The bond order of $O_{2}^{+}$ is x. The bond order of $O_{2}^{-}$ and $O_{2}^{2 +}$ are respectively.

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$\frac{2}{5} X, \frac{3}{5} X$

$\frac{5}{3} X, \frac{5}{6} X$

$\frac{5}{2} X, \frac{5}{3} X$

$\frac{3}{5} X, \frac{6}{5} X$

answer is B.

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Detailed Solution

Bond order (BO)
$\frac{1}{2} [\begin{array}{l} N u m b e r o f b o n d i n g e^{-} - \\ N u m b e r o f a n t i b o n d i n g e^{-} \end{array}]$
Electronic configuration
$\begin{array}{l} {O_{2}}^{+} = σ l s^{2} σ^{•} l s^{2}, σ 2 s^{2}, σ^{•} 2 s^{2}, σ 2 {p_{z}}^{2} \\ π 2 {p_{x}}^{2} = π 2 {p_{y}}^{2}, π^{•} 2 {p_{x}}^{1} = π 2 {p_{y}}^{0} \end{array}$
$∴ B O = \frac{1}{2} [10 - 5] = \frac{5}{2}$
$\begin{array}{l} B O o f {O_{2}}^{+} = x (g i v e n) \\ ∴ \frac{5}{2} = x . . . . (1) \end{array}$
$\begin{array}{l} {O_{2}}^{-} = σ l s^{2}, σ^{•} l s^{2}, σ 2 s^{2}, σ^{•} 2 s^{2}, σ 2 {p_{z}}^{2} \\ π 2 {p_{x}}^{2} = π 2 {p_{y}}^{2}, π^{*} 2 {p_{x}}^{2} \approx π^{*} 2 {p_{y}}^{1} \end{array}$
$\begin{array}{l} ∴ B O = \frac{1}{2} [10 - 7] = \frac{3}{2} \\ O r B O o f {O_{2}}^{-} = \frac{3 x}{5} . . . . [F r o m (1)] \end{array}$
$\begin{array}{l} {O_{2}}^{2 +} = σ l s^{2}, σ^{•} l s^{2}, σ 2 s^{2}, σ^{•} 2 s^{2}, σ 2 {p_{z}}^{2} \\ π 2 {p_{x}}^{2} = π^{*} 2 {p_{y}}^{1} \end{array}$
$∴ B O = \frac{1}{2} [10 - 4] = \frac{6}{2} O R \frac{6 x}{5} [F r o m (1)]$