Question

Question 19

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

(a) $B{e}_2$ is not a stable molecule

(b) $H{e}_2$ is not stable but $H{e}_2^{+}$ is expected to exist.

(c) Bond strength of N_2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(d) The order of energies of molecular orbitals in $N_2$ molecule is

$\sigma 2s<{\sigma }^{\ast }2s<\sigma 2p_z<({\pi }^{\ast }2p_x\simeq {\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$

Answer 1

Answer:

(d) The order of energies of molecular orbitals in N2 molecule is

σ2s<σ∗2s<σ2pz<(π∗2px≃π∗2py)<σ∗2pz

The existence of molecule, bonding nature and energy order of molecular orbitals can be explained on the basis of molecular orbital theory as follows

(i) Molecules having zero bond order never exists while molecular having the non-zero bond is either exists or expected to exist.

(ii) Higher the value of the bond order higher will be its bond strength.

Electrons present in bonding molecular orbital are known as bonding electrons $\left(N_b\right)$ and electrons present on anti-bonding molecular orbital are knows as anti-bonding electrons $\left(N_a\right)$ and half of their difference is knows as bond order i.e.,

(a)$B{e}_2(4+4=8)=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 1s^2,{\sigma }^{\ast }2s^2$

Bond order (BO) =$\frac{1}{2}$

[Number of bonding electrons $\left(N_b\right)$-Number of anti-bonding electrons $\left(N_a\right)$]

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

Here, bond order of $B{e}_2$ is zero. it does not exist.

(b) $H{e}_2(2+2=4)=\sigma 1s^2,{\sigma }^{\ast }1s^2$

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

Here, bond order of $B{e}_2$ is zero. Hence, it does not exist,

$H{e}_2^{+}(2+2-1=3)=\sigma 1s^2,{\sigma }^{\ast }1s^1$

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

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

(c) $N_2(7+7=14)=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2$

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

Thus, dinitrogen $\left(N_2\right)$ molecule contain triple bond and no any molecule of second period have more than double bond. Hence, bond strength of $\left(N_2\right)$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(d) It is incorrect order of energies of molecular orbitals in $\left(N_2\right)$ molecule is

$\sigma 2s<{\sigma }^{\ast }2s<({\pi }^{\ast }2p_x\simeq {\pi }^{\ast }2p_y)<\sigma 2p_z<{\pi }^{\ast }2p_x\approx {\pi }^{\ast }2p_y<{\sigma }^{\ast }2p_z$