Question

Arrange the following in increasing order of stability: $O_2$, $N_2$, $Li_2$, $H_2$.

• A. $H_2 < Li_2 < O_2 < N_2$
• B. $Li_2 < H_2 < O_2 < N_2$
• C. $N_2 < O_2 < Li_2 < H_2$
• D. $H_2 < Li_2 < N_2 < O_2$

Answer 1

Answer: (B)

(b)

Answer 2

To arrange the given molecules ($O_2$, $N_2$, $Li_2$, $H_2$) in increasing order of stability, we use the concept of bond order from Molecular Orbital Theory (MOT). Stability is generally proportional to bond strength, which in turn is related to the bond order. Higher the bond order, stronger is the bond, and thus greater is the stability.

Let's calculate the bond order for each molecule:

1. $H_2$:

   • Total electrons = 2 (1 from each H atom).
   • Molecular orbital configuration: $(\sigma_{1s})^2$.
   • Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2 = (2 - 0) / 2 = 1.

2. $Li_2$:

   • Total electrons = 6 (3 from each Li atom).
   • Molecular orbital configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2$.
   • Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2 = (4 - 2) / 2 = 1.

3. $N_2$:

   • Total electrons = 14 (7 from each N atom).
   • Molecular orbital configuration (for molecules with $\le$ 14 electrons): $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2$.
   • Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2 = (10 - 4) / 2 = 3.

4. $O_2$:

   • Total electrons = 16 (8 from each O atom).
   • Molecular orbital configuration (for molecules with > 14 electrons): $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1$.
   • Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2 = (10 - 6) / 2 = 2.

The bond orders are:

• $H_2$: 1
• $Li_2$: 1
• $N_2$: 3
• $O_2$: 2

Based on bond order, we expect the stability order to be $N_2 > O_2 > H_2 \approx Li_2$.
However, $H_2$ and $Li_2$ both have a bond order of 1. To compare their stability, we need to consider the actual bond strength (bond dissociation energy) or bond length. The bond in $H_2$ is formed by the overlap of small 1s orbitals, resulting in a short bond length (74 pm) and a high bond dissociation energy (436 kJ/mol). The bond in $Li_2$ is formed by the overlap of larger 2s orbitals, resulting in a much longer bond length (267 pm) and a lower bond dissociation energy (105 kJ/mol). Therefore, $H_2$ is significantly more stable than $Li_2$.

Combining the bond order trend with the comparison of $H_2$ and $Li_2$, the increasing order of stability is determined by the increasing bond strength:
$Li_2$ (BO=1, weak bond) < $H_2$ (BO=1, stronger bond) < $O_2$ (BO=2) < $N_2$ (BO=3).

Thus, the increasing order of stability is $Li_2 < H_2 < O_2 < N_2$.