Question

Why is bond order of Be₂ molecule zero?

(N_b = bonding electrons, N_a = antibonding electrons)

पर्याय

• N_b > N_a

• N_b = N_a 

• ${\displaystyle \frac{{\text{N}}_{\text{b}}}{{\text{N}}_{\text{a}}}}$ = Zero

• N_b < N_a

Answer 1

N_b = N_a 

Explanation:

In Be₂ molecule,

${\displaystyle {\sigma }_{1{\text{s}}^2} {\sigma }_{1{\text{s}}^2}^{\cdot } {\sigma }_{2{\text{s}}^2}{\sigma }_{2{\text{s}}^2}^{\cdot }}$

N_b = 4, N_a = 4

Bond order = ${\displaystyle \frac{{\text{N}}_{\text{b}}-{\text{N}}_{\text{a}}}{2}}$ = 0