Chemistry Notes

Chemical Bonding and Molecular Structure — NEET Notes

Almost nothing on earth exists as an isolated atom. Salt is a lattice of ions, water is a network of polar molecules, oxygen drifts about as a paramagnetic diatomic, and the cytochromes that ferry electrons in your mitochondria are held together by every kind of bond chemistry has named. This chapter is the architecture of matter — why some atoms join, why others refuse, and why the molecules they form have the shapes they do. NEET asks 2–4 questions every year from this unit; VSEPR geometry, hybridisation, MOT bond order, and paramagnetism of O₂ are the recurring blockbusters. By the end of this chapter you should be able to look at a formula like SF₆, BrF₅, or N₂ and read off its hybridisation, geometry, bond order, and magnetic character in one breath.

Kossel-Lewis approach — the octet idea

In 1916, working independently, the German chemist Walther Kossel and the American chemist Gilbert N. Lewis arrived at the same insight: atoms combine in order to attain the electronic configuration of the nearest noble gas. Lewis pictured the atom as a positively charged kernel (nucleus plus inner electrons) surrounded by an outer shell that could hold up to eight valence electrons. He drew these as dots at the corners of an imaginary cube — the Lewis dot symbol. Kossel pointed out that the highly electropositive alkali metals and the highly electronegative halogens sit on opposite sides of the noble gases in the periodic table; sodium loses an electron to become Na⁺ with the Ne configuration, chlorine gains one to become Cl⁻ with the Ar configuration, and the resulting ions stabilise each other by electrostatic attraction. From this came the octet rule: atoms gain, lose, or share electrons to attain eight in their outer shell. Hydrogen and helium are the exception — they aim for a duplet (two electrons, the He configuration).

Langmuir refined the picture in 1919 by introducing the term covalent bond — a shared pair of electrons. A single bond is one shared pair; a double bond two; a triple bond three. Lewis dot structures became the universal shorthand for showing bonding pairs and lone pairs.

"Octet rule with exceptions" — useful as a sketch, false as a law.

The compromise that runs through this whole chapter

The octet rule is a sketch, not a law. It has three classes of well-known exceptions:

Incomplete octet

< 8 e⁻

central atom

LiCl, BeH₂, BCl₃, BF₃, AlCl₃. Li, Be, B start with too few valence electrons to ever reach an octet.

PYQ pattern: NEET 2023 Q.56 — AlCl₃, BeCl₂

Odd-electron molecules

unpaired e⁻

no octet possible

NO (11 valence e⁻), NO₂ (17 e⁻), ClO₂. With an odd total, one atom must end up with seven.

Expanded octet

> 8 e⁻

uses empty d-orbitals

PCl₅ (10 e⁻), SF₆ (12 e⁻), XeF₂, H₂SO₄. Third-period elements onwards have empty 3d orbitals available.

PYQ pattern: NEET 2023 Q.56 — PCl₅

Formal charge is a bookkeeping device for Lewis structures. For an atom in a molecule, formal charge = (valence electrons of free atom) − (non-bonding electrons) − ½ (bonding electrons). It does not represent real charge separation; it helps you pick the most plausible Lewis structure among alternatives. The structure with the smallest formal charges on each atom is usually the lowest in energy.

Ionic bond — the lattice that holds it together

An ionic bond forms when one atom transfers electrons to another. The classic example is sodium chloride: Na (1s²2s²2p⁶3s¹) loses one electron to become Na⁺; Cl (1s²2s²2p⁶3s²3p⁵) gains it to become Cl⁻; both attain noble-gas configurations and the resulting ions are bound by Coulombic attraction. Crucially, the bond does not exist in isolation — in the solid, Na⁺ and Cl⁻ pack into a three-dimensional lattice in which every Na⁺ has six Cl⁻ neighbours and every Cl⁻ has six Na⁺ neighbours. This is the rock-salt structure.

The arithmetic of ionic-bond formation looks alarming at first. The ionisation enthalpy of Na is +495.8 kJ/mol and the electron-gain enthalpy of Cl is −348.7 kJ/mol; the sum is positive at +147.1 kJ/mol. So why does NaCl form spontaneously? The answer is that we have not yet added the term that dominates the whole process — the lattice enthalpy, which is −788 kJ/mol for NaCl. Once the gaseous ions snap into the crystal lattice, far more energy is released than was needed to make them. The ionic compound is stable not because of the octet but because of the lattice.

Lattice enthalpy — the master variable

Lattice enthalpy is defined as the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions. It depends on two simple factors:

Kapustinskii's intuition: lattice enthalpy is large when ionic charges are large and ionic radii are small. MgO (Mg²⁺O²⁻, +2,−2) has a lattice enthalpy of about −3795 kJ/mol; NaCl (+1,−1) only −788 kJ/mol. This is why MgO melts at 2852 °C while NaCl melts at 801 °C.

Charge on ions (z⁺ z⁻)

↑ charge → ↑ |U|

strongest factor

Doubling the charge roughly quadruples the lattice energy. The (+2,−2) MgO is much more strongly bound than the (+1,−1) NaCl.

Inter-ionic distance (r)

↓ r → ↑ |U|

U ∝ 1 / (r⁺ + r⁻)

Smaller ions sit closer and feel stronger Coulombic pull. LiF (small ions) has a much larger lattice enthalpy than CsI (large ions).

Born-Haber cycle — measuring what we cannot weigh

Lattice enthalpy cannot be measured directly — you cannot scoop a mole of gaseous Na⁺ and a mole of gaseous Cl⁻ onto a balance. The Born-Haber cycle gets around this by applying Hess's law: the enthalpy change for the overall reaction Na(s) + ½ Cl₂(g) → NaCl(s) equals the sum of the enthalpy changes of any pathway connecting reactants and products. Break the reaction into measurable steps, leave lattice enthalpy as the only unknown, and solve.

Bond parameters — length, angle, enthalpy, order

Four numbers characterise any covalent bond. Bond length is the equilibrium internuclear distance, measured in picometres — about 74 pm for H–H, 154 pm for C–C, 134 pm for C=C, 120 pm for C≡C. Each atom contributes its covalent radius to this length. Bond angle is the angle between two bonds at a central atom — 104.5° in H₂O, 107° in NH₃, 109.5° in CH₄, 120° in BF₃, 180° in BeCl₂. Bond enthalpy is the energy required to break one mole of a particular bond in the gas phase: 436 kJ/mol for H–H, 498 kJ/mol for O=O, 946 kJ/mol for N≡N — that last figure being one of the strongest known and the reason atmospheric nitrogen is so inert. Bond order is simply the number of shared electron pairs: 1 for a single, 2 for a double, 3 for a triple.

Three patterns recur across NEET problems and are worth memorising. First, for the same pair of atoms, bond order ↑ → bond length ↓ → bond enthalpy ↑. Hence C–C (154 pm, 348 kJ/mol) < C=C (134 pm, 614 kJ/mol) < C≡C (120 pm, 839 kJ/mol). Second, isoelectronic species have identical bond orders: N₂, CO, CN⁻, NO⁺ all carry triple bonds with bond order 3 — NEET 2017 Q.6 tested exactly this. Third, resonance averages bond lengths: all three N–O bonds in NO₃⁻ are equal (between single and double), and all three C–O bonds in CO₃²⁻ are equal too.

Bond polarity and dipole moment

No bond is purely covalent or purely ionic. When the two atoms are different, the more electronegative one pulls the shared pair toward itself, creating a polar covalent bond. The result is a dipole moment µ = Q × r, measured in Debye units (1 D = 3.34 × 10⁻³⁰ C m). For diatomic molecules the bond dipole is the molecular dipole. For polyatomic molecules, the molecular dipole is the vector sum of the bond dipoles, and symmetry can cancel everything out: BeF₂, BF₃, CCl₄, CO₂, and 1,4-dichlorobenzene all have µ = 0 despite having polar bonds. NEET 2020 Q.150 and NEET 2021 Q.91 both turn on this distinction.

A subtle case worth noticing: NH₃ has µ = 1.47 D but NF₃ has µ = 0.23 D, even though fluorine is more electronegative than hydrogen. The reason is the lone pair on nitrogen. In NH₃ the N–H bond dipoles point toward N and add to the lone-pair dipole. In NF₃ the N–F bond dipoles point away from N and cancel the lone-pair dipole. Geometry decides the answer, not electronegativity alone.

The covalent bond — sharing as architecture

Covalent bonds are formed by sharing one or more pairs of electrons between atoms. Each atom contributes at least one electron to the shared pair, and each ends up with a stable noble-gas configuration. The hallmark of a covalent compound is the molecule — a discrete unit, often with definite shape, soluble in non-polar solvents, with low melting points and poor electrical conductivity. Contrast this with ionic compounds, which form three-dimensional lattices, dissolve in water, conduct in the molten or aqueous state, and melt only at very high temperatures.

The Kossel-Lewis picture says nothing about the actual mechanism of covalent bonding — why does H₂ form at all? What does "sharing" mean physically? Three theories of increasing sophistication answer these questions: VSEPR (which predicts geometry), valence bond theory (which explains bonding via orbital overlap), and molecular orbital theory (which treats electrons as belonging to the whole molecule). We will take them in turn.

VSEPR theory — geometry from electron-pair repulsion

The Valence Shell Electron Pair Repulsion theory was proposed by Sidgwick and Powell in 1940 and refined by Nyholm and Gillespie in 1957. It is the most-tested idea in this chapter. The core postulate is brutally simple: electron pairs in the valence shell of the central atom repel each other and arrange themselves to maximise the distance between them. Whether the pair is bonded or unbonded does not matter for placement; both kinds count.

A second postulate sharpens the prediction: lone pairs occupy more space than bonded pairs because a lone pair is held by only one nucleus while a bonded pair is shared between two. The order of repulsion strength is therefore:

Lone pair – Lone pair > Lone pair – Bond pair > Bond pair – Bond pair

The Nyholm-Gillespie repulsion order — NEET 2016 Q.2 tested this directly

This is why H₂O (two lone pairs on O) has an H–O–H angle of 104.5°, NH₃ (one lone pair on N) has 107°, and CH₄ (no lone pairs on C) keeps the ideal tetrahedral 109.5°. The lone pairs squeeze the bond pairs together. NEET 2016 Q.19 — "the H–O–H bond angle in H₂O is larger than the H–C–H bond angle in CH₄" — turns on this single chain.

To apply VSEPR to any molecule, follow a four-step process:

In trigonal bipyramidal and octahedral cases, lone pairs preferentially occupy the positions that minimise lp–bp repulsions. In a trigonal bipyramid, lone pairs go to equatorial positions (only 2 lp–bp repulsions at 90° instead of 3). This explains the see-saw shape of SF₄ (sp³d, 1 lp, equatorial), the T-shape of ClF₃ (sp³d, 2 lp, both equatorial), and the linear shape of XeF₂ (sp³d, 3 lp, all equatorial). NEET 2022 Q.62 asked which of IF₅, SF₄, XeF₂, ClF₃ has maximum lp–lp repulsion — the answer is XeF₂ with three lone pairs in the equatorial plane.

Valence bond theory — the orbital overlap picture

VSEPR predicts where the atoms sit but says nothing about why a bond exists or how strong it is. Valence Bond Theory, introduced by Heitler and London in 1927 and developed by Linus Pauling, fills that gap. Its starting point is the formation of H₂. When two H atoms approach, two new attractive forces appear (each nucleus attracts the other's electron) and two new repulsive forces (electron–electron and nucleus–nucleus). The attractive forces dominate until the atoms reach an equilibrium separation of 74 pm, where the potential energy hits a minimum. At that minimum, the two 1s orbitals have partially merged — overlapped — and the electrons, with opposite spins, are shared between the nuclei. The energy released, 436 kJ/mol, is the bond enthalpy.

From this single picture, three properties of covalent bonds emerge naturally. Directionality: p-orbitals point along axes, so p–p overlap is directional. Strength scales with overlap: bigger overlap, stronger bond. Bond order matters: when two atoms share more than one pair, additional sideways overlaps form on top of the axial overlap.

Sigma and pi bonds

VBT distinguishes two kinds of covalent bond by the geometry of the overlap. A sigma (σ) bond is formed by end-on, head-on overlap along the internuclear axis — s–s, s–p, or p–p axial overlap. A pi (π) bond is formed by sideways overlap of two parallel p-orbitals perpendicular to the internuclear axis, producing two saucer-shaped clouds above and below the bond. Because head-on overlap concentrates electron density between the nuclei, sigma bonds are stronger than pi bonds. The structural recipe for multiple bonds is now clear: a single bond = 1 σ, a double bond = 1 σ + 1 π, a triple bond = 1 σ + 2 π. NEET 2023 Q.53 asked you to count σ-bonds, π-bonds, and lone pairs in pyridine — the answer (11 σ, 3 π, 1 lp) flows from this recipe.

Hybridisation — Pauling's geometry-from-orbitals fix

Plain VBT runs into trouble with polyatomic molecules. Carbon in its ground state is 2s² 2p², with only two unpaired electrons — yet it forms four equivalent C–H bonds in methane, all pointing to the corners of a tetrahedron at 109.5°. Plain p–p overlap would predict 90° between any two C–H bonds and would leave one H attached anywhere. Pauling's resolution was hybridisation: atomic orbitals of slightly different energy on the same atom can mix to produce a new set of equivalent orbitals — hybrid orbitals — that point in specific directions.

The number of hybrid orbitals always equals the number of atomic orbitals mixed. Each hybrid orbital has a fixed s-character and p-character. Hybridisation lets you read off geometry directly: count the steric number of the central atom (bond pairs + lone pairs), assign the matching hybridisation, and the geometry follows.

sp hybridisation

Linear

180° · 50% s, 50% p

One s + one p → 2 hybrids, both 180° apart.

Examples: BeCl₂, CO₂, C₂H₂, HCN, BeF₂.

NEET 2017 Q.37: IBr₂⁻ and XeF₂ — both sp³d but linear with 3 lp

sp² hybridisation

Trigonal planar

120° · 33% s, 67% p

One s + two p → 3 hybrids in a plane, 120° apart.

Examples: BCl₃, BF₃, C₂H₄, CH₂O, SO₃, graphite.

NEET 2021 Q.59: BF₃ is sp², 6 e⁻ around B (electron-deficient)

sp³ hybridisation

Tetrahedral

109.5° · 25% s, 75% p

One s + three p → 4 hybrids at tetrahedron corners.

Examples: CH₄, NH₃, H₂O, SiCl₄, CCl₄, diamond.

NEET 2016 Q.19: CH₄ 109.5° > NH₃ 107° > H₂O 104.5°

sp³d hybridisation

Trigonal bipyramidal

90° + 120° · uses 1 d-orbital

One s + three p + one d → 5 hybrids. 3 equatorial (120°) + 2 axial (90° to equatorial).

Examples: PCl₅, PF₅, SF₄ (see-saw), ClF₃ (T), XeF₂ (linear).

NEET 2021 Q.85: PCl₅ → trigonal bipyramidal

sp³d² hybridisation

Octahedral

90° · uses 2 d-orbitals

One s + three p + two d → 6 hybrids, all 90° apart.

Examples: SF₆, [CrF₆]³⁻, [Co(NH₃)₆]³⁺, IF₅ (square pyramidal, 1 lp), XeF₄ (square planar, 2 lp), BrF₅.

NEET 2021 Q.85: SF₆ → octahedral; BrF₅ → square pyramidal

A small caution: the choice of hybridisation is read off the steric number, not the molecular shape. SF₄ and CH₄ are both sp³d... no, SF₄ is sp³d (steric number 5, one lone pair, see-saw shape). CH₄ is sp³ (steric number 4, no lone pairs, tetrahedral). The steric number tells you the hybrid set; the lone-pair count tells you the molecular shape.

Molecular orbital theory — electrons belong to the whole molecule

Hybridisation rescues VBT for polyatomic molecules, but even VBT runs into problems with simple diatomics. The most famous failure: VBT predicts O₂ to be diamagnetic (all electrons paired in two double-bond electron pairs), yet liquid O₂ is observably paramagnetic — it sticks to a magnet. The theory that explains this, and much more, is Molecular Orbital Theory, developed by Hund and Mulliken in 1932.

MOT abandons the idea that bonding electrons belong to particular atoms. Instead, the atomic orbitals of all atoms in a molecule combine — by linear combination of atomic orbitals (LCAO) — to produce a new set of molecular orbitals that extend over the whole molecule. Two atomic orbitals always combine to make two molecular orbitals: a bonding MO (lower in energy, electron density between the nuclei) by addition of wavefunctions, and an antibonding MO (higher in energy, with a node between the nuclei) by subtraction. We label them σ, π, σ*, π*; the asterisk marks the antibonding orbital.

For atomic orbitals to mix, three conditions must be met: similar energy (1s with 1s, not 1s with 2s), same symmetry about the bond axis (2pz with 2pz if the bond is along z), and appreciable overlap. Once filled, MOs follow the same rules as atomic orbitals — Aufbau, Pauli, Hund.

Bond order, stability, magnetic character

From the MO diagram you read off three quantities:

  • Bond order = ½ (Nb − Na), where Nb is the number of electrons in bonding MOs and Na in antibonding MOs. Bond order > 0 → stable molecule; bond order = 0 → does not exist.
  • Bond length and enthalpy: higher bond order → shorter, stronger bond.
  • Magnetic character: any unpaired electron makes the molecule paramagnetic; all paired makes it diamagnetic.

MO diagrams — H₂, He₂, N₂, O₂, F₂

Now to the second-row diatomics that NEET asks every year. There are two energy orderings of MOs to remember, and the difference is critical.

For B₂, C₂, N₂: σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz. The π2p MOs lie below the σ2pz because of s–p mixing.

For O₂, F₂, Ne₂: σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz. The σ2pz drops below the π2p MOs.

Now the per-molecule rundown — exactly what NEET asks:

H₂ (2 electrons)

Bond order 1

diamagnetic, exists

Configuration: (σ1s)². Nb = 2, Na = 0.

Bond length 74 pm; bond enthalpy 436 kJ/mol.

He₂ (4 electrons)

Bond order 0

does NOT exist

Configuration: (σ1s)² (σ*1s)². Nb = 2, Na = 2.

Antibonding cancels bonding — no net bond.

NEET 2020 Q.151: He₂ doesn't exist

N₂ (14 electrons)

Bond order 3

diamagnetic, very stable

Configuration: (KK)(σ2s)²(σ*2s)²(π2px)²(π2py)²(σ2pz)². Nb = 10, Na = 4.

Bond enthalpy 946 kJ/mol — the highest among diatomics.

NEET 2023 Q.64: π2p below σ2pz for N₂

O₂ (16 electrons)

Bond order 2

PARAMAGNETIC

Configuration: ...(σ2pz)²(π2px)²(π2py)²(π*2px)¹(π*2py)¹. Nb = 10, Na = 6.

Two unpaired electrons in π* — sticks to magnet.

F₂ (18 electrons)

Bond order 1

diamagnetic, weak bond

Configuration: ...(σ2pz)²(π2px)²(π2py)²(π*2px)²(π*2py)². Nb = 10, Na = 8.

Bond enthalpy only 155 kJ/mol — weakest single bond in the second row.

Isoelectronic check

Same BO

if same electron count

N₂, CO, CN⁻, NO⁺: all 14 valence e⁻, bond order 3.

F₂, O₂²⁻: bond order 1.

NEET 2017 Q.6: CN⁻ and CO have same BO

A useful family of related calculations to memorise: O₂ → O₂⁺ → O₂⁻ → O₂²⁻. Each electron added goes into the π* antibonding MOs, so bond order drops: O₂⁺ has BO 2.5, O₂ has BO 2, O₂⁻ (superoxide) has BO 1.5, O₂²⁻ (peroxide) has BO 1. Bond length therefore increases in the same order, and stability decreases. NEET 2018 Q.61 — "which has highest bond order among NO, CN⁻, CN⁺, CN?" — turns on counting electrons and reading off bond order from the MO diagram.

Hydrogen bonding — the bond that runs biology

The last bond in this chapter is by far the most consequential one in the world outside the chemistry lab. A hydrogen bond is the attractive force between a hydrogen atom covalently bonded to a small, highly electronegative atom (F, O, or N) and another electronegative atom — in the same molecule (intramolecular) or in a different one (intermolecular). The shared electron pair in the H–X bond is pulled toward X, leaving H with a partial positive charge δ+ and X with a partial negative charge δ−. The δ+ hydrogen of one molecule is then attracted to the δ− atom of the next.

Hydrogen bonds are weaker than covalent bonds (typically 10–40 kJ/mol vs 200–800 kJ/mol) but much stronger than ordinary van der Waals dispersion forces (2–5 kJ/mol). They are responsible for:

  • The anomalously high boiling point of water (100 °C; H₂S boils at −60 °C despite being heavier).
  • The fact that ice floats on water — H-bonds force water molecules into an open lattice that is less dense than liquid water.
  • The dimerisation of carboxylic acids and the high boiling points of alcohols.
  • The alpha-helix and beta-sheet structures of proteins, and the double helix of DNA — every base pair held by 2 or 3 H-bonds.
  • The mutarotation of sugars and the solubility of NH₃, ethanol, glucose in water.

Intramolecular H-bonds (e.g. in o-nitrophenol, salicylaldehyde) lower boiling points by reducing intermolecular association, while intermolecular H-bonds (e.g. in p-nitrophenol) raise boiling points. NEET 2023 Q.74 grouped hydrogen bonding with dipole–dipole, dipole–induced dipole, and dispersion forces as "intermolecular forces" — explicitly excluding covalent bonds, which act within a molecule.

NEET PYQ Snapshot

Five high-yield previous-year questions — solve before moving on.

NEET 2023

The correct order of energies of molecular orbitals of N₂ molecule is —

  1. σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < (π*2px = π*2py) < σ2pz < σ*2pz
  2. σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz
  3. σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz
  4. σ1s < σ*1s < σ2s < σ*2s < σ2pz < σ*2pz < (π2px = π2py) < (π*2px = π*2py)
Answer: (2)

Why: For B₂, C₂, and N₂, s–p mixing pushes σ2pz above the π2p MOs. The π2p orbitals therefore lie below σ2pz. From O₂ onwards the ordering reverses.

NEET 2022

Amongst the following, which one will have maximum 'lone pair – lone pair' electron repulsions?

  1. IF₅
  2. SF₄
  3. XeF₂
  4. ClF₃
Answer: (3) XeF₂

Why: Count lone pairs on the central atom. SF₄ (sp³d) has 1 lp; ClF₃ (sp³d) has 2 lp; IF₅ (sp³d²) has 1 lp; XeF₂ (sp³d) has 3 lp — all three in equatorial positions of the trigonal bipyramid. More lone pairs → more lp–lp repulsion.

NEET 2021

Match List-I with List-II. (a) PCl₅ (i) Square pyramidal (b) SF₆ (ii) Trigonal planar (c) BrF₅ (iii) Octahedral (d) BF₃ (iv) Trigonal bipyramidal

  1. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
  2. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
  3. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
  4. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Answer: (2)

Why: PCl₅ is sp³d, trigonal bipyramidal. SF₆ is sp³d², octahedral. BrF₅ is sp³d² with one lone pair → square pyramidal. BF₃ is sp², trigonal planar.

NEET 2020

Which of the following set of molecules will have zero dipole moment?

  1. BF₃, HF, CO₂, 1,3-dichlorobenzene
  2. NF₃, BeF₂, H₂O, 1,3-dichlorobenzene
  3. BF₃, BeF₂, CO₂, 1,4-dichlorobenzene
  4. NH₃, BeF₂, H₂O, 1,4-dichlorobenzene
Answer: (3)

Why: µ = 0 requires the molecular geometry to make the vector sum of bond dipoles zero. BF₃ is trigonal planar (3 equal vectors at 120°); BeF₂ and CO₂ are linear with opposing bond dipoles; 1,4-dichlorobenzene has the two C–Cl dipoles pointing in exactly opposite directions across the ring.

NEET 2018

Consider the species CN⁺, CN⁻, NO, and CN. Which one will have the highest bond order?

  1. NO
  2. CN⁻
  3. CN⁺
  4. CN
Answer: (2) CN⁻

Why: Count valence electrons. CN⁻ has 4 + 5 + 1 = 10 valence e⁻ — isoelectronic with N₂, so bond order 3. NO has 11 valence e⁻ (BO 2.5). CN has 9 (BO 2.5). CN⁺ has 8 (BO 2.0). CN⁻ wins.

Expert FAQs

Questions NEET has asked from this chapter, answered straight.

What is the octet rule and what are its limitations?
The octet rule states that atoms combine by gaining, losing, or sharing electrons so that each attains eight electrons in its valence shell — the noble-gas configuration. It has three classes of exceptions: (i) incomplete octets in molecules of Li, Be and B such as LiCl, BeH₂, BCl₃, BF₃, AlCl₃; (ii) odd-electron molecules such as NO and NO₂ where an octet is impossible; and (iii) expanded octets in third-period and heavier elements such as PCl₅ (10 e⁻), SF₆ (12 e⁻) and XeF₂ — these use empty d-orbitals to accommodate more than eight electrons.
What is lattice enthalpy and why is it important?
Lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions — for NaCl it is +788 kJ/mol. It is the single most important quantity in deciding whether an ionic compound will form: even when the sum of ionisation enthalpy and electron-gain enthalpy is positive, the large negative lattice enthalpy can still drive the overall process and stabilise the crystal. Lattice enthalpy is computed indirectly through the Born-Haber cycle using Hess's law.
How do you predict molecular geometry using VSEPR theory?
Count total valence electrons; work out bond pairs and lone pairs around the central atom; read off the electron-pair geometry from the steric number (2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral); adjust for lone pairs because the repulsion order is lp–lp > lp–bp > bp–bp, which compresses bond angles. CH₄ stays at 109.5°; NH₃ drops to 107°; H₂O drops to 104.5°.
What is the difference between sigma and pi bonds?
A sigma bond is formed by end-on (axial) overlap of orbitals along the internuclear axis — s–s, s–p, or p–p head-on. A pi bond is formed by sideways overlap of two parallel p-orbitals perpendicular to the internuclear axis, producing two saucer-shaped electron clouds above and below the bond axis. Sigma bonds are stronger because overlap is greater; pi bonds are weaker. A single bond is one sigma; a double bond is one sigma + one pi; a triple bond is one sigma + two pi.
What is hybridisation and why is it needed?
Hybridisation is the intermixing of atomic orbitals of slightly different energies on the same atom to produce a new set of equivalent orbitals — the hybrid orbitals — with definite directional properties. It was introduced by Pauling to explain the observed shapes and bond angles of polyatomic molecules, which simple orbital overlap could not. The five common types are sp (linear, 180°, BeCl₂), sp² (trigonal planar, 120°, BCl₃), sp³ (tetrahedral, 109.5°, CH₄), sp³d (trigonal bipyramidal, PCl₅) and sp³d² (octahedral, SF₆).
Why is O₂ paramagnetic and how does MOT explain it?
Molecular orbital theory explains O₂ paramagnetism in a way Lewis theory cannot. The 16 electrons of O₂ fill MOs in order (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px = π2py)⁴ (π*2px = π*2py)². The last two electrons occupy two degenerate π* antibonding orbitals — one in π*2px and one in π*2py by Hund's rule — leaving two unpaired electrons. Unpaired electrons make O₂ paramagnetic. Bond order = (10 − 6)/2 = 2, which matches the experimental double bond.
Why is the MO energy ordering different for N₂ versus O₂?
For B₂, C₂ and N₂ the order is σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz — the π2p MOs lie below the σ2pz MO because 2s–2p mixing pushes the σ2pz up. For O₂ and F₂ this mixing is small (the 2s and 2p are too far apart in energy) so the σ2pz drops back below the π2p MOs. NEET 2023 tested the N₂ ordering directly.
What is a hydrogen bond and when does it form?
A hydrogen bond is the attractive force between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) and another electronegative atom in the same or a different molecule. The H acquires a δ+ partial charge and the electronegative atom a δ−. Hydrogen bonds are weaker than covalent bonds but stronger than van der Waals forces, and they account for the high boiling point of water, the ice-floats anomaly, the alpha-helix of proteins, and the double helix of DNA. They are intermolecular (HF, H₂O) or intramolecular (o-nitrophenol).

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