JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

Consider the following species:

BrF $_5$ , XeF $_5^-$ , BF $_4^-$ , ICl $_4^-$ , XeF $_4$ , SF $_4$ , NH $_4^+$ , ClF $_3$ , XeF $_2$ , ICl $_2^-$

Number of species having sp $^3$ d hybridized central atom is __________.

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

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Solution

The steric number $H$ is calculated as $H = \dfrac{1}{2}(V + M - C + A)$ , where $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms, $C$ is the cationic charge, and $A$ is the anionic charge.

For $sp^3d$ hybridization, $H = 5$ .

BrF $_5$ : $H = \dfrac{1}{2}(7 + 5) = 6$ ( $sp^3d^2$ )

XeF $_5^-$ : $H = \dfrac{1}{2}(8 + 5 + 1) = 7$ ( $sp^3d^3$ )

BF $_4^-$ : $H = \dfrac{1}{2}(3 + 4 + 1) = 4$ ( $sp^3$ )

ICl $_4^-$ : $H = \dfrac{1}{2}(7 + 4 + 1) = 6$ ( $sp^3d^2$ )

XeF $_4$ : $H = \dfrac{1}{2}(8 + 4) = 6$ ( $sp^3d^2$ )

SF $_4$ : $H = \dfrac{1}{2}(6 + 4) = 5$ ( $sp^3d$ )

NH $_4^+$ : $H = \dfrac{1}{2}(5 + 4 - 1) = 4$ ( $sp^3$ )

ClF $_3$ : $H = \dfrac{1}{2}(7 + 3) = 5$ ( $sp^3d$ )

XeF $_2$ : $H = \dfrac{1}{2}(8 + 2) = 5$ ( $sp^3d$ )

ICl $_2^-$ : $H = \dfrac{1}{2}(7 + 2 + 1) = 5$ ( $sp^3d$ )

The species with $sp^3d$ hybridization are SF $_4$ , ClF $_3$ , XeF $_2$ , and ICl $_2^-$ .

The number of such species is $4$ .

Answer: $4$

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JEE Main Inorganic Chemistry in other years

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JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

Consider the following species:

BrF $_5$ , XeF $_5^-$ , BF $_4^-$ , ICl $_4^-$ , XeF $_4$ , SF $_4$ , NH $_4^+$ , ClF $_3$ , XeF $_2$ , ICl $_2^-$

Number of species having sp $^3$ d hybridized central atom is __________.

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For $sp^3d$ hybridization, $H = 5$ .

BrF $_5$ : $H = \dfrac{1}{2}(7 + 5) = 6$ ( $sp^3d^2$ )

XeF $_5^-$ : $H = \dfrac{1}{2}(8 + 5 + 1) = 7$ ( $sp^3d^3$ )

BF $_4^-$ : $H = \dfrac{1}{2}(3 + 4 + 1) = 4$ ( $sp^3$ )

ICl $_4^-$ : $H = \dfrac{1}{2}(7 + 4 + 1) = 6$ ( $sp^3d^2$ )

XeF $_4$ : $H = \dfrac{1}{2}(8 + 4) = 6$ ( $sp^3d^2$ )

SF $_4$ : $H = \dfrac{1}{2}(6 + 4) = 5$ ( $sp^3d$ )

NH $_4^+$ : $H = \dfrac{1}{2}(5 + 4 - 1) = 4$ ( $sp^3$ )

ClF $_3$ : $H = \dfrac{1}{2}(7 + 3) = 5$ ( $sp^3d$ )

XeF $_2$ : $H = \dfrac{1}{2}(8 + 2) = 5$ ( $sp^3d$ )

ICl $_2^-$ : $H = \dfrac{1}{2}(7 + 2 + 1) = 5$ ( $sp^3d$ )

The species with $sp^3d$ hybridization are SF $_4$ , ClF $_3$ , XeF $_2$ , and ICl $_2^-$ .

The number of such species is $4$ .

Answer: $4$

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JEE Main Inorganic Chemistry in other years

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JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

Did you get this right?

Solution

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JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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