NEETChemistryChemical Bonding and Molecular Structure
Match List I with List II. List I (Molecules) List II (Geometry and Dipole moment) (A) XeF ₄ (i) Square planar, = 0 (B) SF ₄ (ii) See-saw, 0 (C) PCl ₅ (iii) Trigonal bipyramidal, = 0 (D) H ₂ O (iv) Bent, 0 Choose the correct answer from the options given below:
Options
- A(A)-(ii), (B)-(i), (C)-(iii), (D)-(iv)
- B(A)-(i), (B)-(iii), (C)-(ii), (D)-(iv)
- C(A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
- D(A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)
Correct answer
C. (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
Step-by-step solution
According to VSEPR theory: XeF ₄ has 4 bond pairs and 2 lone pairs on the central Xe atom. Its geometry is square planar. The lone pairs are opposite to each other, and the four Xe-F bond dipoles cancel each other out, resulting in a net dipole moment = 0 . SF ₄ has 4 bond pairs and 1 lone pair on the central S atom. Its geometry is see-saw. Due to its asymmetrical shape, the bond dipoles do not cancel out, resulting in 0 . PCl ₅ has 5 bond pairs and 0 lone pairs on the central P atom. Its geometry is trigonal bipy