To determine if a pair of compounds is isoelectronic and isostructural, we need to analyze their total number of valence electrons and their molecular geometry, respectively.

Concept:

• Isoelectronic: Species having the same number of electrons.
• Isostructural: Species having the same molecular geometry and hybridization.

Why (C) is correct:

Let's analyze option (C) ${\mathrm{IBr}}_2^{-},{\mathrm{XeF}}_2$.

1. Isoelectronic Check:

• For ${\mathrm{IBr}}_2^{-}$: Valence electrons = 7 (I) + 2 \times 7 (Br) + 1 (charge) = 7 + 14 + 1 = 22 valence electrons.
• For ${\mathrm{XeF}}_2$: Valence electrons = 8 (Xe) + 2 \times 7 (F) = 8 + 14 = 22 valence electrons.

Since both have 22 valence electrons, they are isoelectronic.

2. Isostructural Check (using VSEPR theory):

• For ${\mathrm{IBr}}_2^{-}$:
  Central atom: I
  Number of bond pairs = 2 (with Br)
  Number of lone pairs = (Total valence electrons - electrons used in bonding) / 2 = (22 - 4) / 2 = 18 / 2 = 9. This is incorrect. Let's use the steric number method.
  Steric number = (Number of surrounding atoms) + (Number of lone pairs)
  Number of valence electrons on I = 7. Add 1 for the negative charge = 8.
  Number of electrons used in bonding with 2 Br atoms = 2.
  Number of lone pair electrons = 8 - 2 = 6.
  Number of lone pairs = 6 / 2 = 3.
  Steric number = 2 (bond pairs) + 3 (lone pairs) = 5.
  Hybridization = ${\mathrm{sp}}^3\mathrm{d}$.
  Geometry (electron domain) = Trigonal bipyramidal.
  Molecular geometry (due to 3 lone pairs in equatorial positions) = Linear.
• For ${\mathrm{XeF}}_2$:
  Central atom: Xe
  Number of valence electrons on Xe = 8.
  Number of electrons used in bonding with 2 F atoms = 2.
  Number of lone pair electrons = 8 - 2 = 6.
  Number of lone pairs = 6 / 2 = 3.
  Steric number = 2 (bond pairs) + 3 (lone pairs) = 5.
  Hybridization = ${\mathrm{sp}}^3\mathrm{d}$.
  Geometry (electron domain) = Trigonal bipyramidal.
  Molecular geometry (due to 3 lone pairs in equatorial positions) = Linear.

Since both have a linear molecular geometry, they are isostructural.

Option Analysis:

• A) ${\mathrm{BeCl}}_2,{\mathrm{XeF}}_2$
  ${\mathrm{BeCl}}_2$: 2 (Be) + 2 \times 7 (Cl) = 16 valence electrons. Linear geometry.
  ${\mathrm{XeF}}_2$: 22 valence electrons. Linear geometry.
  Not isoelectronic (16 vs 22 electrons).
• B) ${\mathrm{TeI}}_2,{\mathrm{XeF}}_2$
  ${\mathrm{TeI}}_2$: 6 (Te) + 2 \times 7 (I) = 20 valence electrons. (Steric number = 2 bond pairs + 2 lone pairs = 4, Bent geometry).
  ${\mathrm{XeF}}_2$: 22 valence electrons. Linear geometry.
  Not isoelectronic (20 vs 22 electrons) and not isostructural.
• D) ${\mathrm{IF}}_3,{\mathrm{XeF}}_2$
  ${\mathrm{IF}}_3$: 7 (I) + 3 \times 7 (F) = 28 valence electrons. (Steric number = 3 bond pairs + 2 lone pairs = 5, T-shaped geometry).
  ${\mathrm{XeF}}_2$: 22 valence electrons. Linear geometry.
  Not isoelectronic (28 vs 22 electrons) and not isostructural.

Correct Answer: (C)