Looking for classes? Ksquare Career Institute, Bengaluru →
The correct bond order in the following species is:
A
$O_2^{2+} < O_2^+ < O_2^-$
B
$O_2^{2+} < O_2^- < O_2^+$
C
$O_2^+ < O_2^- < O_2^{2+}$
D
$O_2^- < O_2^+ < O_2^{2+}$
Detailed Solution
Bond order $= \frac{N_b - N_a}{2}$, with 10 bonding electrons in each species.
$O_2^{2+}$ (14 e⁻): $\frac{10 - 4}{2} = 3$
$O_2^+$ (15 e⁻): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (17 e⁻): $\frac{10 - 7}{2} = 1.5$
So $O_2^- < O_2^+ < O_2^{2+}$.
$O_2^{2+}$ (14 e⁻): $\frac{10 - 4}{2} = 3$
$O_2^+$ (15 e⁻): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (17 e⁻): $\frac{10 - 7}{2} = 1.5$
So $O_2^- < O_2^+ < O_2^{2+}$.