Question

When burnt in excess of oxygen, sodium forms a compound X and potassium forms a compound Y. The magnetic natures of X and Y respectively are

• A. Both X and Y are paramagnetic in nature
• B. X is diamagnetic and Y is paramagnetic in nature
• C. X is paramagnetic and Y is diamagnetic in nature
• D. Both X and Y are diamagnetic in nature

Hint:

Superoxides ( \( \text{KO}_2, \text{RbO}_2, \text{CsO}_2 \) ) are colored and paramagnetic. Peroxides ( \( \text{Na}_2\text{O}_2 \) ) are colorless and diamagnetic.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We identify the oxide products formed by alkali metals with excess oxygen and determine their magnetic nature based on the presence of unpaired electrons in the anions.
Step 2: Key Formula or Approach:

- Sodium (Na) + Excess \( \text{O}_2 \to \text{Na}_2\text{O}_2 \) (Peroxide). - Potassium (K) + Excess \( \text{O}_2 \to \text{KO}_2 \) (Superoxide). - Magnetic nature depends on Molecular Orbital configuration of \( \text{O}_2^{2-} \) and \( \text{O}_2^{-} \).
Step 3: Detailed Explanation:

Compound X:
Sodium Peroxide (\( \text{Na}_2\text{O}_2 \)) contains the peroxide ion \( \text{O}_2^{2-} \). Total valence electrons = \( 6+6+2 = 14 \) (for the O-O part in MOT diagram context, usually considered 18 total e-). Configuration: \( \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p^4 \pi^* 2p^4 \). All electrons are paired. Hence, Diamagnetic. Compound Y:
Potassium Superoxide (\( \text{KO}_2 \)) contains the superoxide ion \( \text{O}_2^{-} \). Total valence electrons = \( 6+6+1 = 13 \) (or 17 total e-). Configuration: \( \dots \pi^* 2p^3 \). There is one unpaired electron in the antibonding \( \pi \) orbital. Hence, Paramagnetic.
Step 4: Final Answer:

X is diamagnetic and Y is paramagnetic.