Chemistry - CHEMICAL BONDING Question with Solution | TestHub

(P) ClF₃: Chlorine has 7 valence electrons; it forms 3 bonds and has 2 lone pairs ($A{X}_3{E}_2$ system), resulting in a T-shape (Match: 3).

(Q) CF₄: Carbon has 4 valence electrons and forms 4 bonds with no lone pairs ($A{X}_4$ system), resulting in a Tetrahedral shape (Match: 1).

(R) SF₄: Sulfur has 6 valence electrons; it forms 4 bonds and has 1 lone pair ($A{X}_4{E}_1$ system), resulting in a see-saw shape (Match: 2).

(S) NF₃: Nitrogen has 5 valence electrons; it forms 3 bonds and has 1 lone pair ($A{X}_3{E}_1$ system), resulting in a Trigonal pyramidal shape (Match: 4).

The correct code is (A) P $\to$ 3; Q $\to$ 1; R $\to$ 2; S $\to$ 4.