The electronic configuration of the N_2 molec | vedclass

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Question

(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{1}{2}(N_b - N_a)$.Here,$N_b$ is the number of electrons in bonding mole

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{1}{2}(N_b - N_a)$.Here,$N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.From the given configuration: $N_b = 2 + 2 + 2 + 2 = 8$ (excluding $KK$ shell) and $N_a = 2$.Therefore,$	ext{Bond Order} = \frac{1}{2}(8 - 2) = \frac{6}{2} = 3$.

The electronic configuration of the $N_2$ molecule in its ground state can be written as: $KK[\sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2 \sigma 2p_z^2]$. What is the bond order of the $N_2$ molecule?

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