Chemical Bonding and Molecular Structure Part 3

Concept:The dipole moment depends on molecular geometry and bond polarity. Bond order is determined using molecular orbital theory. Formal charge and octet rule help assess stability.Explanation:Statement A: $\mathrm{NF}_3$ and $\mathrm{NH}_3$ are trigonal pyramidal. In $\mathrm{NH}_3$, the lone pair dipole and bond dipoles add, giving a larger net dipole. In $\mathrm{NF}_3$, the N–F bond dipoles oppose the lone pair direction, reducing the net dipole. Hence, $\mu(\mathrm{NF}_3) < \mu(\mathrm{NH}_3)$. So A is incorrect.Statement B: $\mathrm{BeH}_2$ is linear (sp hybridised). The two Be–H bond dipoles are equal and opposite, so they cancel. Dipole moment is zero. B is correct.Statement C: $\mathrm{O}_2^{2-}$ has 14 valence electrons, same as $\mathrm{F}_2$. Using MO theory, bond order for $\mathrm{O}_2^{2-}$ is 1, and for $\mathrm{F}_2$ it is also 1. They are identical. C is correct.Statement D: In a resonance form of ozone, $\mathrm{O} = \mathrm{O} - \mathrm{O}$, the central oxygen has one lone pair, one double bond, and one single bond. Formal charge = $6 - (2 + \frac{6}{2}) = +1$. It is $+1$, not $-1$. D is incorrect.Statement E: $\mathrm{NO}_2$ has 17 valence electrons, an odd number. The nitrogen atom does not complete its octet. It is an odd-electron molecule and quite reactive (dimerises to $\mathrm{N}_2\mathrm{O}_4$). E is incorrect.Answer:B and C are correct. Option A: B & C Only.