Question

Bond order of $$\text{B}{\text{e}}_{\text{2}}$$ is
A. 1
B. 2
C. 3
D. 0

Answer 1

Hint: According to the molecular orbital theory, the bond order is defined as the number of covalent bonds in a molecule. Bond order is equal to half of the difference between the number of electrons in bonding ($$N_b$$) and antibonding molecular orbitals ($$N_a$$).

Complete Solution :
$$\text{B}{\text{e}}_{\text{2}}$$ molecule will be formed by the overlapping of atomic orbitals of two beryllium atoms.
A Be atom has four electrons. It has two valence electrons and its electronic configuration is $$1s^{2}2s^2$$. Therefore, $$\text{B}{\text{e}}_{\text{2}}$$ molecule has eight electrons which are to be filled in four molecular orbitals.

Thus, electronic configuration of $$\text{B}{\text{e}}_{\text{2}}$$ is $${\left(\sigma 1s\right)}^2{\left({\sigma }^{\ast }1s\right)}^2{\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2$$
Here, bonding electrons, $$N_b$$ = 4 and anti-bonding electrons, $$N_a$$= 4
Therefore, bond order (B.O.) of $$\text{B}{\text{e}}_{\text{2}}$$ molecule is
$$\begin{array}{rl}& \text{B}\text{.O}\text{.=}\frac{1}{2}(N_b-N_a)\\ & \text{B}\text{.O}\text{.}=\frac{1}{2}(4-4)=0\end{array}$$
 - Zero value of bond order corresponds to non-existence of $$\text{B}{\text{e}}_{\text{2}}$$ molecule.
So, the correct answer is “Option D”.

Note: The bond order of a molecule conveys the following information:
1. The stability of a molecule can also be expressed in terms of bond order. Higher the bond order, more stable is the molecule.
2. Bond length: Bond order and bond length are inversely related. Thus, higher the bond order, shorter is the bond length and vice-versa.
3. Bond dissociation energy: Bond order in a molecule is directly proportional to its bond dissociation energy. Greater the bond order, more will be the value of bond dissociation energy.