Question

Which of the following statement is not correct from the view point of molecular orbital theory?

विकल्प

• $$\mathrm{Be}{\phantom{A}}_2$$ is not a stable molecule.

• $$\mathrm{He}{\phantom{A}}_2$$ is not stable but $$\mathrm{He}{\phantom{A}}_2^{+}$$ is expected to exist.

• Bond strength of $$\mathrm{N}{\phantom{A}}_2$$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

• The order of energies of molecular orbitals in $$\mathrm{N}{\phantom{A}}_2$$ molecule is σ2s < σ^*2s < σ2p_z < (π2p_x = π2p_y) < (π^*2p_x = π^*2p_y) < σ^*2p_z 

Answer 1

The order of energies of molecular orbitals in $$\mathrm{N}{\phantom{A}}_2$$ molecule is σ2s < σ^*2s < σ2p_z < (π2p_x = π2p_y) < (π^*2p_x = π^*2p_y) < σ^*2p_z 

Explanation:

(i) $$\mathrm{Be}{\phantom{A}}_2$$ (4 + 4 = 8) = σ1s², σ*1s², σ1s², σ*2s²

Bond order $$(\mathrm{BO})=\frac{1}{2}$$ [Number of bonding electron (N_b) – Number of anti-bonding electrons N_a]

= ${\displaystyle \frac{4-4}{2}}$ = 0

Here, bond order of $$\mathrm{He}{\phantom{A}}_2$$ is zero. THus. it does not exist.

(ii) $$\mathrm{NHe}{\phantom{A}}_2$$ (2 + 2 = 4) = σ1s², σ*1s² 

BO = ${\displaystyle \frac{2-2}{2}}$ = 0

Here, bond order of $$\mathrm{He}{\phantom{A}}_2$$ is zero. Thus, it does not exist.

$$\mathrm{He}{\phantom{A}}_2^{+}$$ (2 + 2 – 1 = 3) = σ1s², σ*1s¹  

BO = ${\displaystyle \frac{2-1}{2}}$ = 0.5

Since, the bond order is not zero, this molecule is expected to exist.

(iii) $$\mathrm{N}{\phantom{A}}_2$$ (7 + 7 = 14) = $$\mathrm{\sigma }{\phantom{A}}_1\mathrm{s}{\phantom{A}}^2,\mathrm{\sigma }{\phantom{A}}^{\ast }1\mathrm{s}{\phantom{A}}^2,\mathrm{\sigma }{\phantom{A}}_2\mathrm{s}{\phantom{A}}^2,\mathrm{\sigma }{\phantom{A}}^{\ast }2\mathrm{s}{\phantom{A}}^2,\mathrm{\pi }{\phantom{A}}_2\mathrm{p}{\phantom{A}}_x^2\approx \mathrm{\pi }{\phantom{A}}_2\mathrm{p}{\phantom{A}}_y^2,\mathrm{\pi }{\phantom{A}}_2\mathrm{p}{\phantom{A}}_z^2$$

BO = ${\displaystyle \frac{10-4}{2}}$ = 3   

Thus, dinitrogen $$(\mathrm{N}{\phantom{A}}_2)$$ molecule contains triple bond and no other molecule of second period have more than double bond. Hence, bond strength of $$\mathrm{N}{\phantom{A}}_2$$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(iv) It is incorrect. The correct order of energies of molecular orbitals in $$\mathrm{N}{\phantom{A}}_2$$ molecule is

σ2s < σ^*2s < σ2p_z < (π2p_x = π2p_y) < π^*2p_x = π^*2p_y < σ^*2p_z