NCERT Solutions · Class 11 Chemistry Chemical Bonding and Molecular Structure

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 electronsWhen 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, \(\displaystyle \text{H}_2 \), each H atom brings one electron: \[\text{H}\cdot \; + \; \cdot\text{H} \;\longrightarrow\; \text{H}:\text{H} \;\; (\text{or } \text{H–H}) \] 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 electronsNot 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, \(\displaystyle \text{NH}_3 \), nitrogen has $\displaystyle 5$ 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 $\displaystyle 5$ valence electrons distribute as $\displaystyle 3$ (used, one each, in the $\displaystyle 3$ N–H bond pairs) + $\displaystyle 2$ (left over as $\displaystyle 1$ pair on nitrogen). So the Lewis structure of \(\displaystyle \text{NH}_3 \) has 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 \(\displaystyle \text{NH}_4^+ \)) — 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 \(\displaystyle \text{H–H} \), i.e. \(\displaystyle \text{H}_2 \). 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 \(\displaystyle \text{NH}_3 \), in addition to its $\displaystyle 3$ N–H bond pairs.

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 formedTake the internuclear axis to be the straight line joining the two bonded nuclei (call it the \(\displaystyle z\)-axis). 2. Shape of the electron cloud 3. Strength of overlapHead-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 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.

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 \(\displaystyle N_A \) and \(\displaystyle N_B \), and the electrons \(\displaystyle e_A \) and \(\displaystyle e_B \). 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: 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 \(\displaystyle r \) (the distance between \(\displaystyle N_A \) and \(\displaystyle N_B \)): 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 \(\displaystyle \mathrm{H_2} \), measured as \(\displaystyle 74 \) pm. The depth of the minimum below the zero reference — how much energy was released in getting there — is the bond enthalpy, measured as \(\displaystyle 435.8\ \mathrm{kJ\ mol^{-1}} \). 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 \(\displaystyle \mathrm{H_2} \) 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 \(\displaystyle r = 74 \) pm, releasing \(\displaystyle 435.8\ \mathrm{kJ\ mol^{-1}} \); and that minimum-energy, bonded arrangement is the stable \(\displaystyle \mathrm{H_2} \) molecule.Answer: In valence bond theory, \(\displaystyle \mathrm{H_2} \) 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 $\displaystyle 74$ pm, releasing the bond enthalpy of $\displaystyle 435.8$ kJ mol⁻¹ — this lower-energy, bonded state is the stable H₂ molecule.

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 \(\displaystyle \phi_A \) and \(\displaystyle \phi_B \) are the wavefunctions of atomic orbitals on atoms A and B, the molecular orbital wavefunction is written as\[\psi_{MO} = c_A \, \phi_A \pm c_B \, \phi_B \]where \(\displaystyle c_A \) and \(\displaystyle c_B \) 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 $\displaystyle 1$: 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 \(\displaystyle \mathrm{N_2} \) or \(\displaystyle \mathrm{O_2} \), 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 $\displaystyle 2$: 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: 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 \(\displaystyle \sigma \) and bonding perpendicular to it is labelled \(\displaystyle \pi \): the labels track which symmetry class the combination belongs to.Condition $\displaystyle 3$: 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 \(\displaystyle \sigma \) bonding) overlap more effectively than orbitals overlapping sideways (as in \(\displaystyle \pi \) bonding), which is one reason a \(\displaystyle \sigma \) bond is generally stronger than a \(\displaystyle \pi \) 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 ($\displaystyle 1$) have the same or nearly the same energy, ($\displaystyle 2$) have the same symmetry about the internuclear (molecular) axis, and ($\displaystyle 3$) 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.

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 \(\displaystyle \text{Be}_2 \) exists, write its full MO electron configuration and compute the bond order from it.Step $\displaystyle 1$: Count the electrons.Beryllium (\(\displaystyle Z = 4\)) has the atomic configuration \(\displaystyle 1s^2\, 2s^2\), so each Be atom brings $\displaystyle 4$ electrons. The molecule \(\displaystyle \text{Be}_2 \) is built from two Be atoms, so it has\[4 + 4 = 8 \text{ electrons total.} \]Step $\displaystyle 2$: 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\[\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s \]where \(\displaystyle \sigma\) orbitals are bonding (formed by in-phase overlap, lower energy, hold the nuclei together) and \(\displaystyle \sigma^*\) 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 $\displaystyle 2$ electrons (Pauli exclusion), so filling the $\displaystyle 8$ electrons of \(\displaystyle \text{Be}_2 \) gives:\[\text{KK}\,(\sigma 2s)^2 (\sigma^*2s)^2 \]or written out fully:\[(\sigma 1s)^2 (\sigma^*1s)^2 (\sigma 2s)^2 (\sigma^*2s)^2 \]Here "KK" is shorthand for the filled \(\displaystyle 1s\) shell contribution, \(\displaystyle (\sigma1s)^2(\sigma^*1s)^2\), which is common to both atoms and cancels in bonding just like a filled inner shell does in Lewis structures.Step $\displaystyle 3$: Compute the bond order.The bond order formula is\[\text{Bond order} = \frac{1}{2}\left(N_b - N_a\right) \]where \(\displaystyle N_b\) is the number of electrons in bonding MOs and \(\displaystyle N_a\) is the number of electrons in antibonding MOs.From the configuration above:\[N_b = 2 \,(\sigma 1s) + 2\,(\sigma 2s) = 4 \] \[N_a = 2\,(\sigma^*1s) + 2\,(\sigma^*2s) = 4 \]So\[\text{Bond order} = \frac{1}{2}(4 - 4) = 0 \]Step $\displaystyle 4$: Interpret the zero.A bond order of zero means that for every pair of electrons pulling the two nuclei together (in \(\displaystyle \sigma\) orbitals), there is exactly one pair pushing them apart (in \(\displaystyle \sigma^*\) orbitals) — the bonding and antibonding contributions cancel completely, so no net bond forms. This is the same situation as \(\displaystyle \text{He}_2 \), 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 \(\displaystyle \text{Be}_2 \) is \(\displaystyle (\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2\), giving bond order \(\displaystyle = \tfrac{1}{2}(4-4) = 0\). 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 \(\displaystyle \text{Be}_2 \) does not exist.

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 \[\text{Bond order} = \frac{N_b - N_a}{2} \] where \(\displaystyle N_b\) is the number of electrons sitting in bonding molecular orbitals and \(\displaystyle N_a\) 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 \(\displaystyle O_2\)Each oxygen atom brings $\displaystyle 8$ electrons, so neutral \(\displaystyle O_2\) has $\displaystyle 16$ electrons in all. For \(\displaystyle O_2\) (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: \[\sigma 1s,\ \sigma^*1s,\ \sigma 2s,\ \sigma^*2s,\ \sigma 2p_z,\ (\pi 2p_x = \pi 2p_y),\ (\pi^*2p_x = \pi^*2p_y),\ \sigma^*2p_z \] Filling $\displaystyle 16$ electrons into this ladder gives the configuration of neutral \(\displaystyle O_2\): \[\sigma 1s^2\ \sigma^*1s^2\ \sigma 2s^2\ \sigma^*2s^2\ \sigma 2p_z^2\ \pi 2p_x^2\ \pi 2p_y^2\ \pi^*2p_x^1\ \pi^*2p_y^1 \] The step people skip: the \(\displaystyle 1s\) core pair \(\displaystyle \sigma1s^2\sigma^*1s^2\) always contributes \(\displaystyle 2-2=0\) 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, \(\displaystyle \pi^*2p_x,\pi^*2p_y\): by Hund's rule the two electrons go into the two degenerate \(\displaystyle \pi^*\) orbitals singly before either pairs up, which is exactly why ordinary \(\displaystyle O_2\) 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 \(\displaystyle \pi^*2p\) level, because that is the highest (least stable, most easily changed) occupied MO. Removing an electron from \(\displaystyle O_2\) removes it from \(\displaystyle \pi^*\); adding an electron to \(\displaystyle O_2\) adds it to \(\displaystyle \pi^*\).\(\displaystyle O_2\) ($\displaystyle 16$ electrons, neutral): Bonding electrons \(\displaystyle N_b = 2+2+2+2+2 = 10\) (from \(\displaystyle \sigma1s,\sigma2s,\sigma2p_z,\pi2p_x,\pi2p_y\)). Antibonding electrons \(\displaystyle N_a = 2+2+1+1 = 6\) (from \(\displaystyle \sigma^*1s,\sigma^*2s,\pi^*2p_x,\pi^*2p_y\)). \[\text{Bond order} = \frac{10-6}{2} = 2 \] Two unpaired electrons in \(\displaystyle \pi^*2p_x\) and \(\displaystyle \pi^*2p_y\) → paramagnetic.\(\displaystyle O_2^{+}\) ($\displaystyle 15$ electrons — one electron removed from \(\displaystyle O_2\)'s \(\displaystyle \pi^*\) level): \(\displaystyle N_b = 10\) (unchanged), \(\displaystyle N_a = 2+2+1+0 = 5\). \[\text{Bond order} = \frac{10-5}{2} = 2.5 \] One unpaired electron left in \(\displaystyle \pi^*\) → paramagnetic.\(\displaystyle O_2^{-}\), superoxide ($\displaystyle 17$ electrons — one electron added to \(\displaystyle O_2\)'s \(\displaystyle \pi^*\) level): \(\displaystyle N_b = 10\), \(\displaystyle N_a = 2+2+2+1 = 7\). \[\text{Bond order} = \frac{10-7}{2} = 1.5 \] One \(\displaystyle \pi^*\) orbital is now full ($\displaystyle 2$ electrons) and the other has $\displaystyle 1$ → one unpaired electron → paramagnetic.\(\displaystyle O_2^{2-}\), peroxide ($\displaystyle 18$ electrons — two electrons added to \(\displaystyle O_2\)'s \(\displaystyle \pi^*\) level, filling it completely): \(\displaystyle N_b = 10\), \(\displaystyle N_a = 2+2+2+2 = 8\). \[\text{Bond order} = \frac{10-8}{2} = 1 \] Both \(\displaystyle \pi^*\) orbitals are now completely full — no unpaired electrons → diamagnetic.Ranking the stabilitySince a larger bond order means a stronger, more stable O–O bond, the bond orders \(\displaystyle 2.5 > 2 > 1.5 > 1\) give the stability order directly: \[O_2^{+} \;>\; O_2 \;>\; O_2^{-}\,(\text{superoxide}) \;>\; O_2^{2-}\,(\text{peroxide}) \] This tracks with measured bond lengths too: the O–O bond gets progressively longer (weaker) as you go from \(\displaystyle O_2^{+}\) (shortest) to \(\displaystyle O_2^{2-}\) (longest), exactly matching the falling bond order — each extra electron beyond \(\displaystyle O_2\) goes into an antibonding orbital and stretches/weakens the bond, while removing an electron from \(\displaystyle O_2\) empties an antibonding orbital and strengthens it.**Answer: Bond orders are \(\displaystyle O_2^{+}=2.5\), \(\displaystyle O_2=2\), \(\displaystyle O_2^{-}\)(superoxide)\(\displaystyle =1.5\), \(\displaystyle O_2^{2-}\)(peroxide)\(\displaystyle =1\), so stability decreases in the order \(\displaystyle O_2^{+} > O_2 > O_2^{-} > O_2^{2-}\). Magnetically, \(\displaystyle O_2^{+}\), \(\displaystyle O_2\), and \(\displaystyle O_2^{-}\) each have unpaired electrons in the \(\displaystyle \pi^*2p\) orbitals and are paramagnetic, while \(\displaystyle O_2^{2-}\) has its \(\displaystyle \pi^*2p\) orbitals completely filled and is diamagnetic.

The plus and minus signs on an orbital are the sign (phase) of the wave function \(\displaystyle \psi \) in that region of space — they are not electrical charges.An atomic orbital is a plot of the wave function \(\displaystyle \psi \), which is a mathematical solution of the Schrödinger wave equation for an electron. Like any wave, \(\displaystyle \psi \) 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 \(\displaystyle \psi \) 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. Where the sign actually matters — the physics behind the labeling. The sign of \(\displaystyle \psi \) 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:\[\psi_{MO} = \psi_A \pm \psi_B \]Here \(\displaystyle \psi_A \) and \(\displaystyle \psi_B \) are the atomic orbital wave functions on the two combining atoms. 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 \(\displaystyle \psi \) 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.

A bond gets longer when its electron pair is pushed harder from the sides — and in \(\displaystyle PCl_5\) the two axial bonds are pushed harder than the three equatorial ones.Step $\displaystyle 1$ — where five bonds come from.Phosphorus (\(\displaystyle Z=15\)) has the ground-state configuration \[1s^2\,2s^2\,2p^6\,3s^2\,3p^3 \] Only three electrons here are unpaired (in the three \(\displaystyle 3p\) orbitals), which would let P form just three bonds. To form five bonds, one of the paired \(\displaystyle 3s\) electrons is promoted into an empty \(\displaystyle 3d\) 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 \(\displaystyle s\), three \(\displaystyle p\), one \(\displaystyle d\).Step $\displaystyle 2$ — hybridisation. These five orbitals ($\displaystyle 1$ s-orbital + $\displaystyle 3$ p-orbitals + $\displaystyle 1$ d-orbital) mix to give five new, equivalent-in-energy orbitals. Because the mix uses one \(\displaystyle s\), three \(\displaystyle p\), and one \(\displaystyle d\) orbital, this is called \(\displaystyle sp^3d\) hybridisation. Each of the five \(\displaystyle sp^3d\) hybrid orbitals overlaps end-on with a singly filled \(\displaystyle 3p\) orbital of a chlorine atom, giving five P–Cl \(\displaystyle \sigma\) bonds. The five orbitals point toward the corners of a trigonal bipyramid: three of them lie in one plane through P, \(\displaystyle 120^\circ\) 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 \(\displaystyle 90^\circ\) with every equatorial bond and \(\displaystyle 180^\circ\) with each other.Common slip: "\(\displaystyle sp^3d\) 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 \(\displaystyle 120^\circ\) and two at \(\displaystyle 90^\circ\); the other two see three bonds at \(\displaystyle 90^\circ\). That difference in neighbours is the whole answer to the second part.Step $\displaystyle 3$ — 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 \(\displaystyle 90^\circ >\) repulsion at \(\displaystyle 120^\circ >\) repulsion at \(\displaystyle 180^\circ\). So the bonds that suffer the most \(\displaystyle 90^\circ\) repulsions are pushed the hardest.Count the \(\displaystyle 90^\circ\) neighbours of each type of bond pair: Because each axial bond pair is squeezed by three \(\displaystyle 90^\circ\) 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 \(\displaystyle \approx 219\ \text{pm}\) versus equatorial P–Cl \(\displaystyle \approx 204\ \text{pm}\) — the axial bonds are noticeably the longer, weaker pair.)Answer: In \(\displaystyle PCl_5\), one \(\displaystyle 3s\) electron of phosphorus is promoted to an empty \(\displaystyle 3d\) orbital, giving five unpaired electrons in \(\displaystyle 3s,\,3p_x,\,3p_y,\,3p_z,\,3d\); these five orbitals undergo \(\displaystyle sp^3d\) hybridisation and overlap with chlorine \(\displaystyle 3p\) orbitals to give a trigonal bipyramidal molecule with three equatorial P–Cl bonds (\(\displaystyle 120^\circ\) apart) and two axial P–Cl bonds (perpendicular to the equatorial plane, \(\displaystyle 90^\circ\) 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 \(\displaystyle 90^\circ\) angle, whereas each equatorial bond pair experiences only two such \(\displaystyle 90^\circ\) repulsions (its other neighbours being at the weaker \(\displaystyle 120^\circ\)); this larger repulsion on the axial pairs pushes them, and the axial bonds, farther out, so they end up longer than the equatorial bonds.

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 bondA 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 \[\text{X–H}\cdots \text{Y} \] where X and Y are F, O, or N; the solid line X–H is the ordinary covalent bond, and the dotted line H\(\displaystyle \cdots\)Y is the hydrogen bond itself.Why it forms — and why it is unusually strongFluorine, 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, \(\displaystyle \delta+\).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 \(\displaystyle \delta+\) end of a polar molecule could, so the electrostatic attraction it feels is much stronger than a normal dipole–dipole interaction.Comparing the strengths 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\(\displaystyle _2\)O, and NH\(\displaystyle _3\) 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\(\displaystyle \cdots\)Y. It is stronger than van der Waals forces (hydrogen bond energies \(\displaystyle \sim 10\)–\(\displaystyle 40\ \text{kJ mol}^{-1}\) versus \(\displaystyle \sim 0.4\)–\(\displaystyle 4\ \text{kJ mol}^{-1}\) 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.

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\[\text{Bond order} = \frac{1}{2}\left(N_b - N_a\right) \]where \(\displaystyle N_b\) is the total number of electrons occupying bonding molecular orbitals and \(\displaystyle N_a\) 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 \(\displaystyle \mathrm{Li_2}\) through \(\displaystyle \mathrm{N_2}\) (fewer than $\displaystyle 15$ electrons, where the \(\displaystyle 2s\) and \(\displaystyle 2p\) orbitals mix and push \(\displaystyle \sigma 2p_z\) upward), the order is\[\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^*2p_x = \pi^*2p_y < \sigma^*2p_z \]For \(\displaystyle \mathrm{O_2}, \mathrm{F_2}, \mathrm{Ne_2}\) (heavier, where that mixing is too weak to matter), \(\displaystyle \sigma 2p_z\) drops back below the \(\displaystyle \pi 2p\) pair:\[\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^*2p_x = \pi^*2p_y < \sigma^*2p_z \]Using the \(\displaystyle \mathrm{N_2}\) order for \(\displaystyle \mathrm{O_2}\) (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.\(\displaystyle \mathrm{N_2}\) — $\displaystyle 14$ electrons ($\displaystyle 7$ + $\displaystyle 7$)Filling the first order above:\[\sigma 1s^2\ \sigma^*1s^2\ \sigma 2s^2\ \sigma^*2s^2\ (\pi 2p_x^2\ \pi 2p_y^2)\ \sigma 2p_z^2 \]Bonding electrons: \(\displaystyle \sigma1s(2) + \sigma2s(2) + \pi2p(4) + \sigma2p_z(2) = 10\) Antibonding electrons: \(\displaystyle \sigma^*1s(2) + \sigma^*2s(2) = 4\)\[\text{Bond order} = \frac{1}{2}(10-4) = 3 \]A bond order of $\displaystyle 3$ is exactly the triple bond \(\displaystyle \mathrm{N \equiv N}\) you'd draw from the Lewis structure — MO theory and Lewis theory agree here.\(\displaystyle \mathrm{O_2}\) — $\displaystyle 16$ electrons ($\displaystyle 8$ + $\displaystyle 8$)Now use the second (heavier-molecule) order, with \(\displaystyle \sigma2p_z\) filled before \(\displaystyle \pi2p\):\[\sigma 1s^2\ \sigma^*1s^2\ \sigma 2s^2\ \sigma^*2s^2\ \sigma 2p_z^2\ (\pi 2p_x^2\ \pi 2p_y^2)\ (\pi^*2p_x^1\ \pi^*2p_y^1) \]The last two electrons go into the two degenerate \(\displaystyle \pi^*2p\) orbitals singly (Hund's rule), which is exactly why \(\displaystyle \mathrm{O_2}\) is paramagnetic.Bonding electrons: \(\displaystyle \sigma1s(2) + \sigma2s(2) + \sigma2p_z(2) + \pi2p(4) = 10\) Antibonding electrons: \(\displaystyle \sigma^*1s(2) + \sigma^*2s(2) + \pi^*2p(2) = 6\)\[\text{Bond order} = \frac{1}{2}(10-6) = 2 \]A bond order of $\displaystyle 2$ matches the double bond \(\displaystyle \mathrm{O=O}\).\(\displaystyle \mathrm{O_2^+}\) — $\displaystyle 15$ electronsThis is \(\displaystyle \mathrm{O_2}\) with one electron removed. Electrons are always removed from the highest-energy occupied orbital, which for \(\displaystyle \mathrm{O_2}\) is \(\displaystyle \pi^*2p\). So one of the two singly-occupied \(\displaystyle \pi^*2p\) electrons is taken away, leaving only one electron in \(\displaystyle \pi^*2p\):Bonding electrons: still \(\displaystyle 10\) Antibonding electrons: \(\displaystyle \sigma^*1s(2) + \sigma^*2s(2) + \pi^*2p(1) = 5\)\[\text{Bond order} = \frac{1}{2}(10-5) = 2.5 \]Removing an antibonding electron raises the bond order above that of neutral \(\displaystyle \mathrm{O_2}\) — this is why \(\displaystyle \mathrm{O_2^+}\) has a shorter, stronger bond than \(\displaystyle \mathrm{O_2}\) itself, even though it has fewer electrons overall.\(\displaystyle \mathrm{O_2^-}\) — $\displaystyle 17$ electronsThis is \(\displaystyle \mathrm{O_2}\) with one extra electron added. It goes into the same \(\displaystyle \pi^*2p\) set, pairing up with one of the two electrons already there:Bonding electrons: still \(\displaystyle 10\) Antibonding electrons: \(\displaystyle \sigma^*1s(2) + \sigma^*2s(2) + \pi^*2p(3) = 7\)\[\text{Bond order} = \frac{1}{2}(10-7) = 1.5 \]Adding an antibonding electron lowers the bond order below that of \(\displaystyle \mathrm{O_2}\), so \(\displaystyle \mathrm{O_2^-}\) (superoxide) has a weaker, longer bond than \(\displaystyle \mathrm{O_2}\).Putting the four together, bond order falls as more electrons enter the antibonding \(\displaystyle \pi^*2p\) set:\[\mathrm{N_2}\ (3) > \mathrm{O_2^+}\ (2.5) > \mathrm{O_2}\ (2) > \mathrm{O_2^-}\ (1.5) \]Answer: Bond order is \(\displaystyle \dfrac{1}{2}(N_b - N_a)\), half the difference between electrons in bonding and antibonding molecular orbitals. Bond order of \(\displaystyle \mathrm{N_2} = 3\); \(\displaystyle \mathrm{O_2} = 2\); \(\displaystyle \mathrm{O_2^+} = 2.5\); \(\displaystyle \mathrm{O_2^-} = 1.5\).