Let's analyze each option:

Bond length generally increases down a group as the atomic size increases. Chlorine is larger than fluorine, so the H-Cl bond length is greater than the H-F bond length. Therefore, H-F < H-Cl. This statement is incorrect.

Let's determine the bond order using Molecular Orbital Theory:

• He\u2082: Total electrons = 4. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}$. Bond order = $\frac{1}{2}(2-2)=0$.

• He\u2082\u207a: Total electrons = 3. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 1}$. Bond order = $\frac{1}{2}(2-1)=0.5$.

Since 0 < 0.5, He\u2082 < He\u2082\u207a in terms of bond order. This statement is incorrect.

Let's determine the number of unpaired electrons using Molecular Orbital Theory:

• O\u2082: Total electrons = 16. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2)({\pi }_{2p_x}^{\ast 1}={\pi }_{2p_y}^{\ast 1})$. It has 2 unpaired electrons in the ${\pi }^{\ast }$ antibonding orbitals.

• O\u2082\u00b2\u207b (Peroxide ion): Total electrons = 18. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2)({\pi }_{2p_x}^{\ast 2}={\pi }_{2p_y}^{\ast 2})$. All electrons are paired. It has 0 unpaired electrons.

Since 2 > 0, O\u2082 has more unpaired electrons than O\u2082\u00b2\u207b. This statement is correct.

Let's determine the bond order first, as bond length is inversely proportional to bond order.

• O\u2082: Bond order = $\frac{1}{2}(10-6)=2$.

• O\u2082\u00b2\u207b: Bond order = $\frac{1}{2}(10-8)=1$.

Since the bond order of O\u2082 (2) is greater than O\u2082\u00b2\u207b (1), the bond length of O\u2082 will be shorter than O\u2082\u00b2\u207b. Therefore, O\u2082 < O\u2082\u00b2\u207b in terms of bond length. This statement is incorrect.

Correct Answer: (C)