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NEET · Chemistry · STD 11 - 4. Chemical bonding and molecular structure

The correct order of energies of molecular orbitals of \(N _2\) molecule, is

  1. A \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right)\)\( < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma 2 p_{ z } < \sigma^* 2 p_{ z }\)
  2. B \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) \)\(< \sigma 2 p_{ z } < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }\)
  3. C \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \sigma 2 p_{ z } \)\(< \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }\)
  4. D \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \sigma 2 p_{ z } < \sigma^* 2 p_{ z } \)\(< \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) \)\(< \sigma 2 p_{ z } < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }\)

Step-by-step Solution

Detailed explanation

For molecules like \(B _2, C _2, N _2\) etc. the increasing order of energies of various molecular orbitals is \(\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) \)
\(< \sigma 2 p_{ z } < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }\)
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