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 hybridization of the central atom in each molecule is: (a) BCl3: sp^2, (b) AlCl4-: sp^3, (c) CS2: sp, and (d) GeH4: sp^3.

Step-by-step solution

Determine the Lewis structure of each molecule.

For each molecule, first draw the Lewis structure. This will help in identifying the central atom and determining the number of electron pairs surrounding it. (a) BCl3: B (3 valence electrons) and Cl (7 valence electrons) The Lewis structure is B surrounded by three Cl atoms, with each B-Cl bond having a single bond and Cl atoms having three lone pairs. (b) AlCl4-: Al (3 valence electrons) and Cl (7 valence electrons), with an additional electron due to the negative charge The Lewis structure is Al surrounded by four Cl atoms, with each Al-Cl bond having a single bond and Cl atoms having three lone pairs. (c) CS2: C (4 valence electrons) and S (6 valence electrons) The Lewis structure is C bonded with two S atoms, with each C-S bond having a double bond. (d) GeH4: Ge (4 valence electrons) and H (1 valence electron) The Lewis structure is Ge surrounded by four H atoms, with each Ge-H bond having a single bond.

Determine the total number of electron groups around the central atom.

Counting the number of electron groups is essential for determining the hybridization. Each bond and lone pair counts as one electron group. (a) BCl3: 3 bonds (B-Cl) (b) AlCl4-: 4 bonds (Al-Cl) (c) CS2: 2 bonds (C-S) (d) GeH4: 4 bonds (Ge-H)

Determine the hybridization of the central atom.

Based on the number of electron groups surrounding the central atom, determine the hybridization. (a) BCl3: 3 electron groups → sp^2 hybridization (b) AlCl4-: 4 electron groups → sp^3 hybridization (c) CS2: 2 electron groups → sp hybridization (d) GeH4: 4 electron groups → sp^3 hybridization So, the hybridization of the central atom in each molecule is as follows: (a) BCl3: sp^2 (b) AlCl4-: sp^3 (c) CS2: sp (d) GeH4: sp^3

Key concepts

Lewis Structures

Lewis structures are crucial for visualizing molecules. They show the arrangement of atoms and the distribution of electrons in a molecule. This helps determine the molecular geometry and electron configuration. For each molecule, you start by counting the total number of valence electrons. Consider each atom's electrons and the overall charge of the molecule. It’s key to draw these structures properly.

• BCl extsubscript{3}: Boron (B) has 3 valence electrons. Each Chlorine (Cl) has 7 valence electrons. There are no extra electrons from a charge, so we arrange B in the center bonded to 3 Cls, completing the Lewis structure with 3 single bonds.


• AlCl extsubscript{4} extsuperscript{-}: Aluminum (Al) has 3 valence electrons. Each Chlorine (Cl) has 7 valence electrons. Due to the negative charge, add one more electron, forming a structure where Al bonds with 4 Cls, using single bonds.


• CS extsubscript{2}: Carbon (C) holds 4 valence electrons. Each Sulphur (S) brings 6 valence electrons. Here, C is in the center with double bonds to each S, reflecting the need for every atom to gain a full outer shell.


• GeH extsubscript{4}: Germanium (Ge) comes with 4 valence electrons. Hydrogen (H) supplies 1 electron each. Therefore, Ge shares single bonds with all 4 H atoms, completing the Lewis structure.

Ultimately, Lewis structures set the groundwork for understanding molecule shape.

Electron Groups

Electron groups are collective entities surrounding an atom, comprising bonds and lone pairs. They dictate molecular shape, as each group repels others, influencing geometry and hybridization.

• In BCl extsubscript{3}, there are three groups—a single bond with each Cl. No lone pairs on B means three electron groups.


• For AlCl extsubscript{4} extsuperscript{-}, four single bonds exist due to Al's connections with Cls. These four bonds count as four electron groups.


• CS extsubscript{2} uses two double bonds between C and each S atom, making two distinct electron groups.


• In GeH extsubscript{4}, there are four single bonds between Ge and each H, equating to four electron groups.

Grouping electrons in this way helps to predict hybridization. The total number of groups guides which hybridization state is most suitable.

Molecular Geometry

Molecular geometry pertains to the three-dimensional shape of a molecule and is determined by the spatial arrangement of electron groups around the central atom. This geometry influences physical and chemical properties.
For BCl extsubscript{3}, with three electron groups, it adopts a trigonal planar shape. This allows the groups to be as far apart as possible.
In AlCl extsubscript{4} extsuperscript{-}, the four electron groups lead to a tetrahedral geometry, ensuring optimal spacing.
With CS extsubscript{2}, having two electron groups results in a linear geometry, as these groups align in opposite directions.
In GeH extsubscript{4}, the four bonds push for a tetrahedral shape, similar to AlCl4-.

• The geometry can be predicted more accurately using the VSEPR theory, which states electron groups repel each other, trying to maintain maximum distance.
• This variation affects intermolecular forces and reactivity in chemical interactions.

Understanding molecular geometry provides insight into a compound's potential interactions and behaviors.