NEET Corner — Chemical Bonding the NEET Way

How NEET Tests This Chapter — and the 40-Second Mindset

Chemical Bonding is a reliable scorer: 2 to 4 questions every year, almost all of them recall plus one decision. No Born-Haber cycles, no percentage ionic character calculations, no Bent's rule essays. Know the tables in this section cold and each of those questions is a 40-second $+4$.

What actually comes up:

Question style	What it looks like	Time you should spend
Shape or hybridisation (the big one)	"The shape of $\mathrm{XeF_4}$ / $\mathrm{SF_4}$ / $\mathrm{ClF_3}$ is…" / "The hybridisation of the central atom in $\mathrm{I_3^-}$ is…"	30 seconds
MO bond order and magnetism	"Which species has the highest bond order?" / "Which is paramagnetic?" / "Bond order of $\mathrm{O_2^-}$ is…"	40 seconds
Bond-angle ordering	"Correct order of bond angle: $\mathrm{CH_4}$, $\mathrm{NH_3}$, $\mathrm{H_2O}$" / "$\mathrm{NH_3}$ vs $\mathrm{PH_3}$"	30 seconds
Dipole moment	"Which molecule has zero dipole moment?" / "$\mathrm{NH_3}$ vs $\mathrm{NF_3}$"	25 seconds
Sigma and pi count	"Number of $\sigma$ and $\pi$ bonds in $\mathrm{CH_2{=}C{=}CH_2}$"	20 seconds
Hydrogen bonding	"Correct boiling-point order of $\mathrm{H_2O}$, $\mathrm{H_2S}$, $\mathrm{H_2Se}$, $\mathrm{H_2Te}$" / "intramolecular H-bond is present in…"	25 seconds
Resonance and bond order	"Bond order of C-O in $\mathrm{CO_3^{2-}}$" / "which has all equal bond lengths"	30 seconds
Lewis and formal charge	"Formal charge on the central O in ozone" / "which does not obey the octet rule"	30 seconds
Assertion-Reason / statement-type	"Assertion: $\mathrm{O_2}$ is paramagnetic. Reason: it has two unpaired electrons in $\pi^*$ orbitals"	40 seconds

All nine styles live in this section.

Favourite topics, ranked

If you have time for only five things:

1. The shape-and-hybridisation table — steric number 2 to 7, with and without lone pairs, angles and the stock examples ($\mathrm{XeF_2}$, $\mathrm{XeF_4}$, $\mathrm{SF_4}$, $\mathrm{ClF_3}$, $\mathrm{BrF_5}$, $\mathrm{I_3^-}$, $\mathrm{IF_7}$).
2. The MO bond-order card — $\mathrm{H_2}$ to $\mathrm{F_2}$ and the oxygen and nitrogen ions, with magnetism.
3. Bond-angle orders — lone-pair squeeze, electronegativity effects, and the hydride series.
4. Dipole moment — zero or not, and the $\mathrm{NH_3}$ / $\mathrm{NF_3}$ story.
5. Hydrogen bonding — boiling-point orders and the intramolecular cases.

The 40-second mindset

NEET is 180 questions in 200 minutes. A bonding question is where you bank time for Biology. The target is to answer in 40 seconds and move on.

1. Read the last line first. It names the property — shape, hybridisation, bond order, angle, dipole, magnetism, boiling point. Fix the property before looking at the species.
2. Count before you think. Shape: bond pairs plus lone pairs on the central atom. MO: total electrons. Sigma/pi: single bonds plus one per multiple bond. Counting is faster and safer than remembering.
3. Apply the plain rule, then check whether a famous exception is sitting there — $\mathrm{NF_3}$ in a dipole list, $\mathrm{SbH_3}$ in a boiling-point list, $\mathrm{B_2}$ in a magnetism list, $\mathrm{N_2^+}$ next to $\mathrm{N_2^-}$.
4. Watch the word. "Geometry" versus "shape", "increasing" versus "decreasing", "stability" versus "bond length" (opposite orders), "paramagnetic" versus "diamagnetic".
5. Match to the options and go. If your answer appears exactly once, bubble it. If it appears nowhere, you miscounted — recount once, then move on.

Key Point: This chapter rewards counting — electron pairs, total electrons, sigma bonds — rather than reasoning from first principles under the clock. Treat it as a counting chapter with a short list of famous exceptions.

[NEET] Negative marking ($-1$) makes a 40-second wrong answer worse than a 40-second skip. If a shape question matches no option after one recount, mark it for review and come back.

What NEET does NOT ask from this chapter

• Born-Haber cycle arithmetic — that is JEE territory.
• Percentage ionic character from dipole moments — rare, and never with awkward numbers.
• MO diagrams of heteronuclear molecules beyond bond order and magnetism of CO, NO, $\mathrm{NO^+}$, $\mathrm{CN^-}$.
• Bent's rule, Drago's rule by name, or $d$-orbital participation debates — you only need the orders they produce.

If a question here takes more than 90 seconds, you are using the wrong tool. Move.

The Verbatim-Recall Table — Definitions NEET Asks Word for Word

NEET lifts a definition, changes one word, and asks "which statement is correct". Learn these as flash-cards; the bold words are the ones a paper-setter changes.

Table 1: The definitions

Term	Definition as NEET wants it	Instant example or number
Octet rule (Kössel-Lewis, 1916)	Atoms combine by transfer or sharing of valence electrons so as to have eight electrons in their outermost shell	Exceptions: $\mathrm{BeH_2}$, $\mathrm{BCl_3}$ (incomplete), NO, $\mathrm{NO_2}$ (odd electron), $\mathrm{PF_5}$, $\mathrm{SF_6}$ (expanded)
Electrovalent (ionic) bond	Bond formed by complete transfer of one or more electrons from one atom to another, held by electrostatic attraction between the ions	NaCl, $\mathrm{CaF_2}$; favoured by low $\Delta_i H$ of the metal, high negative $\Delta_{eg} H$ of the non-metal, high lattice enthalpy
Covalent bond (Lewis; the word is Langmuir's, 1919)	Bond formed by mutual sharing of an electron pair, each atom contributing one electron	$\mathrm{Cl_2}$, $\mathrm{H_2O}$, $\mathrm{CH_4}$
Coordinate (dative) bond	A covalent bond in which both shared electrons come from one atom	$\mathrm{NH_4^+}$, $\mathrm{H_3O^+}$, $\mathrm{NH_3 \rightarrow BF_3}$, one of the three bonds in CO
Lattice enthalpy	The energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions	NaCl: 788 kJ/mol
Bond length	The equilibrium distance between the nuclei of two bonded atoms in a molecule; $R = r_A + r_B$	$\mathrm{H_2}$ 74 pm, $\mathrm{N_2}$ 109 pm, $\mathrm{Cl_2}$ 199 pm
Bond angle	The angle between the orbitals containing bonding electron pairs around the central atom	$\mathrm{H{-}O{-}H}$ 104.5°
Bond enthalpy	The energy required to break one mole of bonds of a particular type between two atoms in the gaseous state	$\mathrm{H_2}$ 435.8, $\mathrm{O_2}$ 498, $\mathrm{N_2}$ 946.0 kJ/mol
Bond order (Lewis)	The number of bonds between the two atoms in a molecule	$\mathrm{H_2}$ 1, $\mathrm{O_2}$ 2, $\mathrm{N_2}$ 3
Bond order (MO)	Half the difference between the number of electrons in bonding and antibonding molecular orbitals	$\mathrm{O_2}$: $\frac{1}{2}(10 - 6) = 2$
Resonance	When a single Lewis structure cannot describe a molecule accurately, several structures with similar energy, the same positions of nuclei, and the same number of bonding and non-bonding pairs are taken as canonical structures of the resonance hybrid	$\mathrm{O_3}$ (both O-O 128 pm), $\mathrm{CO_3^{2-}}$, $\mathrm{NO_3^-}$, benzene
Dipole moment	The product of the magnitude of charge and the distance between the centres of positive and negative charge; a vector quantity; unit Debye	$\mu = Q \times r$; $1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$
VSEPR postulates (Sidgwick and Powell 1940; Nyholm and Gillespie 1957)	Shape depends on the number of valence-shell electron pairs (bonded and non-bonded) around the central atom; pairs repel and take positions that minimise repulsion; a multiple bond counts as one super pair; repulsion order lp-lp $>$ lp-bp $>$ bp-bp	$\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ 107°, $\mathrm{H_2O}$ 104.5°
Hybridisation	The intermixing of atomic orbitals of slightly different energies to give an equal number of new orbitals of equivalent energy and shape	one 2s + three 2p of C gives four $sp^3$
Sigma ($\sigma$) bond	Formed by end-to-end (head-on) overlap of orbitals along the internuclear axis; s-s, s-p or p-p	Every single bond is a sigma bond
Pi ($\pi$) bond	Formed by sidewise overlap of p orbitals whose axes are parallel and perpendicular to the internuclear axis; weaker than a sigma bond	Second bond of $\mathrm{C{=}C}$; second and third of $\mathrm{N{\equiv}N}$
Bonding MO	Formed by addition (constructive interference) of atomic wave functions; lower energy, high electron density between the nuclei	$\sigma 1s$, $\pi 2p_x$
Antibonding MO	Formed by subtraction (destructive interference); higher energy, a node between the nuclei	$\sigma^* 1s$, $\pi^* 2p_x$
Hydrogen bond	The attractive force that binds the hydrogen atom of one molecule with a highly electronegative atom (F, O or N) of another molecule (or another part of the same molecule); shown by a dotted line; weaker than a covalent bond, stronger than van der Waals forces	HF, $\mathrm{H_2O}$, $\mathrm{NH_3}$, alcohols, DNA base pairs

Key Point (Definition): Three phrases NEET deletes to make a statement false — "gaseous" in bond enthalpy and lattice enthalpy, "same positions of nuclei" in resonance, and "one atom supplies both electrons" in coordinate bond. An option missing the phrase is the wrong option.

Table 2: The three formulas NEET actually uses

Formula	Statement	Worked one-liner
Formal charge	$\text{F.C.} = V - L - \frac{1}{2}S$, where $V$ = valence electrons of the free atom, $L$ = lone-pair (non-bonding) electrons, $S$ = shared (bonding) electrons	Central O in $\mathrm{O_3}$: $6 - 2 - \frac{1}{2}(6) = +1$; end O with double bond: $6 - 4 - \frac{1}{2}(4) = 0$; end O with single bond: $6 - 6 - \frac{1}{2}(2) = -1$
MO bond order	$\text{B.O.} = \frac{1}{2}(N_b - N_a)$	$\mathrm{N_2}$: $\frac{1}{2}(10 - 4) = 3$; $\mathrm{O_2^-}$: $\frac{1}{2}(10 - 7) = 1.5$
Resonance bond order	$\text{B.O.} = \dfrac{\text{total bonds between the pair in all structures}}{\text{number of resonance structures}}$, or equivalently $\dfrac{\text{total bonds to the central atom}}{\text{number of bonded atoms}}$	$\mathrm{CO_3^{2-}}$: $\frac{4}{3} = 1.33$; $\mathrm{O_3}$: $\frac{3}{2} = 1.5$; $\mathrm{NO_3^-}$: 1.33; $\mathrm{SO_4^{2-}}$: 1.5; benzene C-C: 1.5

[NEET] Formal charge is a book-keeping number, not a real charge; the formal charges sum to the charge on the species (zero for $\mathrm{O_3}$, $-2$ for $\mathrm{CO_3^{2-}}$). The structure with the lowest formal charges is the most stable, and the negative formal charge sits preferably on the more electronegative atom.

Bond-order rule of thumb: higher bond order means shorter bond length and higher bond enthalpy. Isoelectronic species share a bond order — $\mathrm{N_2}$, CO, $\mathrm{NO^+}$, $\mathrm{CN^-}$ all have bond order 3; $\mathrm{F_2}$ and $\mathrm{O_2^{2-}}$ both have bond order 1.

The Shape-and-Hybridisation Master Table

This table answers more NEET questions than any other in Chapter 4. Learn it row by row.

The steric-number one-liner

Key Point: Steric number $=$ (atoms bonded to the central atom) $+$ (lone pairs on the central atom). Steric number 2 is $sp$, 3 is $sp^2$, 4 is $sp^3$, 5 is $sp^3d$, 6 is $sp^3d^2$, 7 is $sp^3d^3$. A double or triple bond counts as one bonded atom.

For a quick lone-pair count on the central atom: lone pairs $= \frac{1}{2}(V - B \pm \text{charge})$, where $V$ is the group valence electrons of the central atom and $B$ is the number of electrons it uses in bonds (one per single bond to a monovalent atom, two for a double bond to O). Add the charge for an anion, subtract for a cation. Faster still is the hybrid-orbital count $H = \frac{1}{2}(V + M - c + a)$, with $M$ the number of monovalent atoms attached (H, halogens), $c$ the cation charge and $a$ the anion charge; doubly bonded oxygen atoms are not counted in $M$:

Species	$V$	$M$	Charge	$H$	Hybridisation
$\mathrm{SF_4}$	6	4	0	$\frac{1}{2}(10) = 5$	$sp^3d$
$\mathrm{XeF_4}$	8	4	0	$\frac{1}{2}(12) = 6$	$sp^3d^2$
$\mathrm{NH_4^+}$	5	4	$-1$	$\frac{1}{2}(8) = 4$	$sp^3$
$\mathrm{I_3^-}$	7	2	$+1$	$\frac{1}{2}(10) = 5$	$sp^3d$
$\mathrm{ClO_4^-}$	7	0	$+1$	$\frac{1}{2}(8) = 4$	$sp^3$
$\mathrm{NO_3^-}$	5	0	$+1$	$\frac{1}{2}(6) = 3$	$sp^2$
$\mathrm{NO_2^+}$	5	0	$-1$	$\frac{1}{2}(4) = 2$	$sp$
$\mathrm{SO_2}$	6	0	0	$\frac{1}{2}(6) = 3$	$sp^2$

The master table

Steric no.	Hybridisation	Lone pairs	Electron geometry	Molecular shape	Bond angle(s)	Examples
2	$sp$	0	Linear	Linear	180°	$\mathrm{BeCl_2}$, $\mathrm{BeF_2}$, $\mathrm{BeH_2}$, $\mathrm{HgCl_2}$, $\mathrm{CO_2}$, $\mathrm{HCN}$, $\mathrm{C_2H_2}$, $\mathrm{NO_2^+}$
3	$sp^2$	0	Trigonal planar	Trigonal planar	120°	$\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{AlCl_3}$ (monomer), $\mathrm{SO_3}$, $\mathrm{CO_3^{2-}}$, $\mathrm{NO_3^-}$, $\mathrm{C_2H_4}$, $\mathrm{HCHO}$
3	$sp^2$	1	Trigonal planar	Bent (V-shaped)	slightly less than 120°	$\mathrm{SO_2}$, $\mathrm{O_3}$, $\mathrm{NO_2^-}$, $\mathrm{SnCl_2}$ (gas)
4	$sp^3$	0	Tetrahedral	Tetrahedral	109.5°	$\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{SiCl_4}$, $\mathrm{NH_4^+}$, $\mathrm{BF_4^-}$, $\mathrm{SO_4^{2-}}$, $\mathrm{PO_4^{3-}}$, $\mathrm{ClO_4^-}$
4	$sp^3$	1	Tetrahedral	Trigonal pyramidal	107° in $\mathrm{NH_3}$	$\mathrm{NH_3}$, $\mathrm{NF_3}$, $\mathrm{PCl_3}$, $\mathrm{PH_3}$, $\mathrm{H_3O^+}$, $\mathrm{ClO_3^-}$, $\mathrm{SO_3^{2-}}$, $\mathrm{XeO_3}$
4	$sp^3$	2	Tetrahedral	Bent (V-shaped)	104.5° in $\mathrm{H_2O}$	$\mathrm{H_2O}$, $\mathrm{H_2S}$, $\mathrm{OF_2}$, $\mathrm{SCl_2}$, $\mathrm{NH_2^-}$, $\mathrm{ClO_2^-}$
5	$sp^3d$	0	Trigonal bipyramidal	Trigonal bipyramidal	120° (equatorial), 90° (axial)	$\mathrm{PCl_5}$, $\mathrm{PF_5}$, $\mathrm{SbCl_5}$, $\mathrm{AsF_5}$
5	$sp^3d$	1	Trigonal bipyramidal	See-saw	less than 120° and 90°	$\mathrm{SF_4}$, $\mathrm{SeF_4}$, $\mathrm{XeO_2F_2}$
5	$sp^3d$	2	Trigonal bipyramidal	T-shaped	slightly less than 90°	$\mathrm{ClF_3}$, $\mathrm{BrF_3}$, $\mathrm{ICl_3}$
5	$sp^3d$	3	Trigonal bipyramidal	Linear	180°	$\mathrm{XeF_2}$, $\mathrm{I_3^-}$, $\mathrm{ICl_2^-}$
6	$sp^3d^2$	0	Octahedral	Octahedral	90°	$\mathrm{SF_6}$, $\mathrm{SeF_6}$, $\mathrm{PF_6^-}$, $\mathrm{SiF_6^{2-}}$
6	$sp^3d^2$	1	Octahedral	Square pyramidal	slightly less than 90°	$\mathrm{BrF_5}$, $\mathrm{IF_5}$, $\mathrm{XeOF_4}$
6	$sp^3d^2$	2	Octahedral	Square planar	90°	$\mathrm{XeF_4}$, $\mathrm{ICl_4^-}$, $\mathrm{BrF_4^-}$
7	$sp^3d^3$	0	Pentagonal bipyramidal	Pentagonal bipyramidal	72° (equatorial), 90° (axial)	$\mathrm{IF_7}$
7	$sp^3d^3$	1	Pentagonal bipyramidal	Distorted octahedral	—	$\mathrm{XeF_6}$

Where the lone pairs sit

• Trigonal bipyramidal (steric number 5): lone pairs always go equatorial — an equatorial position has only two 90° neighbours, an axial one has three. So $\mathrm{SF_4}$ is see-saw, $\mathrm{ClF_3}$ is T-shaped and $\mathrm{XeF_2}$ is linear (three lone pairs fill the whole equatorial plane).
• Octahedral (steric number 6): the first lone pair can go anywhere (square pyramidal, $\mathrm{BrF_5}$); the second goes opposite to it to minimise lp-lp repulsion (square planar, $\mathrm{XeF_4}$).
• PCl5 detail: the three equatorial bonds are at 120° in one plane; the two axial bonds are at 90° to that plane. Axial bond pairs suffer more repulsion from the equatorial pairs, so the axial P-Cl bonds are slightly longer and weaker, which is why $\mathrm{PCl_5}$ is reactive.

The orbital and character facts that ride along

Hybridisation	Orbitals mixed	Percentage s-character	Which d orbitals
$sp$	one s + one p	50 percent	—
$sp^2$	one s + two p	33.3 percent	—
$sp^3$	one s + three p	25 percent	—
$sp^3d$	one s + three p + one d	20 percent	$d_{z^2}$
$sp^3d^2$	one s + three p + two d	16.7 percent	$d_{z^2}$ and $d_{x^2 - y^2}$
$sp^3d^3$	one s + three p + three d	14.3 percent	$d_{z^2}$, $d_{x^2 - y^2}$, $d_{xy}$

More s-character means a shorter, stronger bond and a more electronegative carbon. C-H bond length: ethyne (sp) $<$ ethene ($sp^2$) $<$ ethane ($sp^3$); C-C bond length: $\mathrm{C{\equiv}C}$ 120 pm $<$ $\mathrm{C{=}C}$ 133 pm $<$ $\mathrm{C{-}C}$ 154 pm.

Carbon in organic molecules at a glance: four single bonds means $sp^3$; one double bond means $sp^2$; one triple bond or two double bonds means $sp$. In allene $\mathrm{CH_2{=}C{=}CH_2}$ the end carbons are $sp^2$ and the middle carbon is $sp$; the two $\mathrm{CH_2}$ planes are perpendicular.

[NEET] Three hybridisation conditions that appear as statements: the orbitals must belong to the valence shell; the orbitals must have nearly the same energy; promotion of an electron is not a necessary condition; and filled orbitals of the valence shell can also take part (that is how the lone pair of $\mathrm{NH_3}$ sits in an $sp^3$ orbital). Hybridisation is a model applied once the shape is known, not the cause of the shape.

Lone-Pair Squeeze, the Bond-Angle Order Sets, and the Sigma/Pi Counting Rule

Rule 1: lone pairs squeeze bond angles

Repulsion order: lp-lp $>$ lp-bp $>$ bp-bp. A lone pair is held by only one nucleus, so it spreads out and pushes the bond pairs together. Each extra lone pair on a tetrahedral centre knocks the angle down:

Molecule	Lone pairs on central atom	Angle	Shape
$\mathrm{CH_4}$	0	109.5°	tetrahedral
$\mathrm{NH_3}$	1	107°	trigonal pyramidal
$\mathrm{H_2O}$	2	104.5°	bent

So $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$, and by the same argument $\mathrm{NH_4^+}$ (109.5°) $>$ $\mathrm{NH_3}$ (107°) $>$ $\mathrm{NH_2^-}$ (about 104°), and $\mathrm{H_3O^+}$ $>$ $\mathrm{H_2O}$.

Rule 2: down a group, the angle falls (bigger, less electronegative central atom)

The bond pairs sit farther from a big, less electronegative central atom, so they repel each other less and the lone pair wins more of the space. Some students remember this as "more p-character in the bonds".

Series	Order	Angles to quote
Group 15 hydrides	$\mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3}$	107°, about 94°, about 92°, about 91°
Group 16 hydrides	$\mathrm{H_2O} > \mathrm{H_2S} > \mathrm{H_2Se} > \mathrm{H_2Te}$	104.5°, about 92°, about 91°, about 90°

Rule 3: a more electronegative outer atom pulls the bond pair away and shrinks the angle

Series	Order	Why
$\mathrm{NH_3}$ vs $\mathrm{NF_3}$	$\mathrm{NH_3}$ (107°) $>$ $\mathrm{NF_3}$ (about 102°)	F pulls bond pairs away from N; bond pairs repel less, the lone pair squeezes more
Phosphorus trihalides	$\mathrm{PF_3} < \mathrm{PCl_3} < \mathrm{PBr_3} < \mathrm{PI_3}$	Same reason, reversed as the halogen becomes less electronegative
Oxygen series	$\mathrm{OF_2}$ (about 103°) $<$ $\mathrm{H_2O}$ (104.5°) $<$ $\mathrm{Cl_2O}$ (about 111°)	F shrinks it; the big Cl atoms crowd each other and open it
Boron trihalides	$\mathrm{BF_3} = \mathrm{BCl_3} = \mathrm{BBr_3} = 120^\circ$	All $sp^2$, no lone pair — no change

Rule 4: multiple bonds and charge

A double bond takes more room than a single bond. In ethene $\mathrm{H{-}C{-}H}$ is a little less than 120° and $\mathrm{H{-}C{=}C}$ a little more. For the nitrogen dioxide family:

$$\mathrm{NO_2^+}\ (180^\circ) > \mathrm{NO_2}\ (\text{about } 134^\circ) > \mathrm{NO_2^-}\ (\text{about } 115^\circ)$$

$\mathrm{NO_2^+}$ is $sp$ with no lone pair; $\mathrm{NO_2}$ has a single odd electron on N (a half lone pair, small squeeze); $\mathrm{NO_2^-}$ has a full lone pair (big squeeze).

Key Point: Three questions in order: how many lone pairs on the central atom (more means smaller), is the central atom bigger or less electronegative (smaller angle), are the outer atoms more electronegative (smaller angle). Ask them in sequence and any angle-order item takes 30 seconds.

[NEET] The ready-made orders, largest angle first: $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$; $\mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3}$; $\mathrm{H_2O} > \mathrm{H_2S}$; $\mathrm{NH_3} > \mathrm{NF_3}$; $\mathrm{Cl_2O} > \mathrm{H_2O} > \mathrm{OF_2}$; $\mathrm{NO_2^+} > \mathrm{NO_2} > \mathrm{NO_2^-}$; $\mathrm{BF_3}$ (120°) $>$ $\mathrm{NH_3}$ (107°) $>$ $\mathrm{H_2O}$ (104.5°).

The sigma/pi counting rule

Key Point: Every single bond is one sigma. Every double bond is one sigma plus one pi. Every triple bond is one sigma plus two pi. The number of sigma bonds equals the total number of bonds drawn between atoms; the number of pi bonds equals (number of double bonds) $+ 2 \times$ (number of triple bonds).

Two speed shortcuts: for a neutral, acyclic molecule, sigma bonds $=$ (total atoms) $- 1$; for benzene and other rings, sigma bonds $=$ total atoms, because the ring adds one bond.

Molecule	Structure in words	Sigma	Pi	Total
$\mathrm{C_2H_2}$ (ethyne)	$\mathrm{H{-}C{\equiv}C{-}H}$	3 (two C-H, one C-C)	2	5
$\mathrm{C_2H_4}$ (ethene)	$\mathrm{H_2C{=}CH_2}$	5 (four C-H, one C-C)	1	6
$\mathrm{C_2H_6}$ (ethane)	$\mathrm{H_3C{-}CH_3}$	7	0	7
$\mathrm{C_6H_6}$ (benzene)	ring of six C, six C-H	12 (six C-C, six C-H)	3	15
$\mathrm{CO_2}$	$\mathrm{O{=}C{=}O}$	2	2	4
HCN	$\mathrm{H{-}C{\equiv}N}$	2	2	4
$\mathrm{CH_2{=}C{=}CH_2}$ (allene)	two double bonds, four C-H	6	2	8
$\mathrm{CH_2{=}CH{-}C{\equiv}CH}$ (vinylacetylene)	one double, one triple, four C-H, one C-C single	7	3	10
$\mathrm{N_2}$	$\mathrm{N{\equiv}N}$	1	2	3
$\mathrm{SO_2}$ (one Lewis structure)	$\mathrm{O{=}S{-}O}$ with one double bond drawn	2	1	3
$\mathrm{HCOOH}$ (formic acid)	$\mathrm{H{-}C({=}O){-}O{-}H}$	4	1	5
$\mathrm{CH_3{-}C{\equiv}N}$ (acetonitrile)	three C-H, one C-C, one C-N triple	5	2	7

Check ethyne with the shortcut: 4 atoms, acyclic, so $4 - 1 = 3$ sigma. Benzene: 12 atoms in a ring, so 12 sigma. Allene: 7 atoms, so 6 sigma.

The sigma bond is stronger than the pi bond (head-on overlap beats sidewise), a pi bond cannot exist alone between two atoms — it always rides on a sigma bond — and rotation about a double bond is restricted because it would break the pi overlap. The $\mathrm{N{\equiv}N}$ bond enthalpy of 946 kJ/mol is among the highest for a diatomic, which is why nitrogen is so unreactive.

[NEET] In a bond-line or resonance question, count pi bonds in one canonical structure — the hybrid does not have "one and a half" pi bonds for counting purposes. Benzene has 3 pi bonds, $\mathrm{CO_3^{2-}}$ has 1, $\mathrm{NO_3^-}$ has 1, $\mathrm{SO_4^{2-}}$ (with two S=O drawn) has 2.

The MO Bond-Order and Magnetism Card

Fill the molecular orbitals in energy order, count bonding and antibonding electrons, and bond order, stability, bond length and magnetism all fall out.

The two energy orders

For $\mathrm{O_2}$, $\mathrm{F_2}$ (and $\mathrm{Ne_2}$) — 16 or more electrons, $\sigma 2p_z$ below the $\pi 2p$ pair:

$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

For $\mathrm{Li_2}$ to $\mathrm{N_2}$ — 14 or fewer electrons, where $2s$-$2p$ mixing pushes $\sigma 2p_z$ above the $\pi 2p$ pair:

$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

The difference matters only for the magnetism of $\mathrm{B_2}$ and $\mathrm{C_2}$: with the second order, $\mathrm{B_2}$ puts its last two electrons one each into the degenerate $\pi 2p_x$ and $\pi 2p_y$ (paramagnetic), and $\mathrm{C_2}$ fills both (diamagnetic). Bond orders come out the same either way.

The card

Species	Total electrons	Valence MO configuration (after $\sigma 2s^2\, \sigma^* 2s^2$ where applicable)	$N_b$	$N_a$	Bond order	Unpaired electrons	Magnetism
$\mathrm{H_2^+}$	1	$\sigma 1s^1$	1	0	0.5	1	paramagnetic
$\mathrm{H_2}$	2	$\sigma 1s^2$	2	0	1	0	diamagnetic
$\mathrm{He_2^+}$	3	$\sigma 1s^2\, \sigma^* 1s^1$	2	1	0.5	1	paramagnetic
$\mathrm{He_2}$	4	$\sigma 1s^2\, \sigma^* 1s^2$	2	2	0	0	does not exist
$\mathrm{Li_2}$	6	$\sigma 2s^2$	4	2	1	0	diamagnetic
$\mathrm{Be_2}$	8	$\sigma 2s^2\, \sigma^* 2s^2$	4	4	0	0	does not exist
$\mathrm{B_2}$	10	$\pi 2p_x^1\, \pi 2p_y^1$	6	4	1	2	paramagnetic
$\mathrm{C_2}$	12	$\pi 2p_x^2\, \pi 2p_y^2$	8	4	2	0	diamagnetic
$\mathrm{N_2^+}$	13	$\pi 2p_x^2\, \pi 2p_y^2\, \sigma 2p_z^1$	9	4	2.5	1	paramagnetic
$\mathrm{N_2}$	14	$\pi 2p_x^2\, \pi 2p_y^2\, \sigma 2p_z^2$	10	4	3	0	diamagnetic
$\mathrm{N_2^-}$	15	$\ldots \sigma 2p_z^2\, \pi^* 2p_x^1$	10	5	2.5	1	paramagnetic
$\mathrm{O_2^+}$	15	$\sigma 2p_z^2\, \pi 2p_x^2\, \pi 2p_y^2\, \pi^* 2p_x^1$	10	5	2.5	1	paramagnetic
$\mathrm{O_2}$	16	$\ldots \pi^* 2p_x^1\, \pi^* 2p_y^1$	10	6	2	2	paramagnetic
$\mathrm{O_2^-}$ (superoxide)	17	$\ldots \pi^* 2p_x^2\, \pi^* 2p_y^1$	10	7	1.5	1	paramagnetic
$\mathrm{O_2^{2-}}$ (peroxide)	18	$\ldots \pi^* 2p_x^2\, \pi^* 2p_y^2$	10	8	1	0	diamagnetic
$\mathrm{F_2}$	18	same as $\mathrm{O_2^{2-}}$	10	8	1	0	diamagnetic
$\mathrm{Ne_2}$	20	$\ldots \sigma^* 2p_z^2$	10	10	0	0	does not exist
CO	14	isoelectronic with $\mathrm{N_2}$	10	4	3	0	diamagnetic
$\mathrm{NO^+}$	14	isoelectronic with $\mathrm{N_2}$	10	4	3	0	diamagnetic
$\mathrm{CN^-}$	14	isoelectronic with $\mathrm{N_2}$	10	4	3	0	diamagnetic
NO	15	isoelectronic with $\mathrm{O_2^+}$	10	5	2.5	1	paramagnetic

The counts include the four $1s$ electrons of $\sigma 1s^2\, \sigma^* 1s^2$ for second-period species, which cancel; drop them and the bond order is unchanged.

The speed trick — count total electrons, read the bond order

For any second-period diatomic (or isoelectronic ion), the bond order depends only on the total electron count:

Total electrons	10	11	12	13	14	15	16	17	18
Bond order	1	1.5	2	2.5	3	2.5	2	1.5	1

Bond order rises by 0.5 per electron up to 14 and falls by 0.5 per electron after 14. 14 electrons is the peak ($\mathrm{N_2}$, CO, $\mathrm{NO^+}$, $\mathrm{CN^-}$, $\mathrm{C_2^{2-}}$). Magnetism: 13, 15 and 17 electrons have one unpaired electron; 16 has two ($\mathrm{O_2}$); 10 has two ($\mathrm{B_2}$); 12, 14 and 18 are diamagnetic.

The ordered sets NEET asks

Question	Order
Bond order of oxygen species	$\mathrm{O_2^+}\ (2.5) > \mathrm{O_2}\ (2) > \mathrm{O_2^-}\ (1.5) > \mathrm{O_2^{2-}}\ (1)$
Stability of oxygen species	same as bond order: $\mathrm{O_2^+} > \mathrm{O_2} > \mathrm{O_2^-} > \mathrm{O_2^{2-}}$
Bond length of oxygen species	reverse: $\mathrm{O_2^{2-}} > \mathrm{O_2^-} > \mathrm{O_2} > \mathrm{O_2^+}$
Bond order of nitrogen species	$\mathrm{N_2}\ (3) > \mathrm{N_2^+} = \mathrm{N_2^-}\ (2.5)$
Stability of $\mathrm{N_2^+}$ vs $\mathrm{N_2^-}$	$\mathrm{N_2^+} > \mathrm{N_2^-}$ — same bond order, but $\mathrm{N_2^-}$ carries an extra antibonding electron
Bond length of nitrogen species	$\mathrm{N_2^-} > \mathrm{N_2^+} > \mathrm{N_2}$
Which gains stability on losing an electron	$\mathrm{O_2}$ (2 to 2.5), NO (2.5 to 3), $\mathrm{F_2}$ (1 to 1.5) — anything with electrons in antibonding orbitals; $\mathrm{N_2}$ and CO lose stability (3 to 2.5)
Which gains stability on gaining an electron	$\mathrm{C_2}$ (2 to 2.5), $\mathrm{B_2}$ (1 to 1.5); $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{F_2}$ all lose
Bond order 3 club	$\mathrm{N_2}$, CO, $\mathrm{NO^+}$, $\mathrm{CN^-}$, $\mathrm{C_2^{2-}}$
Bond order 1 club	$\mathrm{H_2}$, $\mathrm{Li_2}$, $\mathrm{B_2}$, $\mathrm{F_2}$, $\mathrm{O_2^{2-}}$
Zero bond order (do not exist)	$\mathrm{He_2}$, $\mathrm{Be_2}$, $\mathrm{Ne_2}$

Key Point: Paramagnetic means at least one unpaired electron; diamagnetic means all paired. Among the neutral second-period diatomics only $\mathrm{B_2}$ and $\mathrm{O_2}$ are paramagnetic. Every odd-electron species ($\mathrm{H_2^+}$, $\mathrm{He_2^+}$, $\mathrm{N_2^+}$, $\mathrm{O_2^+}$, $\mathrm{O_2^-}$, NO) is automatically paramagnetic. The paramagnetism of $\mathrm{O_2}$ is what MO theory explains and valence bond theory cannot.

[NEET] Three conditions for atomic orbitals to combine into molecular orbitals: same or nearly the same energy, same symmetry about the molecular axis, and maximum overlap. The number of MOs formed equals the number of AOs combined; $N$ atomic orbitals give $\frac{N}{2}$ bonding and $\frac{N}{2}$ antibonding MOs.

The Dipole-Moment Checklist and the Hydrogen-Bonding Orders

Dipole moment — zero or not, in three ticks

A molecule has zero dipole moment if the individual bond dipoles cancel by symmetry. The checklist:

1. Is it a symmetrical shape with identical outer atoms? Linear $\mathrm{AB_2}$, trigonal planar $\mathrm{AB_3}$, tetrahedral $\mathrm{AB_4}$, trigonal bipyramidal $\mathrm{AB_5}$, octahedral $\mathrm{AB_6}$, square planar $\mathrm{AB_4}$ and linear $\mathrm{AB_2}$ with three lone pairs all give zero.
2. Does the central atom have lone pairs that break the symmetry? Bent, pyramidal, see-saw, T-shaped and square pyramidal molecules are polar.
3. Are the outer atoms different? $\mathrm{CHCl_3}$, $\mathrm{CH_3Cl}$, HCl, COS are polar even with a symmetrical shape.

Zero dipole moment ($\mu = 0$)	Non-zero dipole moment ($\mu \neq 0$)
$\mathrm{CO_2}$, $\mathrm{CS_2}$, $\mathrm{BeCl_2}$, $\mathrm{HgCl_2}$, $\mathrm{C_2H_2}$ (linear)	$\mathrm{H_2O}$ (1.85 D), $\mathrm{H_2S}$ (0.95 D), $\mathrm{SO_2}$, $\mathrm{O_3}$, $\mathrm{NO_2}$ (bent)
$\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{SO_3}$, $\mathrm{AlCl_3}$ (trigonal planar)	$\mathrm{NH_3}$ (1.47 D), $\mathrm{NF_3}$ (0.23 D), $\mathrm{PCl_3}$, $\mathrm{PH_3}$ (pyramidal)
$\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{SiF_4}$, $\mathrm{SnCl_4}$ (tetrahedral)	$\mathrm{CHCl_3}$ (1.04 D), $\mathrm{CH_3Cl}$, $\mathrm{CH_2Cl_2}$ (unequal outer atoms)
$\mathrm{PCl_5}$, $\mathrm{PF_5}$ (trigonal bipyramidal)	$\mathrm{SF_4}$ (see-saw), $\mathrm{ClF_3}$ (T-shaped)
$\mathrm{SF_6}$ (octahedral), $\mathrm{XeF_4}$ (square planar), $\mathrm{XeF_2}$ (linear)	$\mathrm{BrF_5}$, $\mathrm{IF_5}$ (square pyramidal), $\mathrm{XeOF_4}$
$\mathrm{H_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{Cl_2}$ (homonuclear)	HF (1.78 D), HCl (1.07 D), HBr (0.79 D), HI (0.38 D), CO (small, about 0.1 D)
trans-1,2-dichloroethene, p-dichlorobenzene, benzene, $\mathrm{C_2H_4}$	cis-1,2-dichloroethene, o- and m-dichlorobenzene, chlorobenzene

The ordered sets

Set	Order	Reason
Hydrogen halides	HF (1.78) $>$ HCl (1.07) $>$ HBr (0.79) $>$ HI (0.38) D	Electronegativity difference falls down the group
Group 16 hydrides	$\mathrm{H_2O}$ (1.85) $>$ $\mathrm{H_2S}$ (0.95) $>$ $\mathrm{H_2Se}$ $>$ $\mathrm{H_2Te}$	Same, plus the angle closes toward 90°
Group 15 hydrides	$\mathrm{NH_3}$ (1.47) $>$ $\mathrm{PH_3}$ $>$ $\mathrm{AsH_3}$	Same
$\mathrm{NH_3}$ vs $\mathrm{NF_3}$	$\mathrm{NH_3}$ (1.47) $\gg$ $\mathrm{NF_3}$ (0.23) D	In $\mathrm{NH_3}$ the lone-pair dipole and the three N-H dipoles point the same way (toward N) and add; in $\mathrm{NF_3}$ the N-F dipoles point toward F, opposite to the lone pair, and nearly cancel it
Chloromethanes	$\mathrm{CH_3Cl} > \mathrm{CH_2Cl_2} > \mathrm{CHCl_3} > \mathrm{CCl_4}$ (0)	Vector addition of C-Cl dipoles; $\mathrm{CHCl_3}$ is 1.04 D
Dichlorobenzenes	ortho $>$ meta $>$ para (0)	Two C-Cl dipoles at 60°, 120°, 180°
$\mathrm{H_2O}$ vs $\mathrm{H_2S}$ vs $\mathrm{CO_2}$	$\mathrm{H_2O} > \mathrm{H_2S} > \mathrm{CO_2}$ (0)	$\mathrm{CO_2}$ is linear; its two C=O dipoles cancel

Key Point: Dipole moment tells you shape. If a question says "$\mathrm{XY_2}$ has zero dipole moment", the molecule is linear; if it says "$\mathrm{XY_3}$ has a non-zero dipole moment", it is pyramidal, not planar.

[NEET] Percentage ionic character grows with dipole moment for the same bond length; at an electronegativity difference of about 1.7 the bond is roughly 50 percent ionic, and larger differences mean mostly ionic. Fajans' rules give trends, not laws. Ionic character orders: $\mathrm{LiCl} < \mathrm{NaCl} < \mathrm{KCl} < \mathrm{RbCl} < \mathrm{CsCl}$ (bigger cation, less polarising); $\mathrm{NaI} < \mathrm{NaBr} < \mathrm{NaCl} < \mathrm{NaF}$ (smaller anion, less polarisable); $\mathrm{BeCl_2} < \mathrm{MgCl_2} < \mathrm{CaCl_2} < \mathrm{BaCl_2}$; $\mathrm{AlCl_3}$ is largely covalent (small, highly charged $\mathrm{Al^{3+}}$).

Hydrogen bonding — the boiling-point orders

Hydrogen bonding needs H attached to F, O or N. It raises boiling points, melting points and water solubility, and it lowers the density of ice.

Hydride series	Boiling-point order	The story
Group 16	$\mathrm{H_2O} \gg \mathrm{H_2Te} > \mathrm{H_2Se} > \mathrm{H_2S}$	Water is abnormally high (100 °C) because of extensive H-bonding; the rest rise with molar mass (van der Waals)
Group 17	$\mathrm{HF} > \mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}$	HF is abnormally high (zig-zag H-bonded chains); the rest rise with size
Group 15	$\mathrm{SbH_3} > \mathrm{NH_3} > \mathrm{AsH_3} > \mathrm{PH_3}$	$\mathrm{NH_3}$ is high because of H-bonding, but $\mathrm{SbH_3}$ is so heavy that its van der Waals forces win; $\mathrm{NH_3}$ is second, not first
Group 14	$\mathrm{SnH_4} > \mathrm{GeH_4} > \mathrm{SiH_4} > \mathrm{CH_4}$	No H-bonding at all; plain molar-mass order

Other pairs: ethanol ($\mathrm{C_2H_5OH}$, about 78 °C) $\gg$ dimethyl ether ($\mathrm{CH_3OCH_3}$, about $-24$ °C), same formula but only the alcohol H-bonds; $\mathrm{H_2O_2}$ $>$ $\mathrm{H_2O}$; carboxylic acids exist as H-bonded dimers (acetic acid in benzene shows twice the expected molar mass).

H-bond strength order: $\mathrm{F{-}H \cdots F} > \mathrm{O{-}H \cdots O} > \mathrm{N{-}H \cdots N}$, following electronegativity of the acceptor. Typical strength is a few tens of kJ/mol, far below a covalent bond (hundreds) but above van der Waals.

Intermolecular versus intramolecular

Type	Where	Effect	Stock examples
Intermolecular	Between two different molecules (same or different compounds)	Raises boiling point, melting point, viscosity, solubility in water	HF, $\mathrm{H_2O}$, ice, $\mathrm{NH_3}$, alcohols, carboxylic-acid dimers, p-nitrophenol, p-hydroxybenzoic acid, DNA base pairs, proteins
Intramolecular	Between two groups inside the same molecule, usually forming a 5- or 6-membered ring (chelation)	Lowers boiling point and water solubility relative to the para isomer; makes the ortho isomer steam-volatile	o-nitrophenol, salicylaldehyde, salicylic acid (o-hydroxybenzoic acid), o-chlorophenol, chloral hydrate

Key Point: Ortho-nitrophenol is more volatile than para-nitrophenol — intramolecular H-bonding in the ortho isomer means no bonds between molecules, so it escapes easily. The para isomer is locked by intermolecular H-bonds and boils much higher.

Ice is less dense than water because H-bonding holds each water molecule in an open tetrahedral cage (four H-bonds per molecule); on melting, some bonds break, the molecules pack closer, and density rises up to 4 °C. Solid and liquid HF is a zig-zag chain, and $\mathrm{KHF_2}$ (containing $\mathrm{HF_2^-}$) exists because of the very strong $\mathrm{F{-}H \cdots F}$ bond — the reason HF is a weak acid while HCl is strong.