Question

Oxidation state of hydrogen in compound X is -1 and in compound Y is +1. X and Y are respectively

• A. \( \text{LiAlH}_4, \text{H}_2\text{O} \)
• B. \( \text{NH}_3, \text{NaH} \)
• C. \( \text{CH}_4, \text{H}_2\text{O} \)
• D. \( \text{H}_2\text{S}, \text{NaBH}_4 \)

Hint:

Hydrides of group 1, 2, and 13 (like LiH, NaH, LiAlH4) always have H with -1 oxidation state.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

The oxidation state of hydrogen is determined by the electronegativity of the element it bonds with. - If bonded to a metal (less electronegative), H is -1 (hydride ion). - If bonded to a non-metal (more electronegative), H is +1.
Step 2: Key Formula or Approach:

Identify the nature of the bond (ionic vs covalent) and the partner element in each compound.
Step 3: Detailed Explanation:

Compound X (H is -1):
We need a metal hydride.
- \( \text{LiAlH}_4 \) (Lithium Aluminum Hydride): H is bonded to Al/Li (metals). Oxidation state = -1.
- \( \text{NH}_3 \): H bonded to N (non-metal). Oxidation state = +1.
- \( \text{CH}_4 \): H bonded to C. Oxidation state = +1.
- \( \text{H}_2\text{S} \): H bonded to S. Oxidation state = +1. So, X must be \( \text{LiAlH}_4 \) or \( \text{NaBH}_4 \). Compound Y (H is +1):
We need a covalent compound with a non-metal.
- \( \text{H}_2\text{O} \): H bonded to O. Oxidation state = +1.
- \( \text{NaH} \): H bonded to Na. Oxidation state = -1. Comparing options: (A) \( \text{LiAlH}_4 \) (-1) and \( \text{H}_2\text{O} \) (+1). Correct. (B) \( \text{NH}_3 \) (+1) and \( \text{NaH} \) (-1). Incorrect order. (C) \( \text{CH}_4 \) (+1) and \( \text{H}_2\text{O} \) (+1). Incorrect X. (D) \( \text{H}_2\text{S} \) (+1) and \( \text{NaBH}_4 \) (-1). Incorrect order.
Step 4: Final Answer:

X is \( \text{LiAlH}_4 \) and Y is \( \text{H}_2\text{O} \).