Stability of the species `Li_(2), Li_(2)^(-), Li_(2)^(+)` increases in the order of

Options

A. `Li_ 2 lt Li_2^(+) lt Li_2^(-)`
B. `Li_2^(-) lt Li_2 lt Li_2^(+)`
C. `Li_2 lt Li_2^(-) lt Li_2^(+)`
D. `Li_2^(-) lt Li_2^(+) lt Li_2`

Answer

D

Explanation

Atomic number of Li is 3. Thus, the total number of electrons in `Li_2` is 6.
`Li_2(6)sigma1s^2sigma^**1s^2sigma2s^2`
Bond order `=(N_b-N_a)/(2)=(4-2)/(2)=1`
`Li_2^(+)` is formed by the loss of one `e^(-)` from `sigma2s` molecular orbital.
`Li_2^(+)(5)sigma1s^2sigma^**2s^2sigma2s^1`
Bond order `=(N_b-N_a)/(2)=(3-2)/(2)=0.5`
`Li_2^(-)` is formed by the gain of one `e^(-)` into `sigma^**2s` orbital.
`Li_2^(-)(7)=(N_b-N_a)/(2)=(4-3)/(2)=0.5`
Both `Li_2^(+)` and `Li_2^(-)` have identical bond orders but `Li_2^(+)` is more stable than `Li_2^(-)` because `Li_2^(-)` has more antibonding electrons. (Note that antibonding influence is greater than bonding influence.) Thus, the correct order is
`Li_2^(-) lt Li_2^(+) lt Li_2`