Question

Why is bond order of $Be_2$ molecule zero?
(N$_b$ = bonding electrons, N$_a$ = antibonding electrons)

• A. $N_b>N_a$
• B. $N_b<N_a$
• C. $\dfrac{N_b}{N_a} = 0$
• D. $N_b = N_a$

Hint:

If bonding and antibonding electrons are equal, the molecule is unstable and has zero bond order.

Answer 1

The Correct Option is D

Step 1: Write the bond order formula.
Bond order is given by:
\[ \text{Bond order} = \frac{N_b - N_a}{2} \]
Step 2: Electronic configuration of $Be_2$.
Each Be atom has electronic configuration $1s^2\,2s^2$.
In $Be_2$, the molecular orbital configuration becomes:
\[ (\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2 \]
Step 3: Count bonding and antibonding electrons.
Number of bonding electrons $N_b = 4$
Number of antibonding electrons $N_a = 4$
Step 4: Calculate bond order.
\[ \text{Bond order} = \frac{4 - 4}{2} = 0 \]
Step 5: Conclusion.
Since the number of bonding electrons is equal to the number of antibonding electrons, the bond order of $Be_2$ is zero.