Question

Species having same bond order are:

• A. $O_2^{-}$
• B. $F_2^{+}$
• C. $N_2^{-}$
• D. $N_2$

Answer 1

Answer: (A), (B)

$F_2^{+}$

Calculation of bond order


Bond order can be calculated by using the given formula:

$\text{Bond order}=\frac{1}{2}(N_b-N_a)$

$N_b$ = Number of bonding electrons

$N_a$ = Number of antibonding electrons

Option (A):

Number of electrons in $N_2=14$

Electronic configuration of $N_2$ molecule:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2$

$N_b=10,N_a=4$

$\text{Bond order}=\frac{1}{2}(10-4)=3$

Option (B)

Number of electrons in $N_2^{-}=15$

Electronic configuration of $N_2^{-}$ molecule:
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$

$N_b=10,N_a=5$

$\text{Bond order}=\frac{1}{2}(10-5)=2.5$

Option (C)

Number of electrons in $F_2^{+}=17$

Electronic configuration of $F_2^{+}$ molecule:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

$N_b=10,N_a=7$

$\text{Bond order}=\frac{1}{2}(10-7)=1.5$

Option (D):

Number of electrons in $0_2^{-}=17$

Electronic configuration of $0_2^{-}$ molecule:

(\sigma1s^{2}~\sigma^\ast1s^{2}~\sigma2s^{2}~~\sigma^\ast2s^{2}~\sigma2p^{2}_{z}~\pi2p^{2}_{x}=\pi2p^{2}_{y}~\pi^\ast2p^{2}_{x}=\pi^\ast2p^{1}_{y}\)

$N_b=10,N_a=7$

$\text{Bond order}=\frac{1}{2}(10-7)=1.5$

So, molecules having same bond order are $F_2^{+}\text{~}and{0}_2^{-}$.

Hence,the correct options are (C) and (D).