Question: 69: Match List-I with List-II.
| List – I (Hydrides) | List – II (Nature) |
|---|---|
| (a) \mathrm{MgH}_{2} | (i) Electron precise |
| (b) \mathrm{GeH}_{4} | (ii) Electron deficient |
| (c) \mathrm{B}_{2}\mathrm{H}_{6} | (iii) Electron rich |
| (d) HF | (iv) Ionic |
Choose the correct answer from the options given below
(1) (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii)
(2) (a) – (ii), (b) – (iii), (c) – (iv), (d) – (i)
(3) (a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
(4) (a) – (iii), (b) – (i), (c) – (ii), (d) – (iv)
Answer: Option (3)
Explanation:
\mathrm{MgH}_{2} is a hydride of an alkaline earth metal.
It consists of \mathrm{Mg}^{2+} and \mathrm{H}^{-} ions,
so it is classified as an ionic hydride.
\mathrm{GeH}_{4} is a covalent hydride where the number of valence electrons is exactly sufficient to form normal covalent bonds,
hence it is an electron precise hydride.
\mathrm{B}_{2}\mathrm{H}_{6} has fewer electrons than required for conventional two-centre two-electron bonds and contains three-centre two-electron bonds,
so it is an electron deficient hydride.
HF contains lone pairs on fluorine in excess of bonding requirements,
making it an electron rich hydride.
Thus, the correct matching is (a) – (iv), (b) – (i), (c) – (ii), (d) – (iii).