The hybridization scheme for the central atom includes a \(d\) orbital contribution in (a) \(\mathrm{I}_{3}^{-} ;\) (b) \(\mathrm{PCl}_{3}\) (c) \(\mathrm{NO}_{3}^{-} ;\) (d) \(\mathrm{H}_{2}\) Se.

Short Answer

Expert verified
Only \( \mathrm{I}_{3}^{-} \) involves a d orbital in its hybridization scheme.

Step by step solution

01

Identify the Central Atom and Electron Pairs for \( \mathrm{I}_{3}^{-} \)

The central atom for \( \mathrm{I}_{3}^{-} \) is iodine (I), with two other iodine atoms and one lone pair of electrons. This gives it a total of 3 electron pairs.
02

Identify the Central Atom and Electron Pairs for \( \mathrm{PCl}_{3} \)

The central atom for \( \mathrm{PCl}_{3} \) is phosphorus (P), with three other chlorine atoms and one lone pair of electrons. This gives it a total of 4 electron pairs.
03

Identify the Central Atom and Electron Pairs for \( \mathrm{NO}_{3}^{-} \)

The central atom for \( \mathrm{NO}_{3}^{-} \) is nitrogen (N), with three other oxygen atoms. This gives it a total of 3 electron pairs.
04

Identify the Central Atom and Electron Pairs for \( \mathrm{H}_{2} Se \)

The central atom for \( \mathrm{H}_{2} Se \) is selenium (Se), with two hydrogen atoms and two pairs of lone electrons. This gives it a total of 4 electron pairs.
05

Identify Which Molecules Involve a d Orbital

Looking at the total number of electron pairs, the only molecule with more than four electron pairs is \( \mathrm{I}_{3}^{-} \). Therefore, \( \mathrm{I}_{3}^{-} \) involves a d orbital in its hybridization scheme.

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