Question

Match List - I with List- II. Choose the correct answer from the options given below:

	List - I (Hydrides)		List - II (Nature)
(a)	MgH2	(i)	Electron precise
(b)	GrH4	(ii)	Electron deficient
(c)	B2H6	(iii)	Electron rich
(d)	HF	(iv)	Ionic

• A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
• B. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
• C. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
• D. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Hydrides are binary compounds of hydrogen with other elements.
They are classified based on bonding and electronic distribution into ionic, covalent, and metallic types.
Detailed Explanation:
(a) $MgH_{2$:} Magnesium is an alkaline earth metal. It forms stoichiometric saline or ionic hydrides.
Matching: (a) $\rightarrow$ (iv).
(b) $GeH_{4$:} Germanium belongs to Group 14. These hydrides have exactly enough electrons to form normal covalent bonds (octet complete).
These are "electron-precise" hydrides.
Matching: (b) $\rightarrow$ (i).
(c) $B_{2H_{6}$:} Diborane (Group 13) has fewer electrons than needed for conventional bonds ($3c-2e$ bonds).
These are "electron-deficient" hydrides.
Matching: (c) $\rightarrow$ (ii).
(d) $HF$: Hydrogen fluoride (Group 17) has three lone pairs on Fluorine.
Hydrides with lone pairs are "electron-rich".
Matching: (d) $\rightarrow$ (iii).
Step 2: Final Answer:
The correct combination is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
This matches option (1).