Question

Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.

Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals.



Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order :

$\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<(\pi 2p_x\approx \pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x\approx {\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$ and for oxygen and fluorine order of energy of molecular orbitals is given below :
$\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<\sigma 2p_z<(\pi 2p_x\approx \pi 2p_y)<({\pi }^{\ast }2p_x\approx {\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$
Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called 'Sigma', $\left(\sigma \right)$ and if the overlap is lateral, the molecular orbital is called 'pi', $\left(\pi \right)$. The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.

Which of the following pairs is expected to have the same bond order?

• A. $O_2^{+},N_2^{-}$
  
  
• B. $O_2^{-},N_2^{-}$
• C. $O_2^{-},N_2^{+}$
  
  
• D. $O_2{N}_2$
  
  

Answer 1

The correct option is A $O_2^{+},N_2^{-}$



Formula to calculate bond order

Bond order (BO)$=\left|\frac{\left(N_b\right)-\left(N_a\right)}{2}\right|$

$N_b=$ Number of electrons in bonding molecular orbital

$N_a=$ Number of electrons in anti bonding molecular orbitals

Calculating bond order in molecules

Electron configuration of $O_2^{+}$:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$

BO of $O_2^{+}=\left|\frac{10-5}{2}\right|=2.5$

Electronic configuration of $N_2$:

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

BO of $N_2=\left|\frac{10-4}{2}\right|=3$


Electronic conifiguration of $N_2^{+}$:

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

BO of $N_2^{+}=\left|\frac{9-4}{2}\right|=2.5$

Electronic configuration of $N_2^{-}$:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_x^1=\pi 2p_y^0$

Bo of $N_2^{-}=\left|\frac{10-5}{2}\right|=2.5$

Species	BO
$O_2$	2
$N_2$	3
$O_2^{+}$	2.5
$N_2^{-}$	2.5
$O_2^{-}$	1.5
$N_2^{+}$	2.5

$O_2^{+}\text{and}{N}_2^{+}$ pair have same bond order. So, correct answer is option B.