Among the following, the CORRECT combinations are
$$\text{A. IF}_{3} \rightarrow \text{T-shaped}(\text{sp}^{3}\text{d})$$
$$\text{B. IF}_{5} \rightarrow \text{Square pyramidal(sp}^{3}\text{d}^{2})$$
$$\text{C. IF}_{7} \rightarrow \text{Pentagonal bipyrnmidal(sp}_{3}\text{d}^{3})$$
$$\text{D. ClO}_{4}^{-} \rightarrow \text{Square planar(sp}^{3}\text{d})$$
Choose the correct answer from the options given below:

We need to identify the correct combinations of molecule/ion, shape, and hybridisation.

A. $$IF_3$$ → T-shaped ($$sp^3d$$)

I has 7 valence electrons, bonds with 3 F atoms, leaving 2 lone pairs.

Electron geometry: trigonal bipyramidal ($$sp^3d$$), molecular geometry: T-shaped. CORRECT ✓

B. $$IF_5$$ → Square pyramidal ($$sp^3d^2$$)

I has 7 valence electrons, bonds with 5 F atoms, leaving 1 lone pair.

Electron geometry: octahedral ($$sp^3d^2$$), molecular geometry: square pyramidal. CORRECT ✓

C. $$IF_7$$ → Pentagonal bipyramidal ($$sp^3d^3$$)

I has 7 valence electrons, bonds with 7 F atoms, no lone pairs.

Hybridisation: $$sp^3d^3$$, geometry: pentagonal bipyramidal. CORRECT ✓

D. $$ClO_4^-$$ → Square planar ($$sp^3d$$)

Cl has 7 valence electrons + 1 from charge = 8. It bonds with 4 oxygen atoms.

$$ClO_4^-$$ has a tetrahedral geometry with $$sp^3$$ hybridisation, not square planar with $$sp^3d$$. INCORRECT ✗

The correct combinations are A, B and C only, which corresponds to Option 2.