Easy Practice — Set 1

Quick Recap — The p-Block (Groups 13-18)

• Families: 13 boron, 14 carbon, 15 nitrogen (pnictogens), 16 oxygen (chalcogens), 17 halogens, 18 noble gases.
• Inert pair effect: heavier members favour a lower oxidation state (Tl$^+$, Pb$^{2+}$, Bi$^{3+}$).
• Catenation is strongest in carbon; allotropes: C (diamond, graphite, fullerene), P (white/red/black), S (rhombic/monoclinic), O (O$_2$/O$_3$).
• Anomalous first members (small size, no d-orbitals): B, C, N, O, F differ from their heavier congeners.

Beyond-NCERT JEE Essentials

1. Anomaly of the first element (B, C, N, O, F)

The head of each group breaks the family pattern: it is very small, highly electronegative, has a high charge-to-size ratio and, decisively, has NO valence d-orbitals, so its maximum covalency is only 4. Its compact 2p orbitals also overlap sideways well, giving strong $p\pi-p\pi$ multiple bonds that heavier congeners avoid.

• N forms N2 (a $p\pi-p\pi$ triple bond, about 941 kJ/mol) and O forms O2 (a double bond), so both are gases; heavier P and S must use single bonds, building solid P4 and S8 instead.
• With no d-orbitals nitrogen cannot expand its octet: it makes NF3 but never a pentahalide, whereas P, As and Sb form PCl5 and PF5 using 3d orbitals.
• Boron's covalency caps at 4, so [BF4]- exists but [BF6]3- cannot, even though [AlF6]3- forms freely.

[JEE Tip] The head element resembles the SECOND element of the next group, the diagonal relationship (Li~Mg, Be~Al, B~Si). So boron behaves like silicon (acidic oxide, volatile hydrides, easily hydrolysed halides), not like the aluminium directly below it.

2. Inert pair effect (groups 13-15)

Going down a group the ns2 pair is shielded poorly by the intervening d and f electrons, so it is gripped tightly and stops bonding. The LOWER oxidation state becomes stable and the HIGHER state turns strongly oxidising.

• Group 13: Tl+ is more stable than Tl3+, so Tl(III) (e.g. TlCl3) is oxidising.
• Group 14: Pb2+ beats Pb4+, so PbO2 and PbCl4 are strong oxidisers while Sn2+ (SnCl2) is a reducing agent. PbI4 does not exist at all, since Pb4+ would oxidise the iodide.
• Group 15: Bi3+ is favoured over Bi5+, so Bi(V) is strongly oxidising.

[JEE Tip] Both PbO2 (Pb +4) and NaBiO3 (Bi +5) oxidise Mn2+ to purple MnO4- in acid, a favourite "spot the oxidiser" cue.

3. Oxoacids: acidity, basicity, reducing power

• Chlorine oxoacids: acid strength climbs with the oxidation state of Cl (+1, +3, +5, +7) and the number of terminal (non-OH) oxygens, which spread the conjugate base's charge, so HOCl < HClO2 < HClO3 < HClO4. Oxidising power runs the opposite way: HOCl > HClO2 > HClO3 > HClO4.
• Basicity = number of ionisable O-H protons, NOT total H. In the phosphorus acids any H bonded directly to P is non-ionisable:
• H3PO2 = one P-OH, two P-H: monobasic, P is +1
• H3PO3 = two P-OH, one P-H: dibasic, P is +3
• H3PO4 = three P-OH, no P-H: tribasic, P is +5
• A P-H bond makes the acid reducing, so reducing power falls H3PO2 > H3PO3 > H3PO4 (H3PO4 is not reducing). H3PO2, with two P-H bonds, precipitates metals such as silver from their salts.

[JEE Tip] Count P-OH for basicity and P-H for reducing power. The trap is boric acid B(OH)3: three OH groups yet only a weak MONOBASIC acid, because it accepts OH- from water (a Lewis acid) to give [B(OH)4]- rather than donating its own H+.

4. Allotropes, catenation and pi-bonding

• Key allotropes: boron (B12 icosahedra), carbon (diamond $sp^3$ network, graphite $sp^2$ layers with a delocalised pi cloud, fullerene C60), phosphorus (white P4 strained and glowing, red polymeric, black layered and most stable), sulphur (rhombic and monoclinic, both built from S8 crown rings).
• Catenation strength: C >> Si > Ge in group 14, and S > Se in group 16, tracking the falling E-E bond strength; it dies out down each group as atoms swell.
• $p\pi-p\pi$ bonding is effective only for the small 2p elements, so CO2 is a discrete O=C=O gas while SiO2 is a giant Si-O single-bonded solid, and N2 is triple-bonded while P4 is only single-bonded. Heavier atoms instead use $d\pi-p\pi$ bonding (as in phosphates and R3P=O).

5. Shapes by VSEPR: interhalogens, polyhalides, xenon

Add the bonding pairs (bp) and lone pairs (lp) on the central atom, then read the shape:

• ClF3: 3 bp + 2 lp, $sp^3d$, T-shaped.
• BrF5: 5 bp + 1 lp, $sp^3d^2$, square pyramidal.
• I3- (polyhalide): 2 bp + 3 lp, $sp^3d$, linear.
• XeF2: 2 bp + 3 lp, $sp^3d$, linear (the three lone pairs sit equatorial).
• XeF4: 4 bp + 2 lp, $sp^3d^2$, square planar (the two lone pairs go trans).
• XeF6: 6 bp + 1 lp, $sp^3d^3$, distorted octahedral, because the lone pair prevents a regular octahedron.
• XeO3: 3 bp + 1 lp, $sp^3$, pyramidal; XeOF4: 5 bp + 1 lp, $sp^3d^2$, square pyramidal.

[JEE Tip] Every xenon fluoride is a strong oxidiser and fluorinator. Complete hydrolysis of XeF6 gives explosive XeO3 plus HF, whereas partial hydrolysis gives XeOF4, a classic product-prediction trap.

6. High-value anomalies (the "explain why")

• NF3 has a tiny dipole (0.24 D) but NH3 a large one (1.47 D), even though N-F is more polar than N-H. In NH3 the bond dipoles point toward N and add to the lone-pair moment; in NF3 they point away toward F and oppose it, so they nearly cancel.
• Nitrogen forms no pentahalide because it has no valence d-orbitals; phosphorus, one period down, forms PCl5 easily.
• BF3 is a WEAKER Lewis acid than BCl3 or BBr3: the well-matched $2p\pi-2p\pi$ back-donation from F fills boron's empty orbital and eases its electron deficiency (acidity BF3 < BCl3 < BBr3).
• Drying agents must not react with the gas dried: never dry NH3 with acidic P4O10, conc. H2SO4 or CaCl2 (which forms CaCl2.8NH3); use CaO. P4O10 is powerful enough to strip water from HNO3 (to N2O5) and H2SO4 (to SO3).
• N2 is inert because its triple bond (about 941 kJ/mol) is among the strongest known, which is why it serves as an inert atmosphere.

Worked Examples — Beyond-NCERT Essentials

Example 1 (order oxoacid acidity). Arrange HOCl, HClO2, HClO3 and HClO4 by acid strength and name the best oxidiser. Acidity grows with the oxidation state of chlorine (+1, +3, +5, +7) and with the number of terminal (non-OH) oxygens, which delocalise the conjugate base's negative charge. More terminal O means a more stable anion and an easier loss of H+, so acidity is HOCl < HClO2 < HClO3 < HClO4. Oxidising power runs the other way; the electron-rich +1 acid is the strongest oxidant. Answer: acidity HOCl < HClO2 < HClO3 < HClO4; strongest oxidiser is HOCl.

Example 2 (predict an interhalogen shape). Give the hybridisation and shape of BrF5. Bromine brings 7 valence electrons; five form Br-F bonds, leaving one lone pair. So 5 bp + 1 lp = 6 domains, giving $sp^3d^2$ and an octahedral electron geometry. The single lone pair takes one octahedral vertex and pushes the five F atoms into a square pyramid. Answer: $sp^3d^2$, square pyramidal.

Example 3 (predict a xenon shape). Why is XeF4 square planar rather than tetrahedral? Xenon has 8 valence electrons; four bond to F, leaving two lone pairs. So 4 bp + 2 lp = 6 domains, giving $sp^3d^2$ and an octahedral electron geometry. The two lone pairs take opposite (trans) corners to minimise repulsion, leaving the four F atoms coplanar. Answer: $sp^3d^2$, square planar with the lone pairs trans.

Example 4 (explain an anomaly). N-F bonds are more polar than N-H, yet NF3 (0.24 D) has a far smaller dipole than NH3 (1.47 D). Explain. Both molecules are pyramidal with a lone pair on nitrogen. In NH3 the bond dipoles point from H toward the more electronegative N, the same direction as the lone-pair moment, so the two add. In NF3 the more electronegative F pulls the bond dipoles away from N, opposing the lone-pair moment, so they largely cancel. Answer: the resultant bond dipole opposes the lone pair in NF3 but reinforces it in NH3, so NF3 is much less polar.

Example 5 (strongest reducing or most acidic hydride). Among HF, HCl, HBr and HI, identify the strongest acid and the strongest reducing agent. Both trends follow the H-X bond enthalpy, which falls down the group. The weakest bond (H-I) ionises most readily (strongest acid) and gives up its hydrogen most easily (strongest reducer). HF, with the strongest bond plus extensive hydrogen bonding, is weakest on both counts. Answer: HI is both the strongest acid and the strongest reducing agent; HF is the weakest.

Example 6 (basicity of an oxoacid). H3PO3 has three hydrogens; why is it dibasic, and is it a reducing agent? Its structure is HP(O)(OH)2: two hydrogens sit on oxygen (ionisable) and one sits directly on phosphorus (not ionisable). Only the two O-H protons are released, so the basicity is 2. The single P-H bond makes H3PO3 a reducing agent. Answer: dibasic (two P-OH groups), and yes, the P-H bond makes it reducing.

Example 7 (inert pair effect). Explain why PbCl4 is a strong oxidiser while PbCl2 is stable, and why PbI4 is unknown. Down group 14 the inert pair effect makes Pb2+ (+2) the stable state and Pb4+ (+4) unstable, so Pb4+ readily accepts electrons and PbCl4 acts as an oxidiser. Iodide is a good reducer, so the redox is spontaneous (Pb4+ + 2I- gives Pb2+ + I2); PbI4 therefore cannot be isolated, although PbCl4 exists as an unstable oxidising liquid. Answer: the inert pair effect makes Pb4+ oxidising; it oxidises iodide, so PbI4 does not exist.

Example 8 (maximum covalency). PCl5 exists but NCl5 does not. Explain, and predict whether nitrogen or phosphorus uses d-orbitals in its oxoanions. Nitrogen (period 2) has no valence d-orbitals, so its covalency is capped at 4 and it cannot bind five chlorines. Phosphorus (period 3) has empty 3d orbitals, expands its octet to five bonds in PCl5 ($sp^3d$), and also uses $d\pi-p\pi$ bonding in oxoanions such as phosphate. Answer: N lacks d-orbitals (maximum covalency 4), so no NCl5; P uses 3d orbitals for PCl5 and for $d\pi-p\pi$ bonding.