Consider the transition from N 2 to O 2 . In N 2 , the σ 2 pz is higher in energy than π 2 px , y due to a mixing parameter λ . As we move to O 2 , the 2 s − 2 p gap Δ E increases. Let the c

Question

Consider the transition from N 2 ​ to O 2 ​ . In N 2 ​ , the σ 2 pz is higher in energy than π 2 px , y ​ due to a mixing parameter λ . As we move to O 2 ​ , the 2 s − 2 p gap Δ E increases. Let the crossover occur at a critical gap Δ E c ​ . If a molecule M 2 ​ has Δ E < Δ E c ​ and 13 electrons, which of the following is true regarding its ion M 2 ​ − ?

TestHub · Accepted Answer

Question Explanation: Molecular Orbital Theory and the effect of s-p mixing on energy levels.

Concept: Molecular Orbital Theory (s-p mixing)

Solution:

The problem states $Δ E < Δ E_{c}$ , which means the molecule follows the $N_{2}$ type energy level sequence where s-p mixing occurs. In this sequence, the $^{π} 2 p_{x, y}$ orbitals are lower in energy than the $σ_{2 p_{z} orbital.}$

The molecule $M_{2}$ has 13 electrons, and its ion $M_{2}^{-}$ will have 14 electrons. Following the $N_{2}$ configuration:

$σ_{1 s}^{2}, σ_{1 s}^{* 2}, σ_{2 s}^{2}, σ_{2 s}^{* 2}, (π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2}), σ_{2 p_{z}}^{2}$ .

Bond Order $= \frac{10 - 4}{2} = 3.0$

Since all electrons are paired, $M_{2}^{-}$ is diamagnetic.

Adding an electron to a bonding orbital ( $M_{2} \to M_{2}^{-}$ ) decreases the bond length (increases bond strength).

Final Answer: Option (c)

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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