Indicate the hybridization of the central atom in (a) \(\mathrm{BCl}_{3}\), (b) \(\mathrm{AlCl}_{4}^{-}\) (c) \(\mathrm{CS}_{2}\), (d) \(\mathrm{GeH}_{4}\).

In each given compound, the first atom is the central atom: (a) BCl3: central atom - B (Boron) (b) AlCl4-: central atom - Al (Aluminum) (c) CS2: central atom - C (Carbon) (d) GeH4: central atom - Ge (Germanium)

(a) B - 3 valence electrons (b) Al - 3 valence electrons (c) C - 4 valence electrons (d) Ge - 4 valence electrons

To find the hybridization, add the number of bonding electrons and lone pairs around the central atom, then use the formula: \(Hybridization = \frac{1}{2} \times [Valence \: electrons \: of \: central \: atom + atoms \: bonded \: to \: central \: atom - charge]\) (a) BCl3: B has 3 valence electrons, 3 atoms bonded to it (3 Cl), and no charge. Hybridization for BCl3 = \(\frac{1}{2} \times [3 + 3 - 0]\) = \(\frac{1}{2} \times [6]\) = 3, so it's sp² hybridization. (b) AlCl4-: Al has 3 valence electrons, 4 atoms bonded to it (4 Cl), and a charge of -1. Hybridization for AlCl4- = \(\frac{1}{2} \times [3 + 4 - (-1)]\) = \(\frac{1}{2} \times [8]\) = 4, so it's sp³ hybridization. (c) CS2: C has 4 valence electrons, 2 atoms bonded to it (2 S), and no charge. Hybridization for CS2 = \(\frac{1}{2} \times [4 + 2 - 0]\) = \(\frac{1}{2} \times [6]\) = 3, so it's sp² hybridization. (d) GeH4: Ge has 4 valence electrons, 4 atoms bonded to it (4 H), and no charge. Hybridization for GeH4 = \(\frac{1}{2} \times [4 + 4 - 0]\) = \(\frac{1}{2} \times [8]\) = 4, so it's sp³ hybridization.

(a) In BCl3, the central atom B is sp² hybridized. (b) In AlCl4-, the central atom Al is sp³ hybridized. (c) In CS2, the central atom C is sp² hybridized. (d) In GeH4, the central atom Ge is sp³ hybridized.