According to MO theory,the bond orders for O_ | vedclass

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

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$1, 3, 3$

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(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.$1$. For $O_2^{2-}$ ($18$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $\frac{1}{2} (10 - 8) = 1$.$2$. For $CO$ ($14$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $\frac{1}{2} (10 - 4) = 3$.$3$. For $NO^{+}$ ($14$ electrons): Being isoelectronic with $CO$,it also has a bond order of $3$.

According to $MO$ theory,the bond orders for $O_2^{2-}$,$CO$,and $NO^{+}$ respectively,are:

Similar Questions

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