NCERT Solutions · Class 11 Chemistry Chemical Bonding and Molecular Structure

Resonance means one Lewis structure cannot be drawn for \(\displaystyle \text{CO}_3^{2-} \); the real ion is an average of several, and that average is what actually exists.Step $\displaystyle 1$: Try to draw a single Lewis structure.Carbon sits at the centre, bonded to three oxygen atoms. Carbon has $\displaystyle 4$ valence electrons and needs to complete an octet on itself and on each oxygen, while the total charge on the ion is \(\displaystyle -2\). 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:\[\text{O}=\text{C}(-\text{O}^-)(-\text{O}^-) \]Check the formal charge, using \[\text{Formal charge} = (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons}) \] Total charge \(\displaystyle = 0 + 0 + (-1) + (-1) = -2\), 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 \(\displaystyle \text{CO}_3^{2-}\).Step $\displaystyle 2$: 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:\[\text{O}=\text{C}(-\text{O}^-)(-\text{O}^-) \;\longleftrightarrow\; ^-\text{O}-\text{C}(=\text{O})(-\text{O}^-) \;\longleftrightarrow\; ^-\text{O}-\text{C}(-\text{O}^-)(=\text{O}) \]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 $\displaystyle 3$: 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: \[\text{Bond order} = \frac{1+1+2}{3} = \frac{4}{3} \approx 1.33 \] 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.Step $\displaystyle 4$: 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 \(\displaystyle \text{CO}_3^{2-}\). 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 (~$\displaystyle 1.28$ Å, bond order \(\displaystyle 4/3\), between single and double bond lengths), the \(\displaystyle -2\) charge is spread equally over all three oxygens (average \(\displaystyle -2/3\) each) rather than localized on two of them, and the hybrid is more stable (lower in energy) than any single contributing structure.

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 $\displaystyle 1$ and $\displaystyle 2$ actually showStructure $\displaystyle 1$ puts all three hydrogens on oxygen, so phosphorus carries three P–OH bonds and keeps its lone pair:\[\text{P(OH)}_3 \quad\text{(lone pair on P; P is trivalent, all three H atoms sit on O)} \]Structure $\displaystyle 2$ puts only two hydrogens on oxygen and moves the third hydrogen onto phosphorus itself, with a P=O double bond replacing the lone pair:\[\text{HP}(=\!\text{O})(\text{OH})_2 \quad\text{(one P–H bond, one P=O bond, two P–OH bonds, no lone pair on P)} \]Why these cannot be canonical forms of one resonance hybridCanonical structures of a real resonance hybrid are required to have an identical arrangement of nuclei (the same skeleton of \(\displaystyle \sigma\)-bonds connecting the same atoms to the same atoms); what differs between them is only the placement of \(\displaystyle \pi\)-electrons or lone pairs, and you get from one structure to the next just by pushing electrons, never by breaking a \(\displaystyle \sigma\)-bond or relocating a nucleus.Compare the two P(OH)\(\displaystyle _3\)-type structures here: in structure $\displaystyle 1$ the third hydrogen is bonded to an oxygen atom, while in structure $\displaystyle 2$ that same hydrogen is bonded directly to the phosphorus atom. That is a change in which atom the hydrogen's \(\displaystyle \sigma\)-bond is attached to — a genuine difference in atomic connectivity, not a shift of \(\displaystyle \pi\)-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 \(\displaystyle \sigma\)-bond. Since the nuclear framework itself differs between structures $\displaystyle 1$ and $\displaystyle 2$, 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 realOnly structure $\displaystyle 2$ correctly represents phosphorous acid. This is confirmed experimentally: \(\displaystyle \text{H}_3\text{PO}_3\) behaves as a dibasic (diprotic) acid — only two of its three hydrogens ionize in water. That fits structure $\displaystyle 2$, where two hydrogens sit on oxygen (ionizable as \(\displaystyle \text{O–H}\)) and the third sits directly on phosphorus (a non-ionizable P–H bond). If structure $\displaystyle 1$ were correct, all three hydrogens would be on oxygen and \(\displaystyle \text{H}_3\text{PO}_3\) would have to be tribasic, which it is not.**Answer: No, structures $\displaystyle 1$ and $\displaystyle 2$ 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 $\displaystyle 1$ but directly to phosphorus in structure $\displaystyle 2$, a difference in atomic connectivity, not electron placement. So the two are different structures (tautomers), and only structure $\displaystyle 2$, \(\displaystyle \text{HP}(=\!\text{O})(\text{OH})_2\), is the real structure of \(\displaystyle \text{H}_3\text{PO}_3\) — consistent with its known dibasic character.

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 \[\text{FC} = (\text{valence electrons of the free atom}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons}). \] 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.\(\displaystyle \text{SO}_3 \) (sulfur trioxide)Valence electrons: S contributes $\displaystyle 6$, each O contributes $\displaystyle 6$, so \(\displaystyle 6 + 3(6) = 24 \) electrons \(\displaystyle = 12 \) pairs. With S at the centre bonded to three O atoms, the only way to use exactly these $\displaystyle 24$ 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: Sum: \(\displaystyle 0+(-1)+(-1)+2 = 0 \), 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($\displaystyle 1$), single (−$\displaystyle 1$) bonds to O($\displaystyle 2$) and O($\displaystyle 3$) Structure II — double bond to O($\displaystyle 2$), single (−$\displaystyle 1$) bonds to O($\displaystyle 1$) and O($\displaystyle 3$) Structure III — double bond to O($\displaystyle 3$), single (−$\displaystyle 1$) bonds to O($\displaystyle 1$) and O($\displaystyle 2$)with resonance arrows (\(\displaystyle \leftrightarrow\)) connecting I, II and III. A common mistake here is picturing \(\displaystyle \mathrm{SO_{3}}\) 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 \(\displaystyle \dfrac{2+1+1}{3} = \dfrac{4}{3} \) (double in one out of three structures, single in the other two).\(\displaystyle \text{NO}_2 \) (nitrogen dioxide)Valence electrons: \(\displaystyle 5 + 2(6) = 17 \). This is an odd number, so the electrons cannot all be paired up — \(\displaystyle \mathrm{NO_{2}}\) is necessarily a free radical, with one unpaired electron left over. (This is exactly why \(\displaystyle \mathrm{NO_{2}}\) 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: Sum: \(\displaystyle 1+0-1=0 \), 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 firstconnected by \(\displaystyle \leftrightarrow\). 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.\(\displaystyle \text{NO}_3^- \) (nitrate ion)Valence electrons: \(\displaystyle 5 + 3(6) + 1(\text{for the negative charge}) = 24 \) electrons \(\displaystyle = 12 \) pairs — the extra "+$\displaystyle 1$" is because a negative ion carries one more electron than the sum of the neutral atoms. The structure works out exactly like \(\displaystyle \mathrm{SO_{3}}\): one N=O double bond and two N–O single bonds, since that is again the only arrangement of $\displaystyle 24$ electrons that gives every atom an octet.Formal charges: Sum: \(\displaystyle 1+0-1-1 = -1 \), 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($\displaystyle 1$), N–O⁻ to O($\displaystyle 2$) and O($\displaystyle 3$) Structure II — N=O to O($\displaystyle 2$), N–O⁻ to O($\displaystyle 1$) and O($\displaystyle 3$) Structure III — N=O to O($\displaystyle 3$), N–O⁻ to O($\displaystyle 1$) and O($\displaystyle 2$)again joined by \(\displaystyle \leftrightarrow\). The nitrate hybrid has three N–O bonds of identical length, each with bond order \(\displaystyle \dfrac{4}{3} \) — this is exactly why the N–O bond length measured experimentally in \(\displaystyle \text{NO}_3^- \) is one single value, intermediate between a genuine single and double bond, the same way it is for every S–O bond in \(\displaystyle \mathrm{SO_{3}}\).Answer: \(\displaystyle \mathrm{SO_{3}}\) has three resonance structures (double bond rotating among the three equivalent oxygens, each singly-bonded O carrying \(\displaystyle -1\) and S carrying \(\displaystyle +2\)); the hybrid gives every S–O bond order \(\displaystyle 4/3 \). \(\displaystyle \mathrm{NO_{2}}\), being an odd-electron ($\displaystyle 17$-valence-electron) free radical, has two resonance structures — mirror images with the unpaired electron and \(\displaystyle +1\) formal charge on N, one O neutral (double-bonded) and the other \(\displaystyle -1\) (single-bonded). \(\displaystyle \mathrm{NO_{3}}\)⁻ has three resonance structures exactly like \(\displaystyle \mathrm{SO_{3}}\) (N carries \(\displaystyle +1\), two O's carry \(\displaystyle -1\) each, one O is neutral, double bond rotating among the three O's), giving every N–O bond order \(\displaystyle 4/3 \) and identical experimental bond lengths.

An ionic bond forms when one atom empties its outer shell and another fills its outer shell to $\displaystyle 8$ 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 ($\displaystyle 8$ electrons in the outer shell, except for atoms very close to helium). A metal with only $\displaystyle 1$–$\displaystyle 3$ valence electrons finds it easier to lose them (small ionization enthalpy) than to gain $\displaystyle 5$–$\displaystyle 7$ more; a non-metal with $\displaystyle 5$–$\displaystyle 7$ 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 SPotassium, \(\displaystyle Z = 19 \), configuration \(\displaystyle [\text{Ar}]4s^{1} \), has $\displaystyle 1$ valence electron (group $\displaystyle 1$). Its Lewis symbol is a single dot: \[\overset{\displaystyle \bullet}{\text{K}} \] Sulphur, \(\displaystyle Z = 16 \), configuration \(\displaystyle [\text{Ne}]3s^{2}3p^{4} \), has $\displaystyle 6$ valence electrons (group $\displaystyle 16$) and needs $\displaystyle 2$ more to reach $\displaystyle 8$: \[\substack{\bullet\,\bullet \\ \bullet\,\text{S}\,\bullet \\ \bullet\,\bullet} \] One potassium atom can only give away $\displaystyle 1$ electron, but sulphur needs $\displaystyle 2$ — this is the mismatch that trips people up. It is fixed by using two potassium atoms per sulphur atom, so the electrons lost ($\displaystyle 1$ + $\displaystyle 1$ = $\displaystyle 2$) exactly equal the electrons gained ($\displaystyle 2$): \[\overset{\displaystyle \bullet}{\text{K}} \;+\; \overset{\displaystyle \bullet}{\text{K}} \;+\; \substack{\bullet\,\bullet \\ \bullet\,\text{S}\,\bullet \\ \bullet\,\bullet} \;\longrightarrow\; \text{K}^{+} \;+\; \text{K}^{+} \;+\; \left[\,\substack{\bullet\bullet \\ \bullet\bullet\,\text{S}\,\bullet\bullet \\ \bullet\bullet}\,\right]^{2-} \] Each \(\displaystyle \text{K}^{+} \) 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. \(\displaystyle \text{S}^{2-} \) now carries $\displaystyle 8$ dots (a full octet, also argon-like: $\displaystyle 18$ electrons). The compound formed is \(\displaystyle \text{K}_2\text{S} \).(b) Ca and OCalcium, \(\displaystyle Z = 20 \), configuration \(\displaystyle [\text{Ar}]4s^{2} \), has $\displaystyle 2$ valence electrons (group $\displaystyle 2$): \[\substack{\bullet \\ \text{Ca} \\ \bullet} \] Oxygen, \(\displaystyle Z = 8 \), configuration \(\displaystyle [\text{He}]2s^{2}2p^{4} \), has $\displaystyle 6$ valence electrons (group $\displaystyle 16$) and needs $\displaystyle 2$ more: \[\substack{\bullet\,\bullet \\ \bullet\,\text{O}\,\bullet \\ \bullet\,\bullet} \] Here the numbers already match — calcium has exactly the $\displaystyle 2$ electrons oxygen needs — so the transfer is one atom to one atom: \[\substack{\bullet \\ \text{Ca} \\ \bullet} \;+\; \substack{\bullet\,\bullet \\ \bullet\,\text{O}\,\bullet \\ \bullet\,\bullet} \;\longrightarrow\; \text{Ca}^{2+} \;+\; \left[\,\substack{\bullet\bullet \\ \bullet\bullet\,\text{O}\,\bullet\bullet \\ \bullet\bullet}\,\right]^{2-} \] \(\displaystyle \text{Ca}^{2+} \) is drawn with no dots (its remaining shell is the complete, argon-like core, $\displaystyle 18$ electrons); \(\displaystyle \text{O}^{2-} \) has a full octet of $\displaystyle 8$ dots (neon-like, $\displaystyle 10$ electrons). The compound formed is \(\displaystyle \text{CaO} \).(c) Al and NAluminium, \(\displaystyle Z = 13 \), configuration \(\displaystyle [\text{Ne}]3s^{2}3p^{1} \), has $\displaystyle 3$ valence electrons (group $\displaystyle 13$): \[\substack{\bullet \\ \text{Al}\,\bullet \\ \bullet} \] Nitrogen, \(\displaystyle Z = 7 \), configuration \(\displaystyle [\text{He}]2s^{2}2p^{3} \), has $\displaystyle 5$ valence electrons (group $\displaystyle 15$) and needs $\displaystyle 3$ more: \[\substack{\bullet\,\bullet \\ \bullet\,\text{N}\,\bullet \\ \bullet} \] The numbers match again — aluminium has exactly the $\displaystyle 3$ electrons nitrogen is short of — so it is a one-to-one transfer: \[\substack{\bullet \\ \text{Al}\,\bullet \\ \bullet} \;+\; \substack{\bullet\,\bullet \\ \bullet\,\text{N}\,\bullet \\ \bullet} \;\longrightarrow\; \text{Al}^{3+} \;+\; \left[\,\substack{\bullet\bullet \\ \bullet\bullet\,\text{N}\,\bullet\bullet \\ \bullet\bullet}\,\right]^{3-} \] \(\displaystyle \text{Al}^{3+} \) is left with no dots (neon-like core, $\displaystyle 10$ electrons); \(\displaystyle \text{N}^{3-} \) is drawn with a full octet of $\displaystyle 8$ dots (also neon-like, $\displaystyle 10$ electrons). The compound formed is \(\displaystyle \text{AlN} \).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 $\displaystyle 0$ or $\displaystyle 8$ dots in their outer shell (never anything in between).Answer: (a) \(\displaystyle 2\text{K} + \text{S} \rightarrow 2\text{K}^{+} + \text{S}^{2-} \) (forming \(\displaystyle \text{K}_2\text{S} \)); (b) \(\displaystyle \text{Ca} + \text{O} \rightarrow \text{Ca}^{2+} + \text{O}^{2-} \) (forming \(\displaystyle \text{CaO} \)); (c) \(\displaystyle \text{Al} + \text{N} \rightarrow \text{Al}^{3+} + \text{N}^{3-} \) (forming \(\displaystyle \text{AlN} \)) — 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.

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 ideaEvery polar bond has a bond dipole moment, \[\mu = q \times d \] where \(\displaystyle q\) is the magnitude of charge separated at the two ends of the bond and \(\displaystyle d\) 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: 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 \(\displaystyle \mu\) separated by the bond angle \(\displaystyle \theta\) at the central atom, the resultant is \[\mu_{\text{net}} = 2\mu\cos\left(\frac{\theta}{2}\right) \] Here \(\displaystyle \theta\) 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 CO₂CO₂ is experimentally found to have zero net dipole moment. Putting \(\displaystyle \mu_{\text{net}} = 0\) into the formula above: \[0 = 2\mu_{C=O}\cos\left(\frac{\theta}{2}\right) \] Since the C=O bond itself is polar, \(\displaystyle \mu_{C=O} \neq 0\), so the only way the equation holds is \[\cos\left(\frac{\theta}{2}\right) = 0 \quad\Rightarrow\quad \frac{\theta}{2} = 90^\circ \quad\Rightarrow\quad \theta = 180^\circ \] A bond angle of \(\displaystyle 180^\circ\) means the two C=O bond dipoles point in exactly opposite directions along the same line, so they cancel each other out perfectly: \[\text{O} \xleftarrow{\;\mu\;} \text{C} \xrightarrow{\;\mu\;} \text{O} \] 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 H₂OH₂O is experimentally found to have a substantial net dipole moment, \(\displaystyle \mu_{\text{net}} = 1.84\ \text{D}\) — clearly not zero.If water had the same linear shape as CO₂ (\(\displaystyle \theta = 180^\circ\)), the same algebra as above would force \(\displaystyle \mu_{\text{net}} = 0\), because the two O–H bond dipoles would again point in exactly opposite directions and cancel. That directly contradicts the measured $\displaystyle 1.84$ D.So \(\displaystyle \theta\) cannot be \(\displaystyle 180^\circ\). 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 \(\displaystyle 180^\circ\) — 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: \[\text{H} \nwarrow^{\mu} \;\text{O}\; \nearrow^{\mu} \text{H} \quad\longrightarrow\quad \mu_{\text{net}} = 2\mu_{O-H}\cos\left(\frac{\theta}{2}\right) \neq 0 \] (The actual H–O–H angle is close to \(\displaystyle 104.5^\circ\), 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	$\displaystyle 2$ 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

Answer: CO₂ is linear (O=C=O, bond angle \(\displaystyle 180^\circ\)) 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 \(\displaystyle \approx 104.5^\circ\)) because its net dipole moment is not zero (\(\displaystyle 1.84\ \text{D}\)); 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 \(\displaystyle 180^\circ\), i.e., if the molecule is non-linear.

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 \[\mu = q \times d \] where \(\displaystyle q\) is the magnitude of charge separated at each end of the bond and \(\displaystyle d\) 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 polarA 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. 2. Using dipole moment to confirm molecular shapeBecause dipole moment is a vector sum of the individual bond dipoles, its value (especially when it is zero) tells you the geometry. 3. Distinguishing cis and trans isomersFor 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.This gives a physical way to tell the two isomers apart even before doing any other test.4. Distinguishing ortho, meta, and para disubstituted benzenesFor a disubstituted benzene \(\displaystyle \mathrm{C_6H_4XY}\), the angle between the two substituent dipoles is different in each isomer (about $\displaystyle 60$° for ortho, $\displaystyle 120$° for meta, $\displaystyle 180$° for para). Since a smaller angle between two dipole vectors gives a larger vector sum, the dipole moments follow \[\mu_{ortho} > \mu_{meta} > \mu_{para} \] and when \(\displaystyle X = Y\) (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 bondNo real bond is $\displaystyle 100$% 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: \[\% \text{ ionic character} = \frac{\mu_{observed}}{\mu_{calculated \, (100\% \, ionic)}} \times 100 \] Here \(\displaystyle \mu_{observed}\) is the dipole moment measured experimentally, and \(\displaystyle \mu_{calculated}\) is the dipole moment you would get from \(\displaystyle \mu = q \times d\) if one full electronic charge were transferred completely (a purely ionic bond) over the same bond length \(\displaystyle d\). 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 $\displaystyle 17$%, 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 ($\displaystyle 1$) to tell whether a molecule is polar or nonpolar, ($\displaystyle 2$) to work out or confirm a molecule's shape (a zero dipole moment forces a symmetric geometry such as linear \(\displaystyle \mathrm{BeF_2}\), while a nonzero one forces an asymmetric geometry such as bent \(\displaystyle \mathrm{H_2O}\)), ($\displaystyle 3$) to distinguish cis (higher \(\displaystyle \mu\)) from trans (lower \(\displaystyle \mu\)) isomers, ($\displaystyle 4$) to distinguish ortho (highest \(\displaystyle \mu\)), meta, and para (lowest/zero \(\displaystyle \mu\)) disubstituted benzenes, and ($\displaystyle 5$) to calculate the percentage ionic character of a bond via \(\displaystyle \%\text{ ionic character} = (\mu_{observed}/\mu_{calculated\ for\ 100\%\ ionic\ bond}) \times 100\), as for HCl (~$\displaystyle 17$% ionic).

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 meansElectronegativity 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, \(\displaystyle \overset{\delta+}{\text{H}} - \overset{\delta-}{\text{Cl}} \).Two things about electronegativity that trip students up: What electron gain enthalpy meansElectron gain enthalpy, \(\displaystyle \Delta_{eg}H \), 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 \(\displaystyle -1\):\[X(g) + e^- \rightarrow X^-(g) \qquad \Delta_{eg}H \]Here \(\displaystyle X(g)\) is the atom in the gas phase (isolated, no bonding partner) and \(\displaystyle \Delta_{eg}H\) 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, \(\displaystyle \text{kJ mol}^{-1}\) — for example \(\displaystyle \Delta_{eg}H\) for chlorine is about \(\displaystyle -349\ \text{kJ mol}^{-1}\) (energy is released, hence negative, because the incoming electron is attracted by the nucleus).The core contrasts, side by side 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 \(\displaystyle \text{kJ mol}^{-1}\)) when one mole of isolated gaseous atoms each gains one electron to form gaseous anions, \(\displaystyle X(g)+e^-\rightarrow X^-(g)\). 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).

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 \(\displaystyle (\delta^-) \) and the other atom with an equal partial positive charge \(\displaystyle (\delta^+) \).Example: the H–Cl bond in hydrogen chlorideElectronegativity (Pauling scale) is a measure of how strongly an atom in a bond attracts the shared electron pair toward itself: \[\chi(\mathrm{H}) = 2.1, \qquad \chi(\mathrm{Cl}) = 3.0 \]Because \(\displaystyle \chi(\mathrm{Cl}) > \chi(\mathrm{H})\), 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: \[\overset{\delta^+}{\mathrm{H}} - \overset{\delta^-}{\mathrm{Cl}} \]Here \(\displaystyle \delta^+\) and \(\displaystyle \delta^-\) 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 \(\displaystyle \mathrm{Na}^+\) and \(\displaystyle \mathrm{Cl}^-\). 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, \[\mu = Q \times d \] where \(\displaystyle Q\) is the magnitude of the partial charge separated and \(\displaystyle d\) is the distance between the charge centres (essentially the bond length). For HCl, \(\displaystyle \mu \approx 1.03\ \mathrm{D}\) (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 \(\displaystyle \mu = 0\) 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 ( \(\displaystyle \Delta\chi\) 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 \(\displaystyle (\delta^-)\) and the other atom an equal partial positive charge \(\displaystyle (\delta^+)\), without any electron being fully transferred. HCl is the standard example: since \(\displaystyle \chi(\mathrm{Cl}) = 3.0 > \chi(\mathrm{H}) = 2.1\), the bond is written \(\displaystyle \overset{\delta^+}{\mathrm{H}}-\overset{\delta^-}{\mathrm{Cl}}\), and this charge separation gives HCl a measurable dipole moment (\(\displaystyle \mu \approx 1.03\ \mathrm{D}\)), unlike a nonpolar bond such as H–H where both atoms share the pair equally and \(\displaystyle \mu = 0\).

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, \(\displaystyle \chi \), 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\[\Delta\chi = |\chi_A - \chi_B| \]\(\displaystyle \Delta\chi = 0 \) means the pair is shared perfectly equally — a purely covalent bond with zero ionic character. As \(\displaystyle \Delta\chi \) 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 \(\displaystyle \Delta\chi \).Using Pauling's electronegativity values for the atoms involved:\[\chi_{Li}=0.98,\ \chi_F=3.98,\ \chi_K=0.82,\ \chi_O=3.44,\ \chi_N=3.04,\ \chi_S=2.58,\ \chi_{Cl}=3.16 \]Now take each bond in turn.LiF (Li–F bond): \[\Delta\chi = 3.98 - 0.98 = 3.00 \]K₂O (K–O bond): \[\Delta\chi = 3.44 - 0.82 = 2.62 \]N₂ (N–N bond): the two atoms are identical, so there is no pull-imbalance at all — \[\Delta\chi = 3.04 - 3.04 = 0 \] 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): \[\Delta\chi = 3.44 - 2.58 = 0.86 \]ClF₃ (Cl–F bond): \[\Delta\chi = 3.98 - 3.16 = 0.82 \]The step that trips people up here is reaching for the rounded, whole-number electronegativities many textbooks print for quick use (Li = $\displaystyle 1.0$, F = $\displaystyle 4.0$, K = $\displaystyle 0.8$, O = $\displaystyle 3.5$, S = $\displaystyle 2.5$, Cl = $\displaystyle 3.0$, N = $\displaystyle 3.0$). Those round off SO₂ and ClF�$\displaystyle 3$ to the same difference, \(\displaystyle \Delta\chi = 1.0 \), and make it look like a tie. Using Pauling's actual decimal values, as above, the tie breaks: the Cl–F gap ($\displaystyle 0.82$) is very slightly smaller than the S–O gap ($\displaystyle 0.86$), because chlorine ($\displaystyle 3.16$) sits a bit closer to fluorine ($\displaystyle 3.98$) than sulfur ($\displaystyle 2.58$) sits to oxygen ($\displaystyle 3.44$).Putting the five \(\displaystyle \Delta\chi \) 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 \(\displaystyle \Delta\chi \).Answer: N₂ < ClF₃ < SO₂ < K₂O < LiF, in order of increasing ionic character — N₂ is purely covalent (Δχ = $\displaystyle 0$, identical atoms), while LiF is the most ionic (Δχ = $\displaystyle 3.00$, the largest electronegativity gap of the five).

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 $\displaystyle 8$ 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 $\displaystyle 1$ — count the total valence electrons. Acetic acid is \(\displaystyle \text{CH}_3\text{COOH} \), i.e. \(\displaystyle \text{C}_2\text{H}_4\text{O}_2 \): $\displaystyle 2$ carbons, $\displaystyle 4$ hydrogens, $\displaystyle 2$ oxygens. Each atom contributes electrons equal to its group number in the old (A) notation — carbon (Group $\displaystyle 14$) gives $\displaystyle 4$, hydrogen (Group $\displaystyle 1$) gives $\displaystyle 1$, oxygen (Group $\displaystyle 16$) gives 6.\[\text{Total valence electrons} = 2(4) + 4(1) + 2(6) = 8 + 4 + 12 = 24 \text{ electrons} = 12 \text{ electron pairs} \]Step $\displaystyle 2$ — fix the skeleton, then ask who is short a bond. Call the two carbons \(\displaystyle C_1\) (the one carrying the three H's) and \(\displaystyle C_2\). The connectivity that any correct sketch of acetic acid must show is: \(\displaystyle C_1\) bonded to $\displaystyle 3$ H's and to \(\displaystyle C_2\); \(\displaystyle C_2\) bonded to two different oxygens, \(\displaystyle O_a\) and \(\displaystyle O_b\); and \(\displaystyle O_b\) 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: 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 $\displaystyle 3$ — the missing bond must be a C=O double bond, and it can only go on one of the two oxygens. \(\displaystyle C_2\) 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 \(\displaystyle C_2\)–\(\displaystyle O_b\) bond, because \(\displaystyle O_b\) already has two bonds (to \(\displaystyle C_2\) and to H); oxygen stops at two bonds and two lone pairs, so pushing a third bond onto \(\displaystyle O_b\) 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 \(\displaystyle O_a\), the oxygen attached to nothing but \(\displaystyle C_2\):\[C_2 : \; 1(\text{C–}C_1) + 2(\text{C=}O_a) + 1(\text{C–}O_b) = 4 \text{ bonds}, \; 8 \text{ electrons — a full octet, four bonds, exactly as carbon requires.} \]Step $\displaystyle 4$ — give both oxygens their lone pairs so each reaches an octet. Step $\displaystyle 5$ — check the electron count balances, and that no atom needs a formal charge. Bonding pairs in the finished structure: $\displaystyle 3$ (\(\displaystyle C_1\)–H) + $\displaystyle 1$ (\(\displaystyle C_1\)–\(\displaystyle C_2\)) + $\displaystyle 2$ (\(\displaystyle C_2\)=\(\displaystyle O_a\)) + $\displaystyle 1$ (\(\displaystyle C_2\)–\(\displaystyle O_b\)) + $\displaystyle 1$ (\(\displaystyle O_b\)–H) = $\displaystyle 8$ pairs. Lone pairs: $\displaystyle 2$ on \(\displaystyle O_a\) + $\displaystyle 2$ on \(\displaystyle O_b\) = $\displaystyle 4$ pairs. Total = $\displaystyle 8$ + $\displaystyle 4$ = $\displaystyle 12$ pairs — exactly the $\displaystyle 24$ electrons counted in Step $\displaystyle 1$, with nothing left over and nothing missing.Formal charge, using \(\displaystyle \text{FC} = V - N - \tfrac{1}{2}B \) (\(\displaystyle V\) = valence electrons of the free atom, \(\displaystyle N\) = its non-bonding electrons, \(\displaystyle B\) = its bonding electrons), comes out to zero on every atom: 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:``` H | H — C — C = O | \ H O — H ```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 \(\displaystyle \text{CH}_3\text{-CO-OH} \), 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 \(\displaystyle C_1\) (methyl carbon) singly bonded to three H atoms and to \(\displaystyle C_2\); \(\displaystyle C_2\) doubly bonded to one oxygen (the carbonyl O, carrying $\displaystyle 2$ lone pairs) and singly bonded to the other oxygen (the hydroxyl O, carrying $\displaystyle 2$ 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 ($\displaystyle 2$ bonds + $\displaystyle 2$ lone pairs, or $\displaystyle 1$ double bond + $\displaystyle 2$ 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.