Ncert Exercises & Solutions

Use the molecular orbital energy level diagram to show that N₂ would be expected to have a triple bond. F₂, a single bond and Ne₂, no bond.

Formation of N₂ molecule Electronic configuration of N-atom

{}_{7}N = 1 s^{2}, 2 s^{2}, 2 p_{x}^{1}, 2 p_{y}^{1}, 2 p_{z}^{1}

N_{2}

molecule

= σ 1 s^{2}, σ * 1 s^{2}, σ 2 s^{2}, σ * 2 s^{2}, π 2 {p_{x}}^{2} \approx π 2 {p_{y}}^{2}, σ 2 {p_{z}}^{2}

Bond order

= \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 4) = 3 .

Bond order value of 3 means that N₂ contains a triple bond.

Formation of F₂ molecule

{}_{9}F = 1 s^{2}, 2 s^{2}, 2 {p_{x}}^{2}, 2 {p_{y}}^{2}, 2 {p_{z}}^{2}

F_{2}

molecule

= σ 1 s^{2}, σ * 1 s^{2}, σ 2 s^{2}, σ * 2 s^{2}, σ 2 {p_{z}}^{2}, π 2 {p_{x}}^{2} \approx π 2 {p_{y}}^{2}, π * 2 {p^{2}}_{x} \approx π * 2 {p^{2}}_{y}

Bond order

= \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 8) = 1

Bond order value 1 means that F₂ contains single bond.

Formation of Ne₂ molecule

{}_{10}{Ne} = 1 s^{2}, 2 s^{2}, 2 {p_{x}}^{2}, 2 {p_{y}}^{2}, 2 {p_{z}}^{2}

Ne₂ molecule

= σ 1 s^{2}, σ * 1 s^{2}, σ 2 s^{2}, σ * 2 s^{2}, σ 2 {p_{z}}^{2}, π 2 {p_{x}}^{2} \approx π 2 {p_{y}}^{2}, π * 2 {p^{2}}_{x} \approx π * 2 {p^{2}}_{y}, σ * 2 {p_{z}}^{2}

Bond order

= \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} (10 - 10) = 0

Bond order value zero means that there is no formation of bond between two Ne-atoms.

Hence, Ne₂ molecule does not exist.