Question

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

Short answer

The hybridizations of the central atoms in the given compounds are as follows: (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.

Step-by-step solution

Identify the central atom in each compound.

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)

Calculate the number of valence electrons for each central atom.

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

Determine the number of bonding electrons and lone pairs around each central atom and find the hybridization.

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.

Write the final answer.

(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.

Key concepts

Valence Electrons

Valence electrons are the electrons present in the outermost shell of an atom and are crucial for understanding chemical bonding. They are responsible for the formation of chemical bonds since they can be gained, lost, or shared by atoms. For each atom, the number of valence electrons can be determined from its group number in the periodic table.

For instance:

• Boron (B) in group 13 has 3 valence electrons.
• Aluminum (Al) in group 13 also has 3 valence electrons.
• Carbon (C) in group 14 has 4 valence electrons.
• Germanium (Ge) in group 14 has 4 valence electrons.

Understanding the number of valence electrons helps in predicting the type and number of bonds an atom can form, which is essential for determining its hybridization state.

Central Atom

The central atom in a molecule is typically the one that can form the most bonds. It is usually the atom that has the least valence electrons but can form a variety of bonds with surrounding atoms. Knowing which atom serves as the central atom is key to understanding the molecule's geometry and hybridization.

In the compounds:

• BCl₃: The central atom is Boron (B), which forms three bonds with chlorine atoms.
• AlCl₄^-: The central atom is Aluminum (Al), bonding with four chlorine atoms.
• CS₂: The central atom is Carbon (C), bonding with two sulfur atoms.
• GeH₄: The central atom is Germanium (Ge), bonding with four hydrogen atoms.

By identifying the central atom, we can then calculate hybridization and predict molecular shapes.

sp² Hybridization

sp² Hybridization occurs when a central atom forms three bonds and involves mixing one s orbital and two p d orbitals from the same atom to form three equivalent hybrid orbitals. This type of hybridization allows the formation of trigonal planar geometry with bond angles of approximately 120 degrees.

Let's look at examples where sp² hybridization is observed:

• In BCl₃, Boron (B) with three valence electrons forms three bonds with chlorine atoms, adopting a trigonal planar structure typical of sp² hybridization.
• In CS₂, Carbon (C) forms two double bonds with two sulfur atoms, leading to a linear structure. Though linear in nature, carbon adopts sp hybridization as it utilizes two π bonds out of its available electrons. However, typically for three areas of electron density, sp² would be used if steric concerns demanded it.

Whenever a molecule has three regions of electron density, it most likely undergoes sp² hybridization.

sp³ Hybridization

sp³ Hybridization describes a scenario where one s orbital and three p orbitals from the same atom combine to form four equivalent hybrid orbitals. It allows the central atom to form four bonds, often resulting in a tetrahedral geometry with bond angles of around 109.5 degrees.

Examples of sp³ hybridization include:

• In AlCl₄^-, Aluminum (Al), with an extra electron due to its charge, forms four bonds with chlorine atoms, adopting a tetrahedral geometry and thus, sp³ hybridization.
• In GeH₄, Germanium (Ge) forms four bonds with hydrogen atoms, each adopting a tetrahedral structure, indicative of sp³ hybridization.

Compounds with a central atom bonded to four atoms typically utilize sp³ hybridization, demonstrating a common occurrence in larger molecules with single bonds.