Question

Which of the following statements is not correct from the view point of molecular orbital theory?
A) $B{e_2}$ is not a stable molecule.
B) $H{e_2}$ is not stable but $He_2^+ $ is expected to exist.
C) The bond strength of ${N_2}$ is maximum amongst the mononuclear diatomic molecules belonging to the second period.
D) The order of energies of the molecular orbitals in ${N_2}$ molecule is
$\sigma_{2s} \;<\; \sigma^*_{2s} \;<\;\sigma_{2p_z}\;<\;(\pi_{2p_x}\;=\;\pi_{2p_y)}\;<\;(\pi^*_{2p_x}\;=\;\pi_{2p_y}\;)\;<\;\sigma^*_{2p_z}$

Answer 1

Hint: Molecular orbital theory was proposed for chemical bonding. According to the theory the atomic orbitals participating in molecule formation get disturbed when the nuclei approach near, all of which get mixed up to rise to new orbitals which are called molecular orbitals.

Complete step by step answer:
Molecular orbitals are polycentric and are arranged in the increasing energy levels in the atomic orbitals. If the overlap is additive in phase it forms bonding orbitals which are lower in energy than the atomic orbitals and if the the overlap is out of the phase then it forms anti bonding ($\sigma^*,\pi^*,\delta^*$ etc.) orbital, its energy is higher than the energy of atomic orbital.
Stability, magnetic property, bond strength and energy, bond order of molecule orbitals can be explained on the basis of molecular orbital theory as follows.
i) Molecules having zero bond order never exist while molecules having non-zero bond order either exist or are expected to exist.
ii) The higher the value of the bond order higher will be its bond strength.
iii) Magnetic property of molecules can be decided from its molecular orbital. Paired electrons makes it diamagnetic and unpaired electrons make it paramagnetic.
Let us analyze the given options above:-
Electrons present in bonding molecular orbital are known as bonding electrons and electrons present in an antibonding molecular orbital are known as antibonding electrons and half of their difference between the electrons present in bonding and antibonding molecular orbital is known as bond order. If the number of electrons present in bonding molecular orbital $(N_b)$ is more when compared to antibonding molecular orbital $(N_a)$ then the molecule is said to be stable.
a) $Be_2$ = (No of electrons in $Be_2$ is (4 + 4 = 8) = \[(\sigma_{1s})^2,(\sigma^* _{1s})^2,(\sigma_{2s})^2,(\sigma^*_{2s})^2\]
Bond order (BO)=$\dfrac{1}{2}({N_{_b}} - {N_a}) = \dfrac{1}{2}(4 - 4) = 0$
Ignoring the inner shell it is apparent that the effects of bonding \[(\sigma_{2s})^2 \;and\; (\sigma^*_{2s})^2\] anti bonding levels cancel so there is no stabilization and the molecule doesn’t exist.

b) $He_2$ (no of electrons in $He_2$ is 2+2=4) = \[(\sigma_{1s})^2,(\sigma^* _{1s})^2\]
BO=$\dfrac{1}{2}({N_b} - {N_a}) = \dfrac{1}{2}(2 - 2) = 0$
Here, the bond order of $Be_2$ is zero, thus, it does not exist.
$He^{2+}$ (no of electrons in $He^{2+}$ is 2+1 = 3)
Electronic Configuration = \[(\sigma_{1s})^2,(\sigma^* _{1s})^1\]
BO= $\dfrac{1}{2}({N_b} - {N_a}) = \dfrac{1}{2}(2 - 1) = 0.5$
Since the bond order is not zero, this molecule is expected to exist. Also we observe that in He atom the energy gets canceled out which does not happen in its ion it does exist but is not very stable.
c) $N_2$ (No of electrons in $N_2$ is 7+7=14) = \[(\sigma_{1s})^2,(\sigma^* _{1s})^2, (\sigma_{2s})^2,(\sigma^* _{2s})^2,[(\pi2_{p_x})^2\;=\;\pi2_{p_y})^2], (\sigma_{2p_z})^2\]

BO= $\dfrac{1}{2}({N_b} - {N_a}) = \dfrac{1}{2}(10 - 4) = 3$
Thus, the dinitrogen $(N_2)$ molecule contains a triple bond as even after cancelation of energies it is still left with one sigma and two pi-bonds . Hence, the bond strength of $N_2$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

d) The correct order of energies of molecular orbitals in $N_2$ molecule is
$\sigma_{2s}\;<\;\sigma^*_{2s}\;<\;(\pi_{2p_x}\;=\;\pi_{2p_y)}\;<\;\sigma_{2p_z}\;<\;(\pi^*_{2p_x}\;=\;\pi^*_{2p_y})\;<\;\sigma^*_{2p_z}$
Since we know that antibonding orbitals have higher energy than bonding orbitals.

Hence the correct option is option A,B and C.

Note: Stability depends on the number of bonding and antibonding electrons present in orbitals. If the number of electrons present in bonding is more when compared to antibonding then the molecule is said to be stable. Molecular orbitals are polycentric and are of varying energies.