Question 1

Explain the formation of a chemical bond.

Accepted Answer

A chemical bond forms when bringing two atoms together lowers the total energy of the system below the energy of the two separate, isolated atoms — and this energy drop happens because electrons rearrange. Everything else (octet rule, shared pairs, ionic transfer) is a bookkeeping device for tracking how the electrons rearrange to make that energy drop happen. 1. The Kössel–Lewis picture: why electrons move at all Kössel and Lewis noticed that the noble gases (He, Ne, Ar, …) are chemically inert, and that every one of them has 8 electrons in its outermost shell (2 for He). They proposed that atoms of other elements are also "trying" to reach that same stable outer-shell arrangement, called the octet rule: an atom is most stable when its valence shell holds 8 electrons (a duplet, 2 electrons, for hydrogen and helium). An isolated atom, other than a noble gas, does not have this arrangement, so it is energetically restless. A bond forms because combining with another atom lets both reach a stable octet — and there are only two ways electrons can move to achieve this: - Complete transfer of electrons — the ionic (electrovalent) bond. An electropositive atom (few electrons beyond a stable core, e.g. Na with configuration 1s^2 2s^2 2p^6 3s^1 ) loses its outer electron(s) entirely to an electronegative atom (nearly a full octet, e.g. Cl with 1s^2 2s^2 2p^6 3s^2 3p^5 ). Na becomes Na^+ (octet complete at 1s^2 2s^2 2p^6 ), Cl becomes Cl^- (octet complete at 3s^2 3p^6 ). The bond itself is then just the electrostatic (Coulombic) attraction between these two oppositely charged ions — that attraction is the ionic bond in NaCl. - Sharing of electrons — the covalent bond. When neither atom can afford to give up an electron outright (e.g. two identical H atoms, or two Cl atoms), each contributes one electron to a shared pair that sits between both nuclei and is counted toward both atoms' octets simultaneously. In H 2, each H "sees" 2 electrons (a filled duplet, like He) instead of its lone 1; in Cl 2, each Cl "sees" 8 electrons around it instead of 7. Lewis represented this with dot structures, e.g. H:H or H"-"H for the shared pair. A short aside on the common mix-up: it is not that atoms have some literal desire to be full — the octet rule is only a convenient rule of thumb that happens to track configurations of low energy for elements near the top of the periodic table. The real criterion, always, is energy. 2. The energetics picture: why H₂ actually forms Take two isolated H atoms, each with one proton and one electron, infinitely far apart — define this separated state as having potential energy 0. Now let them approach each other. Two kinds of forces act simultaneously as the internuclear distance decreases: - Attractive forces: each nucleus attracts the other atom's electron. - Repulsive forces: the two nuclei repel each other, and the two electrons repel each other. At large separation, attraction wins (the electrons can spread over both nuclei), so potential energy keeps decreasing as the atoms approach — energy is being released. This continues until a particular internuclear distance is reached at which the attractive and repulsive forces exactly balance and the potential energy hits a minimum. This distance is the bond length (74 pm for H 2 ), and the energy released in reaching it (relative to the separated atoms) is the bond enthalpy (~435 kJ/mol for H 2 ). If the atoms are pushed closer than this equilibrium distance, nucleus–nucleus and electron–electron repulsion dominate and the potential energy rises sharply again — so the minimum is a genuinely stable point, not an arbitrary stopping place. The molecule H 2 is more stable than two separate H atoms precisely because its potential energy at bond length is lower than the potential energy of the two isolated atoms; the difference is the energy that must be supplied back to break the bond. Putting the two pictures together: the octet rule tells you which electron arrangement (transferred or shared) an atom will settle into, and the potential-energy argument tells you why settling into that arrangement is favourable — it is the arrangement at which attractive and repulsive forces between the combining atoms balance to give the lowest possible energy for the system. Answer: A chemical bond forms because it lowers the potential energy of the combining atoms below that of the isolated atoms. Atoms achieve this either by completely transferring electrons from an electropositive atom to an electronegative one, producing oppositely charged ions held together by electrostatic attraction (ionic bond, e.g. Na⁺Cl⁻), or by sharing a pair of electrons between two atoms so that each attains a stable noble-gas-like octet (or duplet, for H) around it (covalent bond, e.g. H–H). In either case, the driving force is that attractive forces (nucleus–electron) come to outweigh repulsive forces (nucleus–nucleus, electron–electron) up to a particular internuclear distance — the bond length — at which the system's potential energy is at a minimum; the energy released in reaching this minimum is the bond enthalpy, and it is this energy lowering, not any innate "desire" for a full shell, that constitutes bond formation.

Question 2

Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Accepted Answer

A Lewis dot symbol shows only the valence electrons of an atom — the electrons in the outermost shell that take part in bonding — placed as dots around the element's symbol. The inner, filled shells are never shown; they are chemically inert and play no part in bonding. The number of dots equals the number of valence electrons. To find that number, write the electron configuration and count the electrons outside the last noble-gas core (for a transition-metal-adjacent element like Br, the filled inner d subshell is not valence — a common mistake is to drag those ten electrons in as well). Notice each count matches the element's old-style group number (IA, IIA, IIIA, VA, VIA, VIIA) — that shortcut is a fast check, but the configuration is what actually justifies it. Placing the dots: fill the four sides singly first (Hund's rule), then start pairing. There are only four positions around a symbol — top, right, bottom, left. Electrons are added one per side going around, exactly as electrons occupy separate orbitals singly before pairing up; only once all four sides already carry one dot does a fifth electron double up on a side. This is why boron (3 electrons) still has three lone, unpaired dots, while nitrogen's fifth electron is forced to pair with one already there. Filling right → top → left → bottom, then pairing in the same order: Na: Mg: B: N: O: Br: Checking each dot count against the valence-electron totals above: Na has 1 dot, Mg has 2 (both unpaired), B has 3 (all unpaired — the fourth side is still empty), N has 5 (one paired side, three unpaired — this pair-plus-three-singles pattern is exactly why nitrogen forms three bonds and keeps one lone pair), O has 6 (two paired sides, two unpaired — matching oxygen's two bonds), and Br has 7 (three paired sides, one lone unpaired dot — matching the single bond every halogen forms). Answer: Na · (1 e⁻), · Mg · (2 e⁻, unpaired), · B · with a third single dot (3 e⁻, unpaired), N with one lone pair and three single dots (5 e⁻), O with two lone pairs and two single dots (6 e⁻), Br with three lone pairs and one single dot (7 e⁻) — as drawn above.

Question 3

Write Lewis symbols for the following atoms and ions: S and S 2 –; Al and Al 3 +; H and H^-

Accepted Answer

A Lewis symbol shows only the electrons in an atom's (or ion's) outermost, valence shell as dots placed around the symbol — the inner, core electrons never appear, because they play no part in bonding. To know how many dots to draw, find the valence-shell electron count from the electron configuration, not from the atomic number itself — the atomic number gives the total number of electrons, and only the ones sitting in the outermost shell get drawn. The dot-placement rule is what decides whether dots stand alone or in pairs: starting from a bare symbol, one dot is placed on each of the four sides in turn, and only once all four sides already carry a single dot does the next electron start pairing up with one of them. So 1–4 valence electrons show up as that many single dots on different sides, while 5–8 valence electrons show up as some sides paired and the rest single, up to a full octet of four pairs. S and S^2- Sulphur, Z=16, has the configuration 1s^2 2s^2 2p^6 3s^2 3p^4, so its valence shell ( n=3 ) holds 2+4=6 electrons. By the placement rule, all four sides get a dot first and two of those sides then take a second dot, leaving 2 lone pairs and 2 single electrons: S: · · · · S ·· The step people skip with anions: an anion isn't just "a charge label" — it physically adds electrons to the same valence shell, so the dot count goes up. S^2- forms when S gains 2 electrons to complete its octet, 3s^2 3p^6, i.e. 6+2=8 valence electrons — a full set of 4 lone pairs. Being a charged species, the symbol goes inside square brackets with the charge written as a right superscript outside the bracket: S^2-: [ · · · · ·· S ·· ]^2- Al and Al^3+ Aluminium, Z=13, has the configuration 1s^2 2s^2 2p^6 3s^2 3p^1, so its valence shell holds 2+1=3 electrons — three single dots on three of the four sides, the fourth side left empty (three lone electrons have nothing to pair with yet): Al: · · Al · The step people skip with cations: a cation forms by losing electrons from the outermost shell entirely, not by "cancelling" some of the dots and leaving the rest. Al^3+ loses all 3 of those valence electrons at once — that is exactly what it takes to bare the filled, stable n=2 shell underneath; losing only 1 or 2 would leave an unstable, half-emptied outer shell instead. With zero electrons left in the valence shell, the Lewis symbol carries no dots at all, just the symbol and the charge in brackets: Al^3+: [ Al ]^3+ H and H^- Hydrogen, Z=1, has the configuration 1s^1 — a single valence electron, drawn as one dot beside the symbol: H: H · The hydride ion H^- forms when H gains 1 electron, reaching 1s^2 — the same two-electron shell as helium. Both electrons now sit together as a single lone pair beside the symbol, and the ion is bracketed with a 1- charge: H^-: [ H ·· ]^- Answer: S has 6 valence electrons (2 lone pairs + 2 single dots); S^2- has 8, written [ S ]^2- with 4 lone pairs; Al has 3 valence electrons, drawn as 3 single dots; Al^3+ has 0, written [ Al ]^3+ with no dots at all; H has 1 valence electron, H·; H^- has 2, written [ H ·· ]^- as one lone pair.

Question 4

Draw the Lewis structures for the following molecules and ions: H 2S, SiCl 4, BeF 2, CO 3^2-, HCOOH

Accepted Answer

A Lewis structure is just the valence-electron count turned into a picture: every atom gets its full outer shell, and nothing is invented beyond what the arithmetic gives you. The same four steps apply to every species below: 1. Add up the valence electrons contributed by each atom (group number: 1 for H, 4 for C, 6 for O and S, 7 for Cl and F, 2 for Be), and add one extra electron for every unit of negative charge on the species. Forgetting the charge electrons is the single most common mistake on questions like this. 2. Put the atom that can form the most bonds in the centre — H can only ever form one bond, so it is never central. 3. Join the centre to the outer atoms with single bonds, then use the remaining electrons to give every outer atom a full octet (a duet, for H). 4. Whatever is left goes on the central atom as lone pairs. If the central atom still doesn't reach eight electrons after that, take a lone pair off a neighbouring atom and turn it into a second bond (a double bond) to the centre. H 2 S Valence electrons: 2(1) + 6 = 8, i.e. 4 pairs. S is central. Two S–H single bonds use 2 pairs (4 electrons); the remaining 2 pairs both go on S. Structure: H - S - H, with two lone pairs on sulfur (not on the hydrogens — a hydrogen atom's shell is full with just the 2 electrons of its one bond). That's 2 bond pairs + 2 lone pairs = 4 pairs on S, an octet. SiCl 4 Valence electrons: 4 + 4(7) = 32, i.e. 16 pairs. Si is central (it can form four bonds; each Cl can only form one). Four Si–Cl single bonds use 4 pairs, leaving 12 pairs. Give each Cl 3 lone pairs to complete its octet (1 bond pair + 3 lone pairs = 4 pairs = 8 electrons per Cl): 4 × 3 = 12 lone pairs, which exactly uses up what's left. Structure: Si at the centre, singly bonded to four Cl atoms (tetrahedral), each Cl carrying 3 lone pairs, and none left over for Si — its 4 bond pairs already give it an octet. BeF 2 Valence electrons: 2 + 2(7) = 16, i.e. 8 pairs. Be is central. Two Be–F single bonds use 2 pairs, leaving 6 pairs — exactly 3 lone pairs on each F (1 bond pair + 3 lone pairs = octet for each F). Structure: F - Be - F (linear), each F with 3 lone pairs, and none on Be. Do not try to force an octet onto Be here — Be contributes only 2 valence electrons in total, so after it forms its 2 bonds it has just 4 electrons around it, and there is no spare lone pair anywhere in the molecule that could be shared into a second bond to Be without stripping an F below its own octet. BeF 2 (like BeH 2, BeCl 2 and BF 3 ) is one of the genuine, well-known exceptions to the octet rule: Be is simply electron-deficient. CO 3^2- Valence electrons: 4 + 3(6) + 2 = 24 (the +2 is for the ion's two negative charges — leaving it out is exactly the trap step 1 above warns about). That's 12 pairs. C is central, bonded to three O atoms. If all three C–O bonds were single bonds, that uses only 3 pairs, leaving 9 pairs to spread over the three O's as 3 lone pairs each — but that leaves C with only 3 bond pairs (6 electrons), short of an octet. So one bond is upgraded: a lone pair on one oxygen is shared with C, turning that C–O single bond into a C=O double bond. Now: - C: 2 single bonds + 1 double bond = 4 bond pairs → octet. - The doubly-bonded O: 2 shared pairs + 2 lone pairs → octet. - Each of the other two O's: 1 bond pair + 3 lone pairs → octet, and each carries a formal negative charge. To check this is consistent, use formal charge FC = V - L - 1/2B, where V is the atom's own valence electrons, L is the number of electrons sitting in its lone pairs, and B is the number of electrons it shares in bonds: C: 4-0-1/2(8)=0, O(double): 6-4-1/2(4)=0, O(single): 6-6-1/2(2)=-1 Summing over the ion: 0+0+(-1)+(-1) = -2, matching the ion's actual charge, so the structure is self-consistent. Structure: one C=O and two C–O⁻, with C at the centre. But nothing singles out which of the three oxygens should carry the double bond — the three C–O bonds are chemically indistinguishable (confirmed experimentally: all three C–O bond lengths in CO 3^2- are equal, in between a normal single and double bond). So the drawing above is only one of three equivalent resonance structures (the double bond could equally be drawn to any one of the three oxygens); the real ion is the resonance hybrid of all three, with every C–O bond of order 4/3. HCOOH (formic acid) Valence electrons: 2(1) + 4 + 2(6) = 18, i.e. 9 pairs. Read the formula carefully — one H sits on C, and the other sits on an O, i.e. the molecule is H–C(=O)–O–H, not two hydrogens both attached to carbon. C is central, bonded to: one H, one O by a double bond, and a second O by a single bond, with that second O then singly bonded to the second H. Electron count check: bonds used = C–H (1 pair) + C=O (2 pairs) + C–O (1 pair) + O–H (1 pair) = 5 pairs (10 electrons). Remaining: 9-5=4 pairs (8 electrons), split as 2 lone pairs on the doubly-bonded O and 2 lone pairs on the –O–H oxygen. Octet check: C has 1+2+1=4 bond pairs → octet. The =O has 2 shared pairs + 2 lone pairs → octet. The –O– has 2 shared pairs (one to C, one to H) + 2 lone pairs → octet. Both hydrogens have 1 bond pair each, a full duet. So the full structure is: H singly bonded to C; C doubly bonded to one O (2 lone pairs on that O); C singly bonded to a second O (2 lone pairs on that O), which is itself singly bonded to the remaining H. Answer: - H 2 S: H–S–H, with 2 lone pairs on S. - SiCl 4: Si singly bonded to four Cl atoms (tetrahedral), 3 lone pairs on each Cl, none on Si. - BeF 2: F–Be–F (linear), 3 lone pairs on each F, none on Be — an accepted exception to the octet rule. - CO 3^2-: C bonded to three O atoms, one C=O (2 lone pairs on that O) and two C–O⁻ (3 lone pairs on each); the three C–O bonds are equivalent by resonance, bond order 4/3 each. - HCOOH: H–C(=O)–O–H, 2 lone pairs on the doubly-bonded O and 2 lone pairs on the –O–H oxygen.

Question 5

Define octet rule. Write its significance and limitations.

Accepted Answer

The octet rule says atoms react to end up with eight electrons in their outermost (valence) shell — the same stable arrangement as a noble gas. Statement of the rule Atoms of different elements combine — by losing, gaining, or sharing electrons — in such a way that each atom acquires eight electrons in its valence shell, giving it the electronic configuration of the nearest noble gas. This is called Kössel–Lewis's octet rule. (Hydrogen and a few very light elements are content with just two electrons, a "duet," since their nearest noble gas is helium.) Take common examples: - Na → Na^+ + e^-: sodium loses its one valence electron, so Na^+ has the argon-like configuration 2, 8 — 8 electrons in the outer shell. - Cl + e^- → Cl^-: chlorine gains one electron to reach 2, 8, 8, also 8 in the outer shell. - In H 2 O, oxygen shares one electron pair with each of the two hydrogen atoms. Counting both its own and the shared electrons, oxygen ends up surrounded by 8 electrons (2 lone pairs + 2 bond pairs, each bond pair counted twice). Significance of the octet rule - It gives a simple, quick way to predict how atoms of the representative (s- and p-block) elements will combine, and in what ratio — for example, why sodium and chlorine combine 1: 1 (NaCl) and not 1: 2. - It correctly rationalises why noble gases are chemically inert: their valence shell already has 8 electrons (2 for He), so they have no tendency to lose, gain, or share more. - It underlies the drawing of Lewis structures, from which oxidation states, formal charges, and rough molecular shapes (via VSEPR, built on the same electron-counting idea) can be worked out. - The step people skip: the octet rule only counts electrons around each atom — a shared (bonded) pair is counted once for each of the two atoms it belongs to. Missing that double-counting is the most common error when checking whether an atom "has 8." Limitations of the octet rule The rule is a useful first approximation, not a law, and it breaks down in several well-known ways: 1. Incomplete octet. Some central atoms are stable with fewer than 8 electrons around them. In BeCl 2, beryllium has only 4 electrons around it; in BF 3, boron has only 6. Both molecules exist and are stable despite the "shortfall." 2. Odd-electron molecules. Species with an odd total number of valence electrons can never satisfy the octet rule for every atom, because electrons can only be arranged in pairs to make a full octet. Examples: nitric oxide, NO (11 valence electrons), and nitrogen dioxide, NO 2 (17 valence electrons). 3. Expanded octet. Elements from period 3 onward have accessible d orbitals, so their atoms can accommodate more than 8 electrons. In PF 5, phosphorus is surrounded by 10 electrons; in SF 6, sulphur is surrounded by 12; in H 2 SO 4 and PO 4^3- the central atom likewise exceeds an octet. This is the most frequently tested limitation, so it is worth remembering by period: expansion needs empty low-lying d orbitals, which only period-3-and-beyond atoms (P, S, Cl, …) have — period 2 atoms (like N or O) can never expand their octet. 4. It does not explain the shape of molecules. The octet rule only counts electrons; it says nothing about geometry — why H 2 O is bent while CO 2 is linear, for instance. (That is instead explained by VSEPR theory.) 5. It does not account for the relative stability or relative energy of molecules. The rule is based purely on electron pairing and ignores the actual energy considerations that determine which arrangement is genuinely more stable. 6. Noble gases were once thought totally unreactive, "confirming" the rule — but compounds like XeF 2, XeF 4, and XeF 6 exist, in which xenon (which already has a full octet on its own) forms real bonds, something the rule cannot accommodate. Answer: The octet rule states that atoms combine so as to have 8 electrons (a noble-gas configuration) in their valence shell. Its significance is that it gives a simple basis for predicting combining ratios and for drawing Lewis structures. Its limitations are: it fails for molecules with an incomplete octet ( BeCl 2, BF 3 ), for odd-electron molecules ( NO, NO 2 ), for molecules with an expanded octet ( PF 5, SF 6, H 2 SO 4 ); and it does not explain the shape of molecules, does not account for the relative energy/stability of molecules, and cannot explain noble-gas compounds such as XeF 2 and XeF 4.

Question 6

Write the favourable factors for the formation of ionic bond.

Accepted Answer

An ionic bond forms only when the whole three-step energy balance comes out favourable — not just when one atom "wants" to lose an electron and another "wants" to gain one. Making an ion always costs energy first; the bond only exists because the ions then release far more energy by coming together. So the favourable factors are exactly the quantities that control each step of that balance. Step 1: Take an electron off the metal atom — low ionization enthalpy Ionization enthalpy, Δ i H, is the energy needed to remove an electron from a gaseous atom: M(g) → M^+(g) + e^-, Δ i H 0 This step always absorbs energy, so the lower this cost is, the easier it becomes to generate the cation. Elements with low ionization enthalpy — the alkali and alkaline earth metals (Na, K, Ca, Mg,...) sit at one end of the periodic table for exactly this reason — are the ones that readily form positive ions. Step 2: Put that electron onto the non-metal atom — high (negative) electron gain enthalpy Electron gain enthalpy, Δ eg H, is the energy change when a gaseous atom accepts an electron: X(g) + e^- → X^-(g) For most non-metals this step releases energy ( Δ egH is negative), and the more negative it is, the more favourable ion formation becomes. Halogens and oxygen, which have a strong pull for one extra electron, are the typical anion-formers. The step people mix up: ionization enthalpy is always endothermic (energy in), while electron gain enthalpy is usually exothermic (energy out) — do not treat them as the same sign or you will conclude ionic bonding can never be favourable. Step 3: Pack the ions into a crystal — high lattice enthalpy Even after Steps 1 and 2, the net energy change for forming isolated M^+ and X^- ions from neutral atoms can still be positive (unfavourable), because Δ iH is usually bigger in magnitude than Δ egH. What makes the ionic solid form anyway is the third quantity: lattice enthalpy, Δ latticeH, the energy released when the gaseous ions come together and pack into the crystal lattice: M^+(g) + X^-(g) → MX(s), Δ latticeH 0 Lattice enthalpy is large and negative because of the strong electrostatic (Coulombic) attraction between oppositely charged ions arranged in a regular repeating array, and it easily outweighs the earlier energy cost. The higher the magnitude of the lattice enthalpy, the more favourable the overall formation of the ionic bond — this is why small, highly charged ions (which pack closely and attract strongly) tend to give the most stable ionic solids. Putting the three together The overall enthalpy of formation of the ionic solid is, schematically, Δ H f = Δ i H + Δ eg H + Δ lattice H For the ionic bond to form spontaneously, Δ H f must be negative overall, which needs Δ i H to be as small as possible, Δ eg H to be as negative as possible, and Δ lattice H to be as large (negative) as possible. Answer: The formation of an ionic bond is favoured by (i) low ionization enthalpy of the metal atom, so that the cation forms easily; (ii) high (more negative) electron gain enthalpy of the non-metal atom, so that the anion forms readily; and (iii) high (more negative) lattice enthalpy, released when the oppositely charged gaseous ions pack into a crystal lattice — this last term is usually large enough to outweigh the energy cost of steps (i) and (ii) and make the overall process energetically favourable.

Question 7

Discuss the shape of the following molecules using the VSEPR model: BeCl 2, BCl 3, SiCl 4, AsF 5, H 2S, PH 3

Accepted Answer

The shape of a molecule under VSEPR is decided by ALL the electron pairs around the central atom — lone pairs included — not just the atoms you can see bonded to it. A molecule that "looks" like it should be linear because it has two atoms attached can still be bent, if the central atom is also hiding a couple of lone pairs. So the method is always the same four steps, and only the count changes from molecule to molecule. The method, once, so you're not re-deriving it six times: 1. Count the valence electrons of the central atom (its group number). 2. Work out how many single (σ) bonds it makes to the surrounding atoms, and how many of its valence electrons are left over as lone pairs. 3. Add them: steric number SN = ( bond pairs ) + ( lone pairs ). This number fixes the electron-pair geometry — how the electron clouds arrange themselves to stay as far apart as possible. 4. The molecular shape is what you see when you look only at where the atoms sit, ignoring the lone pairs (even though the lone pairs are still there, pushing on the bonds). The one repulsion fact you need throughout: lone pair–lone pair repulsion is strongest, lone pair–bond pair is next, bond pair–bond pair is weakest ( lp-lp lp-bp bp-bp ). A lone pair takes up more space than a bonding pair, so wherever a lone pair is present it squeezes the bond angle to a little less than the "ideal" geometric value — that's the detail students most often forget, treating every SN = 4 species as if it must show the full 109.5°. BeCl 2: Be (group 2) has 2 valence electrons and forms 2 single bonds to the two Cl atoms, using both electrons — nothing is left over, so there are 0 lone pairs on Be (Be ends up with only 4 electrons around it; this is one of the well-known exceptions to the octet rule). SN = 2 ( bond pairs ) + 0 ( lone pairs ) = 2 Two electron domains with no lone pairs arrange themselves exactly opposite each other: linear shape, Cl–Be–Cl = 180^°. BCl 3: B (group 13) has 3 valence electrons and forms 3 single bonds, again using up all of them — 0 lone pairs on B (only 6 electrons around it, another octet exception). SN = 3 + 0 = 3 Three domains, no lone pairs, spread out in a plane: trigonal planar shape, all Cl–B–Cl = 120^°. SiCl 4: Si (group 14) has 4 valence electrons and forms 4 single bonds, using all 4 — 0 lone pairs. SN = 4 + 0 = 4 Four domains, no lone pairs: tetrahedral shape, every Cl–Si–Cl = 109.5^° (this is the one case here where the electron-pair geometry and molecular shape are identical, because there's nothing hidden away as a lone pair). AsF 5: As (group 15) has 5 valence electrons and forms 5 single bonds to the 5 F atoms, using all 5 — 0 lone pairs. As is in Period 4, so it can expand its octet and accommodate 5 bond pairs (unlike N, which cannot form NF 5 ). SN = 5 + 0 = 5 Five domains with no lone pairs give the shape unique to SN = 5: trigonal bipyramidal shape — three F atoms in an equatorial plane at 120^° to each other, and two F atoms axial, each at 90^° to every equatorial F and 180^° to each other. (This is also why trigonal bipyramidal is the odd one out among these shapes: it has two different bond angles instead of one.) H 2 S: S (group 16) has 6 valence electrons and forms only 2 single bonds (to the 2 H atoms), using 2 of its electrons — the remaining 4 electrons sit as 2 lone pairs on S. SN = 2 ( bond pairs ) + 2 ( lone pairs ) = 4 The electron-pair geometry is tetrahedral, but with 2 of those 4 corners occupied by lone pairs instead of atoms, the molecular shape is bent (angular/V-shaped) — this is the trap: H 2S has the same number of bonded atoms as BeCl 2 (two), but it is not linear, precisely because those two lone pairs are still there repelling the bonds. With two lone pairs pushing inward (lp-lp repulsion, the strongest kind), the H–S–H angle is squeezed well below the tetrahedral 109.5°, down to about 92^° experimentally — close to a plain 90^°, which tells you sulfur is bonding using orbitals with very little s-character. PH 3: P (group 15) has 5 valence electrons and forms 3 single bonds (to the 3 H atoms), using 3 electrons — the remaining 2 electrons sit as 1 lone pair on P. SN = 3 ( bond pairs ) + 1 ( lone pair ) = 4 Again the electron-pair geometry is tetrahedral, but with one corner taken by a lone pair, the molecular shape is trigonal pyramidal. Here only lp-bp repulsion (not lp-lp) is doing the squeezing, so the angle isn't pushed down as far as in H 2 S — but it is still well under 109.5°, coming out to about 93.5^° experimentally (even smaller than the 107^° seen in NH 3, because P is a larger, less electronegative central atom and bonds using orbitals with even less s-character than N does). Answer: BeCl 2 — linear ( 180^°, 0 lone pairs on Be). BCl 3 — trigonal planar ( 120^°, 0 lone pairs on B). SiCl 4 — tetrahedral ( 109.5^°, 0 lone pairs on Si). AsF 5 — trigonal bipyramidal ( 90^° axial–equatorial, 120^° equatorial–equatorial, 180^° axial–axial; 0 lone pairs on As). H 2 S — bent/angular (2 lone pairs on S squeeze the angle to ≈ 92^° ). PH 3 — trigonal pyramidal (1 lone pair on P squeezes the angle to ≈ 93.5^° ).

Question 8

Although geometries of NH 3 and H 2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Accepted Answer

The bond angle is squeezed by how many lone pairs are pushing on it — water has two, ammonia has only one. Both molecules are built on the same starting point. The central atom (N in NH 3, O in H 2 O ) is sp^3 hybridised, which gives four hybrid orbitals arranged toward the corners of a regular tetrahedron, where the ideal angle between any two orbitals is 109.5^°. If all four positions were occupied by bonding pairs (as in CH 4 ), the molecule would be a perfect tetrahedron with exactly this angle. In NH 3 and H 2 O, some of those four positions are filled by lone pairs instead of bonds, and that is what "distorts" the tetrahedron and pulls the angle down. Count the pairs around each central atom. - N in NH 3: 5 valence electrons, 3 used in N–H bonds, leaving 1 lone pair. So the arrangement is 3 bond pairs (BP) + 1 lone pair (LP). - O in H 2 O: 6 valence electrons, 2 used in O–H bonds, leaving 2 lone pairs. So the arrangement is 2 bond pairs (BP) + 2 lone pairs (LP). Both are "distorted tetrahedral" in the sense that the electron-pair geometry is tetrahedral but the molecular shape (the arrangement of atoms you'd actually draw) is pyramidal for NH 3 and bent (angular) for H 2 O — the lone pairs are there but not "seen" in the molecular shape, even though they still take up space and push on the bonds. Apply the VSEPR repulsion order. The rule (from VSEPR theory) for how strongly two electron pairs push each other apart is: lone pair–lone pair (lp–lp) lone pair–bond pair (lp–bp) bond pair–bond pair (bp–bp) This ordering comes from the fact that a lone pair is held only by the central atom's nucleus, so its electron cloud is less confined and spreads out more than a bond pair, which is pulled toward two nuclei at once. A more spread-out cloud repels neighbouring electron pairs more strongly. This is the step people skip past: it is not enough to say "lone pairs repel more" — you have to say why (they aren't shared between two nuclei, so they occupy more angular space around the central atom), otherwise the comparison between two different molecules doesn't follow from it. Now compare the two molecules pair by pair. - In NH 3: there is only one lone pair, so the only repulsions distorting the tetrahedron are 3 (lp–bp) interactions (the lone pair pushing on each of the three N–H bonds), plus the weaker (bp–bp) interactions among the bonds themselves. - In H 2 O: there are two lone pairs, so on top of the (lp–bp) and (bp–bp) interactions, there is now also an (lp–lp) interaction between the two lone pairs — the strongest repulsion in the whole list, and NH 3 does not have this term at all. Because H 2 O carries this extra, stronger lone pair–lone pair repulsion, its two lone pairs push each other (and, through them, the bonding pairs) further apart than a single lone pair can in NH 3. The two O–H bonds get squeezed closer together in response, closing the H–O–H angle more than the H–N–H angle is closed in ammonia. The numbers confirm the direction. Starting from the ideal tetrahedral angle of 109.5^°: - NH 3 (1 lone pair, 3 lp–bp repulsions): bond angle compressed to 107.8^° - H 2 O (2 lone pairs, plus 1 lp–lp repulsion): bond angle compressed further, to 104.5^° So both angles are less than 109.5^° because lone pairs are present at all, but water's angle is smaller than ammonia's specifically because water has one more lone pair, and that second lone pair introduces the strongest kind of repulsion (lp–lp) that ammonia's single lone pair cannot produce. Answer: Both NH 3 and H 2 O start from a tetrahedral ( sp^3 ) arrangement of electron pairs, but NH 3 has only 1 lone pair (giving lp–bp and bp–bp repulsions) while H 2 O has 2 lone pairs (adding a stronger lp–lp repulsion as well). Since lp–lp repulsion lp–bp repulsion bp–bp repulsion, the extra lone pair in water pushes the bonding pairs closer together than the single lone pair in ammonia does. This is why the H–O–H angle ( 104.5^° ) is smaller than the H–N–H angle ( 107.8^° ), even though both are less than the ideal tetrahedral angle of 109.5^°.

Question 9

How do you express the bond strength in terms of bond order?

Accepted Answer

Bond order and bond strength move together: the higher the bond order, the stronger the bond. Bond order comes out of Molecular Orbital Theory. It is defined as Bond order = 1/2(N b - N a) where N b is the number of electrons sitting in bonding molecular orbitals and N a is the number of electrons sitting in antibonding molecular orbitals. Why this connects to strength: electrons in a bonding orbital hold the two nuclei together, while electrons in an antibonding orbital pull them apart. A larger N b - N a means the "holding together" effect wins by a bigger margin, so the two atoms are bound more tightly. That tighter binding shows up in a real, measurable quantity — the bond dissociation energy, i.e. the energy you must supply to break the bond. Bond order and bond dissociation energy (bond strength) are directly proportional: Bond order ∝ Bond dissociation energy (Bond strength) So as bond order goes up, bond strength goes up in step with it, and the molecule is more stable. (The same rise in bond order also pulls the bonded atoms closer, which is why bond order is inversely proportional to bond length — a separate but related consequence of the same electron count.) A quick numerical check with H 2, He 2^+, and He 2 makes the trend concrete: - H 2: N b = 2, N a = 0 ⇒ bond order = 1/2(2-0) = 1 - He 2^+: N b = 2, N a = 1 ⇒ bond order = 1/2(2-1) = 0.5 - He 2: N b = 2, N a = 2 ⇒ bond order = 1/2(2-2) = 0 H 2 has the highest bond order and is indeed the strongest, most stable of the three (it exists as a normal stable molecule); He 2^+ is weaker and less stable; He 2 has zero bond order, meaning no bond forms at all — helium does not exist as He 2. The step people get wrong here: bond order is not itself an energy — it is a pure number (a ratio of electron counts) that tells you how bond strength compares across molecules or ions. To get an actual strength value in kJ/mol you still need the measured bond dissociation energy; bond order only tells you the direction and rough scale of that comparison. Answer: Bond strength is expressed through bond order via the relation Bond order =1/2(N b-N a) ( N b, N a = electrons in bonding and antibonding molecular orbitals). Bond order is directly proportional to bond dissociation energy, so a higher bond order means a stronger, more stable bond (and, correspondingly, a shorter bond length); a bond order of zero means no bond exists.

Question 10

Define the bond length.

Accepted Answer

Bond length is the equilibrium distance between the centres (nuclei) of two bonded atoms in a molecule. When two atoms come together to form a covalent bond, they do not stop at any random separation. As the atoms approach each other, the attractive forces (nucleus–electron attraction, and the sharing of the bonding electron pair) and the repulsive forces (nucleus–nucleus repulsion, and electron–electron repulsion between the two atoms' electron clouds) both grow, but not at the same rate. - At large separation, attraction dominates and the potential energy of the system falls as the atoms move closer. - At very short separation, repulsion dominates and the potential energy rises sharply. - In between, there is exactly one internuclear distance at which the potential energy of the system is a minimum — the system is most stable here. That distance of minimum potential energy, at which attractive and repulsive forces are balanced, is defined as the bond length. It is usually measured as the distance between the nuclei of the two bonded atoms, using techniques such as X-ray diffraction, electron diffraction, or spectroscopic methods, and is expressed in picometres (pm) or angstroms (Å); 1 Å = 100 pm. A short aside people often miss: bond length is not the sum of some arbitrarily chosen atomic radii — it is an experimentally measured, characteristic value for a given pair of bonded atoms in a given bond order (a C–C single bond, C=C double bond, and C≡C triple bond each have their own distinct, shorter-with-higher-order bond length). It is also convenient to express bond length as the sum of the covalent radii of the two bonded atoms (and, where one atom is more electronegative, a small correction term), because covalent radii are themselves derived from measured bond lengths. Answer: Bond length is the equilibrium (average) distance between the nuclei of two atoms bonded together in a molecule — the internuclear distance at which the potential energy of the system is minimum, i.e., where the attractive and repulsive forces between the atoms are balanced. It is measured experimentally (by X-ray/electron diffraction or spectroscopy) and expressed in picometres (pm) or angstroms (Å), where 1 Å = 100 pm.

Question 11

Explain the important aspects of resonance with reference to the CO 3^2- ion.

Accepted Answer

Resonance means one Lewis structure cannot be drawn for CO 3^2-; the real ion is an average of several, and that average is what actually exists. Step 1: Try to draw a single Lewis structure. Carbon sits at the centre, bonded to three oxygen atoms. Carbon has 4 valence electrons and needs to complete an octet on itself and on each oxygen, while the total charge on the ion is -2. The only way to satisfy carbon's octet with three C–O bonds is to make one C=O double bond and two C–O single bonds: O = C (- O^-)(- O^-) Check the formal charge, using Formal charge = ( valence electrons ) - ( lone-pair electrons ) - 1/2( bonding electrons ) - Carbon (4 bonds, no lone pairs): 4 - 0 - 8/2 = 0 - The doubly-bonded O (2 lone pairs + a double bond): 6 - 4 - 4/2 = 0 - Each singly-bonded O (3 lone pairs + a single bond): 6 - 6 - 2/2 = -1 Total charge = 0 + 0 + (-1) + (-1) = -2, which matches the ion — so this structure is chemically valid. The mistake people make here is stopping at this one structure and treating it as "the" structure of CO 3^2-. Step 2: Notice the structure is not unique. Nothing distinguishes the three oxygen atoms from one another — any one of them could equally well be the one holding the double bond. So there are three ways to place that double bond, giving three canonical (resonance) structures: O = C (- O^-)(- O^-) ↔^- O - C (= O )(- O^-) ↔^- O - C (- O^-)(= O ) These three structures satisfy the requirements for genuine resonance forms: the positions of all the nuclei (C and the three O's) are identical in each one, only the placement of the double bond (i.e., of electrons) differs, and all three have the same energy by symmetry. Step 3: The real ion is the resonance hybrid, not any one canonical form and not a mixture that flips between them. None of the three drawn structures is the actual molecule — carbon–oxygen bonds do not switch back and forth between single and double. The true structure is a single, unchanging hybrid that blends all three contributing structures. Two consequences follow directly from this, and both are confirmed experimentally: - All three C–O bonds are identical in length, each about 1.28 Å — intermediate between a pure C–O single bond (about 1.43 Å) and a pure C=O double bond (about 1.20 Å). Averaging over the three structures, each bond is single in two of them and double in one, giving a bond order of Bond order = (1+1+2)/3 = 4/3 ≈ 1.33 which matches a length between single and double — exactly what is seen, and something a single Lewis structure (which would predict one short C=O and two longer C–O bonds) cannot explain. - The -2 charge is delocalized equally over all three oxygen atoms, not sitting on two specific ones as any single structure would suggest. Each oxygen carries an average formal charge of -2/3, rather than two oxygens carrying -1 each and one carrying 0. Step 4: Resonance also lowers the energy. The hybrid is more stable than any individual canonical structure would be if it existed alone — this extra stability is called the resonance (or delocalization) energy, and it is why the carbonate ion is far less reactive than a hypothetical ion with a truly localized double bond would be. Answer: No single Lewis structure fits CO 3^2-. Three equivalent canonical structures can be drawn, differing only in which oxygen carries the C=O double bond; the real ion is the resonance hybrid of these three. As a result, all three C–O bonds are equal in length (~1.28 Å, bond order 4/3, between single and double bond lengths), the -2 charge is spread equally over all three oxygens (average -2/3 each) rather than localized on two of them, and the hybrid is more stable (lower in energy) than any single contributing structure.

Question 12

H 3PO 3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H 3PO 3? If not, give reasons for the same.

Accepted Answer

Resonance (canonical) structures must have every atom in exactly the same place — only the electrons are allowed to move. If two Lewis structures show a hydrogen atom bonded to two different atoms in the two pictures, they are not resonance forms of one molecule; they are two different molecules (or tautomers) drawn side by side. What structures 1 and 2 actually show Structure 1 puts all three hydrogens on oxygen, so phosphorus carries three P–OH bonds and keeps its lone pair: P(OH) 3 (lone pair on P; P is trivalent, all three H atoms sit on O) Structure 2 puts only two hydrogens on oxygen and moves the third hydrogen onto phosphorus itself, with a P=O double bond replacing the lone pair: HP (= O )( OH ) 2 (one P–H bond, one P=O bond, two P–OH bonds, no lone pair on P) Why these cannot be canonical forms of one resonance hybrid Canonical structures of a real resonance hybrid are required to have an identical arrangement of nuclei (the same skeleton of σ -bonds connecting the same atoms to the same atoms); what differs between them is only the placement of π -electrons or lone pairs, and you get from one structure to the next just by pushing electrons, never by breaking a σ -bond or relocating a nucleus. Compare the two P(OH) 3 -type structures here: in structure 1 the third hydrogen is bonded to an oxygen atom, while in structure 2 that same hydrogen is bonded directly to the phosphorus atom. That is a change in which atom the hydrogen's σ -bond is attached to — a genuine difference in atomic connectivity, not a shift of π -electrons over a fixed skeleton. This is exactly the point students blur: moving a whole atom (and its bond) between two different positions is tautomerism, not resonance — resonance never breaks or remakes a σ -bond. Since the nuclear framework itself differs between structures 1 and 2, they fail the basic requirement for canonical forms of a resonance hybrid; they are two distinct structural formulas (tautomers of each other), not two contributors to a single delocalized structure. Which one is real Only structure 2 correctly represents phosphorous acid. This is confirmed experimentally: H 3 PO 3 behaves as a dibasic (diprotic) acid — only two of its three hydrogens ionize in water. That fits structure 2, where two hydrogens sit on oxygen (ionizable as O–H ) and the third sits directly on phosphorus (a non-ionizable P–H bond). If structure 1 were correct, all three hydrogens would be on oxygen and H 3 PO 3 would have to be tribasic, which it is not. Answer: No, structures 1 and 2 are not canonical (resonance) forms of one hybrid. A true resonance hybrid needs the same positions of all atoms in every canonical structure, with only electrons redistributed; here the third hydrogen is bonded to oxygen in structure 1 but directly to phosphorus in structure 2, a difference in atomic connectivity, not electron placement. So the two are different structures (tautomers), and only structure 2, HP (= O )( OH ) 2, is the real structure of H 3 PO 3 — consistent with its known dibasic character.

Question 13

Write the resonance structures for SO 3, NO 2 and NO 3^-.

Accepted Answer

Resonance means one real electron distribution, drawn as several "contributing" Lewis structures that differ only in where the electrons sit — the atoms never move. For each of these three species you first count all the valence electrons available, then place them so every atom gets an octet (or, for the odd-electron case, as close as you can). Whenever more than one placement is equally valid, all of them are resonance structures, and the true molecule is their average (hybrid). The bookkeeping tool used throughout is the formal charge, worked out from FC = ( valence electrons of the free atom ) - ( lone-pair electrons ) - 1/2( bonding electrons ). The formal charges on all atoms of a structure must add up to the actual charge on the species — that check is what tells you a structure is drawn correctly, not guesswork. SO 3 (sulfur trioxide) Valence electrons: S contributes 6, each O contributes 6, so 6 + 3(6) = 24 electrons = 12 pairs. With S at the centre bonded to three O atoms, the only way to use exactly these 24 electrons and still give every atom an octet is: one S–O bond as a double bond, and the other two as single bonds, with a negative charge sitting on the two singly-bonded oxygens and a compensating positive charge on S. Checking formal charges: - O in the S=O bond (2 lone pairs, 4 bonding electrons): 6-4- 12(4)=0 - O in an S–O single bond (3 lone pairs, 2 bonding electrons): 6-6- 12(2)=-1 (there are two of these) - S (no lone pairs, 8 bonding electrons around it): 6-0- 12(8)=+2 Sum: 0+(-1)+(-1)+2 = 0, matching the neutral molecule. Since all three oxygens are identical, the double bond is equally entitled to sit on any one of them. That gives three resonance structures: Structure I — double bond to O(1), single (−1) bonds to O(2) and O(3) Structure II — double bond to O(2), single (−1) bonds to O(1) and O(3) Structure III — double bond to O(3), single (−1) bonds to O(1) and O(2) with resonance arrows ( ↔ ) connecting I, II and III. A common mistake here is picturing SO 3 as flipping between three different molecules in time — it is not; it is one molecule whose real bonding is the average of the three drawings, so all three S–O bonds come out experimentally identical, each with bond order (2+1+1)/3 = 4/3 (double in one out of three structures, single in the other two). NO 2 (nitrogen dioxide) Valence electrons: 5 + 2(6) = 17. This is an odd number, so the electrons cannot all be paired up — NO 2 is necessarily a free radical, with one unpaired electron left over. (This is exactly why NO 2 dimerises to N2O4.) Put N in the centre, joined to one O by a double bond and to the other O by a single bond, with the extra unpaired electron residing on N: - N: nonbonding electrons owned = 1 (the unpaired electron), bonding electrons = 4 (double bond) + 2 (single bond) = 6, so FC = 5-1- 12(6) = +1 - O (double-bonded, 2 lone pairs): 6-4- 12(4)=0 - O (single-bonded, 3 lone pairs): 6-6- 12(2)=-1 Sum: 1+0-1=0, correct for the neutral molecule. Because the two oxygens are equivalent, there are two resonance structures, mirror images of each other: Structure I: unpaired electron on N, N=O to one oxygen, N–O⁻ to the other Structure II: unpaired electron on N, N=O to the other oxygen, N–O⁻ to the first connected by ↔. The aside worth remembering: whenever the total valence-electron count for a species comes out odd, stop trying to force every atom into a full octet with all electrons paired — the molecule genuinely has an unpaired electron, and that is the correct structure, not a failure to balance it. NO 3^- (nitrate ion) Valence electrons: 5 + 3(6) + 1( for the negative charge ) = 24 electrons = 12 pairs — the extra "+1" is because a negative ion carries one more electron than the sum of the neutral atoms. The structure works out exactly like SO 3: one N=O double bond and two N–O single bonds, since that is again the only arrangement of 24 electrons that gives every atom an octet. Formal charges: - N (no lone pairs, 8 bonding electrons): 5-0- 12(4+2+2) = 5-4 = +1 - O (double-bonded): 6-4- 12(4)=0 - O (single-bonded, ×2): 6-6- 12(2)=-1 each Sum: 1+0-1-1 = -1, matching the charge on the nitrate ion. With three equivalent oxygens, three resonance structures result, related by moving the double bond: Structure I — N=O to O(1), N–O⁻ to O(2) and O(3) Structure II — N=O to O(2), N–O⁻ to O(1) and O(3) Structure III — N=O to O(3), N–O⁻ to O(1) and O(2) again joined by ↔. The nitrate hybrid has three N–O bonds of identical length, each with bond order 4/3 — this is exactly why the N–O bond length measured experimentally in NO 3^- is one single value, intermediate between a genuine single and double bond, the same way it is for every S–O bond in SO3. Answer: SO 3 has three resonance structures (double bond rotating among the three equivalent oxygens, each singly-bonded O carrying -1 and S carrying +2 ); the hybrid gives every S–O bond order 4/3. NO 2, being an odd-electron (17-valence-electron) free radical, has two resonance structures — mirror images with the unpaired electron and +1 formal charge on N, one O neutral (double-bonded) and the other -1 (single-bonded). NO 3 ⁻ has three resonance structures exactly like SO 3 (N carries +1, two O's carry -1 each, one O is neutral, double bond rotating among the three O's), giving every N–O bond order 4/3 and identical experimental bond lengths.

Question 14

Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.

Accepted Answer

An ionic bond forms when one atom empties its outer shell and another fills its outer shell to 8 electrons — the number of electrons transferred is fixed by each atom's valence-electron count, which you read straight off its group number, not guessed. The rule being used throughout is the octet rule: main-group atoms react so as to end up with the same electron count as the nearest noble gas (8 electrons in the outer shell, except for atoms very close to helium). A metal with only 1–3 valence electrons finds it easier to lose them (small ionization enthalpy) than to gain 5–7 more; a non-metal with 5–7 valence electrons finds it easier to gain the few it is missing (favourable electron gain enthalpy) than to lose many. Whenever the electrons lost by the metal don't equal the electrons gained by one non-metal atom, you need more than one atom of one of them — that balancing is the step people skip. (a) K and S Potassium, Z = 19, configuration [ Ar ]4s^1, has 1 valence electron (group 1). Its Lewis symbol is a single dot: K Sulphur, Z = 16, configuration [ Ne ]3s^23p^4, has 6 valence electrons (group 16) and needs 2 more to reach 8: S One potassium atom can only give away 1 electron, but sulphur needs 2 — this is the mismatch that trips people up. It is fixed by using two potassium atoms per sulphur atom, so the electrons lost (1 + 1 = 2) exactly equal the electrons gained (2): K + K + S → K^+ + K^+ + [ S ]^2- Each K^+ is written with no dots at all — the electron removed came from the outermost (4s) shell, so the shell left behind is the already-complete argon-like core, and Lewis symbols only ever put dots on the outermost shell. S^2- now carries 8 dots (a full octet, also argon-like: 18 electrons). The compound formed is K 2 S. (b) Ca and O Calcium, Z = 20, configuration [ Ar ]4s^2, has 2 valence electrons (group 2): Ca Oxygen, Z = 8, configuration [ He ]2s^22p^4, has 6 valence electrons (group 16) and needs 2 more: O Here the numbers already match — calcium has exactly the 2 electrons oxygen needs — so the transfer is one atom to one atom: Ca + O → Ca^2+ + [ O ]^2- Ca^2+ is drawn with no dots (its remaining shell is the complete, argon-like core, 18 electrons); O^2- has a full octet of 8 dots (neon-like, 10 electrons). The compound formed is CaO. (c) Al and N Aluminium, Z = 13, configuration [ Ne ]3s^23p^1, has 3 valence electrons (group 13): Al Nitrogen, Z = 7, configuration [ He ]2s^22p^3, has 5 valence electrons (group 15) and needs 3 more: N The numbers match again — aluminium has exactly the 3 electrons nitrogen is short of — so it is a one-to-one transfer: Al + N → Al^3+ + [ N ]^3- Al^3+ is left with no dots (neon-like core, 10 electrons); N^3- is drawn with a full octet of 8 dots (also neon-like, 10 electrons). The compound formed is AlN. In every case the check that makes the diagram correct is the same: count the dots you started with on each side, make sure the number of electrons the metal atom(s) give away exactly equals the number the non-metal atom needs, and confirm both resulting ions end up with either 0 or 8 dots in their outer shell (never anything in between). Answer: (a) 2 K + S → 2 K^+ + S^2- (forming K 2 S ); (b) Ca + O → Ca^2+ + O^2- (forming CaO ); (c) Al + N → Al^3+ + N^3- (forming AlN ) — in each pair the metal loses electrons equal to its group-derived valence-electron count and the non-metal gains exactly enough to complete its octet, so both ions attain a noble-gas electron configuration.

Question 15

Although both CO 2 and H 2O are triatomic molecules, the shape of H 2O molecule is bent while that of CO 2 is linear. Explain this on the basis of dipole moment.

Accepted Answer

A molecule's dipole moment is a vector sum of its individual bond dipole moments — so the net value depends on the SHAPE of the molecule, not only on how polar each bond is. Setting up the idea Every polar bond has a bond dipole moment, μ = q × d where q is the magnitude of charge separated at the two ends of the bond and d is the bond length. The direction of this vector runs along the bond, toward the more electronegative atom. In both CO₂ and H₂O the central atom is joined to two atoms by polar bonds: - In CO₂, each C=O bond is polar because O is more electronegative than C. - In H₂O, each O–H bond is polar because O is more electronegative than H. So both molecules are built from two identical, individually polar bonds. The only thing that decides whether the molecule ends up polar or non-polar is the angle between those two bonds — because the net dipole moment is the vector sum of the two bond dipoles, not their arithmetic sum. For two bond dipoles of equal magnitude μ separated by the bond angle θ at the central atom, the resultant is μ net = 2μ cos ((θ)/2) Here θ is the angle between the two bonds (O=C=O or H–O–H), since each bond dipole vector points outward along its own bond from the central atom. A common trap: assuming that if the bonds are polar, the molecule must be polar. It is not automatic — geometry can make two (or more) equally strong bond dipoles cancel completely, even though neither individual bond is non-polar. Applying it to CO₂ CO₂ is experimentally found to have zero net dipole moment. Putting μ net = 0 into the formula above: 0 = 2μ C=O cos ((θ)/2) Since the C=O bond itself is polar, μ C=O ≠ 0, so the only way the equation holds is cos ((θ)/2) = 0 ⇒ (θ)/2 = 90^° ⇒ θ = 180^° A bond angle of 180^° means the two C=O bond dipoles point in exactly opposite directions along the same line, so they cancel each other out perfectly: O μ C μ O This cancellation is only geometrically possible if O=C=O is linear. So the observed zero dipole moment is direct evidence that CO₂ must be linear — if it were bent even slightly, the two bond dipoles would no longer be antiparallel and a net moment would appear. Applying it to H₂O H₂O is experimentally found to have a substantial net dipole moment, μ net = 1.84 D — clearly not zero. If water had the same linear shape as CO₂ ( θ = 180^° ), the same algebra as above would force μ net = 0, because the two O–H bond dipoles would again point in exactly opposite directions and cancel. That directly contradicts the measured 1.84 D. So θ cannot be 180^°. The only way the two O–H bond dipoles can add up to a non-zero resultant is if the H–O–H angle is less than 180^° — i.e., the molecule is bent (angular). In that shape the two bond dipole vectors no longer point in opposite directions; they combine along the bisector of the H–O–H angle, reinforcing rather than cancelling: H^μ O^μ H → μ net = 2μ O-H cos ((θ)/2) ≠ 0 (The actual H–O–H angle is close to 104.5^°, consistent with two lone pairs on oxygen pushing the bonding pairs together from the ideal tetrahedral angle — but the dipole-moment argument alone already tells us, without needing VSEPR, that the shape cannot be linear.) So the reasoning, side by side: Bonds Net dipole moment (measured) What that forces the shape to be --- --- --- --- CO₂ 2 polar C=O bonds 0 Linear ( 180^° ) — only way two equal dipoles cancel H₂O 2 polar O–H bonds 1.84 D ≠ 0 Bent ( θ < 180^° ) — dipoles must add, not cancel Answer: CO₂ is linear (O=C=O, bond angle 180^° ) because its two equal C=O bond dipole moments point in exactly opposite directions and cancel, giving the experimentally observed zero net dipole moment — this cancellation is possible only in a linear geometry. H₂O is bent/angular (bond angle ≈ 104.5^° ) because its net dipole moment is not zero ( 1.84 D ); the two O–H bond dipoles can only add up to give this non-zero resultant if the H–O–H angle is less than 180^°, i.e., if the molecule is non-linear.

Question 16

Write the significance/applications of dipole moment.

Accepted Answer

A dipole moment tells you how unevenly charge sits across a molecule, and that single number lets you settle four different structural questions — polarity, shape, isomer identity, and how ionic a bond really is. Dipole moment is defined as μ = q × d where q is the magnitude of charge separated at each end of the bond and d is the distance between the charge centres. It is a vector — it has both a magnitude and a direction, pointing from the positive end to the negative end of the charge separation. That "it's a vector" fact is the one people skip, and it is exactly what makes dipole moment useful for shape, not just for "is this molecule polar." 1. Deciding whether a molecule is polar A nonzero dipole moment means the molecule is polar; a dipole moment of exactly zero means the individual bond dipoles have cancelled by symmetry. This is more informative than looking at individual bonds, because a molecule can have polar bonds and still be nonpolar overall. - H 2O: μ = 1.85 D (nonzero) — the two O–H bond dipoles do not cancel. - CO 2: μ = 0 — the two C=O bond dipoles are equal and opposite, so they cancel exactly. 2. Using dipole moment to confirm molecular shape Because dipole moment is a vector sum of the individual bond dipoles, its value (especially when it is zero) tells you the geometry. - BeF 2 has μ = 0: this is only possible if the molecule is linear, so the two Be–F bond dipoles point in exactly opposite directions and cancel. - H 2O has μ ≠ 0: the O–H bond dipoles do not cancel, which is only consistent with a bent (angular) shape, not a linear one. - NH 3 ( μ = 1.47 D, pyramidal) has a larger dipole moment than NF 3 ( μ = 0.24 D ), even though N–F bonds are individually more polar than N–H bonds. In NH 3 the lone pair's dipole and the three N–H bond dipoles all point the same general way (away from H, toward N) and add up; in NF 3 the lone pair's dipole opposes the three N–F bond dipoles (which point toward the more electronegative F), so they partly cancel. The lesson: dipole moment reflects the vector sum, not the polarity of any one bond taken alone. 3. Distinguishing cis and trans isomers For a pair of geometrical isomers, the cis form has bond dipoles on the same side, so they add up to a larger resultant; the trans form has them on opposite sides, so they partly or fully cancel. - In 1,2-dichloroethene, μ cis μ trans (the trans isomer's C–Cl dipoles point in opposite directions and largely cancel). This gives a physical way to tell the two isomers apart even before doing any other test. 4. Distinguishing ortho, meta, and para disubstituted benzenes For a disubstituted benzene C 6H 4XY, the angle between the two substituent dipoles is different in each isomer (about 60° for ortho, 120° for meta, 180° for para). Since a smaller angle between two dipole vectors gives a larger vector sum, the dipole moments follow μ ortho μ meta μ para and when X = Y (e.g., para-dichlorobenzene), the para isomer's dipole moment is exactly zero because the two identical bond dipoles point in exactly opposite directions. 5. Calculating the percentage ionic character of a bond No real bond is 100% ionic — even in bonds we call "ionic," there is some sharing of electron density, so the measured dipole moment is always lower than what a purely ionic bond of the same bond length would give. Comparing the two lets you quantify how ionic a bond actually is: % ionic character = (μ observed)/(μ calculated (100 % ionic)) × 100 Here μ observed is the dipole moment measured experimentally, and μ calculated is the dipole moment you would get from μ = q × d if one full electronic charge were transferred completely (a purely ionic bond) over the same bond length d. The gap between the two values is the "sharing" of the bond pair that the ionic picture ignores — this is the step people get wrong, treating dipole moment as if it only ever meant "ionic vs covalent" as a yes/no label rather than a measurable percentage. For example, HCl's percentage ionic character comes out to about 17%, showing it is a predominantly covalent bond with partial ionic character, even though chlorine is considerably more electronegative than hydrogen. Answer: Dipole moment is used (1) to tell whether a molecule is polar or nonpolar, (2) to work out or confirm a molecule's shape (a zero dipole moment forces a symmetric geometry such as linear BeF 2, while a nonzero one forces an asymmetric geometry such as bent H 2O ), (3) to distinguish cis (higher μ ) from trans (lower μ ) isomers, (4) to distinguish ortho (highest μ ), meta, and para (lowest/zero μ ) disubstituted benzenes, and (5) to calculate the percentage ionic character of a bond via % ionic character = (μ observed/μ calculated for 100 % ionic bond) × 100, as for HCl (~17% ionic).

Question 17

Define electronegativity. How does it differ from electron gain enthalpy?

Accepted Answer

Electronegativity is the pull a bonded atom exerts on a shared pair of electrons; electron gain enthalpy is the energy change when a free atom grabs a whole extra electron. They are answers to two different questions about two different situations. What electronegativity means Electronegativity is defined as the tendency of an atom in a molecule (i.e., already bonded to another atom) to attract the shared pair of electrons of a covalent bond towards itself. Picture H–Cl. Both atoms share one electron pair in the bond. Chlorine pulls that shared pair closer to itself than hydrogen does — chlorine is said to be more electronegative than hydrogen. That pull is what gives the H–Cl bond its partial charges, δ+ H - δ- Cl. Two things about electronegativity that trip students up: - It is a property of a bonded atom, not a free atom. An isolated Cl atom sitting alone in a container is not "attracting a shared pair" — there is no bond and no partner to pull against. Electronegativity only has meaning once the atom is part of a molecule. - It has no fixed experimental value and no units. It is expressed only as a number on a relative scale — the most common is the Pauling scale, where fluorine is fixed at 4.0 and every other element is ranked relative to it. Because it is relative, the same atom can show a different electronegativity depending on what it is bonded to and even on its hybridisation (for example, carbon's electronegativity rises as its hybridisation goes from sp^3 → sp^2 → sp, because more s-character pulls the bonding electrons in more tightly). What electron gain enthalpy means Electron gain enthalpy, Δ egH, is the enthalpy change when one mole of isolated gaseous atoms, each in its ground state, each gains one electron to become a mole of gaseous anions of charge -1: X(g) + e^- → X^-(g) Δ egH Here X(g) is the atom in the gas phase (isolated, no bonding partner) and Δ egH is the energy released or absorbed for that single, well-defined process. Because it is a measured energy change for a specific reaction, it does have a fixed value and units, kJ mol^-1 — for example Δ egH for chlorine is about -349 kJ mol^-1 (energy is released, hence negative, because the incoming electron is attracted by the nucleus). The core contrasts, side by side - State of the atom: electron gain enthalpy is defined for an isolated gaseous atom; electronegativity is defined for an atom already bonded inside a molecule. One aside worth flagging explicitly, since it's the step most students blur together: electron gain enthalpy describes an atom becoming an ion on its own, while electronegativity describes an atom's pull while already sharing an electron pair — they are not two measurements of the same event. - What happens to the electron: electron gain enthalpy is about the atom fully accepting one whole electron to form an anion; electronegativity is about the atom pulling a shared pair partly towards itself, without fully taking possession of it — the bond does not break, no ion is formed. - Units and fixedness: electron gain enthalpy is a measurable quantity with units of kJ mol^-1 and one defined value per element (in its ground state); electronegativity is a dimensionless, relative number that can change from one compound to another for the same element. Because of this last point, an element does not have "an" electronegativity the way it has "an" electron gain enthalpy — it has a range of electronegativity values depending on context, while its electron gain enthalpy is a single fixed number for the isolated atom. Answer: Electronegativity is the relative tendency of an atom already bonded in a molecule to attract the shared pair of a covalent bond towards itself — a dimensionless number (e.g., Pauling scale) that varies with the atom's bonding partner and hybridisation. Electron gain enthalpy is the measurable energy change (in kJ mol^-1 ) when one mole of isolated gaseous atoms each gains one electron to form gaseous anions, X(g)+e^-→ X^-(g). The two differ in the state of the atom (bonded vs. isolated), in what happens to the electron (partial pull on a shared pair vs. complete gain of an electron), and in whether the value is a fixed, unit-bearing quantity (electron gain enthalpy) or a relative, unit-less, context-dependent number (electronegativity).

Question 18

Explain with the help of suitable example polar covalent bond.

Accepted Answer

A covalent bond is polar when the two bonded atoms have different electronegativities, so the shared pair of electrons is pulled closer to the more electronegative atom — that atom ends up with a partial negative charge (δ^-) and the other atom with an equal partial positive charge (δ^+). Example: the H–Cl bond in hydrogen chloride Electronegativity (Pauling scale) is a measure of how strongly an atom in a bond attracts the shared electron pair toward itself: χ( H ) = 2.1, χ( Cl ) = 3.0 Because χ( Cl ) χ( H ), chlorine attracts the bonding electron pair more strongly than hydrogen does. The pair is not withdrawn all the way to chlorine — that would make it an ionic bond — but it does spend more time closer to Cl than to H. This unequal, but still shared, distribution is written as: δ^+ H - δ^- Cl Here δ^+ and δ^- mean "a small fraction of a full unit charge," not a whole electron transferred — this is the exact step students blur with ionic bonding. In an ionic bond (like NaCl) an electron is essentially fully transferred, giving whole charges Na^+ and Cl^-. In a polar covalent bond (like HCl) the electron pair is still shared — it just isn't shared equally. Why this is called "polar": the separation of a small positive charge and a small negative charge, held apart by the bond length, creates a permanent electric dipole. Its size is the dipole moment, μ = Q × d where Q is the magnitude of the partial charge separated and d is the distance between the charge centres (essentially the bond length). For HCl, μ ≈ 1.03 D (debye), confirming the bond is polar. A truly nonpolar covalent bond — say, H–H or Cl–Cl, where both atoms are identical and so have identical electronegativity — has μ = 0 because the bonding pair is shared exactly equally and no charge separation exists at all. The size of the polarity also scales with how different the electronegativities are: H–F ( Δχ large) is more polar than H–Cl, which is more polar than H–Br, which is more polar than H–I, tracking the falling electronegativity of the halogen down the group. Answer: A polar covalent bond is one in which the shared electron pair is displaced toward the more electronegative atom, giving that atom a partial negative charge (δ^-) and the other atom an equal partial positive charge (δ^+), without any electron being fully transferred. HCl is the standard example: since χ( Cl ) = 3.0 χ( H ) = 2.1, the bond is written δ^+ H - δ^- Cl, and this charge separation gives HCl a measurable dipole moment ( μ ≈ 1.03 D ), unlike a nonpolar bond such as H–H where both atoms share the pair equally and μ = 0.

Question 19

Arrange the bonds in order of increasing ionic character in the molecules: LiF, K 2O, N 2, SO 2 and ClF3.

Accepted Answer

Ionic character comes from unequal sharing of the bonding electron pair — the bigger the difference in electronegativity between the two bonded atoms, the more that pair sits on one atom, and the more "ionic" the bond looks. Electronegativity, χ, measures how strongly an atom pulls the shared pair of a bond towards itself (Pauling scale). For a bond between atoms A and B, the electronegativity difference is Δχ = χ A - χ B Δχ = 0 means the pair is shared perfectly equally — a purely covalent bond with zero ionic character. As Δχ grows, the more electronegative atom hogs the pair, a partial negative charge builds on it and a partial positive charge on the other, and the bond's ionic character rises. So ranking these five bonds by increasing ionic character is the same as ranking them by increasing Δχ. Using Pauling's electronegativity values for the atoms involved: χ Li=0.98, χ F=3.98, χ K=0.82, χ O=3.44, χ N=3.04, χ S=2.58, χ Cl=3.16 Now take each bond in turn. LiF (Li–F bond): Δχ = 3.98 - 0.98 = 3.00 K₂O (K–O bond): Δχ = 3.44 - 0.82 = 2.62 N₂ (N–N bond): the two atoms are identical, so there is no pull-imbalance at all — Δχ = 3.04 - 3.04 = 0 This is the case people get wrong: N₂ isn't "a little ionic," it is the one bond here with genuinely zero ionic character, because a bond between two atoms of the same element cannot be polarised in either direction. SO₂ (S–O bond): Δχ = 3.44 - 2.58 = 0.86 ClF₃ (Cl–F bond): Δχ = 3.98 - 3.16 = 0.82 The step that trips people up here is reaching for the rounded, whole-number electronegativities many textbooks print for quick use (Li = 1.0, F = 4.0, K = 0.8, O = 3.5, S = 2.5, Cl = 3.0, N = 3.0). Those round off SO₂ and ClF�3 to the same difference, Δχ = 1.0, and make it look like a tie. Using Pauling's actual decimal values, as above, the tie breaks: the Cl–F gap (0.82) is very slightly smaller than the S–O gap (0.86), because chlorine (3.16) sits a bit closer to fluorine (3.98) than sulfur (2.58) sits to oxygen (3.44). Putting the five Δχ values in increasing order — 0 (N 2) < 0.82 (ClF 3) < 0.86 (SO 2) < 2.62 (K 2O) < 3.00 (LiF) — gives the same order for increasing ionic character, since ionic character rises monotonically with Δχ. Answer: N₂ < ClF₃ < SO₂ < K₂O < LiF, in order of increasing ionic character — N₂ is purely covalent (Δχ = 0, identical atoms), while LiF is the most ionic (Δχ = 3.00, the largest electronegativity gap of the five).

Question 20

The skeletal structure of CH 3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Accepted Answer

A skeleton only tells you which atoms touch which — it never tells you which of those bonds are single, double, or triple. That is fixed by two hard rules: total valence electrons must be conserved, and every atom except H must end up with 8 electrons around it (the octet rule), while carbon specifically must always show exactly four bonds. Whatever bonds the sketch you were shown has wrong, working from those two rules rebuilds the one correct structure — there is only one, because acetic acid has no resonance or charge to argue about. Step 1 — count the total valence electrons. Acetic acid is CH 3 COOH, i.e. C 2 H 4 O 2: 2 carbons, 4 hydrogens, 2 oxygens. Each atom contributes electrons equal to its group number in the old (A) notation — carbon (Group 14) gives 4, hydrogen (Group 1) gives 1, oxygen (Group 16) gives 6. Total valence electrons = 2(4) + 4(1) + 2(6) = 8 + 4 + 12 = 24 electrons = 12 electron pairs Step 2 — fix the skeleton, then ask who is short a bond. Call the two carbons C 1 (the one carrying the three H's) and C 2. The connectivity that any correct sketch of acetic acid must show is: C 1 bonded to 3 H's and to C 2; C 2 bonded to two different oxygens, O a and O b; and O b further bonded to the fourth H (the acidic one). This is fixed — nothing here is in question. Now check each carbon assuming, for a moment, that every bond drawn is a single bond: - C 1: 3 (C–H) + 1 (C– C 1 – C 2 ) = 4 bonds, 8 electrons around it. This is a complete octet and exactly carbon's required four bonds — C 1 needs nothing changed. - C 2: 1 (C– C 1 ) + 1 (C– O a ) + 1 (C– O b ) = only 3 bonds, 6 electrons around it. Carbon is never left with three bonds and an incomplete octet — this is exactly the kind of error a wrong sketch makes, drawing every bond around the carbonyl carbon as single. C 2 is one bonding pair short of an octet. This is the step most mis-drawn sketches get wrong: they draw both C–O bonds as plain single lines, which quietly leaves the carbonyl carbon with only six electrons around it. Carbon cannot stop at three bonds; the fourth bond has to come from somewhere. Step 3 — the missing bond must be a C=O double bond, and it can only go on one of the two oxygens. C 2 cannot reach any new atom (the skeleton is fixed), so its fourth bond must be a second bond to an atom it already touches — one of the two C–O single bonds has to become a double bond. It cannot be the C 2 – O b bond, because O b already has two bonds (to C 2 and to H); oxygen stops at two bonds and two lone pairs, so pushing a third bond onto O b would over-fill its shell past eight electrons — this is the other common drawing error, putting the double bond on the hydroxyl oxygen instead of the carbonyl one. The double bond can only sit on O a, the oxygen attached to nothing but C 2: C 2: 1( C– C 1) + 2( C= O a) + 1( C– O b) = 4 bonds, 8 electrons — a full octet, four bonds, exactly as carbon requires. Step 4 — give both oxygens their lone pairs so each reaches an octet. - O a (the carbonyl oxygen, double-bonded to C 2 only): the double bond contributes 4 electrons; it needs 4 more to reach 8, so O a carries 2 lone pairs. - O b (the hydroxyl oxygen, singly bonded to C 2 and to H): its two single bonds contribute 4 electrons; it needs 4 more, so O b also carries 2 lone pairs. Step 5 — check the electron count balances, and that no atom needs a formal charge. Bonding pairs in the finished structure: 3 ( C 1 –H) + 1 ( C 1 – C 2 ) + 2 ( C 2 = O a ) + 1 ( C 2 – O b ) + 1 ( O b –H) = 8 pairs. Lone pairs: 2 on O a + 2 on O b = 4 pairs. Total = 8 + 4 = 12 pairs — exactly the 24 electrons counted in Step 1, with nothing left over and nothing missing. Formal charge, using FC = V - N - 1/2B ( V = valence electrons of the free atom, N = its non-bonding electrons, B = its bonding electrons), comes out to zero on every atom: - C 1: 4 - 0 - 1/2(8) = 0 - C 2: 4 - 0 - 1/2(8) = 0 - O a: 6 - 4 - 1/2(4) = 0 - O b: 6 - 4 - 1/2(4) = 0 - each H: 1 - 0 - 1/2(2) = 0 All-zero formal charges on a structure that already satisfies every octet is the signature of the correct, most stable Lewis structure — there is no better arrangement to search for. The correct Lewis structure, drawn out in full: with two lone pairs sitting on the doubly-bonded oxygen (the carbonyl O) and two lone pairs on the oxygen that also carries the H (the hydroxyl O). In condensed form this is CH 3 -CO-OH, with a C–C single bond, a C=O double bond to one oxygen, and single bonds C–O–H to the other — never two single C–O bonds, and never the double bond on the oxygen that already holds the acidic hydrogen. Answer: The correct Lewis structure has C 1 (methyl carbon) singly bonded to three H atoms and to C 2; C 2 doubly bonded to one oxygen (the carbonyl O, carrying 2 lone pairs) and singly bonded to the other oxygen (the hydroxyl O, carrying 2 lone pairs), which is in turn singly bonded to the fourth H. Every carbon shows exactly four bonds and every oxygen ends with an octet (2 bonds + 2 lone pairs, or 1 double bond + 2 lone pairs); a structure with both C–O bonds single, or with the double bond on the O–H oxygen instead, is the kind of error the sketch is testing for.

Question 21

Apart from tetrahedral geometry, another possible geometry for CH 4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH 4 is not square planar?

Accepted Answer

A geometry claim about CH 4 can only be tested by looking at what it predicts for a molecule derived from CH 4^- and here square planar and tetrahedral geometry make different, checkable predictions. Notice first that this cannot be settled by looking at CH 4 itself. All four atoms attached to carbon are identical (H), so both candidate shapes are highly symmetric: - Tetrahedral CH 4 has the point-group symmetry T d. - Square planar CH 4 (four H at the corners of a square, C at the centre) has the symmetry D 4h. In either case, the four identical C–H bond dipoles are arranged symmetrically enough that their vector sum is zero. So "does CH 4 have a dipole moment?" gives the same answer (no) whichever shape is correct — it cannot distinguish them. The aside people miss: you have to break the symmetry by substitution before dipole moment becomes a useful test. Break the symmetry: replace two H atoms by two Cl atoms, giving CH 2Cl 2. Ask how many different ways there are to place 2 Cl and 2 H on each candidate shape. If CH 4 were square planar (H at the four corners of a square), the two Cl atoms could occupy: - Adjacent corners — the Cl-C-Cl angle is 90^°. Call this the cis form. - Diagonally opposite corners — the Cl-C-Cl angle is 180^°. Call this the trans form. These are genuinely different molecules, not the same structure viewed two ways, because a square has two geometrically distinct pairs of corners (adjacent vs. diagonal). So square planar geometry predicts two isomers of CH 2Cl 2, with different properties. In particular, for the diagonal ( trans ) form the two C-Cl bond-dipole vectors point in exactly opposite directions and cancel: μ net = μ C-Cl,1 + μ C-Cl,2 = 0 ( trans, ∠ Cl-C-Cl = 180^°) while for the adjacent ( cis ) form the two dipoles are at 90^° to each other and do not cancel, giving a nonzero resultant. So square planar geometry predicts one isomer with zero dipole moment and one isomer with a nonzero dipole moment. If CH 4 is tetrahedral, look at the four corners of a regular tetrahedron. Every pair of corners is related to every other pair by a symmetry operation of the tetrahedron — a regular tetrahedron has no "adjacent" versus "diagonal" corners the way a square does; all six edges (and hence all 4 2 =6 ways of picking 2 vertices out of 4) are equivalent. So however you choose 2 of the 4 H positions to replace with Cl, you get the same single structure. Tetrahedral geometry predicts exactly one isomer of CH 2Cl 2, with a Cl-C-Cl angle close to 109.5^°. Since 109.5^° ≠ 180^°, the two C–Cl dipoles do not cancel, so this one isomer must have a nonzero net dipole moment. Compare with experiment. Only one compound called dichloromethane ( CH 2Cl 2 ) is known — chemists have never isolated two distinct isomers with different melting points, boiling points, or spectra. And that single compound has a measured, nonzero dipole moment ( μ ≈ 1.60 D ). This matches the tetrahedral prediction exactly: one isomer, nonzero dipole moment. It flatly contradicts the square planar prediction, which requires two isomers to exist, one of them (the trans form) with zero dipole moment. Since only one form is ever found, and it is polar, the square planar hypothesis is falsified by this evidence, while the tetrahedral hypothesis survives it. This same reasoning is why chemists trust sp^3 tetrahedral carbon generally: replacing any two of methane's four equivalent hydrogens must give a single, unique, chemically identical product, and every disubstituted methane derivative ever isolated confirms exactly that pattern. Answer: CH 4 is tetrahedral and not square planar because the two geometries make different, testable predictions about the disubstituted derivative CH 2Cl 2: square planar geometry requires two isomers (a cis form with Cl-C-Cl=90^° and nonzero dipole moment, and a trans form with Cl-C-Cl=180^° and zero dipole moment), whereas tetrahedral geometry — since all pairs of vertices of a regular tetrahedron are equivalent — predicts only one isomer, necessarily with a nonzero dipole moment (as 109.5^° ≠ 180^° ). Experimentally, only one form of CH 2Cl 2 exists and it does have a nonzero dipole moment ( ≈ 1.60 D ), which matches the tetrahedral prediction and rules out the square planar one.

Question 22

Explain why BeH 2 molecule has a zero dipole moment although the Be–H bonds are polar.

Accepted Answer

A molecule's net dipole moment is the vector sum of its individual bond dipoles, not just a list of them — direction matters as much as size. Each Be–H bond in BeH 2 is polar because beryllium and hydrogen have different electronegativities, so each bond carries its own small dipole moment, with the negative end pointing toward the more electronegative atom (H is more electronegative than Be, so each dipole points from Be towards H). Step 1: Find the shape of the molecule. Beryllium has the configuration 1s^2 2s^2. To form two bonds it promotes one 2s electron to 2p, giving 2s^1 2p^1, and these two orbitals mix to form two equivalent sp hybrid orbitals. Two sp hybrid orbitals always point in exactly opposite directions, 180^° apart, because that is the arrangement that keeps them as far apart as possible. So BeH 2 is a linear molecule: H-Be-H with a bond angle of 180^°. This is the step students skip: you cannot decide whether a molecule is polar just by looking at whether its bonds are polar — you must first fix the geometry, because the dipole moment depends on the angle between the bonds. Step 2: Add the two bond-dipole vectors. Let each Be–H bond dipole have magnitude μ, directed from Be towards each H atom. Since the two Be–H bonds point in exactly opposite directions ( 180^° apart), the two dipole vectors are equal in magnitude and opposite in sense along the same line: μ 1 = - μ 2 The resultant dipole moment of the molecule is the vector sum: μ net = μ 1 + μ 2 = μ 1 + (- μ 1) = 0 So the two bond dipoles are equal in magnitude and opposite in direction, and they cancel each other exactly. Why this matters: a "polar bond" is a property of one bond in isolation (it depends only on the electronegativity difference between the two bonded atoms). A "polar molecule" is a property of the whole molecule and depends on the geometry too. BeH 2 has polar bonds but a symmetric linear geometry, so the bond dipoles are positioned to cancel exactly, making the molecule as a whole nonpolar — its net dipole moment is zero. Answer: BeH 2 is linear (Be is sp hybridized, bond angle 180^° ), so its two equal and polar Be–H bond dipoles point in exactly opposite directions along the same line. Being equal in magnitude and opposite in direction, they cancel each other vectorially, giving the molecule a net dipole moment of zero even though each individual Be–H bond is polar.

Question 23

Which out of NH 3 and NF 3 has higher dipole moment and why?

Accepted Answer

The net dipole moment of a pyramidal AX₃E molecule is the vector sum of the three bond dipoles AND the dipole due to the lone pair — and in NH₃ these two contributions add, while in NF₃ they oppose each other. That is why NH₃ (1.47 D) beats NF₃ (0.24 D), even though fluorine is the far more electronegative atom. Setting up the geometry Both NH₃ and NF₃ have nitrogen as sp³ hybridised, with three bond pairs and one lone pair, giving a pyramidal shape (point group C 3v ). By this symmetry, whatever net dipole the molecule has must lie along the C 3 axis — the axis running through N and the lone pair, straight through the centre of the base formed by the three H (or F) atoms. So the total dipole moment is μ net = μ bonds + μ lone pair where μ bonds is the vector sum of the three individual N–H (or N–F) bond dipoles, and μ lone pair is the dipole contributed by the lone pair sitting on nitrogen. A dipole moment vector is drawn pointing from the less electronegative (partially positive, δ^+ ) atom toward the more electronegative (partially negative, δ^- ) atom. The lone pair sits in a hybrid orbital that points away from the three bonding pairs, along the C 3 axis, out of the top of the pyramid (with N at the apex). It is a concentration of electron density with no nucleus at the far end to pull it back, so it acts like a dipole pointing outward from N, away from the three bonds — and this direction is fixed by the geometry alone; it does not depend on whether the atoms at the base are H or F. Case 1: NH₃ Nitrogen (electronegativity ≈ 3.0) is more electronegative than hydrogen (≈ 2.1). So each N–H bond dipole points from H toward N — that is, from the base of the pyramid up toward the apex, the same general direction as the lone pair's dipole. Here μ bonds and μ lone pair point the same way along the C 3 axis, so they reinforce: μ NH 3 = μ bonds + μ lone pair (magnitudes add) This gives NH₃ a comparatively large observed dipole moment, μ NH 3 = 1.47 D. Case 2: NF₃ Now the electronegativity order is reversed: fluorine (≈ 4.0) is more electronegative than nitrogen (≈ 3.0). So each N–F bond dipole points from N toward F — from the apex down toward the base — which is the opposite direction to the lone pair's dipole (still pointing outward from N, away from the base, because that direction is fixed by geometry, not electronegativity). This is the step people miss: they assume "F is more electronegative, so NF₃ should have the bigger dipole moment." It is exactly the opposite, because the lone pair's contribution does not flip with the substituent — only the bond dipoles do, and in NF₃ they now fight the lone pair instead of helping it. Here μ bonds and μ lone pair point in opposite directions along the C 3 axis, so they partly cancel: μ NF 3 = μ lone pair - μ bonds (magnitudes subtract) Even though each individual N–F bond is more polar than an N–H bond, this cancellation leaves NF₃ with a small net dipole moment, μ NF 3 = 0.24 D. Answer: NH₃ has the higher dipole moment (1.47 D vs. 0.24 D for NF₃). In both molecules the lone pair on N contributes a dipole pointing away from the three bonds along the pyramidal axis. In NH₃, N is more electronegative than H, so the N–H bond dipoles point toward N — the same direction as the lone pair — and they add. In NF₃, F is more electronegative than N, so the N–F bond dipoles point toward F — opposite to the lone pair — and they partly cancel. That cancellation, not the individual bond polarities, is why NF₃ ends up with the smaller net dipole moment.

Question 24

What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.

Accepted Answer

Hybridisation is atomic orbitals of an atom mixing together and re‑emerging as a new set of orbitals — same number, same total energy, but now all identical in shape and energy, and pointing in specific directions in space. An atom's pure orbitals (one s, three p ) do not, by themselves, explain the bond angles seen in real molecules. Take methane, CH 4: carbon's ground-state configuration is 1s^2 2s^2 2p x^1 2p y^1, which offers only two half-filled orbitals — not enough to form four bonds, and nowhere near the observed 109.5^° angle if you tried to use unmixed s and p orbitals directly. Hybridisation resolves this: one electron is promoted from 2s to the empty 2p z, giving four half-filled orbitals, and then these four orbitals — one s and three p — blend into four new, identical orbitals called hybrid orbitals. Each carries a fixed share of s and p character, and this share is what fixes the geometry. Rules that govern hybridisation (why the shapes come out fixed, not arbitrary): - The number of hybrid orbitals formed always equals the number of pure orbitals mixed. Mix 2, get 2; mix 4, get 4 — none are lost or created. - The hybrid orbitals are all identical in energy and shape (this is the point of mixing — the original s and p orbitals were not identical). - Hybrid orbitals are more effective at forming stable bonds than pure orbitals: each hybrid orbital has one large lobe and one small lobe (unlike a symmetric pure p orbital), and the large lobe overlaps another atom's orbital more completely, giving a stronger bond. - Hybrid orbitals arrange themselves in space so as to keep as far apart as possible — this is what fixes the bond angle. Everything below follows from this one rule. The type of hybridisation is named after which and how many orbitals were mixed, and the aside people miss is this: the percentage of s -character in a hybrid orbital, not the label "sp/sp²/sp³" by itself, is what determines the bond angle — more s -character pulls the electron density closer to the nucleus and pushes the hybrid orbitals farther apart. sp hybridisation (1 s + 1 p → 2 sp orbitals): linear, 180^°. One s orbital mixes with one p orbital. Two sp orbitals result, each with 50 % s -character and 50 % p -character. To stay as far apart as possible, two orbitals can only point in exactly opposite directions along one axis, so the shape is a straight line: sp orbitals: 180^° apart, linear Example: in BeCl 2, beryllium's two sp hybrid orbitals overlap with chlorine orbitals on either side, giving a linear Cl - Be - Cl molecule. The same hybridisation gives the linear H-C ≡ C-H skeleton in acetylene. sp² hybridisation (1 s + 2 p → 3 sp² orbitals): trigonal planar, 120^°. One s orbital mixes with two p orbitals. Three equivalent sp² orbitals result, each with 33.3 % s -character and 66.7 % p -character (less s -character than sp, so a smaller angle than 180^° ). Three orbitals maximise their mutual separation by lying in one plane, pointing to the corners of an equilateral triangle: sp^2 orbitals: 120^° apart, trigonal planar Example: boron in BCl 3 is sp² hybridised, giving a flat, triangular molecule with all three Cl-B-Cl angles equal to 120^°. The same hybridisation gives the flat C 2H 4 (ethylene) skeleton around each carbon. sp³ hybridisation (1 s + 3 p → 4 sp³ orbitals): tetrahedral, 109.5^°. One s orbital mixes with all three p orbitals. Four equivalent sp³ orbitals result, each with 25 % s -character and 75 % p -character — the least s -character of the three, hence the largest departure from a small angle and the characteristic tetrahedral value. Four orbitals maximise their separation only by pointing to the four corners of a regular tetrahedron, with the central atom at its centre: sp^3 orbitals: 109^° 28' (≈ 109.5^°) apart, tetrahedral Example: carbon in CH 4 is sp³ hybridised; all four H-C-H angles are 109.5^°, reproducing methane's familiar tetrahedral shape. The place students slip is treating hybridisation as something that happens to isolated atoms for its own sake — it is really a bookkeeping device that predicts the geometry an atom's bonds will take once you know how many hybrid orbitals it forms and how many pure p orbitals (versus s ) went into each one. Answer: Hybridisation is the mixing of atomic orbitals of comparable energy on one atom to give an equal number of new, identical hybrid orbitals with fixed directional geometry. sp hybrid orbitals (from 1 s + 1 p) are linear, 180° apart; sp² hybrid orbitals (from 1 s + 2 p) are trigonal planar, 120° apart; sp³ hybrid orbitals (from 1 s + 3 p) are tetrahedral, 109.5° apart — each shape following from the orbitals arranging themselves as far apart as possible in space.

Question 25

Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl + Cl − → AlCl − 3 4

Accepted Answer

A central atom's hybridisation is fixed by how many sigma-bond pairs (plus lone pairs) sit around it — count those first, on each side of the reaction. The reaction is AlCl 3 + Cl^- → AlCl 4^- Step 1: Look at Al in AlCl3. Aluminium has 3 valence electrons (3s^2 3p^1). Each Cl atom takes one of these in a shared pair, so Al forms exactly 3 Al–Cl sigma bonds and is left with no lone pair of its own. Three bond pairs and zero lone pairs around the central atom means 3 electron domains, arranged in a plane at 120^° — a trigonal planar shape. That geometry is built from sp2 hybrid orbitals (one s + two p orbitals mixed to give 3 equivalent orbitals). So in AlCl 3, Al is sp2 hybridised, and because it has an empty, unhybridised p orbital left over, it is electron-deficient — that empty orbital is exactly why AlCl 3 is a strong Lewis acid. Step 2: Look at what the incoming Cl^- does. A chloride ion carries a lone pair it is not using in any bond. This lone pair is donated into that empty p orbital on Al — a coordinate (dative) bond. Al does not lose or gain a proton or rearrange its own electrons; it simply accepts a new electron pair. Al now has 4 shared pairs around it (the original 3 Al–Cl sigma bonds plus the new Al←Cl coordinate bond), still with no lone pair. Four electron domains with no lone pair means a tetrahedral shape at 109.5^° angles, which requires mixing one s and three p orbitals — sp3 hybridisation. A step people skip here: don't try to explain the change by looking at oxidation state or charge — Al's oxidation state stays +3 throughout. The hybridisation change is purely a count of electron domains going from 3 to 4, nothing else. Step 3: State the change. Al (sp^2, trigonal planar, in AlCl 3) + Cl^- Al (sp^3, tetrahedral, in AlCl 4^-) The extra electron pair donated by Cl^- forces one more hybrid orbital to be made available, converting the vacant p orbital's role into a fourth sp^3 hybrid orbital that houses the new bond pair. Answer: Yes, the hybridisation of Al changes. In AlCl 3, Al is sp2 hybridised (3 bond pairs, trigonal planar, with an empty p orbital that makes it electron-deficient). When Cl^- donates its lone pair into that empty orbital to form AlCl 4 −, Al becomes sp3 hybridised (4 bond pairs, tetrahedral).

Question 26

Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF 3 + NH 3 → F 3B.NH 3

Accepted Answer

Hybridisation is fixed by how many electron pairs sit around the atom, not by how many bonds it started with — so you have to recount the pairs on B and on N after the reaction, not just look at the reactants. The reaction referred to is boron trifluoride accepting the lone pair of ammonia to form a coordinate (dative) adduct: BF 3 + NH 3 → F 3 B ← NH 3 Here the arrow ← stands for a coordinate bond: the shared pair of electrons is supplied entirely by nitrogen, not by boron. Step 1 — hybridisation of B and N before the reaction. The hybridisation of a central atom is decided by its steric number, steric number = ( number of σ bonds ) + ( number of lone pairs ), with steric number 2, 3, 4 corresponding to sp, sp^2, sp^3. - In BF 3, boron forms 3 σ bonds (to the three F atoms) and carries no lone pair (boron has only 3 valence electrons, so it is left with an empty 2p orbital and an incomplete octet — this is exactly what makes BF 3 a Lewis acid). Steric number =3+0=3, so boron is sp^2 hybridised, and BF 3 is trigonal planar with 120^° F–B–F angles. - In NH 3, nitrogen forms 3 σ bonds (to the three H atoms) and carries 1 lone pair. Steric number =3+1=4, so nitrogen is sp^3 hybridised, and NH 3 is pyramidal with bond angles close to 107^° (compressed from the ideal 109.5^° by lone-pair–bond-pair repulsion). Step 2 — what the coordinate bond does to each atom's count. Nitrogen's lone pair, which is already sitting in one of its sp^3 orbitals, is donated straight into boron's empty 2p orbital. No bonds break and no new orbitals need to be built on nitrogen — a pair that was non-bonding simply becomes bonding. On boron, though, a brand-new fourth electron pair has appeared where there was none before. A common mistake here is to assume that because nitrogen is the one donating, it must be nitrogen whose hybridisation changes. It is the opposite: donating a lone pair that already occupies a hybrid orbital changes nothing about that atom's orbital count; it is receiving a new pair into a previously empty orbital that forces a re-hybridisation. Step 3 — recount the steric numbers in the product F 3 B ← NH 3. - Boron now has 3 σ bonds to F plus the new coordinate bond from N: steric number =4+0=4, so boron must go from sp^2 to sp^3. Its geometry changes correspondingly from trigonal planar to (distorted) tetrahedral. - Nitrogen still has exactly 4 electron pairs around it — 3 N–H bond pairs plus the pair it now shares with boron (previously a lone pair, now a bond pair). Steric number =4+0=4, same as before, so nitrogen stays sp^3; only the character of that fourth pair (lone → bonding) has changed, not the hybridisation itself. So the change is one-sided: boron is forced to rehybridise to accommodate the new bond it accepts, while nitrogen's hybridisation was already set up to supply that bond without any change to itself. Answer: Yes, but only for boron. In BF 3, boron is sp^2 hybridised (steric number 3, trigonal planar); on forming the coordinate bond with NH 3 it becomes sp^3 hybridised (steric number 4, tetrahedral). Nitrogen remains sp^3 hybridised throughout — it already had steric number 4 in NH 3, and donating its lone pair as the new N→B bond does not change that count, so its hybridisation is unchanged.

Question 27

Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C 2H 4 and C 2H 2 molecules.

Accepted Answer

A double bond is one sigma (σ) bond plus one pi (π) bond, and a triple bond is one sigma bond plus two pi bonds — the σ bond comes from head-on overlap of hybrid orbitals, and each π bond comes from sideways overlap of unhybridized p orbitals. To "draw" this, you have to first fix the hybridization at each carbon, because that tells you how many p orbitals are left over to form π bonds. Ethylene, C 2 H 4 — the double bond Each carbon here is sp^2 hybridized. One s orbital and two p orbitals mix to give three sp^2 hybrid orbitals, lying in one plane at 120^° to each other. The third p orbital of carbon (say 2p z ) is left unhybridized, standing perpendicular to that plane. Picture one carbon with its three sp^2 lobes pointing into a flat triangle, and a single unhybridized p z orbital sticking up and down through the centre of that triangle (a dumb-bell perpendicular to the plane of the three lobes). The second carbon is built the same way. Now bring the two carbons together: - Two of the three sp^2 orbitals on each carbon overlap head-on with the 1s orbital of a hydrogen atom, giving four C–H σ bonds. - The remaining sp^2 orbital on each carbon points at the other carbon and overlaps head-on with it, along the internuclear axis — this end-to-end overlap is the C–C σ bond. (All the atoms so far — 2 C and 4 H — lie in one plane, because sp^2 orbitals are planar.) - The two leftover p z orbitals, one on each carbon, are now parallel to each other and both perpendicular to the molecular plane. They overlap sideways — lobe above the plane overlapping lobe above the plane, lobe below overlapping lobe below — to give one π bond, pictured as two banana-shaped electron clouds, one above and one below the plane of the six atoms. So the C=C double bond in ethylene is: 1 σ bond (head-on sp^2 – sp^2 overlap, along the C–C axis) + 1 π bond (sideways p z – p z overlap, above and below the plane). A step people skip: the π overlap is only possible because the two p z orbitals stay parallel — this is exactly why there is no free rotation about a C=C bond (rotating one CH₂ end would break the sideways overlap). Acetylene, C 2 H 2 — the triple bond Each carbon here is sp hybridized. One s orbital and one p orbital mix to give two sp hybrid orbitals, oriented at 180^° to each other (a straight line). That leaves two unhybridized p orbitals on each carbon — say 2p y and 2p z — mutually perpendicular to each other and both perpendicular to the sp axis. Picture one carbon with its two sp lobes pointing left and right along a straight line, and two p-orbital dumb-bells, one vertical and one horizontal, both centred on the carbon and both at right angles to that line (like a plus-sign cross-section sitting on the axis). Bringing the two carbons together, with hydrogens on the outside: - One sp orbital on each carbon overlaps with a hydrogen 1s orbital, giving two C–H σ bonds, one at each end. - The other sp orbital on each carbon points at the other carbon and overlaps head-on with it — this is the C–C σ bond, and it keeps H–C–C–H perfectly linear. - On carbon 1, the two unhybridized p orbitals (say p y and p z ) are parallel to the identically labelled p y and p z orbitals on carbon 2 (all four are perpendicular to the C–C axis, in two mutually perpendicular planes). Each parallel pair overlaps sideways: p y – p y gives one π bond (electron clouds above and below, say, a horizontal plane), and p z – p z gives a second π bond (electron clouds in front and behind, in a vertical plane perpendicular to the first). So the C≡C triple bond in acetylene is: 1 σ bond (head-on sp – sp overlap) + 2 π bonds (two sideways p – p overlaps in mutually perpendicular planes). The combined π electron density from these two π bonds forms a cylindrical sheath wrapped around the C–C σ bond, which is why the whole HC ≡ CH molecule is linear and symmetric about that axis. The pattern to remember: hybridize the σ-framework first ( sp^2 for a double bond, sp for a triple bond), let whatever p orbitals are left over sit perpendicular to that framework, and every extra bond beyond the first σ bond is one more sideways p – p overlap. Answer: In C 2 H 4, each carbon is sp^2 hybridized; the C=C double bond is one σ bond from head-on sp^2 – sp^2 overlap plus one π bond from sideways overlap of the unhybridized p z orbitals (above and below the planar molecule). In C 2 H 2, each carbon is sp hybridized; the C≡C triple bond is one σ bond from head-on sp – sp overlap plus two π bonds from sideways overlap of the two pairs of unhybridized p orbitals lying in mutually perpendicular planes, giving the linear H–C≡C–H molecule a cylindrical π-electron sheath around the σ bond.

Question 28

What is the total number of sigma and pi bonds in the following molecules? (a) C 2H 2 (b) C 2H 4

Accepted Answer

A single bond between two atoms is always one sigma (σ) bond; a double bond is one σ plus one pi (π); a triple bond is one σ plus two π. Every other bond in the molecule — to hydrogen, or from one multiply-bonded carbon further along the chain — is a plain σ bond formed by head-on (axial) overlap. A π bond, by contrast, comes only from the sideways overlap of unhybridized p-orbitals left over once a carbon has hybridized. So to count bonds correctly, first work out how each carbon is hybridized, because that tells you how many p-orbitals are left over to form π bonds. (a) C 2H 2 — ethyne (acetylene), H-C≡ C-H Each carbon here is bonded to only two other atoms (one H and one C), and the molecule is linear, so each carbon is sp hybridized. Two sp hybrid orbitals point in opposite directions along the molecular axis on each carbon; the two p-orbitals not used in hybridization (say p y and p z ) stay unhybridized on each carbon. Building the bonds: - Each sp orbital on carbon overlaps head-on with the 1s orbital of an H atom → 2 C-H σ bonds. - One remaining sp orbital on each carbon overlaps head-on with the corresponding sp orbital on the other carbon → 1 C-C σ bond. - The unhybridized p y orbitals on the two carbons overlap sideways → 1 π bond. The unhybridized p z orbitals overlap sideways in the perpendicular plane → 1 more π bond. So the triple bond between the carbons is itself 1 σ + 2 π, on top of the two C–H sigma bonds: σ bonds = 2 ( C-H ) + 1 ( C-C ) = 3, π bonds = 2 Where people slip: it is tempting to say a triple bond is "three bonds" and stop there — but only one of those three is σ; the other two are π, and they must be counted separately because they come from a different kind of overlap. (b) C 2H 4 — ethene (ethylene), H 2C=CH 2 Each carbon here is bonded to three other atoms (two H and one C), and the geometry around each carbon is trigonal planar, so each carbon is sp2 hybridized. Three sp2 orbitals lie in a plane on each carbon; one p-orbital (perpendicular to that plane) is left unhybridized on each carbon. Building the bonds: - Two sp2 orbitals on each carbon overlap head-on with 1s orbitals of H atoms → 2 × 2 = 4 C-H σ bonds. - The remaining sp2 orbital on each carbon overlaps head-on with the corresponding sp2 orbital on the other carbon → 1 C-C σ bond. - The unhybridized p-orbitals on the two carbons, both perpendicular to the molecular plane, overlap sideways → 1 π bond. σ bonds = 4 ( C-H ) + 1 ( C-C ) = 5, π bonds = 1 Where people slip: the double bond C=C is 1 σ + 1 π, not two σ bonds — writing "5 sigma, 1 pi" only comes out right if you first hybridize each carbon correctly (sp2, not sp3) and remember that only one p-orbital per carbon is left over to make the π bond. Answer: (a) C 2H 2: 3 sigma bonds and 2 pi bonds (total 5). (b) C 2H 4: 5 sigma bonds and 1 pi bond (total 6).

Question 29

Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s.

Accepted Answer

A sigma (σ) bond forms by head-on (end-to-end) overlap of orbitals along the line joining the two nuclei — the internuclear axis. If the orbitals overlap sideways instead, with their lobes lying parallel to each other above and below the axis, the bond is a pi (π) bond, not a sigma bond. So the test for each pair is: does the overlap happen along the x-axis, or across it? Go through each case, taking the internuclear axis as the x-axis. (a) 1s and 1s. An s orbital is spherically symmetric — it has no preferred direction, so it always presents a lobe pointing straight along whatever axis joins the two nuclei. Two 1s orbitals overlap head-on along the x-axis. ⇒ forms a σ bond. (b) 1s and 2p x. The 2p x orbital's own axis of orientation is the x-axis. Since the x-axis is also the internuclear axis here, the 2p x lobe points directly at the 1s orbital, and they overlap head-on. ⇒ forms a σ bond. (c) 2p y and 2p y. The 2p y orbital is oriented along the y-axis, which is perpendicular to the internuclear (x) axis. Two 2p y orbitals on neighbouring atoms therefore lie side by side, both pointing up-down along y while sitting apart along x. Their lobes overlap sideways (above and below the internuclear axis), not head-on. ⇒ this is exactly how a π bond is defined — lateral overlap of parallel p orbitals whose axes are perpendicular to the internuclear axis. This pair does NOT form a sigma bond; it would give a π bond instead. (d) 1s and 2s. Like 1s, the 2s orbital is spherically symmetric, so it also always overlaps head-on along whatever axis connects the two nuclei. ⇒ forms a σ bond. A common mix-up here is treating "any two p orbitals" as automatically sigma-forming. What matters is not that the orbitals are p-type, but whether each orbital's own lobe is directed along the internuclear axis ( p x is, when x is the internuclear axis) or perpendicular to it ( p y and p z are not). Answer: (c) 2p y and 2p y will not form a sigma bond, because a σ bond needs head-on overlap along the internuclear (x) axis, and the 2p y orbitals are oriented perpendicular to that axis — they can only overlap sideways, which gives a π bond, not a σ bond.

Question 30

Which hybrid orbitals are used by carbon atoms in the following molecules? CH 3 - CH 3; (b) CH 3 - CH = CH 2; (c) CH 3 - CH 2 - OH; (d) CH 3 - CHO (e) CH 3COOH

Accepted Answer

The hybridisation of a carbon atom is decided by how many other atoms are directly (sigma) bonded to it, not by how many total bonds (sigma + pi) it has. A carbon joined to four different atoms by four single (sigma) bonds is sp^3. A carbon that has one double bond (one sigma + one pi, to the same neighbour) has only three sigma-bonded neighbours, so it is sp^2. A carbon with a triple bond, or two double bonds, has only two sigma-bonded neighbours, so it is sp. The double or triple bond's pi part is built from the leftover, unhybridised p orbital(s) — it plays no role in fixing the hybridisation itself. So the working, for each molecule, is simply: draw every carbon, count how many atoms are sigma-bonded to it, and read off the hybridisation from that count. (a) CH 3-CH 3 (ethane) Each carbon is bonded to three H atoms and the other carbon — four sigma bonds, no multiple bond anywhere in the molecule. Both carbon atoms: sp^3. (b) CH 3-CH=CH 2 (propene) - C 1 (the CH 3 carbon): bonded to 3 H's + C 2, all single bonds → 4 sigma bonds → sp^3. - C 2 (the =CH- carbon): bonded to C 1, H, and doubly bonded to C3. That is 3 sigma-bonded neighbours (C1, H, C3) plus one pi bond to C 3 → sp^2. - C 3 (the =CH 2 carbon): bonded to 2 H's and doubly bonded to C 2 → 3 sigma-bonded neighbours (2 H + C2) plus one pi bond → sp^2. So: CH 3 carbon is sp^3; both carbons of the C=C double bond are sp^2. (c) CH 3-CH 2-OH (ethanol) - CH 3 carbon: bonded to 3 H's + the other carbon, all single bonds → sp^3. - CH 2 carbon: bonded to 2 H's, the CH 3 carbon, and the O of -OH, all single bonds → four sigma bonds → sp^3. Here's the aside worth flagging: the oxygen having lone pairs does not change the carbon's hybridisation — only what is sigma-bonded to the carbon itself counts. Both carbons are sp^3. (d) CH 3-CHO (acetaldehyde) - CH 3 carbon: 3 H's + the carbonyl carbon, all single bonds → sp^3. - CHO carbon: bonded to the CH 3 carbon, one H, and doubly bonded to O (the carbonyl C=O ). That is 3 sigma-bonded neighbours (C, H, O) plus one pi bond to O → sp^2. So: CH 3 carbon is sp^3; the carbonyl ( CHO ) carbon is sp^2. (e) CH 3COOH (acetic acid) - CH 3 carbon: 3 H's + the carboxyl carbon, all single bonds → sp^3. - COOH carbon: bonded to the CH 3 carbon, singly bonded to the -OH oxygen, and doubly bonded to the other O. That is 3 sigma-bonded neighbours (C, O of OH, O of C=O) plus one pi bond → sp^2. The step people miss here: the carboxyl carbon has two different oxygens attached (one by a single bond, one by a double bond), but it is still only one pi bond, so it stays 3 sigma-neighbours → sp^2, not something "extra" because there are two C–O connections. So: CH 3 carbon is sp^3; the COOH carbon is sp^2. Answer: (a) both C atoms sp^3; (b) the CH 3 carbon sp^3, the two double-bonded carbons ( CH= and =CH 2 ) both sp^2; (c) both C atoms sp^3; (d) the CH 3 carbon sp^3, the CHO (carbonyl) carbon sp^2; (e) the CH 3 carbon sp^3, the COOH carbon sp^2. In every case the rule is the same: a carbon with four single (sigma) bonds is sp^3; a carbon involved in one double bond (three sigma-bonded neighbours) is sp^2.

Question 31

What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.

Accepted Answer

A bond pair is a pair of electrons that is shared between two atoms and holds them together; a lone pair is a pair of valence electrons that belongs to just one atom and takes no part in bonding. Both ideas come straight out of the Lewis dot structure of a molecule, where every dot represents one valence electron and every pair of dots is either "shared" (a bond) or "unshared" (sitting on one atom only). Bond pair of electrons When two atoms come together to form a covalent bond, each contributes one electron, and the resulting pair of electrons is shared between the two nuclei. This shared pair is called a bond pair. It is usually drawn as a single line (—) between the two atoms in a structural formula, or as a pair of dots placed between them in a Lewis structure. Example: In the hydrogen molecule, H 2, each H atom brings one electron: H · + · H → H: H ( or H–H ) The pair of dots between the two H atoms is one bond pair. It is attracted by both nuclei at once, and that mutual attraction is exactly what holds the molecule together. Lone pair of electrons Not every valence electron pair on an atom is used for bonding. Some atoms have extra valence electrons left over after forming all the bonds they need to — these are grouped into pairs that stay on that one atom alone, without being shared with any other atom. Such a pair is called a lone pair (also called a non-bonding pair). A common mix-up: a lone pair still belongs to the atom's valence shell and still repels other electron pairs (this is the basis of VSEPR theory), even though it does not connect two atoms the way a bond pair does. Example: In ammonia, NH 3, nitrogen has 5 valence electrons. Three of these electrons pair up with the single electron of each of the three H atoms, giving three bond pairs (three N–H bonds). Nitrogen's 5 valence electrons distribute as 3 (used, one each, in the 3 N–H bond pairs) + 2 (left over as 1 pair on nitrogen). So the Lewis structure of NH 3 has - 3 bond pairs — one N–H bond pair for each of the three N–H bonds, and - 1 lone pair — sitting on the nitrogen atom, not shared with any H. That lone pair on nitrogen is why ammonia is pyramidal rather than perfectly symmetric in bond angle, and why nitrogen can donate this pair to form a fourth bond (as in NH 4^+ ) — it is available precisely because it was never shared in the first place. Answer: A bond pair is a pair of electrons shared between two bonded atoms — for example, the single shared pair in H–H, i.e. H 2. A lone pair is a pair of valence electrons that stays on one atom only and is not shared — for example, the one lone pair on the nitrogen atom in NH 3, in addition to its 3 N–H bond pairs.

Question 32

Distinguish between a sigma and a pi bond.

Accepted Answer

A sigma (σ) bond forms by head-on (coaxial) overlap of orbitals along the line joining the two nuclei; a pi (π) bond forms by sideways (lateral) overlap of orbitals lying parallel to each other and perpendicular to that line. Everything else that distinguishes them — shape of the electron cloud, strength, and whether the molecule can rotate freely about the bond — follows from this one difference in how the orbitals meet. 1. How each bond is formed Take the internuclear axis to be the straight line joining the two bonded nuclei (call it the z -axis). - A σ bond comes from orbitals that overlap end-to-end, along this axis: s - s, s - p z, p z - p z, or overlap of hybrid orbitals ( sp, sp^2, sp^3, …) that themselves point along the axis. The overlap region is centred symmetrically on the axis. - A π bond comes from two unhybridized p orbitals (say p x - p x or p y - p y ) on the two atoms that are oriented parallel to each other but perpendicular to the internuclear axis. They overlap sideways, above and below the axis, not along it. 2. Shape of the electron cloud - In a σ bond the electron density is concentrated symmetrically around the internuclear axis — if you look straight down the bond axis, the cross-section looks the same all the way round (cylindrical symmetry). This is why a single bond can be freely rotated: spinning one half of the molecule about the axis does not change the overlap at all. - In a π bond the electron density lies in two lobes, one above and one below the plane that contains the two nuclei, with a nodal plane (zero electron density) containing the internuclear axis itself. Because the overlapping lobes sit off to the sides rather than centred on the axis, rotating one atom about the bond axis would swing the lobes out of alignment and break the overlap — so a π bond prevents free rotation about that axis. (This is exactly why ethene, H 2 C = CH 2, is rigid and planar while ethane, H 3 C - CH 3, rotates freely about its C–C bond.) 3. Strength of overlap Head-on overlap (σ) is more complete/effective than sideways overlap (π) between the same pair of orbitals, so σ bonds are stronger than π bonds. A common slip is to assume a double bond, having two bonds, is simply "twice as strong" as a single bond — it is stronger, but not twice as strong, precisely because its second bond (the π bond) is the weaker overlap type. 4. How many of each kind, and whether a π bond can exist alone - A single bond between two atoms is always one σ bond. - A double bond is one σ + one π bond (e.g. C = C ). - A triple bond is one σ + two π bonds, the two π bonds using the two mutually perpendicular p orbitals not used for the σ bond (e.g. N ≡ N: one σ from head-on p z - p z overlap, plus π bonds from p x - p x and p y - p y sideways overlap). A σ bond can exist by itself between two atoms (as in every single bond), but a π bond never occurs on its own between a given pair of atoms — it is always formed in addition to a σ bond already present between them, because the sideways overlap needs the σ framework to hold the two nuclei at the right distance and orientation first. Summary of the distinction σ bond π bond --- --- --- Orbital overlap head-on, along the internuclear axis sideways, perpendicular to the axis Electron cloud symmetric about the axis (cylindrical) above and below a nodal plane containing the axis Rotation about bond axis free restricted Relative strength stronger weaker Occurs alone? yes (every single bond is one σ) no (always accompanies a σ bond) Present in single, double, triple bonds (always exactly one) double bonds (one π) and triple bonds (two π) Answer: A σ bond arises from head-on overlap of orbitals along the internuclear axis, giving an electron cloud that is symmetric about that axis, permitting free rotation, and it can exist alone (as in every single bond). A π bond arises from sideways overlap of parallel p orbitals perpendicular to the axis, giving electron density in two lobes above and below a nodal plane that contains the axis; it is weaker than a σ bond, it restricts free rotation about the bond, and it never occurs by itself — a double bond is one σ + one π, and a triple bond is one σ + two π.

Question 33

Explain the formation of H 2 molecule on the basis of valence bond theory.

Accepted Answer

Valence bond theory says a covalent bond forms when two atomic orbitals, each holding one unpaired electron, overlap so that the electrons pair up — and a bond forms only if that overlap lowers the system's total energy. Take two isolated hydrogen atoms, A and B, far enough apart that they do not interact. Each has one electron in a 1s orbital, and each atom's electron is attracted only to its own nucleus. Call the nuclei N A and N B, and the electrons e A and e B. Since the atoms are far apart, define the potential energy of this separated system as zero — this is just a convenient reference point, not a claim that the atoms have no energy at all. Now let the atoms approach each other. Two new kinds of force switch on, and they pull in opposite directions: - New attractive forces: nucleus N A is now close enough to attract the other atom's electron e B, and N B attracts e A. These are attractions that did not exist when the atoms were far apart. - New repulsive forces: the two nuclei repel each other ( N A – N B ), and the two electrons repel each other ( e A – e B ). The bond forms or doesn't depending on which set wins. Experimentally (and this is the fact valence bond theory takes as its starting point), the new attractive forces are stronger in magnitude than the new repulsive forces, provided the electrons on the two atoms have opposite spins. Opposite spins matter because two electrons of like spin cannot occupy the same region of overlap (Pauli exclusion) — only opposite-spin electrons can pair and let the orbitals overlap constructively. Because attraction dominates, the potential energy of the system falls as the atoms approach. Plot potential energy against the internuclear distance r (the distance between N A and N B ): - At large r, energy is taken as zero (the reference). - As r decreases, energy keeps falling — the atoms are being pulled together, and the system is becoming more stable. - The energy reaches a minimum at one particular distance. - If the atoms are pushed closer than that, the nucleus–nucleus and electron–electron repulsions take over and energy rises sharply again. That minimum is the point where the net attractive and net repulsive forces exactly balance. It is not an arbitrary stopping point — it is the lowest-energy, most stable arrangement the two atoms can reach, so the system settles there rather than anywhere else. This is the same distance where all the bonding actually happens. The internuclear distance at the minimum is the equilibrium bond length of H 2, measured as 74 pm. The depth of the minimum below the zero reference — how much energy was released in getting there — is the bond enthalpy, measured as 435.8 kJ mol^-1. This is the energy that would have to be put back in to tear the two atoms apart again, which is exactly why it is reported as a positive bond-dissociation energy even though forming the bond releases energy (a sign flip that trips people up: bond formation is exothermic, bond breaking needs that same amount of energy supplied). So the physical picture valence bond theory gives for H 2 is: two 1s orbitals, each with one electron, overlap along the internuclear axis; the electrons pair up (necessarily with opposite spins); the resulting attractive forces outweigh the repulsive ones; the system's potential energy drops to a minimum at r = 74 pm, releasing 435.8 kJ mol^-1; and that minimum-energy, bonded arrangement is the stable H 2 molecule. Answer: In valence bond theory, H 2 forms because the 1s orbitals of two H atoms with oppositely-spinning electrons overlap; the new nucleus–electron attractions this creates outweigh the new nucleus–nucleus and electron–electron repulsions, so the system's potential energy falls as the atoms approach and reaches a minimum at the equilibrium bond length of 74 pm, releasing the bond enthalpy of 435.8 kJ mol⁻¹ — this lower-energy, bonded state is the stable H₂ molecule.

Question 34

Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Accepted Answer

A molecular orbital is built by adding or subtracting atomic orbital wavefunctions — but that addition only produces a meaningful bond if three conditions hold. In the LCAO (Linear Combination of Atomic Orbitals) method, if φ A and φ B are the wavefunctions of atomic orbitals on atoms A and B, the molecular orbital wavefunction is written as ψ MO = c A φ A ± c B φ B where c A and c B are coefficients showing how much each atomic orbital contributes. The + combination gives a bonding molecular orbital (constructive overlap, enhanced electron density between the nuclei); the – combination gives an antibonding molecular orbital (destructive overlap, a node between the nuclei). But this combination is only physically valid — only gives orbitals that actually look and behave like real molecular orbitals — when the atomic orbitals going into it satisfy three conditions. Condition 1: The combining atomic orbitals must have the same or nearly the same energy. This is the most restrictive condition. If two atomic orbitals differ greatly in energy, their combination does not produce two new orbitals that are meaningfully different from the originals — the mixing is negligible. This is why, in a homonuclear diatomic like N 2 or O 2, the 1s orbital of one atom combines with the 1s of the other (comparable energy, deep core levels), and the 2s of one combines with the 2s of the other, and the 2p with the 2p — but a 1s orbital does not combine with a 2s or 2p orbital, because the energy gap is too large. The point people miss here: "same or nearly the same energy" is a statement about the two atomic orbitals relative to each other, not about the atoms being identical — in a heteronuclear molecule like CO, a carbon 2p orbital can still combine with an oxygen 2p orbital because their energies are close enough, even though the atoms are different. Condition 2: The combining atomic orbitals must have the same symmetry about the molecular axis. Take the internuclear axis (the line joining the two nuclei) as the z-axis. An atomic orbital's symmetry about this axis decides whether it can overlap with another orbital without the overlap cancelling itself out. For example: - 2p z on atom A and 2p z on atom B both point along the internuclear axis and have the same symmetry about it — they combine, giving a σ bonding/antibonding pair. - 2p x on atom A and 2p x on atom B are both perpendicular to the axis in the same way — they combine, giving a π bonding/antibonding pair. Likewise 2p y with 2p y. - 2p z on atom A and 2p x on atom B do not combine, even though both are 2p orbitals of comparable energy, because they have different symmetry about the internuclear axis — the positive overlap on one side is exactly cancelled by negative overlap on the other side, so the net overlap is zero. The step people get wrong: being the "same type" of orbital (2p with 2p) is not enough by itself — it is the orientation relative to the internuclear axis that must match. This is exactly why bonding along the axis is labelled σ and bonding perpendicular to it is labelled π: the labels track which symmetry class the combination belongs to. Condition 3: The combining atomic orbitals must overlap to the maximum possible extent. The greater the overlap between two atomic orbitals, the greater the electron density concentrated in the bonding region between the nuclei, and correspondingly the greater the stability of the bonding molecular orbital formed. Maximum overlap is what makes a bond strong; orbitals pointing directly at each other (head-on, as in σ bonding) overlap more effectively than orbitals overlapping sideways (as in π bonding), which is one reason a σ bond is generally stronger than a π bond. Only when all three conditions — comparable energy, matching symmetry about the internuclear axis, and effective overlap — are satisfied together does the linear combination of atomic orbitals produce genuine bonding and antibonding molecular orbitals. Answer: The atomic orbitals combining to form molecular orbitals must (1) have the same or nearly the same energy, (2) have the same symmetry about the internuclear (molecular) axis, and (3) overlap to the maximum possible extent — comparable energy allows effective mixing, matching symmetry allows the overlap not to cancel out, and maximum overlap gives the greatest bonding stabilization.

Question 35

Use molecular orbital theory to explain why the Be 2 molecule does not exist.

Accepted Answer

Bond order tells you whether a molecule can exist at all — a bond order of zero means no net bonding, so no molecule. Molecular orbital theory (MOT) says a molecule is stable only if it has more electrons in bonding molecular orbitals than in antibonding ones. To check whether Be 2 exists, write its full MO electron configuration and compute the bond order from it. Step 1: Count the electrons. Beryllium ( Z = 4 ) has the atomic configuration 1s^2 2s^2, so each Be atom brings 4 electrons. The molecule Be 2 is built from two Be atoms, so it has 4 + 4 = 8 electrons total. Step 2: Fill the molecular orbitals in order of increasing energy. For a light diatomic like this, the MO energy order (up to this electron count) is σ 1s < σ^ 1s < σ 2s < σ^ 2s where σ orbitals are bonding (formed by in-phase overlap, lower energy, hold the nuclei together) and σ^ orbitals are antibonding (formed by out-of-phase overlap, higher energy, push electron density away from between the nuclei — this "star" superscript is the detail people skip past). Each MO holds 2 electrons (Pauli exclusion), so filling the 8 electrons of Be 2 gives: KK (σ 2s)^2 (σ^ 2s)^2 or written out fully: (σ 1s)^2 (σ^ 1s)^2 (σ 2s)^2 (σ^ 2s)^2 Here "KK" is shorthand for the filled 1s shell contribution, (σ1s)^2(σ^ 1s)^2, which is common to both atoms and cancels in bonding just like a filled inner shell does in Lewis structures. Step 3: Compute the bond order. The bond order formula is Bond order = 1/2(N b - N a) where N b is the number of electrons in bonding MOs and N a is the number of electrons in antibonding MOs. From the configuration above: N b = 2 (σ 1s) + 2 (σ 2s) = 4 N a = 2 (σ^ 1s) + 2 (σ^ 2s) = 4 So Bond order = 1/2(4 - 4) = 0 Step 4: Interpret the zero. A bond order of zero means that for every pair of electrons pulling the two nuclei together (in σ orbitals), there is exactly one pair pushing them apart (in σ^ orbitals) — the bonding and antibonding contributions cancel completely, so no net bond forms. This is the same situation as He 2, which is why neither molecule is observed under ordinary conditions: MOT predicts that any species whose bonding and antibonding electron counts are equal has zero bond order and therefore does not exist as a stable molecule. Answer: The MO configuration of Be 2 is (σ1s)^2(σ^ 1s)^2(σ2s)^2(σ^ 2s)^2, giving bond order = 1/2(4-4) = 0. A zero bond order means the bonding and antibonding electrons cancel exactly, so there is no net attractive force holding the two Be atoms together — hence Be 2 does not exist.

Question 36

Compare the relative stability of the following species and indicate their magnetic properties; O 2, O 2^+, O 2^- (superoxide), O 2^2- (peroxide)

Accepted Answer

Stability here is read off the bond order, and bond order comes straight from the molecular‑orbital (MO) electron count — you don't need anything else. The formula to use is Bond order = (N b - N a)/2 where N b is the number of electrons sitting in bonding molecular orbitals and N a is the number sitting in antibonding molecular orbitals. A higher bond order means a shorter, stronger, more stable bond. Whether the species is paramagnetic or diamagnetic is read off the same diagram: if any orbital holds a single, unpaired electron, the species is paramagnetic; if every orbital is either empty or fully paired, it is diamagnetic. Building the MO diagram of O 2 Each oxygen atom brings 8 electrons, so neutral O 2 has 16 electrons in all. For O 2 (and its ions, which are built from the same diagram by adding or removing electrons), the molecular orbitals fill in this order of increasing energy: σ 1s, σ^ 1s, σ 2s, σ^ 2s, σ 2p z, (π 2p x = π 2p y), (π^ 2p x = π^ 2p y), σ^ 2p z Filling 16 electrons into this ladder gives the configuration of neutral O 2: σ 1s^2 σ^ 1s^2 σ 2s^2 σ^ 2s^2 σ 2p z^2 π 2p x^2 π 2p y^2 π^ 2p x^1 π^ 2p y^1 The step people skip: the 1s core pair σ1s^2σ^ 1s^2 always contributes 2-2=0 to the bond order — it's one bonding pair cancelled by one antibonding pair — so it can safely be carried along without doing any real work in the arithmetic. The part that actually decides everything for this question is the highest occupied level, π^ 2p x,π^ 2p y: by Hund's rule the two electrons go into the two degenerate π^ orbitals singly before either pairs up, which is exactly why ordinary O 2 itself is paramagnetic (this is the classic result MO theory gets right and simple Lewis structures miss). Now the four species differ only in how many electrons sit in that π^ 2p level, because that is the highest (least stable, most easily changed) occupied MO. Removing an electron from O 2 removes it from π^; adding an electron to O 2 adds it to π^. O 2 (16 electrons, neutral): Bonding electrons N b = 2+2+2+2+2 = 10 (from σ1s,σ2s,σ2p z,π2p x,π2p y ). Antibonding electrons N a = 2+2+1+1 = 6 (from σ^ 1s,σ^ 2s,π^ 2p x,π^ 2p y ). Bond order = 10-6/2 = 2 Two unpaired electrons in π^ 2p x and π^ 2p y → paramagnetic. O 2^+ (15 electrons — one electron removed from O 2 's π^ level): N b = 10 (unchanged), N a = 2+2+1+0 = 5. Bond order = 10-5/2 = 2.5 One unpaired electron left in π^ → paramagnetic. O 2^-, superoxide (17 electrons — one electron added to O 2 's π^ level): N b = 10, N a = 2+2+2+1 = 7. Bond order = 10-7/2 = 1.5 One π^ orbital is now full (2 electrons) and the other has 1 → one unpaired electron → paramagnetic. O 2^2-, peroxide (18 electrons — two electrons added to O 2 's π^ level, filling it completely): N b = 10, N a = 2+2+2+2 = 8. Bond order = 10-8/2 = 1 Both π^ orbitals are now completely full — no unpaired electrons → diamagnetic. Ranking the stability Since a larger bond order means a stronger, more stable O–O bond, the bond orders 2.5 2 1.5 1 give the stability order directly: O 2^+ O 2 O 2^- ( superoxide ) O 2^2- ( peroxide ) This tracks with measured bond lengths too: the O–O bond gets progressively longer (weaker) as you go from O 2^+ (shortest) to O 2^2- (longest), exactly matching the falling bond order — each extra electron beyond O 2 goes into an antibonding orbital and stretches/weakens the bond, while removing an electron from O 2 empties an antibonding orbital and strengthens it. Answer: Bond orders are O 2^+=2.5, O 2=2, O 2^- (superoxide) =1.5, O 2^2- (peroxide) =1, so stability decreases in the order O 2^+ O 2 O 2^- O 2^2-. Magnetically, O 2^+, O 2, and O 2^- each have unpaired electrons in the π^ 2p orbitals and are paramagnetic, while O 2^2- has its π^ 2p orbitals completely filled and is diamagnetic.

Question 37

Write the significance of a plus and a minus sign shown in representing the orbitals.

Accepted Answer

The plus and minus signs on an orbital are the sign (phase) of the wave function ψ in that region of space — they are not electrical charges. An atomic orbital is a plot of the wave function ψ, which is a mathematical solution of the Schrödinger wave equation for an electron. Like any wave, ψ can have a crest (positive amplitude) or a trough (negative amplitude) in different regions. When chemists draw an orbital with a "+" lobe and a "−" lobe, they are marking where ψ is positive and where it is negative — exactly the way a sound wave or a water wave has crests and troughs. This has nothing to do with the orbital carrying positive or negative electric charge: an electron's charge is negative everywhere, regardless of which lobe it is found in. Why this distinction trips people up: it is tempting to read the "+" as "positive charge here" and "−" as "negative charge here," the way you would read a battery terminal. That reading is wrong for orbitals — both lobes hold the same negatively charged electron; only the mathematical sign of the wave amplitude differs between them. The pattern of signs depends on the shape of the orbital. - An s orbital is spherically symmetric and has ψ with the same sign everywhere (it is usually drawn all "+"), because there is no nodal plane cutting through it. - A p orbital, say 2p z, has two lobes on either side of the nucleus separated by a nodal plane (a plane on which ψ = 0 ). The wave function is positive on one side of this node and negative on the other, so the two lobes of a single p orbital are marked "+" and "−". Where the sign actually matters — the physics behind the labeling. The sign of ψ becomes physically important the moment two atomic orbitals overlap to form a molecular orbital, because wave functions add algebraically, the way two waves add by superposition: ψ MO = ψ A ± ψ B Here ψ A and ψ B are the atomic orbital wave functions on the two combining atoms. - If the overlapping lobes carry the same sign (+ with +, or − with −), the wave functions add constructively — like crest meeting crest. The amplitude of ψ builds up in the region between the two nuclei, so the electron-probability density ψ^2 there increases. This is a bonding molecular orbital: extra electron density between the nuclei screens their mutual repulsion and holds the atoms together, and this MO is lower in energy than the parent atomic orbitals. - If the overlapping lobes carry opposite signs (+ with −), the wave functions add destructively — crest meeting trough — and they cancel exactly at a point (or plane) between the nuclei, creating a node there where ψ = 0 and hence ψ^2 = 0. This is an antibonding molecular orbital: electron density is depleted between the nuclei (pushed instead to the outer sides), the internuclear repulsion is no longer well-screened, and this MO is higher in energy than the parent atomic orbitals. So the "+"/"−" labels are not decoration — they are the bookkeeping that tells you, once two orbitals come together, whether their overlap will reinforce the wave (bonding, lower energy) or cancel it (antibonding, higher energy, with a node between the nuclei). Answer: The plus and minus signs on an orbital represent the sign (phase) of the electron wave function ψ in that lobe, not any electrical charge — both lobes belong to the same negatively charged electron. This sign matters when atomic orbitals combine: overlap of lobes with the same sign interferes constructively, building up electron density between the nuclei and giving a lower-energy bonding molecular orbital, while overlap of lobes with opposite signs interferes destructively, creating a node (zero electron density) between the nuclei and giving a higher-energy antibonding molecular orbital.

Question 38

Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?

Accepted Answer

A bond gets longer when its electron pair is pushed harder from the sides — and in PCl 5 the two axial bonds are pushed harder than the three equatorial ones. Step 1 — where five bonds come from. Phosphorus ( Z=15 ) has the ground-state configuration 1s^2 2s^2 2p^6 3s^2 3p^3 Only three electrons here are unpaired (in the three 3p orbitals), which would let P form just three bonds. To form five bonds, one of the paired 3s electrons is promoted into an empty 3d orbital of the same shell (this costs energy, but it is repaid by forming two extra P–Cl bonds). The excited-state configuration is 3s^1 3p x^1 3p y^1 3p z^1 3d^1 — five singly occupied orbitals: one s, three p, one d. Step 2 — hybridisation. These five orbitals (1 s-orbital + 3 p-orbitals + 1 d-orbital) mix to give five new, equivalent-in-energy orbitals. Because the mix uses one s, three p, and one d orbital, this is called sp^3d hybridisation. Each of the five sp^3d hybrid orbitals overlaps end-on with a singly filled 3p orbital of a chlorine atom, giving five P–Cl σ bonds. The five orbitals point toward the corners of a trigonal bipyramid: three of them lie in one plane through P, 120^° apart (the equatorial bonds), and the other two point straight up and down, perpendicular to that plane (the axial bonds), each making an angle of 90^° with every equatorial bond and 180^° with each other. Common slip: " sp^3d hybrid orbitals are equivalent" only means they are equal in energy and shape when drawn in isolation. Once all five point at real chlorine atoms, they are not in equivalent environments — three see two other bonds at 120^° and two at 90^°; the other two see three bonds at 90^°. That difference in neighbours is the whole answer to the second part. Step 3 — why the axial bonds are longer. In VSEPR, repulsion between two electron pairs falls off sharply as the angle between them opens up: repulsion at 90^° repulsion at 120^° repulsion at 180^°. So the bonds that suffer the most 90^° repulsions are pushed the hardest. Count the 90^° neighbours of each type of bond pair: - An axial P–Cl bond pair sits at 90^° to all three equatorial bond pairs, and at 180^° (negligible repulsion) to the other axial bond. So it faces three strong, close-range repulsions. - An equatorial P–Cl bond pair sits at 90^° to only the two axial bond pairs; its other two equatorial neighbours are at the much gentler 120^°. So it faces only two strong repulsions. Because each axial bond pair is squeezed by three 90^° repulsions while each equatorial bond pair is squeezed by only two, the axial pairs experience the greater net repulsion. An electron pair relieves repulsion by moving farther from the nucleus — so the axial bonding pairs, and with them the axial P–Cl bonds, stretch out longer than the equatorial ones. (Experimentally, axial P–Cl ≈ 219 pm versus equatorial P–Cl ≈ 204 pm — the axial bonds are noticeably the longer, weaker pair.) Answer: In PCl 5, one 3s electron of phosphorus is promoted to an empty 3d orbital, giving five unpaired electrons in 3s, 3p x, 3p y, 3p z, 3d; these five orbitals undergo sp^3d hybridisation and overlap with chlorine 3p orbitals to give a trigonal bipyramidal molecule with three equatorial P–Cl bonds ( 120^° apart) and two axial P–Cl bonds (perpendicular to the equatorial plane, 90^° to each equatorial bond). The axial bonds are longer than the equatorial bonds because each axial bond pair experiences repulsion from three equatorial bond pairs at the unfavourable 90^° angle, whereas each equatorial bond pair experiences only two such 90^° repulsions (its other neighbours being at the weaker 120^° ); this larger repulsion on the axial pairs pushes them, and the axial bonds, farther out, so they end up longer than the equatorial bonds.

Question 39

Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Accepted Answer

A hydrogen bond is a special, strong dipole–dipole attraction that forms because hydrogen bonded to F, O, or N is left almost as a bare proton — and it is stronger than van der Waals forces, not weaker. Defining the hydrogen bond A hydrogen bond forms when a hydrogen atom is covalently bonded to a small, highly electronegative atom — F, O, or N — and this hydrogen also has an attractive interaction with a lone pair of electrons on another highly electronegative atom (F, O, or N), either in a neighbouring molecule or in a different part of the same molecule. It is written as X–H Y where X and Y are F, O, or N; the solid line X–H is the ordinary covalent bond, and the dotted line H Y is the hydrogen bond itself. Why it forms — and why it is unusually strong Fluorine, oxygen, and nitrogen are small atoms with high electronegativity. When one of them is bonded to hydrogen, the shared electron pair is pulled strongly towards X, so the X–H bond becomes highly polar. This leaves the hydrogen end of the bond with a significant partial positive charge, δ+. Here is the point that makes hydrogen bonding special rather than an ordinary dipole–dipole force: hydrogen has only a single electron and no inner shell of electrons at all. Once that electron is pulled towards X, what is left behind is essentially a bare, unshielded proton — an extremely small, concentrated positive charge. This lets the hydrogen approach the lone pair on a nearby Y atom far more closely than any ordinary δ+ end of a polar molecule could, so the electrostatic attraction it feels is much stronger than a normal dipole–dipole interaction. Comparing the strengths - Van der Waals forces (London dispersion forces and ordinary dipole–dipole attractions between molecules) are weak, non-directional interactions with typical bond energies of roughly 0.4 to 4 kJ mol^-1. - Hydrogen bonds are stronger and directional, with typical bond energies of roughly 10 to 40 kJ mol^-1 (still much weaker than a true covalent bond, which is of the order of 150 – 950 kJ mol^-1 ). A short aside on where students slip: it is tempting to lump hydrogen bonding in with the "weak intermolecular forces" category alongside van der Waals forces, since both act between molecules rather than within them. But being intermolecular does not make two forces equally weak — the near-bare-proton character of the hydrogen bonded to F, O, or N is what raises hydrogen bonding well above ordinary van der Waals attraction. This is exactly why hydrogen bonding, and not van der Waals forces, is responsible for the anomalously high boiling points of HF, H 2 O, and NH 3 compared to the other hydrides in their respective groups. Answer: A hydrogen bond is the attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) — X–H — and a lone pair on another highly electronegative atom (F, O, or N) — Y — represented as X–H Y. It is stronger than van der Waals forces (hydrogen bond energies 10 – 40 kJ mol^-1 versus 0.4 – 4 kJ mol^-1 for van der Waals forces), because the hydrogen atom, having lost its only electron towards the electronegative atom it is bonded to, behaves almost like a bare proton and can approach the lone pair on the other electronegative atom very closely, giving a much stronger electrostatic attraction.

Question 40

What is meant by the term bond order? Calculate the bond order of: N 2, O 2, O 2^+ and O 2^-.

Accepted Answer

Bond order tells you how many net bonds hold two atoms together, and it comes straight out of how many electrons molecular orbital (MO) theory puts into bonding orbitals versus antibonding orbitals. The formula is Bond order = 1/2(N b - N a) where N b is the total number of electrons occupying bonding molecular orbitals and N a is the total number occupying antibonding molecular orbitals. An electron in a bonding orbital pulls the two nuclei together; one in an antibonding orbital pushes them apart — so the difference, halved, is the net number of bonds. To use this formula you first need the electrons in each molecule sorted into the right molecular orbitals, in the right energy order. This is the step people get wrong: the energy order of the MOs is not the same for every diatomic. For Li 2 through N 2 (fewer than 15 electrons, where the 2s and 2p orbitals mix and push σ 2p z upward), the order is σ 1s < σ^ 1s < σ 2s < σ^ 2s < π 2p x = π 2p y < σ 2p z < π^ 2p x = π^ 2p y < σ^ 2p z For O 2, F 2, Ne 2 (heavier, where that mixing is too weak to matter), σ 2p z drops back below the π 2p pair: σ 1s < σ^ 1s < σ 2s < σ^ 2s < σ 2p z < π 2p x = π 2p y < π^ 2p x = π^ 2p y < σ^ 2p z Using the N 2 order for O 2 (or vice versa) fills orbitals in the wrong sequence and gives the wrong bond order — so it's worth checking which molecule you're filling before you start. N 2 — 14 electrons (7 + 7) Filling the first order above: σ 1s^2 σ^ 1s^2 σ 2s^2 σ^ 2s^2 (π 2p x^2 π 2p y^2) σ 2p z^2 Bonding electrons: σ1s(2) + σ2s(2) + π2p(4) + σ2p z(2) = 10 Antibonding electrons: σ^ 1s(2) + σ^ 2s(2) = 4 Bond order = 1/2(10-4) = 3 A bond order of 3 is exactly the triple bond N ≡ N you'd draw from the Lewis structure — MO theory and Lewis theory agree here. O 2 — 16 electrons (8 + 8) Now use the second (heavier-molecule) order, with σ2p z filled before π2p: σ 1s^2 σ^ 1s^2 σ 2s^2 σ^ 2s^2 σ 2p z^2 (π 2p x^2 π 2p y^2) (π^ 2p x^1 π^ 2p y^1) The last two electrons go into the two degenerate π^ 2p orbitals singly (Hund's rule), which is exactly why O 2 is paramagnetic. Bonding electrons: σ1s(2) + σ2s(2) + σ2p z(2) + π2p(4) = 10 Antibonding electrons: σ^ 1s(2) + σ^ 2s(2) + π^ 2p(2) = 6 Bond order = 1/2(10-6) = 2 A bond order of 2 matches the double bond O=O. O 2^+ — 15 electrons This is O 2 with one electron removed. Electrons are always removed from the highest-energy occupied orbital, which for O 2 is π^ 2p. So one of the two singly-occupied π^ 2p electrons is taken away, leaving only one electron in π^ 2p: Bonding electrons: still 10 Antibonding electrons: σ^ 1s(2) + σ^ 2s(2) + π^ 2p(1) = 5 Bond order = 1/2(10-5) = 2.5 Removing an antibonding electron raises the bond order above that of neutral O 2 — this is why O 2^+ has a shorter, stronger bond than O 2 itself, even though it has fewer electrons overall. O 2^- — 17 electrons This is O 2 with one extra electron added. It goes into the same π^ 2p set, pairing up with one of the two electrons already there: Bonding electrons: still 10 Antibonding electrons: σ^ 1s(2) + σ^ 2s(2) + π^ 2p(3) = 7 Bond order = 1/2(10-7) = 1.5 Adding an antibonding electron lowers the bond order below that of O 2, so O 2^- (superoxide) has a weaker, longer bond than O 2. Putting the four together, bond order falls as more electrons enter the antibonding π^ 2p set: N 2 (3) O 2^+ (2.5) O 2 (2) O 2^- (1.5) Answer: Bond order is 1/2(N b - N a), half the difference between electrons in bonding and antibonding molecular orbitals. Bond order of N 2 = 3; O 2 = 2; O 2^+ = 2.5; O 2^- = 1.5.

	Bonds	Net dipole moment (measured)	What that forces the shape to be
CO₂	$\displaystyle 2$ polar C=O bonds	\(\displaystyle 0\)	Linear (\(\displaystyle 180^\circ\)) — only way two equal dipoles cancel
H₂O	$\displaystyle 2$ polar O–H bonds	\(\displaystyle 1.84\ \text{D} \neq 0\)	Bent (\(\displaystyle \theta < 180^\circ\)) — dipoles must add, not cancel

	σ bond	π bond
Orbital overlap	head-on, along the internuclear axis	sideways, perpendicular to the axis
Electron cloud	symmetric about the axis (cylindrical)	above and below a nodal plane containing the axis
Rotation about bond axis	free	restricted
Relative strength	stronger	weaker
Occurs alone?	yes (every single bond is one σ)	no (always accompanies a σ bond)
Present in	single, double, triple bonds (always exactly one)	double bonds (one π) and triple bonds (two π)

Exercises 4.1–4.10 (part 1 of 4)

Exercise 4.1

Exercise 4.2

Exercise 4.3

Exercise 4.4

Exercise 4.5

Exercise 4.6

Exercise 4.7

Exercise 4.8

Exercise 4.9

Exercise 4.10