Write the MO electronic configuration of a diatomic molecule having a bond order of three.

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Solution

The molecular orbital configuraion of diatomic molecule with bond order three is
$σ 1 s^{2} σ^{} 1 s^{2} σ 2 s^{2} σ^{} 2 s^{2} π 2 p_{x}^{2} π 2 p_{y}^{2} σ 2 p_{z}^{2}$
= Bond order $= \frac{1}{2}$ (Bonding electrons-Anti Bonding)
$\frac{1}{2} (10 - 4) = 3$
The diatomic molecule is Nitrogen

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σ 1 s^{2} σ^{+} 1 s^{2} σ 2 s^{2} σ^{+} 2 s^{2} σ^{+} 2 s^{2} σ 2 p_{x}^{2}

{\begin{matrix} π 2 p_{y}^{2} π 2 p_{z}^{2} \end{matrix}

X_{2}

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