Concept: Bond order is a measure of the number of chemical bonds between two atoms. It can be calculated using Molecular Orbital Theory (MOT) as half the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals.

Why (B) is correct:

For ${\mathrm{N}}_2$:

Total electrons = 7 (for N) $\times$ 2 = 14 electrons.

The molecular orbital configuration for ${\mathrm{N}}_2$ is: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$

Number of electrons in bonding orbitals (Nb) = 2 + 2 + 2 + 2 = 8

Number of electrons in antibonding orbitals (Na) = 2 + 2 = 4

Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(8-4)=\frac{1}{2}(4)=2$

Wait, there's a mistake in the calculation. Let's re-evaluate for N2.

For ${\mathrm{N}}_2$:

Total electrons = 14.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$

Number of electrons in bonding orbitals (Nb) = 2 (from ${\sigma }_{1s}$) + 2 (from ${\sigma }_{2s}$) + 2 (from ${\pi }_{2p_x}$) + 2 (from ${\pi }_{2p_y}$) + 2 (from ${\sigma }_{2p_z}$) = 10

Number of electrons in antibonding orbitals (Na) = 2 (from ${\sigma }_{1s}^{\ast }$) + 2 (from ${\sigma }_{2s}^{\ast }$) = 4

Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-4)=\frac{1}{2}(6)=3$

Option Analysis:

• A) ${\mathrm{H}}_2$: Total electrons = 2. MO configuration: ${\sigma }_{1s}^2$. Nb = 2, Na = 0. Bond order = $\frac{1}{2}(2-0)=1$.
• B) ${\mathrm{N}}_2$: As calculated above, bond order = 3.
• C) ${\mathrm{O}}_2$: Total electrons = 16. MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$. Nb = 10, Na = 6. Bond order = $\frac{1}{2}(10-6)=2$.
• D) ${\mathrm{He}}^{2+}$: Total electrons = 2 (for He) $\times$ 2 - 2 (for 2+ charge) = 2 electrons. MO configuration: ${\sigma }_{1s}^2$. Nb = 2, Na = 0. Bond order = $\frac{1}{2}(2-0)=1$. (Note: ${\mathrm{He}}_2$ has 4 electrons, MO config ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}$, Nb=2, Na=2, BO=0, so it doesn't exist. ${\mathrm{He}}_2^{+}$ has 3 electrons, MO config ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 1}$, Nb=2, Na=1, BO=0.5). The question asks for ${\mathrm{He}}^{2+}$, which is a single helium atom with a +2 charge, meaning it has lost both electrons, so it's just a nucleus. If it meant ${\mathrm{He}}_2^{2+}$, then it would have 2 electrons and a bond order of 1. Assuming it meant ${\mathrm{He}}_2^{2+}$ as a diatomic species, its bond order is 1.

Correct Answer: (B)