JEE Corner — Born-Haber Cycle, Fajans' Rules, Dipole Vectors, Bent's and Drago's Rules, Back-Bonding and Heteronuclear MOs

The Part of Bonding Your Textbook Left Out

Sections 1 to 11 cover what the Board paper can ask about bonding. JEE Main asks something else: the lattice enthalpy of NaCl from a Born-Haber cycle, which lithium halide is the most covalent, why AgCl is white but AgI yellow, why the bond angle of $\mathrm{PH_3}$ is 93.6° and not 107°, why $\mathrm{N(SiH_3)_3}$ is planar, whether $\mathrm{NO^+}$ has a shorter bond than NO, and above all arrange the following by lattice enthalpy, covalent character, dipole moment, bond angle, bond order, bond length or boiling point.

This section builds that toolkit on top of what you already know.

What "beyond the textbook" means here

Textbook gives you	JEE Main also wants	JEE Advanced adds
NaCl: IE 495.8, $\Delta_{eg}H$ $-348.7$, lattice enthalpy 788 kJ/mol	the full Born-Haber cycle with sublimation and bond dissociation, solving for an unknown	second electron gain enthalpy in MgO, lattice-hydration balance for solubility
Fajans' rules in two lines	pseudo-noble-gas cations, colour, melting point and solubility consequences	ionic potential, ordering mixed sets like $\mathrm{SnCl_2}$ / $\mathrm{SnCl_4}$ / $\mathrm{AgI}$ / $\mathrm{BeCl_2}$
$\mu = Q \times r$, $\mathrm{NH_3}$ vs $\mathrm{NF_3}$	vector addition with $\cos\theta$, disubstituted benzenes, per cent ionic character	bond-dipole back-calculation, $\mu = 0$ from symmetry alone
VSEPR shapes and angles	the electronegativity and size rules for angles, lone-pair positions	Bent's rule, Drago's rule, back-bonding, where the $\tfrac{1}{2}(V + M - C + A)$ formula breaks
MO diagram of $\mathrm{O_2}$, $\mathrm{N_2}$	CO, NO, $\mathrm{NO^+}$, $\mathrm{CN^-}$, HF; the $\pm\tfrac{1}{2}$ rule	s-p mixing and why $\mathrm{N_2^+}$ and $\mathrm{O_2^+}$ behave oppositely; HOMO/LUMO; magnetic moments
resonance in $\mathrm{O_3}$, $\mathrm{CO_3^{2-}}$	fractional bond orders and bond-length orders	s-character and bond length, bond-enthalpy anomalies ($\mathrm{F_2}$)
H-bonds in HF, water, ice	boiling-point orders with the $\mathrm{SbH_3} > \mathrm{NH_3}$ surprise, $\mathrm{HF_2^-}$	acidity puzzles (ortho vs para hydroxybenzoic acid), intermolecular force types

Key Point: Almost everything here is a competition between two numbers. Lattice enthalpy against hydration enthalpy decides solubility. Polarising power against polarisability decides covalent character. Central-atom electronegativity against terminal-atom electronegativity decides a bond angle. A bonding electron against an antibonding electron decides bond order. Name the two quantities that are fighting and which one wins, and you can answer the question.

The Born-Haber Cycle and What Controls Lattice Enthalpy

The cycle for NaCl

Lattice enthalpy cannot be measured directly. Born and Haber built a closed loop of steps whose enthalpies can be measured and let Hess's law do the rest. For sodium chloride, all values in kJ/mol:

Step	Process	$\Delta H$	Why this sign
1	$\mathrm{Na(s)} \rightarrow \mathrm{Na(g)}$	$\Delta_{sub}H = +108$	sublimation: breaking the metallic lattice costs energy
2	$\mathrm{Na(g)} \rightarrow \mathrm{Na^+(g)} + e^-$	$\Delta_iH = +496$	ionization is always endothermic
3	$\tfrac{1}{2}\mathrm{Cl_2(g)} \rightarrow \mathrm{Cl(g)}$	$\tfrac{1}{2}\Delta_{diss}H = +121$	half the Cl-Cl bond enthalpy (242) — one Cl atom is needed
4	$\mathrm{Cl(g)} + e^- \rightarrow \mathrm{Cl^-(g)}$	$\Delta_{eg}H = -349$	electron gain of chlorine is exothermic
5	$\mathrm{Na^+(g)} + \mathrm{Cl^-(g)} \rightarrow \mathrm{NaCl(s)}$	$-\Delta_{lattice}H = -788$	opposite charges coming together release energy
Sum	$\mathrm{Na(s)} + \tfrac{1}{2}\mathrm{Cl_2(g)} \rightarrow \mathrm{NaCl(s)}$	$\Delta_fH$	the enthalpy of formation

$$\Delta_fH = \Delta_{sub}H + \Delta_iH + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H - \Delta_{lattice}H = 108 + 496 + 121 - 349 - 788 = -412\ \mathrm{kJ/mol}$$

The measured value is $-411$ kJ/mol; the 1 kJ gap is rounding. The first three steps cost 725 kJ and the electron gain returns only 349, so it is the lattice — 788 kJ released — that makes the whole thing worthwhile. Sodium chloride does not exist because sodium "wants" to give an electron to chlorine (that transfer alone is uphill by $496 - 349 = +147$ kJ/mol); it exists because the lattice pays.

Sign convention — read the question

Lattice enthalpy is the enthalpy change when one mole of the ionic solid is separated into gaseous ions: $\mathrm{NaCl(s)} \rightarrow \mathrm{Na^+(g)} + \mathrm{Cl^-(g)}$, $\Delta_{lattice}H = +788$ kJ/mol. Many JEE questions instead quote the enthalpy of lattice formation, $-788$ kJ/mol, for the reverse process. Same magnitude, opposite sign. Give each arrow the sign its direction needs; when you compare "lattice enthalpies", compare magnitudes.

Solving for the unknown

Given five of the six numbers, rearrange for the sixth:

$$\boxed{\Delta_{lattice}H = \Delta_{sub}H + \Delta_iH + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H - \Delta_fH}$$

For NaCl with $\Delta_fH = -411$: $108 + 496 + 121 - 349 + 411 = 787$ kJ/mol. The same rearrangement run for KCl is Question 1(b). Four things to watch: halve the $\mathrm{Cl_2}$ bond enthalpy (242 per mole of $\mathrm{Cl_2}$, 121 per mole of Cl atoms); for a bromide or iodide add the vaporisation or sublimation step that turns liquid $\mathrm{Br_2}$ or solid $\mathrm{I_2}$ into gas before the bond is broken; for a divalent metal like Mg put in both ionization enthalpies ($738 + 1451$); for an oxide put in both electron gain enthalpies ($-141$ and $+780$), the second one positive.

What controls the size of the lattice enthalpy

Coulomb's law for a crystal gives, to a good approximation,

$$\Delta_{lattice}H \propto \frac{|z_+ z_-|}{r_+ + r_-}$$

The Kapustinskii equation is good enough for estimates: $U = \dfrac{K\,\nu\,|z_+z_-|}{r_+ + r_-}\left(1 - \dfrac{d}{r_+ + r_-}\right)$, where $\nu$ is the number of ions per formula unit and $d = 34.5$ pm. You need only recognise its shape. Two levers:

Lever	Effect	Data (kJ/mol)
Charge product $\lvert z_+z_- \rvert$	dominant: doubling both charges roughly quadruples the pull, and the ions are smaller too	NaCl 788; MgO about 3900; CaO 3400; $\mathrm{Al_2O_3}$ about 15,000
Sum of radii $r_+ + r_-$	smaller ions, closer nuclei, larger lattice enthalpy	LiF 1037 $>$ NaF 923 $>$ KF 821 $>$ RbF 785 $>$ CsF 740; NaF 923 $>$ NaCl 788 $>$ NaBr 747 $>$ NaI 704

The rule for ordering: compare charges first; if the charges match, the smaller ion pair wins. MgO beats NaCl by a factor of five although $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ are not much smaller than $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ — the charge product 4 against 1 does most of the work. Among the alkali halides, LiF has the highest lattice enthalpy and CsI the lowest (about 600).

Lattice enthalpy against hydration enthalpy — solubility

Dissolving costs $\Delta_{lattice}H$ and returns $\Delta_{hyd}H$ (negative) when water surrounds the ions. The sign of

$$\Delta_{sol}H = \Delta_{lattice}H + \Delta_{hyd}H(\text{cation}) + \Delta_{hyd}H(\text{anion})$$

is a small difference between two big numbers. NaCl: $+788 + (-406) + (-364) = +18$ kJ/mol, slightly endothermic, so it dissolves a little more in hot water. Both terms fall as ions get bigger; which falls faster decides the trend.

• Anion small ($\mathrm{F^-}$, $\mathrm{OH^-}$): the lattice enthalpy is dominated by the cation's size and drops steeply down the group, faster than hydration enthalpy does. Solubility rises down the group: $\mathrm{Be(OH)_2} < \mathrm{Mg(OH)_2} < \mathrm{Ca(OH)_2} < \mathrm{Sr(OH)_2} < \mathrm{Ba(OH)_2}$; same for the group-2 fluorides.
• Anion large ($\mathrm{SO_4^{2-}}$, $\mathrm{CO_3^{2-}}$): the lattice enthalpy is set by the big anion and hardly changes down the group, but the cation's hydration enthalpy keeps falling. Solubility falls: $\mathrm{BeSO_4} > \mathrm{MgSO_4} > \mathrm{CaSO_4} > \mathrm{SrSO_4} > \mathrm{BaSO_4}$ — hence the barium sulfate test for sulfate.
• Size mismatch means soluble: LiF is the least soluble lithium halide (both ions tiny, lattice enthalpy 1037 wins), CsI the least soluble caesium halide (both huge, hydration poor). LiI and CsF are the soluble extremes.

Melting points follow the lattice, with one warning

For genuinely ionic solids, melting point tracks lattice enthalpy: NaF 993 °C $>$ NaCl 801 $>$ NaBr 747 $>$ NaI 661; MgO 2852 °C against NaCl 801. But LiF (845 °C) melts below NaF (993 °C) despite its higher lattice enthalpy, and $\mathrm{BeCl_2}$ (405 °C) far below $\mathrm{MgCl_2}$ (714 °C). The warning is covalent character: a small, hard cation polarises the anion and the solid is no longer purely ionic.

[JEE Main] Write the five steps in a column with signs before the arithmetic, then check your answer against the ballpark: alkali halides 600 to 1050, group-2 oxides 3000 to 4000 kJ/mol. An answer of 200 or 8000 is a sign error.

Fajans' Rules — When an Ionic Bond Is Not Really Ionic

The picture

Put a small, highly charged cation next to a big, soft anion. The cation's field distorts the anion's outer electron cloud until some electron density sits between the two nuclei — which is what a covalent bond is. Fajans (1923) listed what makes this polarisation worse. A cation's ability to distort is its polarising power; an anion's willingness to be distorted is its polarisability. These are trends that rank compounds, not laws that give numbers.

Factor	More covalent character when…	Reason	Example
Cation size	cation is small	field at the surface is $\propto z/r^2$	$\mathrm{LiCl}$ more covalent than $\mathrm{KCl}$
Cation charge	cation charge is high	same reason	$\mathrm{AlCl_3}$ (covalent, sublimes at 180 °C) vs $\mathrm{NaCl}$
Anion size	anion is large	outer electrons far from its nucleus, loosely held	$\mathrm{LiI}$ more covalent than $\mathrm{LiF}$
Anion charge	anion charge is high	extra electrons are loosely held	$\mathrm{Na_2S}$ more covalent than $\mathrm{NaCl}$
Cation configuration	cation has a pseudo-noble-gas ($18$-electron, $d^{10}$) or $(18+2)$ shell	d electrons shield the nuclear charge poorly, so the effective field is higher than a noble-gas cation of the same size	$\mathrm{CuCl}$ vs $\mathrm{NaCl}$; $\mathrm{AgCl}$ vs $\mathrm{KCl}$; $\mathrm{ZnCl_2}$ vs $\mathrm{MgCl_2}$

The first two factors merge into the ionic potential $\phi = z/r$: higher ionic potential, greater polarising power. $\mathrm{Be^{2+}}$ ($2/45$ pm) beats $\mathrm{Ba^{2+}}$ ($2/135$) easily, which is why beryllium chloride is a covalent, polymeric solid soluble in organic solvents while barium chloride is a textbook ionic salt.

Key Point (Definition): Polarising power of a cation increases with its charge and decreases with its size (high $z/r$). Polarisability of an anion increases with its size and charge. High polarising power plus high polarisability means a distorted anion, shared electron density, and covalent character in a nominally ionic bond.

The pseudo-noble-gas cation

$\mathrm{Na^+}$ (102 pm) and $\mathrm{Cu^+}$ (96 pm) match in size and charge, so size and charge alone say NaCl and CuCl should be equally ionic. They are not: CuCl is insoluble in water, NaCl freely soluble. $\mathrm{Na^+}$ is $[\mathrm{Ne}]$, a tight octet; $\mathrm{Cu^+}$ is $[\mathrm{Ar}]3d^{10}$, and its outermost ten d electrons shield the nucleus badly, so a chloride ion next to $\mathrm{Cu^+}$ feels a stronger pull. The same story separates $\mathrm{Ag^+}$ from $\mathrm{K^+}$ (AgCl insoluble, KCl soluble), $\mathrm{Zn^{2+}}$ from $\mathrm{Mg^{2+}}$ (radii 74 and 72 pm; $\mathrm{ZnCl_2}$ melts at 290 °C, $\mathrm{MgCl_2}$ at 714 °C), and $\mathrm{Hg^{2+}}$ from $\mathrm{Ca^{2+}}$ ($\mathrm{HgCl_2}$ is a molecular solid soluble in ethanol). Rule: for the same size and charge, an $18$-electron cation polarises more than an $8$-electron one.

What covalent character does to the compound

Property	More covalent character means…	Evidence
Melting / boiling point	lower — lattice partly molecular	$\mathrm{BeCl_2}$ 405 °C $<$ $\mathrm{MgCl_2}$ 714 $<$ $\mathrm{CaCl_2}$ 772; $\mathrm{AlCl_3}$ sublimes at 180 °C; $\mathrm{SnCl_4}$ is a liquid (b.p. 114 °C) while $\mathrm{SnCl_2}$ is a solid (m.p. 247 °C)
Solubility in water	lower; solubility in organic solvents higher	LiCl dissolves in ethanol and pyridine; AgCl, CuCl, $\mathrm{HgI_2}$ insoluble in water
Electrical conductivity of the melt	lower	molten $\mathrm{BeCl_2}$ is a poor conductor
Colour	deeper — polarisation shrinks the energy gap for moving charge from anion to cation, so the solid absorbs visible light	AgCl white, AgBr pale yellow, AgI yellow; $\mathrm{HgCl_2}$ white, $\mathrm{HgI_2}$ red; $\mathrm{PbCl_2}$ white, $\mathrm{PbI_2}$ yellow; $\mathrm{Ag_2S}$ black
Thermal stability of carbonates, nitrates	lower — the polarising cation pulls $\mathrm{O^{2-}}$ out of the anion	$\mathrm{Li_2CO_3}$ decomposes on gentle heating, $\mathrm{Na_2CO_3}$ does not; $\mathrm{BeCO_3}$ unstable, $\mathrm{BaCO_3}$ stable to 1360 °C

The colour rule deserves a second look. NaCl, KCl and CsCl are all white — no polarisation to speak of. Silver halides run white to yellow as the anion gets bigger and softer: the more polarisable $\mathrm{I^-}$ hands electron density to $\mathrm{Ag^+}$ more readily, the charge-transfer transition drops into the visible, and the solid turns yellow.

Ready-made orderings

Set	Order	Which factor
Ionic character, group 2 chlorides	$\mathrm{BeCl_2} < \mathrm{MgCl_2} < \mathrm{CaCl_2} < \mathrm{SrCl_2} < \mathrm{BaCl_2}$	cation size grows, polarising power falls
Covalent character, lithium halides	$\mathrm{LiF} < \mathrm{LiCl} < \mathrm{LiBr} < \mathrm{LiI}$ — LiI most covalent	anion polarisability grows
Covalent character, period 3 chlorides	$\mathrm{NaCl} < \mathrm{MgCl_2} < \mathrm{AlCl_3} < \mathrm{SiCl_4} < \mathrm{PCl_5}$	cation charge grows, size falls
Same element, two oxidation states	$\mathrm{SnCl_2}$ (ionic, solid) $<$ $\mathrm{SnCl_4}$ (covalent, liquid); $\mathrm{PbCl_2} < \mathrm{PbCl_4}$; $\mathrm{FeCl_2} < \mathrm{FeCl_3}$; $\mathrm{Hg_2Cl_2} < \mathrm{HgCl_2}$	higher charge, smaller cation
Same anion, pseudo-noble-gas vs noble-gas cation	$\mathrm{NaCl} < \mathrm{CuCl}$; $\mathrm{KCl} < \mathrm{AgCl}$; $\mathrm{MgCl_2} < \mathrm{ZnCl_2}$; $\mathrm{CaCl_2} < \mathrm{HgCl_2}$	$d^{10}$ shielding
Melting point, sodium halides	$\mathrm{NaI} < \mathrm{NaBr} < \mathrm{NaCl} < \mathrm{NaF}$	lattice enthalpy, all still ionic
Melting point, alkali chlorides	$\mathrm{LiCl}$ (605 °C) $<$ $\mathrm{NaCl}$ (801) $>$ $\mathrm{KCl}$ (770) $>$ $\mathrm{RbCl}$ (718) $>$ $\mathrm{CsCl}$ (645)	NaCl is the peak: LiCl loses to covalent character, the rest to falling lattice enthalpy
Mixed set	$\mathrm{AgI} > \mathrm{AgBr} > \mathrm{AgCl} > \mathrm{AgF}$ covalent; AgF is the only water-soluble silver halide	anion polarisability

[JEE Main] When a set mixes the factors, decide the winner in this order: (1) cation charge, (2) cation size, (3) anion size, (4) $d^{10}$ vs octet cation. Keep the sign of each consequence straight — covalent character lowers melting point and water solubility, raises solubility in organic solvents, deepens colour, lowers thermal stability of oxo-salts. "Most covalent" and "lowest melting point" are usually the same answer; "most ionic" and "most soluble in water" usually match too, unless lattice enthalpy (LiF, CsI) intervenes.

Dipole Moments as Vectors

The cosine rule

A bond dipole is a vector: magnitude $\mu = Q \times r$, direction along the bond from the positive to the negative end (chemists draw the arrow with its head at the negative end). The molecular dipole is the vector sum. For two bond dipoles $\mu_1$ and $\mu_2$ with angle $\theta$ between them,

$$\boxed{\mu_R = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}} \qquad \text{and for equal dipoles} \qquad \mu_R = 2\mu\cos\frac{\theta}{2}$$

Water. Each O-H bond dipole is about 1.5 D and the angle is 104.5°: $\mu_R = 2 \times 1.5 \times \cos 52.25^\circ = 3.0 \times 0.612 = 1.84$ D, matching the measured 1.85 D. Backwards, from $\mu = 1.85$ D and 104.5°, the bond dipole is $1.85/(2\cos 52.25^\circ) = 1.51$ D. (The lone pairs also contribute; the "bond dipole" you get this way quietly includes their share.)

Ammonia and $\mathrm{NF_3}$. Three N-H dipoles pointing toward N add to a resultant along the pyramid axis, and the lone-pair dipole points the same way — total 1.47 D. In $\mathrm{NF_3}$ the three N-F dipoles point away from N (F is more electronegative) while the lone pair still points away from the fluorines, so the two contributions oppose and the total is only 0.23 D. Same shape, opposite arithmetic. Compare $\mathrm{NH_3}$ 1.47 D against $\mathrm{PH_3}$ 0.58 D: the P-H bonds are almost non-polar (electronegativity 2.1 against 2.1) and the lone pair is in an orbital with high s-character (Drago, later), which is less directional.

Disubstituted benzenes — the ratio question

Take two identical C-Cl bond dipoles $\mu$ on a benzene ring. The angle between them is 60° for ortho, 120° for meta and 180° for para:

Isomer	$\theta$	$\mu_R = 2\mu\cos(\theta/2)$	Ratio	Measured (D)
ortho	60°	$2\mu\cos 30^\circ = \sqrt{3}\,\mu$	$\sqrt{3}$	2.50
meta	120°	$2\mu\cos 60^\circ = \mu$	1	1.72
para	180°	0	0	0

So $\mu_{ortho} : \mu_{meta} : \mu_{para} = \sqrt{3} : 1 : 0$ for two identical substituents, and the meta isomer has the same dipole as the monosubstituted compound (chlorobenzene 1.69 D — close to 1.72). The measured ortho value (2.50) falls short of $\sqrt{3} \times 1.72 = 2.98$ because two chlorines crowded on adjacent carbons partly cancel each other's electron pull; the ideal ratio is what the exam wants unless it hands you data. For 1,3,5-trisubstitution, three equal dipoles at 120° sum to zero. For two different substituents use the full cosine rule, and remember that an electron-donating group's dipole points into the ring: in para-chlorotoluene ($\mathrm{Cl}$ withdrawing, $\mathrm{CH_3}$ donating) the dipoles reinforce, while in para-nitrochlorobenzene they partly cancel — p-chloronitrobenzene 2.6 D $<$ nitrobenzene 4.2 D.

The chloromethanes: $\mathrm{CH_3Cl} > \mathrm{CH_2Cl_2} > \mathrm{CHCl_3} > \mathrm{CCl_4}$

Measured: 1.87, 1.60, 1.04, 0 D. $\mathrm{CCl_4}$ is easy — four equal tetrahedral dipoles cancel exactly. The rest need one geometric fact: in a regular tetrahedron the vector sum of any three bond directions equals minus the fourth, so three C-Cl dipoles at 109.5° add to exactly one C-Cl dipole along the C-H axis. On paper $\mathrm{CHCl_3}$ and $\mathrm{CH_3Cl}$ should then both be worth about one C-Cl dipole plus what the C-H bonds add. They are not equal because three chlorines competing for the same carbon each get less electron density than a lone chlorine does, so each C-Cl bond in $\mathrm{CHCl_3}$ is weaker as a dipole and the resultant (1.04 D) is little more than half of $\mathrm{CH_3Cl}$'s (1.87 D). $\mathrm{CH_2Cl_2}$ sits between: two C-Cl dipoles at 109.5° give $2\mu\cos 54.75^\circ = 1.15\,\mu$, and with the reduced bond polarity the measured 1.60 D lands where you expect.

cis against trans

Two identical polar substituents on a C=C: trans puts them 180° apart in a planar molecule, so the dipoles cancel (trans-1,2-dichloroethene 0 D); cis puts them roughly 60° apart and they add (1.90 D). cis-2-butene has a small dipole (0.33 D), trans-2-butene none. The same rule makes trans-$[\mathrm{Pt(NH_3)_2Cl_2}]$ non-polar and cis-platin polar.

Per cent ionic character

If a bond of length $d$ were 100 % ionic, a full electronic charge would sit at each end: $\mu_{\text{ionic}} = e \times d$. The conversion is $e \times (1\ \text{\AA}) = 1.602 \times 10^{-19} \times 10^{-10} / 3.336 \times 10^{-30} = 4.80$ D, so

$$\boxed{\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{ionic}}} \times 100 = \frac{\mu_{\text{obs}}\ (\mathrm{D})}{4.80 \times d\ (\text{\AA})} \times 100}$$

Molecule	$\mu_{\text{obs}}$ (D)	$d$ (pm)	$\mu_{\text{ionic}} = 4.80 \times d$ (D)	% ionic
HF	1.78	92	4.42	40
HCl	1.07	127	6.10	17.5
HBr	0.79	141	6.77	11.7
HI	0.38	160	7.68	4.9

Constants, if the question supplies SI units: $e = 1.602 \times 10^{-19}$ C, $1\ \mathrm{D} = 3.336 \times 10^{-30}$ C m. Ionic character falls HF $>$ HCl $>$ HBr $>$ HI while acid strength rises the other way — polarity and acidity run opposite down a group because bond strength, not polarity, decides acidity there.

The $\mu = 0$ checklist

A molecule is non-polar when its bond dipoles cancel by symmetry: every position around the central atom holds the same atom and the shape is one of the symmetric ones. Lone pairs are allowed only if they too sit symmetrically.

Shape	Zero-dipole members	Watch out
Linear $\mathrm{AX_2}$	$\mathrm{CO_2}$, $\mathrm{CS_2}$, $\mathrm{BeCl_2}$, $\mathrm{HgCl_2}$, $\mathrm{XeF_2}$ (three lone pairs at 120° in the equatorial plane), $\mathrm{I_3^-}$, $\mathrm{C_2H_2}$	HCN, OCS, $\mathrm{N_2O}$ polar
Trigonal planar $\mathrm{AX_3}$	$\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{SO_3}$, $\mathrm{NO_3^-}$, $\mathrm{CO_3^{2-}}$	$\mathrm{NH_3}$ (pyramidal), $\mathrm{SO_2}$ (bent) polar
Tetrahedral $\mathrm{AX_4}$	$\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{SiF_4}$, $\mathrm{SO_4^{2-}}$, $\mathrm{XeO_4}$	$\mathrm{CHCl_3}$, $\mathrm{CH_3Cl}$ polar; $\mathrm{SF_4}$ (seesaw) polar
Trigonal bipyramidal $\mathrm{AX_5}$	$\mathrm{PCl_5}$, $\mathrm{PF_5}$	$\mathrm{PCl_3F_2}$ zero only if both F are axial; $\mathrm{ClF_3}$, $\mathrm{SF_4}$ polar
Octahedral $\mathrm{AX_6}$	$\mathrm{SF_6}$, $\mathrm{SeF_6}$	$\mathrm{BrF_5}$ (square pyramidal) polar
Square planar $\mathrm{AX_4E_2}$	$\mathrm{XeF_4}$, $[\mathrm{ICl_4}]^-$, trans-$[\mathrm{Pt(NH_3)_2Cl_2}]$	cis isomer polar
Planar organics	benzene, p-dichlorobenzene, trans-alkenes, ethene, 1,3,5-trichlorobenzene	o-, m-, cis- all polar
Homonuclear	$\mathrm{H_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{Cl_2}$, $\mathrm{P_4}$, $\mathrm{S_8}$	$\mathrm{O_3}$ is polar (0.53 D) — bent

[JEE Main] Three checks settle most dipole questions. (1) Are all terminal atoms identical? If not, the molecule is polar (except accidental near-cancellations JEE does not ask). (2) Is the shape one of the symmetric set — linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar? Bent, pyramidal, seesaw, T-shaped and square pyramidal are always polar. (3) Do lone pairs sit symmetrically ($\mathrm{XeF_2}$, $\mathrm{XeF_4}$: yes; $\mathrm{SF_4}$, $\mathrm{ClF_3}$, $\mathrm{BrF_5}$: no)?

VSEPR Fine Print — Bond-Angle Rules, Bent's Rule, Drago's Rule and Back-Bonding

Rule 1: a more electronegative central atom opens the angle

Series	Angles	Trend
Group 15 hydrides	$\mathrm{NH_3}$ 107° $>$ $\mathrm{PH_3}$ 93.6° $>$ $\mathrm{AsH_3}$ 91.8° $>$ $\mathrm{SbH_3}$ 91.3°	angle falls down the group
Group 16 hydrides	$\mathrm{H_2O}$ 104.5° $>$ $\mathrm{H_2S}$ 92.1° $>$ $\mathrm{H_2Se}$ 91° $>$ $\mathrm{H_2Te}$ 90°	same

The usual VSEPR reason: a more electronegative central atom pulls the bond pairs closer to itself, they end up closer to each other, and their mutual repulsion opens the angle. A bigger, less electronegative centre lets the bond pairs drift outward, and the lone pair squeezes the bonds together. But the drop is sharp — 107° to 93.6° — and the values then park near 90°. That "near 90°" is Drago's rule (below), the better explanation.

Rule 2: more electronegative terminal atoms close the angle

Pair	Angles	Why
$\mathrm{NH_3}$ vs $\mathrm{NF_3}$	107° vs 102.3°	F drags the bond pairs away from N; less bp-bp repulsion near N; the angle shrinks
$\mathrm{PCl_3}$ vs $\mathrm{PF_3}$	100.3° vs 97.8°	same
$\mathrm{PF_3} < \mathrm{PCl_3} < \mathrm{PBr_3} < \mathrm{PI_3}$	97.8°, 100.3°, 101°, 102°	electronegativity of X falls, size of X rises — both open the angle
$\mathrm{H_2O}$ vs $\mathrm{OF_2}$	104.5° vs 103.1°	F pulls the bond pairs away from O

Two effects act on the terminal atom: electronegativity (pulls bond density away from the centre, angle shrinks) and size (big atoms bump into each other, angle opens). For H against F only electronegativity matters — H is tiny. For the heavier halogens size takes over. Hence the order JEE likes to test:

$$\mathrm{OF_2}\ (103.1^\circ) < \mathrm{H_2O}\ (104.5^\circ) < \mathrm{Cl_2O}\ (110.9^\circ)$$

$\mathrm{OF_2}$ is below water because fluorine is more electronegative than hydrogen. $\mathrm{Cl_2O}$ is above water because the two chlorines are large and their lone pairs repel each other across a small oxygen; chlorine is also less electronegative than fluorine, so the bond pairs stay nearer oxygen. (Some books add partial delocalisation of oxygen's lone pairs into chlorine's empty 3d orbitals.) F closes, Cl opens. Same pattern in $\mathrm{NF_3}$ 102.3° $<$ $\mathrm{NH_3}$ 107° $<$ $\mathrm{NCl_3}$ 107.1°.

Rule 3: multiple bonds take more room

A double bond holds four electrons and pushes the neighbouring bonds together. In phosgene $\mathrm{COCl_2}$ the Cl-C-Cl angle is 111.8° and each Cl-C=O angle 124.1°; in formaldehyde H-C-H is 116° and H-C=O 122°; in ethene H-C-H is 117.6° and H-C=C 121.2°. Rank the repulsions: triple $>$ double $>$ single, and lone pair $>$ any of them.

$\mathrm{SO_2}$ against $\mathrm{O_3}$. Both are bent, $sp^2$ with one lone pair on the central atom, but ozone's angle is 116.8° and sulfur dioxide's 119.5°. In $\mathrm{SO_2}$ both S-O bonds are full double bonds (sulfur can use its 3d orbitals to hold two $\pi$ bonds at once) and resist the lone pair's squeeze; in $\mathrm{O_3}$ the two bonds share one $\pi$ bond (order 1.5 each) and the small oxygens bring the lone pair closer. The order $\mathrm{O_3} < \mathrm{SO_2}$ is the point; the $d$-orbital story is a common way to argue it.

Lone-pair positions: equatorial in a trigonal bipyramid, trans in an octahedron

An axial position has three neighbours at 90°; an equatorial position only two (the other two at 120°). Lone pairs, being bulkiest, take the fewest 90° contacts — equatorial. So $\mathrm{SF_4}$ is a seesaw (one equatorial lone pair), $\mathrm{ClF_3}$ T-shaped (two), $\mathrm{XeF_2}$ and $\mathrm{I_3^-}$ linear (three). In an octahedron every position is equivalent, so the first lone pair goes anywhere ($\mathrm{BrF_5}$, $\mathrm{IF_5}$: square pyramidal) and the second goes trans to it, 180° away ($\mathrm{XeF_4}$, $[\mathrm{ICl_4}]^-$: square planar). Lone pairs also bend what is left: $\mathrm{SF_4}$'s axial F-S-F closes to 173°, $\mathrm{ClF_3}$'s F-Cl-F to 87.5°.

Bent's rule

Key Point (Bent's rule): s-character concentrates in the orbitals directed toward electropositive substituents, and p-character in the orbitals directed toward electronegative substituents. Lone pairs count as the most electropositive "substituent" of all and take the most s-character.

An electronegative atom pulls the bonding electrons toward itself, so the central atom gains little from spending its low-energy s orbital on that bond; it saves the s-character for bonds where the electrons stay close, and for lone pairs. Three consequences JEE uses:

1. Angles. More p-character means an angle nearer 90°; more s-character, nearer 180°. In $\mathrm{CH_3F}$ the C-F bond takes extra p-character and the three C-H bonds extra s-character, so H-C-H opens to about 110° and each H-C-F closes to about 108.7°. In $\mathrm{CH_2F_2}$ F-C-F is 108.3° and H-C-H 113°.
2. Bond lengths. More p-character means a longer bond. In $\mathrm{CH_3F}$ the single C-F bond soaks up most of carbon's p-character and is long (139 pm); in $\mathrm{CF_4}$ four fluorines share it equally, so each C-F bond is shorter (132 pm).
3. Position in a trigonal bipyramid. The $sp^3d$ set is an $sp^2$ trio (equatorial, 33 % s) plus a $pd$ pair (axial, no s-character), so the most electronegative substituents go axial: $\mathrm{PCl_3F_2}$ has both fluorines axial and three chlorines equatorial, making it non-polar; $\mathrm{PF_3Cl_2}$ has two F axial and one F plus two Cl equatorial; in $\mathrm{PCl_5}$ the axial P-Cl bonds (219 pm) are longer than the equatorial (204 pm). Lone pairs and double bonds go equatorial — the same conclusion VSEPR reached by counting 90° repulsions.

Drago's rule

Drago's rule is Bent's rule pushed to the limit. When the central atom is from period 3 or lower (P, As, Sb, S, Se, Te) and is bonded to atoms whose electronegativity is about 2.1 or less — in practice, hydrogen — the $ns$-$np$ gap is large and the bonds are weak, so hybridisation does not pay. The central atom bonds with almost pure p orbitals, at 90° to each other, and parks its lone pair in the pure s orbital.

Molecule	Angle	Molecule	Angle
$\mathrm{PH_3}$	93.6°	$\mathrm{H_2S}$	92.1°
$\mathrm{AsH_3}$	91.8°	$\mathrm{H_2Se}$	91°
$\mathrm{SbH_3}$	91.3°	$\mathrm{H_2Te}$	90°

That is why the angles crash from 107° ($\mathrm{NH_3}$, real $sp^3$) to 93.6° and then sit near 90°. Two side-effects: a lone pair in a pure s orbital is spherical and held tight, so $\mathrm{PH_3}$ is a far weaker base than $\mathrm{NH_3}$ and does not hydrogen-bond; and the P-H bond, made from a pure 3p orbital, is long and weak. Mind the conditions: Drago does not apply to $\mathrm{NH_3}$ (period 2) or to $\mathrm{PF_3}$ (F far too electronegative — $\mathrm{PF_3}$ is $sp^3$ with the angle pulled to 97.8° by Rule 2 and Bent's rule).

Back-bonding

When an atom with a filled p orbital sits next to an atom with an empty p (or d) orbital of the right symmetry, the lone pair partly delocalises into it: a $p\pi$-$p\pi$ or $p\pi$-$d\pi$ back-bond. It adds partial double-bond character, shortens the bond, and changes shape and basicity.

Case	What is seen	Explanation
$\mathrm{BF_3}$	B-F 130 pm against the 152 pm expected from covalent radii; $\mathrm{BF_3}$ is the weakest Lewis acid of the trihalides: $\mathrm{BF_3} < \mathrm{BCl_3} < \mathrm{BBr_3} < \mathrm{BI_3}$	each F donates filled-2p density into boron's empty 2p ($p\pi$-$p\pi$); boron's vacancy is partly filled, so it accepts an external lone pair less eagerly. Overlap is best for F (2p matches B), worse for bigger halogens
$\mathrm{F_3B}\cdot\mathrm{NH_3}$	boron goes from planar $sp^2$ to tetrahedral $sp^3$, B-F lengthens to about 138 pm, F-B-F closes from 120° to about 109°	the nitrogen lone pair fills boron's empty orbital, switching off the F to B back-bonding, so B-F loses its double-bond character
$\mathrm{N(SiH_3)_3}$ vs $\mathrm{N(CH_3)_3}$	trisilylamine is planar ($sp^2$ N, Si-N-Si 120°) and a very weak base; trimethylamine is pyramidal ($sp^3$, C-N-C 108°) and a good base	N's lone pair delocalises into Si's empty 3d orbitals ($p\pi$-$d\pi$), which needs a pure p lone pair, hence planar N; a delocalised pair cannot be donated. Carbon has no low-lying d orbital, so trimethylamine keeps its pyramid

The hybridisation formula — and when it fails

For a quick steric number, count

$$\boxed{H = \tfrac{1}{2}\left(V + M - C + A\right)}$$

$V$ = valence electrons of the central atom, $M$ = number of monovalent atoms attached (H, halogens), $C$ = positive charge, $A$ = negative charge. Divalent oxygen and sulfur count zero (they bring two electrons and take two). $H = 2, 3, 4, 5, 6, 7$ means $sp, sp^2, sp^3, sp^3d, sp^3d^2, sp^3d^3$.

Species	$\tfrac{1}{2}(V + M - C + A)$	Hybridisation	Shape
$\mathrm{NH_4^+}$	$\tfrac{1}{2}(5 + 4 - 1) = 4$	$sp^3$	tetrahedral
$\mathrm{SF_4}$	$\tfrac{1}{2}(6 + 4) = 5$	$sp^3d$	seesaw (one lone pair)
$\mathrm{XeF_4}$	$\tfrac{1}{2}(8 + 4) = 6$	$sp^3d^2$	square planar (two lone pairs)
$\mathrm{I_3^-}$	$\tfrac{1}{2}(7 + 2 + 1) = 5$	$sp^3d$	linear (three lone pairs)
$\mathrm{ClO_4^-}$	$\tfrac{1}{2}(7 + 0 + 1) = 4$	$sp^3$	tetrahedral
$\mathrm{NO_2^+}$	$\tfrac{1}{2}(5 - 1) = 2$	$sp$	linear

Where it fails. (1) Odd-electron species: $\mathrm{NO_2}$ gives $\tfrac{1}{2}(5) = 2.5$ and $\mathrm{ClO_2}$ gives 3.5 — meaningless; both are really $sp^2$ with the odd electron in the hybrid or p orbital. (2) Drago's molecules: $\mathrm{PH_3}$ and $\mathrm{H_2S}$ score 4, but the central atom is essentially unhybridised. (3) Back-bonded molecules: $\mathrm{N(SiH_3)_3}$ scores 4 ($sp^3$) but its nitrogen is planar $sp^2$; $\mathrm{BF_3}$ scores 3 correctly, yet the formula cannot see the extra $\pi$ bonding. (4) Molecules with no single central atom ($\mathrm{H_2O_2}$, $\mathrm{N_2O}$, $\mathrm{C_2H_4}$ — run it per atom). Use it as a fast first pass, then check against the steric number from a Lewis structure — $\sigma$ bonds plus lone pairs on the central atom — which is the definition and never fails for main-group species.

[JEE Main] For a bond-angle comparison, ask in this order: (1) same central atom, different terminal atoms — terminal electronegativity closes the angle, size opens it (H vs F: electronegativity only; Cl, Br, I: size wins); (2) same terminal atoms, different central atom — a heavy p-block hydride means Drago, about 90°, otherwise a more electronegative centre means a bigger angle; (3) multiple bond or lone pair present — it takes more room; (4) trigonal bipyramid — lone pairs and double bonds equatorial, the most electronegative atoms axial.

Heteronuclear Molecular Orbitals and the Half-Electron Rule

Two atoms, two different energies

In a homonuclear molecule the two atoms contribute equally to every MO. In a heteronuclear one the more electronegative atom has lower-lying atomic orbitals, so the bonding MOs are weighted toward the more electronegative atom and the antibonding MOs toward the less electronegative one. That explains why CO binds metals through carbon and why the HF bond is polar.

CO — the triple bond you did not expect

Carbon monoxide has $4 + 6 = 10$ valence electrons, exactly like $\mathrm{N_2}$. It uses the $\mathrm{N_2}$ ordering (the s-p mixed one, $\pi 2p$ below $\sigma 2p$):

$$\sigma 2s^2\ \sigma^* 2s^2\ \pi 2p_x^2\ \pi 2p_y^2\ \sigma 2p_z^2 \qquad \text{B.O.} = \tfrac{1}{2}(8 - 2) = 3$$

So CO has a triple bond — one $\sigma$ and two $\pi$ — is diamagnetic, and has the shortest (113 pm) and strongest (1072 kJ/mol) bond of any diatomic molecule, stronger even than $\mathrm{N_2}$'s 946. Its Lewis structure, $:\mathrm{C}{\equiv}\mathrm{O}:$ with a formal charge of $-1$ on C and $+1$ on O, says the same thing. The HOMO is $\sigma 2p_z$, concentrated on carbon — essentially the carbon lone pair, which is why CO attaches to a metal through C in carbonyls like $\mathrm{Ni(CO)_4}$. The LUMO is the $\pi^* 2p$ pair, also mostly on carbon, which accepts back-donation from filled metal d orbitals. The measured dipole moment is tiny (0.11 D) and has its negative end on carbon — the lone pair on C outweighs the electronegativity difference.

Isoelectronic with CO (10 valence electrons, bond order 3, diamagnetic): $\mathrm{N_2}$, $\mathrm{CN^-}$, $\mathrm{NO^+}$, $\mathrm{C_2^{2-}}$ (the acetylide ion in $\mathrm{CaC_2}$).

NO — bond order 2.5 and a paramagnetic molecule

Nitric oxide has $5 + 6 = 11$ valence electrons. Filling with the $\mathrm{N_2}$-type ordering (the ordering used for NO in every JEE key; the bond order comes out the same either way):

$$\sigma 2s^2\ \sigma^* 2s^2\ \pi 2p_x^2\ \pi 2p_y^2\ \sigma 2p_z^2\ \pi^* 2p_x^1 \qquad \text{B.O.} = \tfrac{1}{2}(8 - 3) = 2.5$$

One unpaired electron in an antibonding $\pi^*$ orbital: NO is paramagnetic ($\mu = \sqrt{1 \times 3} = 1.73$ BM), has a bond length of 115 pm — between a double (about 122) and a triple (106) N-O bond — and, with its odd electron in a high-energy antibonding orbital, is easily oxidised. Remove that electron and you get the nitrosonium ion $\mathrm{NO^+}$: bond order 3, diamagnetic, bond length 106 pm, found in $\mathrm{NOCl}$, $\mathrm{NOBF_4}$ and $\mathrm{NO^+HSO_4^-}$. Add an electron instead ($\mathrm{NO^-}$): bond order 2, bond length 127 pm, two unpaired electrons like $\mathrm{O_2}$. So

$$\text{bond length: } \mathrm{NO^+}\ (106) < \mathrm{NO}\ (115) < \mathrm{NO^-}\ (127); \qquad \text{bond strength the reverse.}$$

$\mathrm{CN}$ (9 valence electrons) is the mirror case: bond order 2.5, paramagnetic; add an electron to the bonding $\sigma 2p_z$ and $\mathrm{CN^-}$ has bond order 3 and is diamagnetic — adding an electron strengthened the bond, because it went into a bonding orbital.

HF — one bond, three lone pairs

Hydrogen's 1s orbital ($-13.6$ eV) is close in energy only to fluorine's 2p ($-18.6$ eV); fluorine's 2s ($-40$ eV) is far too low to mix, and only the F $2p_z$ pointing at hydrogen has the right symmetry for a $\sigma$ bond. So: one $\sigma$ bonding orbital (mostly F $2p_z$), one $\sigma^*$ (mostly H 1s), and three non-bonding orbitals on fluorine — 2s, $2p_x$, $2p_y$ — the three lone pairs. Eight valence electrons fill $\sigma^2$ and the three non-bonding orbitals; $\sigma^*$ is empty. Bond order $\tfrac{1}{2}(2 - 0) = 1$, diamagnetic, and because the bonding pair lives mostly on fluorine the molecule is strongly polar ($\mathrm{H}^{\delta+}\mathrm{F}^{\delta-}$, 1.78 D). The $2p_x$, $2p_y$ electrons do not count in the bond order — they are neither bonding nor antibonding.

The half-electron rule

Key Point: Adding or removing one electron changes the bond order by exactly $\tfrac{1}{2}$. Which way depends only on which orbital the electron enters or leaves: bonding orbital emptied or antibonding orbital filled, bond order falls by $\tfrac{1}{2}$; antibonding orbital emptied or bonding orbital filled, bond order rises by $\tfrac{1}{2}$.

So an "arrange the species" question is two steps: locate the HOMO (for removal) or LUMO (for addition), then add or subtract $\tfrac{1}{2}$.

Species	Configuration of the last electrons	B.O.	Unpaired	Bond length (pm)
$\mathrm{O_2^{2+}}$	$\ldots \sigma 2p_z^2$	3	0	about 105
$\mathrm{O_2^+}$	$\ldots \pi^* 2p^1$	2.5	1	112
$\mathrm{O_2}$	$\ldots \pi^* 2p_x^1\ \pi^* 2p_y^1$	2	2	121
$\mathrm{O_2^-}$ (superoxide)	$\ldots \pi^* 2p_x^2\ \pi^* 2p_y^1$	1.5	1	128
$\mathrm{O_2^{2-}}$ (peroxide)	$\ldots \pi^* 2p_x^2\ \pi^* 2p_y^2$	1	0	149

$$\text{Bond length: } \mathrm{O_2^{2-}} > \mathrm{O_2^-} > \mathrm{O_2} > \mathrm{O_2^+}; \qquad \text{bond strength / stability the reverse.}$$

For oxygen every added electron goes into $\pi^*$ (antibonding) and every removed electron comes out of $\pi^*$, so the pattern is monotonic. Nitrogen behaves differently:

Species	Last electrons	B.O.	Unpaired	Note
$\mathrm{N_2^+}$	$\ldots \pi 2p^4\ \sigma 2p_z^1$	2.5	1	electron removed from a bonding orbital
$\mathrm{N_2}$	$\ldots \pi 2p^4\ \sigma 2p_z^2$	3	0	110 pm, 946 kJ/mol
$\mathrm{N_2^-}$	$\ldots \sigma 2p_z^2\ \pi^* 2p^1$	2.5	1	electron added to an antibonding orbital

$\mathrm{N_2^+}$ and $\mathrm{N_2^-}$ share bond order 2.5 but are not equally stable: an antibonding electron destabilises a bond a little more than a bonding electron stabilises it, so $\mathrm{N_2^-}$ (3 antibonding electrons) is weaker and longer than $\mathrm{N_2^+}$ (2). Bond length $\mathrm{N_2} < \mathrm{N_2^+} < \mathrm{N_2^-}$; stability $\mathrm{N_2} > \mathrm{N_2^+} > \mathrm{N_2^-}$. The same tie-break makes $\mathrm{H_2^+}$ more stable than $\mathrm{H_2^-}$ (both 0.5).

Why ionising $\mathrm{N_2}$ weakens the bond but ionising $\mathrm{O_2}$ strengthens it — s-p mixing

The answer is the two energy orderings:

• For $\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2}$ (and CO, NO, CN): the 2s and 2p atomic orbitals are close in energy, so the $\sigma 2s$ and $\sigma 2p_z$ MOs, which have the same symmetry, mix. Mixing pushes $\sigma 2p_z$ up — above the $\pi 2p$ pair. Order: $\sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p < \sigma^* 2p_z$.
• For $\mathrm{O_2}$, $\mathrm{F_2}$, $\mathrm{Ne_2}$: the 2s-2p gap is large (the higher nuclear charge pulls 2s down much more than 2p), mixing is negligible, and $\sigma 2p_z$ sits below $\pi 2p$. Order: $\sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p < \sigma^* 2p_z$.

The gap doubles across the period: about 5 eV for B, 8 for C, 12 for N, 15 for O, 21 for F. So the HOMO of $\mathrm{N_2}$ is the bonding $\sigma 2p_z$ — the first electron removed comes from a bonding orbital and the bond weakens ($\mathrm{N_2^+}$, 2.5). The HOMO of $\mathrm{O_2}$ is the antibonding $\pi^* 2p$ — the first electron removed comes from an antibonding orbital and the bond strengthens ($\mathrm{O_2^+}$, 2.5, shorter than $\mathrm{O_2}$). Two more consequences of the mixed ordering: $\mathrm{B_2}$ is paramagnetic (its two $\pi 2p$ electrons occupy $\pi 2p_x$ and $\pi 2p_y$ singly) and $\mathrm{C_2}$ is diamagnetic with two $\pi$ bonds and no $\sigma$ bond. Without s-p mixing $\mathrm{B_2}$ would be diamagnetic — experiment says otherwise, which is the evidence for the mixing.

Molecule	HOMO	LUMO
$\mathrm{N_2}$	$\sigma 2p_z$ (bonding)	$\pi^* 2p$
$\mathrm{O_2}$	$\pi^* 2p$ (half-filled — also the SOMO)	$\sigma^* 2p_z$ (the next empty orbital)
CO	$\sigma 2p_z$ (carbon lone pair)	$\pi^* 2p$ (mostly on C)
NO	$\pi^* 2p$ (singly occupied)	the other $\pi^* 2p$

Magnetic moments

Count unpaired electrons $n$ and use $\mu = \sqrt{n(n + 2)}$ BM: $n = 1$, 1.73; $n = 2$, 2.83; $n = 3$, 3.87.

Species	Unpaired	$\mu$ (BM)	Species	Unpaired	$\mu$ (BM)
$\mathrm{O_2}$, $\mathrm{B_2}$, $\mathrm{NO^-}$	2	2.83	$\mathrm{N_2}$, $\mathrm{C_2}$, $\mathrm{F_2}$, CO, $\mathrm{CN^-}$, $\mathrm{NO^+}$, $\mathrm{O_2^{2-}}$	0	0
$\mathrm{O_2^+}$, $\mathrm{O_2^-}$, NO, CN, $\mathrm{N_2^+}$, $\mathrm{N_2^-}$, $\mathrm{H_2^+}$, $\mathrm{He_2^+}$, $\mathrm{H_2^-}$	1	1.73	$\mathrm{Be_2}$, $\mathrm{He_2}$ (B.O. 0, do not exist)	—	—

[JEE Main] Memorise both orderings and the dividing line ($\mathrm{N_2}$ mixed, $\mathrm{O_2}$ unmixed). Then every question is mechanical: count valence electrons (add for anions, subtract for cations), fill, read off bond order, unpaired electrons and the HOMO. Use the $\mathrm{N_2}$ ordering for CO, NO, $\mathrm{CN^-}$, $\mathrm{NO^+}$. The bond-length orders — $\mathrm{NO^+} < \mathrm{NO} < \mathrm{NO^-}$, $\mathrm{O_2^+} < \mathrm{O_2} < \mathrm{O_2^-} < \mathrm{O_2^{2-}}$, $\mathrm{N_2} < \mathrm{N_2^+} < \mathrm{N_2^-}$ — follow from bond order first and antibonding count second.