Question

Molecular orbital electronic configuration for 'X' anion is- KK σ^*2s² σ*2s² π2p_x² π2p_y² σ2p_z² π*2p_x¹, the anion 'X' is.

Options

• $$\mathrm{N}{\phantom{A}}_2^{-}$$

• $$\mathrm{O}{\phantom{A}}_2^{-}$$

• $$\mathrm{N}{\phantom{A}}_2^{2-}$$

• $$\mathrm{O}{\phantom{A}}_2^{2-}$$

Answer 1

$$\mathrm{N}{\phantom{A}}_2^{-}$$

Explanation:

Given that anion X's electronic configuration is σ1s² σ*1s², σ2s², π2p_x² ≈ π2p_y², σ2p_z², π*2p_x¹

∴ Total amount of anion electrons X = 15

As a result, the anion X is $$\mathrm{N}{\phantom{A}}_2^{-}$$