1. Use VSEPR theory to determine shapes from the steric number:

   $\text{Steric number} = \text{bond pairs} + \text{lone pairs on central atom}$

2. We need the pair whose geometries are, respectively:

   • first species: $T$-shape
   • second species: square pyramidal

Option A: $ClF_3$ and $IO_4^-$

(i) $ClF_3$

• Central atom: $Cl$
• Valence electrons on $Cl = 7$
• It forms 3 bonds with $F$, leaving 2 lone pairs.
• So total electron pairs around $Cl = 5$

Thus electron pair geometry is trigonal bipyramidal, with 2 equatorial lone pairs. Hence molecular shape is:

$T\text{-shape}$

(ii) $IO_4^-$

• Central atom: $I$
• It is analogous to perchlorate-type tetraoxo species.
• Around iodine there are 4 bonded oxygens and no lone pair in the usual VSEPR description.

So geometry is:

$\text{tetrahedral}$

This is not square pyramidal.

❌ Option A is incorrect.

Option B: $ICl_2^-$ and $ICl_5$

(i) $ICl_2^-$

• Central atom: $I$
• Total electron pairs around iodine = 5 (2 bond pairs + 3 lone pairs)
• Electron pair geometry: trigonal bipyramidal
• Molecular shape:

$\text{linear}$

So first species is not $T$-shaped.

(ii) $ICl_5$

• 5 bond pairs and 1 lone pair around iodine
• Steric number = 6
• Electron pair geometry: octahedral
• Molecular shape:

$\text{square pyramidal}$

Second is correct, first is not.

❌ Option B is incorrect.

Option C: $XeOF_2$ and $XeOF_4$

(i) $XeOF_2$

• Central atom: $Xe$
• Xenon has 8 valence electrons.
• It forms 3 sigma bonds total: one with O and two with F.
• In VSEPR, the $Xe=O$ double bond counts as one electron domain.
• So total bonded domains = 3
• Remaining lone pairs on Xe = 2

Thus steric number:

$3 + 2 = 5$

Electron pair geometry: trigonal bipyramidal. With 2 lone pairs occupying equatorial positions, the molecular shape becomes:

$T\text{-shape}$

(ii) $XeOF_4$

• Central atom: $Xe$
• Bonded domains: 5 (one $Xe=O$ and four $Xe-F$ bonds)
• Lone pairs on Xe = 1

Thus steric number:

$5 + 1 = 6$

Electron pair geometry: octahedral. With one lone pair, molecular shape is:

$\text{square pyramidal}$

Both match the required shapes.

✅ Option C is correct.

Option D: $IO_3^-$ and $IO_2F_2^-$

(i) $IO_3^-$

• Central atom: $I$
• 3 bonded domains and 1 lone pair
• Steric number = 4
• Shape:

$\text{trigonal pyramidal}$

So it is not $T$-shaped.

(ii) $IO_2F_2^-$

• Around iodine there are 4 bonded atoms and likely 1 lone pair
• This would lead to trigonal bipyramidal electron arrangement and a seesaw-type/distorted structure, not square pyramidal.

❌ Option D is incorrect.

Final Answer

The required pair is:

$\boxed{\text{C: } XeOF_2 \text{ and } XeOF_4}$

This matches the stored correct answer.