Match List I with List II List I List II A. XeF 4 I.See-saw B. SF 4 II. Square planar C. NH 4 + III. Bent T…

Match List I with List II

List I List II
A. XeF4 I.See-saw
B. SF4 II. Square planar
C. NH4+ III. Bent T- shaped
D.BrF3 IV. Tetrahedral

Choose the correct answer from the options given below :

  1. A-IV, B-III, C-II, D-I
  2. A-II, B-I, C-III, D-IV
  3. A-IV, B-I, C-II, D-III
  4. A-II, B-I, C-IV, D-III

Solution

(A) XeF4:

Xe has 4 bond pairs along with 2 lone pairs in the XeF4. Thus, the hybridisation of Xe is sp3d2. Now since it contains two lone pairs it will show square planar geometry.

(B) SF4:

The lone pair is an equatorial position, and there are two lone-pair—bond pair repulsions. Hence, it is more stable. So, the shape is described as a distorted tetrahedron, a folded square or a see-saw.

(C) NH4+:

In this case steric number = lone pair + sigma bond = 0 + 4 = 4. 

So, it is sp3 hybridisation. There is no lone pair present. Thus, the shape of this ion is tetrahedral.

(D) BrF3:

To decrease repulsion between the lone pairs, the molecule's structure is bent, making it T-shaped.

Asked in: JEE Main 2023 (31 Jan Shift 1)

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