Match List-I with List-II.
List-I (Hydrides) --- List-II (Nature)
(a) MgH2 --- (i) Electron precise

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Question

Match List-I with List-II.
List-I (Hydrides) --- List-II (Nature)
(a) MgH2 --- (i) Electron precise
(b) GeH4 --- (ii) Electron deficient
(c) B2H6 --- (iii) Electron rich
(d) HF --- (iv) Ionic
Choose the correct answer from the options given below:
(1) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(2) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
(4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

Accepted Answer

Correct Answer: B - (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)Step-by-step Solution:MgH2 (Mg is an alkaline earth metal) - (iv) Ionic: Magnesium is electropositive enough to form ionic hydride (Mg2+ and H- ions). Ionic hydrides are formed by s-block metals (Group 1 and 2).GeH4 (Group 14 hydride) - (i) Electron precise: Germanium (Group 14) forms GeH4 with 4 bonding pairs and no lone pairs or electron deficiency - all 8 valence electrons are used in bonds. These are called electron-precise covalent hydrides.B2H6 (diborane) - (ii) Electron deficient: Boron (Group 13) has only 3 valence electrons; it forms 3 bonds but has an empty p orbital. B2H6 has 3c-2e bonds and is electron deficient (fewer electrons than required for normal 2c-2e bonds).HF (Group 17 hydride) - (iii) Electron rich: Fluorine has 3 lone pairs. HF has more electrons than needed for bonding (F has 3 lone pairs), making it electron rich. Group 16 and 17 hydrides are electron rich.Key Concepts:Ionic hydrides: s-block metals + H2 → M+H-; saline hydrides (NaH, MgH2, CaH2).Electron-precise hydrides: Group 14 (CH4, SiH4, GeH4); all electron pairs used in 2c-2e bonds.Electron-deficient hydrides: Group 13 (B2H6); use 3c-2e bonds.Electron-rich hydrides: Group 15, 16, 17 (NH3, H2O, HF); have lone pairs.

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