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(B) The bond order is calculated as: $	ext{Bond order} = \frac{1}{2} [N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the num

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$O_2^{2-}, O_2^{2+}$

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(B) The bond order is calculated as: $	ext{Bond order} = \frac{1}{2} [N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons. For $O_2$ ($16$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 6] = 2.0$. For $O_2^{-}$ ($17$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 7] = 1.5$. For $O_2^{+}$ ($15$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 5] = 2.5$. For $O_2^{2-}$ ($18$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 8] = 1.0$. For $O_2^{2+}$ ($14$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 4] = 3.0$. Calculating the differences: $(A)$ $|2.5 - 1.5| = 1.0$ $(B)$ $|3.0 - 1.0| = 2.0$ $(C)$ $|3.0 - 2.0| = 1.0$ $(D)$ $|3.0 - 2.5| = 0.5$ The maximum difference is $2.0$ for the pair $O_2^{2-}$ and $O_2^{2+}$.

Identify the pair in which the difference in bond order value is maximum.

Similar Questions

$AX$ is a covalent diatomic molecule where $A$ and $X$ are second row elements of the periodic table. Based on Molecular Orbital Theory,the bond order of $AX$ is $2.5$. The total number of electrons in $AX$ is ........... (Round off to the nearest integer).

Among the following species $N_{2}$,$N_{2}^{+}$,$N_{2}^{-}$,$N_{2}^{2-}$,$O_{2}$,$O_{2}^{+}$,$O_{2}^{-}$,$O_{2}^{2-}$,the number of species showing diamagnetism is:

In which of the following ionization processes has the bond energy increased and the magnetic behaviour changed from paramagnetic to diamagnetic?

What is the effect of the following processes on the bond order in $N_{2}$ and $O_{2}$?
$(A)$ $N_{2} \to N_{2}^{+} + e^{-}$
$(B)$ $O_{2} \to O_{2}^{+} + e^{-}$

State whether the bond order increases or decreases when $O_2$ changes to $O_2^+$ and when $N_2$ changes to $N_2^+$.

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