Match List I with List II List I List II A. ${\mathrm{XeF}}_{4}$ I.See-saw B. ${\mathrm{SF}}_{4}$ II. Square planar C. ${\mathrm{NH}}_{4}^{+}$ III. Bent $T-$ shaped D.${\mathrm{BrF}}_{3}$ IV. Tetrahedral Choose the correct answer from the options given below :

JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqeasy

Match List I with List II

List I	List II
A. ${\mathrm{XeF}}_{4}$	I.See-saw
B. ${\mathrm{SF}}_{4}$	II. Square planar
C. ${\mathrm{NH}}_{4}^{+}$	III. Bent $T-$ shaped
D. ${\mathrm{BrF}}_{3}$	IV. Tetrahedral

Choose the correct answer from the options given below :

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Options

A
$A-\mathrm{IV},B-\mathrm{III},C-\mathrm{II},D-I$
B
$A-\mathrm{II},B-I,C-\mathrm{III},D-\mathrm{IV}$
C
$A-\mathrm{IV},B-I,C-\mathrm{II},D-\mathrm{III}$
D
$A-\mathrm{II},B-I,C-\mathrm{IV},D-\mathrm{III}$

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Solution

(A) ${\mathrm{XeF}}_{4}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 1

$\mathrm{Xe}$ has 4 bond pairs along with 2 lone pairs in the ${\mathrm{XeF}}_{4}$ . Thus, the hybridisation of $\mathrm{Xe}$ is ${\mathrm{sp}}^{3}{d}^{2}$ . Now since it contains two lone pairs it will show square planar geometry.

(B) ${\mathrm{SF}}_{4}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 2

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.

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 3

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

So, it is sp $^{3}$ hybridisation. There is no lone pair present. Thus, the shape of this ion is tetrahedral.

(D) ${\mathrm{BrF}}_{3}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 4

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

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Options

$A-\mathrm{IV},B-\mathrm{III},C-\mathrm{II},D-I$

$A-\mathrm{II},B-I,C-\mathrm{III},D-\mathrm{IV}$

$A-\mathrm{IV},B-I,C-\mathrm{II},D-\mathrm{III}$

$A-\mathrm{II},B-I,C-\mathrm{IV},D-\mathrm{III}$

Solution

(A) ${\mathrm{XeF}}_{4}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 1

(B) ${\mathrm{SF}}_{4}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 2

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 3

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

So, it is sp $^{3}$ hybridisation. There is no lone pair present. Thus, the shape of this ion is tetrahedral.

(D) ${\mathrm{BrF}}_{3}$ :

JEE Main 2023 Chemistry, Chemical Bonding and Molecular Structure — solution figure 4

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

JEE Main 2023 — Chemistry Inorganic Chemistry

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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