Identify the pair in which the geometry of the species is T-shape and square pyramidal, respectively:

To decide the geometry of every species we shall apply the VSEPR (Valence-Shell Electron-Pair Repulsion) model, which states that the arrangement of electron-pairs around the central atom is the one that minimises repulsions. The steric number, SN = (number of sigma bonds) + (number of lone pairs), tells us the basic electron-pair geometry, after which the positions of the lone pairs decide the final molecular shape.

Option A ICl$$\_2^-$$ and ICl$$\_5$$

ICl$$\_2^-$$ : The central iodine has two sigma bonds and three lone pairs, so $$\text{SN}=2+3=5$$, giving a trigonal-bipyramidal electron skeleton. Placing the three lone pairs in the equatorial belt leaves the two Cl atoms axial, producing a linear, not T-shaped, molecule.

ICl$$\_5$$ : Iodine has five sigma bonds and one lone pair, so $$\text{SN}=5+1=6$$, i.e. an octahedral skeleton. One lone pair occupies one position, leaving a square pyramidal shape. Thus the first member is not T-shaped, so the pair fails.

Option B IO$$\_3^-$$ and IO$$\_2$$F$$\_2^-$$

IO$$\_3^-$$ : Three sigma bonds + one lone pair ⇒ $$\text{SN}=4$$, giving a tetrahedral skeleton and a trigonal-pyramidal shape.

IO$$\_2$$F$$\_2^-$$ : Four sigma bonds + one lone pair ⇒ $$\text{SN}=5$$, giving a trigonal-bipyramidal skeleton. With one lone pair, the shape is see-saw, not square pyramidal. Hence option B is wrong.

Option C ClF$$\_3$$ and IO$$\_4^-$$

ClF$$\_3$$ : Three sigma bonds + two lone pairs ⇒ $$\text{SN}=5$$, electron skeleton trigonal-bipyramidal. The two lone pairs occupy equatorial positions, so the three F atoms form a T-shape. Good so far.

IO$$\_4^-$$ : Four sigma bonds + zero lone pairs ⇒ $$\text{SN}=4$$, giving a tetrahedral shape, not square pyramidal. Therefore option C is eliminated.

Option D XeOF$$\_2$$ and XeOF$$\_4$$

We now examine both species carefully.

XeOF$$\_2$$

• Valence electrons on Xe = 8. • One O atom and two F atoms each form one sigma bond with Xe, so there are three sigma bonds. • The total steric number around Xe is obtained from $$\text{SN} \;=\; \frac{\text{valence electrons of Xe} + \text{number of monovalent atoms} - \text{charge}}{2} = \frac{8 + 3 - 0}{2} = 5.$$ (Here O is counted once because it provides one sigma bond.) • With $$\text{SN}=5$$ the electron-pair geometry is trigonal bipyramidal. We still have $$5 - 3 = 2$$ lone pairs to place. According to VSEPR, lone pairs prefer equatorial sites to minimise 90° repulsions, so both lone pairs go equatorial. The remaining three bonded atoms (O, F, F) occupy two axial and one equatorial position which produces a right-angled arrangement, i.e. a T-shape. So XeOF$$\_2$$ is T-shaped.

XeOF$$\_4$$

• Valence electrons on Xe = 8. • One O atom + four F atoms give five sigma bonds. • The steric number is $$\text{SN} = \frac{8 + 5 - 0}{2} = 6.$$ • $$\text{SN}=6$$ corresponds to an octahedral electron skeleton. The total lone pairs are $$6 - 5 = 1.$$ • Placing the single lone pair in any one of the six positions of an octahedron leaves a square base of four ligands with the fifth ligand occupying the position trans to the lone pair. This yields a square pyramidal molecular geometry. Hence XeOF$$\_4$$ is square pyramidal.

Thus, in option D the first species (XeOF$$\_2$$) is T-shaped and the second species (XeOF$$\_4$$) is square pyramidal. No other option satisfies the required pair of shapes.

Hence, the correct answer is Option D.