In which of the following pairs do the two mo | vedclass

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Question

(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2}[N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$

Vedclass · Accepted Answer

$N_2, O_2^{2+}$

Vedclass · Answer

(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2}[N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $N_2$ ($14$ electrons): $B.O. = \frac{1}{2}[10 - 4] = 3$.For $O_2^{2+}$ ($14$ electrons): $B.O. = \frac{1}{2}[10 - 4] = 3$.Since both $N_2$ and $O_2^{2+}$ have a bond order of $3$,they have identical bond orders.

Column $I$	Column $II$
$(A)$ $B_2$	$(p)$ Paramagnetic
$(B)$ $N_2$	$(q)$ Undergoes oxidation
$(C)$ $O_2^{-}$	$(r)$ Undergoes reduction
$(D)$ $O_2$	$(s)$ Bond order $\geq 2$
	$(t)$ Mixing of '$s$' and '$p$' orbitals

In which of the following pairs do the two molecules have identical bond order?

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