The molecular orbital configuration of a diat | vedclass

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(A) The bond order is calculated using the formula: $B.O. = \frac{N_b - N_a}{2}$Where $N_b$ is the number of electrons in bonding molecular orbitals a

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$3$

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(A) The bond order is calculated using the formula: $B.O. = \frac{N_b - N_a}{2}$Where $N_b$ 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 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_x) + 2 (\pi 2p_y) + 2 (\pi 2p_z) = 10$$N_a = 2 (\sigma^* 1s) + 2 (\sigma^* 2s) = 4$$B.O. = \frac{10 - 4}{2} = \frac{6}{2} = 3$.

The molecular orbital configuration of a diatomic molecule is
$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$
Its bond order is

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The molecular orbital configuration of a diatomic molecule is$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$Its bond order is