Easy Practice — Set 1

Quick Recap — Bonding Basics & VSEPR

• Ionic bond = transfer of electrons; covalent bond = sharing; coordinate bond = both electrons from one atom.
• Octet rule: atoms aim for 8 valence electrons. VSEPR predicts shape from bond pairs + lone pairs.
• Hybridisation $\to$ shape: $sp$ linear (180$^\circ$), $sp^2$ trigonal planar (120$^\circ$), $sp^3$ tetrahedral (109.5$^\circ$), $sp^3d$ trigonal bipyramidal, $sp^3d^2$ octahedral.
• Lone pairs reduce bond angles ($CH_4\,109.5^\circ>NH_3\,107^\circ>H_2O\,104.5^\circ$). H-bonding needs F, O or N.

Beyond-NCERT JEE Formulae

Fast, high-yield formulae, each with when to reach for it and the trap it hides. (Bonding-type and VSEPR-shape basics are assumed; this is the computational toolkit that actually earns marks.)

1. Formal charge — pick the best Lewis / resonance structure. Use when a species has more than one valid Lewis structure: the dominant contributor keeps formal charges smallest and puts any negative charge on the most electronegative atom.

$$FC = V - L - \tfrac12 B$$

$V$ = valence electrons of the free atom, $L$ = lone-pair (non-bonding) electrons, $B$ = bonding (shared) electrons. Formal charges of all atoms sum to the overall charge — use that as a check.

[JEE Tip] Among resonance contributors, the one with the least charge separation dominates; a structure with like charges on adjacent atoms is negligible.

2. MOT bond order, stability and magnetism. Use for homonuclear diatomics (and NO) and their ions to compare bond length, strength and magnetic nature.

$$\text{Bond order} = \dfrac{N_b - N_a}{2}$$

$N_b$ and $N_a$ = electrons in bonding and antibonding MOs. Any unpaired electron makes the species paramagnetic; all-paired makes it diamagnetic. Energy-level order:

• Light diatomics up to and including N2: $\sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p < \sigma^* 2p_z$ (the two pi bonds fill before sigma2pz).
• From O2 onward (O2, F2): $\sigma 2p_z$ sinks below the $\pi 2p$ pair.

[JEE Tip] Adding an electron to an antibonding MO cuts the bond order (O2 to O2-, 2 to 1.5, bond lengthens); removing one raises it (O2 to O2+, 2 to 2.5, bond shortens).

3. Steric number, shape and hybridisation. Use to fix geometry and hybridisation of any main-group centre, including hypervalent SF4, ClF3, XeF4.

$$SN = (\text{sigma-bonded atoms}) + (\text{lone pairs on central atom})$$

$SN = 2, 3, 4, 5, 6$ gives sp, sp2, sp3, sp3d, sp3d2 respectively. For a one-line count of hybridisation and lone pairs, use

$$H = \tfrac12\,(V + M - c + a)$$

$V$ = valence electrons of the central atom, $M$ = number of monovalent atoms (H and halogens) attached, $c$ = cationic charge, $a$ = anionic charge. $H = 2, 3, 4, 5, 6$ maps to the same sp…sp3d2 series, and lone pairs $= H - (\text{number of sigma bonds})$.

[JEE Tip] Count a double or triple bond as ONE domain for shape. With sp3d, lone pairs take equatorial sites (SF4 see-saw, ClF3 T-shaped, XeF2 linear); with sp3d2 they go trans (XeF4 square planar).

4. Dipole moment and percentage ionic character. Use to judge polarity and how ionic a nominally covalent bond really is.

$$\mu = q \times d$$

measured in debye; $1\ \text{D} = 3.336\times10^{-30}\ \text{C m}$. The molecular dipole is the vector sum of bond dipoles, so two equal bond dipoles $\mu_b$ separated by bond angle $\theta$ give $\mu_{net} = 2\mu_b\cos\tfrac{\theta}{2}$. The ionic character of a bond is

$$\%\ \text{ionic} = \dfrac{\mu_{obs}}{\mu_{ionic}} \times 100$$

where $\mu_{ionic}$ is the moment for a full unit-charge separation; numerically $\mu_{ionic}(\text{in D}) = 4.8 \times d$ with $d$ in angstrom.

[JEE Tip] Symmetric shapes (linear CO2, trigonal planar BF3, tetrahedral CCl4, octahedral SF6) give zero net dipole despite polar bonds; one lone pair (H2O, NH3) breaks the cancellation and the molecule turns polar.

5. Fajans' rules and the bond-order link. Use to compare covalent character of ionic solids and to reason about bond metrics.

Covalent character rises with a small cation, a large anion, a high charge on either ion, and a cation carrying an 18-electron (pseudo-noble-gas) shell. The polarising power of a cation scales as $\propto \dfrac{\text{charge}}{(\text{radius})^2}$. For a bond between the same two atoms, higher bond order means a shorter and stronger bond.

[JEE Tip] This is why AlCl3 is covalent while NaCl is ionic (Al3+ is small and triply charged), and why CuCl is more covalent than NaCl even though Cu+ and Na+ are similar in size — Cu+ has the more polarising 18-electron shell.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Bond order and magnetism of O2, O2+ and O2-. O2 has 16 electrons; in the O2-type ordering ($\sigma 2p_z$ below $\pi 2p$) the bonding MOs hold $N_b = 10$ and antibonding hold $N_a = 6$. $$\text{BO(O2)} = \dfrac{10 - 6}{2} = 2$$ The last two electrons singly occupy the degenerate $\pi^* 2p$ orbitals, so O2 is paramagnetic. Remove one antibonding electron for O2+ ($N_a = 5$): BO $= (10 - 5)/2 = 2.5$. Add one for O2- ($N_a = 7$): BO $= (10 - 7)/2 = 1.5$. Each still has one unpaired electron, so both are paramagnetic. Bond order and bond strength run O2+ (2.5) > O2 (2) > O2- (1.5); bond length is the reverse.

Example 2 — Bond order and magnetism of N2 and NO. N2 (14 electrons): $N_b = 10$, $N_a = 4$. $$\text{BO(N2)} = \dfrac{10 - 4}{2} = 3$$ Every MO is filled or empty, so N2 is diamagnetic. NO has 15 electrons: $N_b = 10$, $N_a = 5$, so BO $= (10 - 5)/2 = 2.5$ with one unpaired electron in $\pi^* 2p$ — NO is paramagnetic. Removing that electron (NO+) gives BO $= 3$, diamagnetic and isoelectronic with N2.

Example 3 — Formal charge in the nitrate ion (NO3-). Use the structure with one N=O double bond and two N-O single bonds. Nitrogen (4 bonds, no lone pair): $V = 5$, $L = 0$, $B = 8$, so $FC = 5 - 0 - \tfrac12(8) = +1$. Doubly bonded O: $V = 6$, $L = 4$, $B = 4$, so $FC = 6 - 4 - \tfrac12(4) = 0$. Each singly bonded O: $V = 6$, $L = 6$, $B = 2$, so $FC = 6 - 6 - \tfrac12(2) = -1$. The charges sum to $(+1) + 0 + (-1) + (-1) = -1$, matching the ion — negative charge sits on oxygen, as expected.

Example 4 — Shape and hybridisation of SF4. Sulfur: $V = 6$, four F so $M = 4$, neutral so $c = a = 0$. $$H = \tfrac12(6 + 4 - 0 + 0) = 5 \Rightarrow sp^3d$$ Sigma bonds $= 4$, so lone pairs $= 5 - 4 = 1$ and $SN = 4 + 1 = 5$. Trigonal-bipyramidal electron geometry with the lone pair equatorial gives a see-saw shape with a non-zero dipole.

Example 5 — Shape and hybridisation of ClF3. Chlorine: $V = 7$, $M = 3$, neutral. $$H = \tfrac12(7 + 3) = 5 \Rightarrow sp^3d$$ Sigma bonds $= 3$, lone pairs $= 5 - 3 = 2$, $SN = 5$. Both lone pairs occupy equatorial sites of the trigonal bipyramid, leaving a T-shaped molecule with bond angles slightly under $90^\circ$.

Example 6 — Shape and hybridisation of XeF4. Xenon: $V = 8$, $M = 4$, neutral. $$H = \tfrac12(8 + 4) = 6 \Rightarrow sp^3d^2$$ Sigma bonds $= 4$, lone pairs $= 6 - 4 = 2$, $SN = 6$. The two lone pairs take trans axial positions of the octahedron, so the four F atoms are square planar; the symmetric arrangement cancels the bond dipoles, making XeF4 non-polar.

Example 7 — Percentage ionic character of HCl. Observed $\mu_{obs} = 1.03$ D and bond length $d = 1.27$ angstrom. A full unit-charge separation would give $\mu_{ionic} = 4.8 \times 1.27 = 6.10$ D, so $$\%\ \text{ionic} = \dfrac{1.03}{6.10} \times 100 \approx 16.9\%$$ The H-Cl bond is about 17% ionic (roughly 83% covalent) — polar covalent, not ionic.

Example 8 — Predicting the more covalent compound (Fajans' rules). AlCl3 versus NaCl: Al3+ is smaller and triply charged, so its polarising power ($\propto \text{charge}/\text{radius}^2$) is far higher — AlCl3 is markedly more covalent (it sublimes as Al2Cl6). CuCl versus NaCl: Cu+ and Na+ are alike in size and charge, but Cu+ has an 18-electron outer shell against Na+ with a noble-gas octet, so Cu+ polarises chloride more and CuCl is the more covalent. General rule: smaller cation, larger anion, higher charge, or an 18-electron cation all push a bond toward covalency.