Chemistry - CHEMICAL BONDING Question with Solution | TestHub

For $A{X}_{n_1}$ (Group 16): A planar and polar molecule is bent-shaped like OF₂, giving $n_1=2$ and $m_1=2$ lone pairs.

For $B{X}_{n_2}$ (Group 15): A trigonal pyramidal molecule like NH₃ has 3 bonds and 1 lone pair, so $n_2=3$ and $m_2=1$.

For $C{X}_{n_3}$ (Group 14): A molecule with zero dipole moment is tetrahedral like CH₄, giving $n_3=4$ and $m_3=0$.

For $D{X}_{n_4}$ (Group 13): A molecule with $120^{\circ }$ angles is trigonal planar like BF₃, giving $n_4=3$ and $m_4=0$.

Substituting these into the expression:

$${\left[\frac{2+3+4+3}{2+1+0+0}\right]}^2={\left[\frac{12}{3}\right]}^2=4^2=16$$