Question

For ion $O_2^{-},$ the bond order is

• A. 2
• B. 1.5
• C. 2.5
• D. 0

Answer 1

Answer: (B)

Number of electrons in $O_2^{-}=8+8+1=17$
MO configuration of
$O_2^{-}=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2$
$\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^1$
Here, $N_b$ (bonding electrons) $=10N_a$ (antibonding electrons)
$=7\because$ Bond order $=\frac{N_b-N_a}{2}$
$\therefore$ Bond order $=\frac{10-7}{2}=1.5$