Question 1Level 2 · ExamSingle correct
Which one of the following equations correctly describes a reaction that actually takes place under the stated conditions? Exactly one option is correct.
A2B(s)+6HCl(aq)→2BCl3+3H2(g), readily at room temperature
BAl(s)+4HNO3(conc.)→Al(NO3)3+NO(g)+2H2O, proceeding steadily at room temperature
C2Al(s)+2NaOH(aq)+6H2O(l)→2Na+[Al(OH)4]−(aq)+3H2(g)
D2B(s)+3H2O(l)→B2O3(s)+3H2(g), rapidly at 298 K
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Answer: C
Check both the chemistry (does it happen under these conditions?) and the balance.
(A) Boron is not attacked by non-oxidising acids such as dilute HCl even at moderate temperatures (its small size and high ionisation enthalpy make it non-metallic). No reaction.
(B) Concentrated nitric acid renders aluminium passive by forming a thin protective oxide layer, so aluminium does not dissolve steadily in it. (The equation is atom-balanced, but the reaction does not proceed.)
(C) Correct. Aluminium is amphoteric and dissolves in aqueous alkali, giving tetrahydroxidoaluminate and hydrogen. Check: Al 2 = 2; Na 2 = 2; O 2 + 6 = 8 = 2 × 4; H 2 + 12 = 14 = 8 + 6; charge 0 = 0.
(D) Crystalline boron is unreactive towards water at room temperature. (Balanced, but it does not occur under these conditions.)
Trap: options (B) and (D) are balanced, so balance alone does not decide; the conditions do.
Answer: (C).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Borax is Na2[B4O5(OH)4]⋅8H2O. For the anion [B4O5(OH)4]2−, let p = number of tetrahedral (sp3) boron atoms, q = number of trigonal-planar (sp2) boron atoms, r = number of bridging oxygen atoms (B–O–B), and s = number of terminal B–OH groups. The ordered set (p,q,r,s) is ______. Exactly one option is correct.
A(4,0,5,4)
B(2,2,5,4)
C(2,2,4,4)
D(2,2,5,2)
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Answer: B
Each B carries one terminal OH, so s=4. Total O = 9; 4 are in OH groups, so r=5 bridging O. Each negative charge of the anion sits on a tetrahedral [BO4]− unit (a trigonal BO3 unit is neutral), so a charge of −2 means p=2 and q=4−2=2. Check by bond count: B–O bonds =2(4)+2(3)=14; each bridging O makes 2 and each OH makes 1: 2(5)+4=14. Consistent.
(A) Would need every B to carry four O; but only 2 B can be tetrahedral when the charge is −2 (see the count above).
(B) Correct (see the count above).
(C) Gives only 4+4=8 O atoms; the formula has 9.
(D) The formula shows four OH groups, one on each boron.
Note: on heating, borax loses water and swells; it then melts to a transparent glass of NaBO2 and B2O3 (the borax bead).
Answer: (B).
Difficulty: Very hardQuestion 3Level 2 · ExamSingle correct
In which one of the following pairs do BOTH species exist as stable, well-characterised species? Exactly one option is correct.
A[SiF6]2− and [GeCl6]2−
B[BF6]3− and [AlF6]3−
CNCl5 and PCl5
D[SiCl6]2− and [SnCl6]2−
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Answer: A
The first member of each p-block group has no valence d orbitals (maximum covalency 4); in heavier members steric size also limits the coordination number.
(A) Both exist. Si and Ge can expand their covalency to 6 using vacant d orbitals; six small F atoms fit around Si, and six Cl atoms fit around the larger Ge.
(B) [AlF6]3− exists (cryolite), but boron has no d orbitals in its valence shell; its maximum covalency is 4, so [BF6]3− does not exist.
(C) PCl5 exists, but N (second period, no d orbitals) cannot exceed covalency 4; NCl5 is unknown.
(D) [SnCl6]2− exists, but six large Cl atoms cannot be accommodated around the small Si atom, so [SiCl6]2− is not known.
Answer: (A).
Difficulty: Hard