Main Group Elements Questions and Answers

Answer: A, C, D

Statements (A), (C) and (D) are correct; (B) is incorrect.

Identifying the compounds. A 1 : 5 mixture of Xe and F₂ at 873 K and 7 bar gives XeF₄ (P); XeF₆ needs a much larger excess of fluorine (about 1 : 20) at high pressure. The strong fluorinating agent O₂F₂ raises XeF₄ to XeF₆: \(\mathrm{XeF_4 + O_2F_2 \to XeF_6 + O_2}\), so Q is XeF₆. Complete hydrolysis of XeF₆ gives \(\mathrm{XeF_6 + 3H_2O \to XeO_3 + 6HF}\), so R is XeO₃.

(A) XeF₄ has four bond pairs and two lone pairs on Xe (square planar): correct.

(B) XeF₆ has six bond pairs and one lone pair, so its geometry is a distorted octahedron, not a perfect one: incorrect.

(C) XeF₆ readily transfers fluorine and is a fluorinating agent: correct.

(D) XeO₃ has three bond pairs and one lone pair on Xe, so it is trigonal pyramidal: correct.

Answer: B, D

The correct statements are (B) and (D).

Orthoboric acid, B(OH)₃, is a weak monobasic acid, but not a proton donor in the usual sense: it is a Lewis acid that accepts OH^- from water, \(\mathrm{B(OH)_{3} + H_{2}O \rightarrow [B(OH)_{4}]^{-} + H^{+}}\). So (A), which attributes its acidity to self-ionisation, is wrong. Because it ionises only slightly, it is a weak electrolyte (D).

Adding a cis-diol such as ethylene glycol forms a stable chelate with the borate ion and pulls the equilibrium to the right, so the solution becomes more acidic (B); this is how boric acid is titrated.

(C) is wrong: hydrogen bonds link the B(OH)₃ units into flat sheets, a layered two-dimensional structure, not a three-dimensional one.

Answer: A, B, C

Paramagnetic species are formed in (A), (B) and (C).

• (A) Sodium dissolves in excess liquid ammonia to give ammoniated electrons, each with an unpaired spin: the blue solution is paramagnetic.
• (B) Potassium burns in excess oxygen to give the superoxide KO₂; the O₂^- ion has one unpaired electron.
• (C) Copper with dilute nitric acid gives nitric oxide, NO, an odd-electron molecule (and Cu²⁺, which also has an unpaired electron).

(D) Oxygen with 2-ethylanthraquinol is the industrial route to hydrogen peroxide, H₂O₂, in which all electrons are paired, so it gives no paramagnetic product.