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The set containing all linear polyhalogen compounds is :

  • A.\(\text{I}_{3}^{-}\), \(\text{Cl}_{3}^{-}\), \(\text{BrCl}_{2}^{-}\)
  • B.\(\text{ICl}_{2}^{-}\), \(\text{I}_{3}^{+}\), \(\text{Cl}_{3}^{+}\)
  • C.\(\text{I}_{3}^{+}\), \(\text{Cl}_{3}^{+}\), \(\text{BrCl}_{2}^{-}\)
  • D.\(\text{Cl}_{3}^{-}\), \(\text{BrCl}_{2}^{+}\), \(\text{ICl}_{2}^{-}\)

Answer: A

The all-linear set is \(\text{I}_{3}^{-}\), \(\text{Cl}_{3}^{-}\), \(\text{BrCl}_{2}^{-}\).

Each of these trihalide anions has a central halogen with two bonding pairs and three lone pairs (AX2E3). The three lone pairs occupy the equatorial positions of a trigonal bipyramid, so the two bonded atoms sit on the axis and the ion is linear. ICl2- is linear for the same reason.

The trihalogen cations I3+, Cl3+ and BrCl2+ have one electron pair fewer on the central atom: two bonding pairs and two lone pairs (AX2E2), which makes them bent, like water. Every other option contains at least one of these bent cations.

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