The corrent bond order in the following species is

Options

A. `O_(2)^(2+) lt O_(2)^(-) lt O_(2)^(+)`
B. `O_(2)^(+) lt O_(2)^(-) lt O_(2)^(2+)`
C. `O_(2)^(-) lt O_(2)^(+) lt O_(2)^(2+)`
D. `O_(2)^(2+) lt O_(2)^(+) lt O_(2)^(-)`

Answer

c

Explanation

`O_(2)^(-)`ion Total number of electron `(16 - 1) = 15`Electron configuration :
`sigma 1s^(2) lt sigma**1s^(2) lt sigma2s^(2) lt sigma** 2s^(2) lt sigma2p_(x)^(2) lt pi 2p_(y)^(2) = pi 2p_(z)^(2) lt pi **2p_(y)^(1)`
Bond order `= (N_(b) - N_(a))/(2) = (10 - 5)/(2) = (5)/(2) = 2(1)/(2)`
`O_(2)` (super oxide ion) Total number of electron `(16 + 1) = 17` Electronic configuration
`sigma 1s^(2) lt sigma**1s^(2) lt sigma2s^(2) lt sigma** 2s^(2) lt sigma2p_(x)^(2) lt pi 2p_(y)^(2) = pi 2p_(z)^(2) lt pi **2p_(y)^(1) lt pi **2p_(x)^(1)`
Bond order `= ((N_(b) - N_(a)))/(2) = (10 - 7)/(2) = (3)/(2) = 2(1)/(2)`
`O_(2)^(+3)` ion Total number of electron `= (16 + 2) = 14` Electronic configuration
`sigma 1s^(2) lt sigma**1s^(2) lt sigma2s^(2) lt sigma** 2s^(2) lt sigma2p_(x)^(2) lt pi 2p_(y)^(2) = pi 2p_(z)^(2) `
Bond order `= ((N_(b) - N_(a)))/(2) = (10 - 4)/(2) = (6)/(2) = 3`
So bond order `O_(2)^(-) lt O_(2)^(+) lt _(2)^(2+)`