JEE Chemistry 2026 Question with Solution

Given: The combinations list the molecular shape and hybridization of $\mathrm{IF_3}$, $\mathrm{IF_5}$, $\mathrm{IF_7}$ and $\mathrm{ClO_4^-}$.

Find: Which combinations are correct.

Hybridization is determined by the total number of sigma bonds and lone pairs around the central atom.

Step 1: Geometry of $\mathrm{IF_3}$

has two lone pairs and three bond pairs with $sp^3d$ hybridization, giving a T-shaped geometry.

Step 2: Geometry of $\mathrm{IF_5}$

has one lone pair and five bond pairs with $sp^3d^2$ hybridization, giving square pyramidal geometry.

Step 3: Geometry of $\mathrm{IF_7}$

has seven bond pairs and no lone pair with $sp^3d^3$ hybridization, giving pentagonal bipyramidal geometry.

Step 4: Geometry of $\mathrm{ClO_4^-}$

has $sp^2d$ hybridization and square planar electron geometry.

Step 5: Conclusion All given combinations are correct. Therefore, the correct option is A.

Given: Four statement-wise combinations of shape and hybridization are provided.

Find: Whether statements A, B, C and D are correct.

From the extracted solution:

• A is stated correct for $\mathrm{IF_3}$ as T-shaped with $sp^3d$ hybridization.
• B is stated correct for $\mathrm{IF_5}$ as square pyramidal with $sp^3d^2$ hybridization.
• C is stated correct for $\mathrm{IF_7}$ as pentagonal bipyramidal with $sp^3d^3$ hybridization.
• D is also stated correct for $\mathrm{ClO_4^-}$ in the provided solution.

Hence all four listed combinations are treated as correct by the source solution, so the answer corresponds to A, B, C and D.

Therefore, the correct option is A.