According to `MO` Thory

Options

A. `O_(2)^(+)` is paramagnetic and bond order is greater than `O_(2)`
B. `O_(2)^(+)` is paramagnetic and bond order is less than `O_(2)`
C. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`
D. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)`

Answer

a

Explanation

In `O_(2)` molecule , the total number of electron `= 16` Electronic distrubution in molencular orbital of
`O_(2) = sigma 1s^(2), sigma 2s^(2), sigma **2s^(2), sigma p1_(s)^(2), pi 2 p _(y), pi 2p_(z), pi **2p_(y)^(1), pi**2p_(z)^(1)` Bond order in `O_(2) = (1)/(2) [N_(b) - N_(a)] = (1)/(2) [10-6] = 2.0`
In `O_(2)^(+) = sigma 1s^(2), sigma **1s^(2), sigma 2s^(2), sigma **2_(s)^(2), sigma p _(s)^(2), pi 2s_(y)^(2), pi 2p_(z)^(2),pi **2p_(y)^(1)` It is paramagnetic due to presence of unpaired orbital.
Bond order in `O_(2)^(+) = (1)/(2) [10-5] = 2.5`
The bond order of `O_(2)^(+)` is `2.5` which is greater than bond order of `O_(2) (2.0)`.
Hence, `O_(2)^(+)` is paramagenetic and order greater than `O_(2)`