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(B) The bond energy is directly proportional to the bond order. The bond order can be calculated using the formula: $	ext{Bond Order} = \frac{N_b - N

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$O_{2}^{+}$

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(B) The bond energy is directly proportional to the bond order. The bond order can be calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$(A)$ $O_{2}^{2-}$: Total electrons = $18$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^2$. Bond order = $\frac{10-8}{2} = 1$.$(B)$ $O_{2}^{+}$: Total electrons = $15$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1$. Bond order = $\frac{10-5}{2} = 2.5$.$(C)$ $O_{2}^{-}$: Total electrons = $17$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^1$. Bond order = $\frac{10-7}{2} = 1.5$.$(D)$ $O_{2}$: Total electrons = $16$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1, (\pi^* 2p_y)^1$. Bond order = $\frac{10-6}{2} = 2$.Comparing the bond orders: $O_{2}^{+} (2.5) > O_{2} (2) > O_{2}^{-} (1.5) > O_{2}^{2-} (1)$.Since $O_{2}^{+}$ has the highest bond order,it has the highest bond energy.

List-$I$	List-$II$
$A$. $\Psi_{MO} = \Psi_{A} - \Psi_{B}$	$I$. Dipole moment
$B$. $\mu = Q \times r$	$II$. Bonding molecular orbital
$C$. $\frac{N_{b} - N_{a}}{2}$	$III$. Anti-bonding molecular orbital
$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$	$IV$. Bond order

Which of the following species has the highest bond energy?

Similar Questions

Total number of electrons present in $(\pi^*)$ molecular orbitals of $O_2$,$O_2^{+}$,and $O_2^{-}$ is ............

What is the bond order of $O_2^-$ (in $.5$)?

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A$. $\Psi_{MO} = \Psi_{A} - \Psi_{B}$ $I$. Dipole moment

$B$. $\mu = Q \times r$ $II$. Bonding molecular orbital

$C$. $\frac{N_{b} - N_{a}}{2}$ $III$. Anti-bonding molecular orbital

$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$ $IV$. Bond order

What is the number of electrons present in antibonding orbitals of $N_2$ molecule according to molecular orbital theory?

Use molecular orbital theory to explain why the $Be_{2}$ molecule does not exist.

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