Bond order of 1.5 is shown by:

Options

A. `O_2^+`
B. `O_2^-`
C. `O_2^(2-)`
D. `O_2`

Answer

B

Explanation

Molecular orbital configuration of `O_2^+` (8+8-1=15)
`=sigma1s^2 , overset***1s^2, sigma2s^2, overset***sigma2s^2, sigma2p_z^2, pi2p_x^2~~pi2p_y^2, overset***pi2p_x^1~~overset***pi2p_y^0`
Bond order (BO) =`(N_b-N_a)/2`
(where, `N_b`=number of electrons in bonding molecular orbital. `N_a`=number of electrons in antibonding molecular orbital)
`therefore BO=(10-5)/2=2.5`
Similarly,
(b)`O_2^-` (8+8+1=17)
so, `BO=(N_b-N_a)/2=(10-7)/2=15`
(c )`O_2^(2-)` (8+8+2=18)
`BO=(N_b-N_a)/2=(10-8)/2=1`
(d)`O_2`(8+8=16)
BO=`(10-6)/2=2`
Thus, `O_2^-` shows the bond order 1.5