Number of compounds with one lone pair of electrons on central atom amongst following is O₃$O_3$, H₂O$H_2O$, SF₄$SF_4$, ClF₃$ClF_3$, NH₃$NH_3$, BrF₅$BrF_5$, XeF₄$XeF_4$

Core Logic

Let us determine the steric number (Z$Z$) and number of lone pairs (LP$LP$) for the central atom in each given molecule. Formula: Z = (1)/(2) (V + M - C + A)$Z = \frac{1}{2} (V + M - C + A)$ Where V$V$ = valence electrons on central atom, M$M$ = number of monovalent atoms, C$C$ = cationic charge, A$A$ = anionic charge. LP = Z - Bond Pairs (B.P.)$LP = Z - \text{Bond Pairs (B.P.)}$

• O₃$O_3$: Central atom O (V=6$V=6$). It forms one double bond and one dative bond. It has 1 lone pair remaining.
• H₂O$H_2O$: Central atom O (V=6$V=6$). Z = (1)/(2)(6 + 2) = 4$Z = \frac{1}{2}(6 + 2) = 4$. LP = 4 - 2 = 2$LP = 4 - 2 = 2$.
• SF₄$SF_4$: Central atom S (V=6$V=6$). Z = (1)/(2)(6 + 4) = 5$Z = \frac{1}{2}(6 + 4) = 5$. LP = 5 - 4 = 1$LP = 5 - 4 = 1$ (See-saw shape).
• ClF₃$ClF_3$: Central atom Cl (V=7$V=7$). Z = (1)/(2)(7 + 3) = 5$Z = \frac{1}{2}(7 + 3) = 5$. LP = 5 - 3 = 2$LP = 5 - 3 = 2$ (T-shape).
• NH₃$NH_3$: Central atom N (V=5$V=5$). Z = (1)/(2)(5 + 3) = 4$Z = \frac{1}{2}(5 + 3) = 4$. LP = 4 - 3 = 1$LP = 4 - 3 = 1$ (Pyramidal).
• BrF₅$BrF_5$: Central atom Br (V=7$V=7$). Z = (1)/(2)(7 + 5) = 6$Z = \frac{1}{2}(7 + 5) = 6$. LP = 6 - 5 = 1$LP = 6 - 5 = 1$ (Square Pyramidal).
• XeF₄$XeF_4$: Central atom Xe (V=8$V=8$). Z = (1)/(2)(8 + 4) = 6$Z = \frac{1}{2}(8 + 4) = 6$. LP = 6 - 4 = 2$LP = 6 - 4 = 2$ (Square Planar).

Step 1: Final Counting

The compounds containing exactly ONE lone pair on the central atom are O₃$O_3$, SF₄$SF_4$, NH₃$NH_3$, and BrF₅$BrF_5$.

Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure