Question

According to molecular orbital theory, which of the following is correct about the molecule $D_2$?

• A. Its bond order is $2.0$
• B. It has two unpaired electrons in 1t-bonding molecular orbital
• C. It is paramagnetic
• D. All of these

Answer 1

The correct option is D All of these

Compound $\left(A\right)$ undergoes hydrolysis slowly to form $\left(C\right),\left(B\right)$, and a diatomic gas $\left(D\right)$ whose $IE$ is almost similar to that of $\left(C\right)$
$\underset{A}{Xe{F}_2}+2H_{2}O\to \underset{C}{2Xe}+\underset{B}{2H_2{F}_2}+\underset{D}{O_2}$
Thus, the compound D is oxygen molecule.
The MO configuration of oxygen molecule is $KK\sigma \left(2s{)}^2{\sigma }^{\ast }\sigma \right(2p_z{)}^2\pi (2p_x{)}^2=\pi (2p_y{)}^2{\pi }^{\ast }(2p_x{)}^1={\pi }^{\ast }(2p_y{)}^1$.
Its bond order is 2.0.
$\text{Bond order}=\frac{n_b-n_a}{2}=\frac{8-4}{2}=2$
It has two unpaired electrons in ${\pi }^{\ast }(2p_x{)}^1={\pi }^{\ast }(2p_y{)}^1$ molecular orbital. It is paramagnetic as it contains unpaired electrons.
Hence, all the three options $A,B$ and $C$ are correct.