Question

Predict the geometry of the following species: (a) \(\mathrm{O}_{3}\) (b) \(\mathrm{OCl}_{2}\) (c) \(\mathrm{SnCl}_{3}^{-}\) (d) \(\mathrm{CS}_{2}\)

Short answer

Question: Predict the molecular geometry in each case using VSEPR theory: (a) \(\mathrm{O}_{3}\), (b) \(\mathrm{OCl}_{2}\), (c) \(\mathrm{SnCl}_{3}^{-}\), (d) \(\mathrm{CS}_{2}\). Answer: (a) In \(\mathrm{O}_{3}\), the molecular geometry is bent or V-shaped. (b) In \(\mathrm{OCl}_{2}\), the molecular geometry is bent or V-shaped. (c) In \(\mathrm{SnCl}_{3}^{-}\), the molecular geometry is trigonal pyramidal. (d) In \(\mathrm{CS}_{2}\), the molecular geometry is linear.

Step-by-step solution

(a) Determine the Central Atom, Bond Pairs, and Lone Pairs for \(\mathrm{O}_{3}\)

The central atom in \(\mathrm{O}_{3}\) (ozone) is Oxygen (O), which is bonded to the other two Oxygen atoms. The central Oxygen atom has one double bond with one Oxygen atom and one single bond with the other, making two bond pairs and one lone pair (since oxygen typically forms two bonds and has two lone pairs).

(a) Predict the geometry for \(\mathrm{O}_{3}\)

With two bond pairs and one lone pair, the central Oxygen atom has three electron groups repelling each other. Hence, the shape will be bent or V-shaped based on a trigonal planar electron-pair geometry.

(b) Determine the Central Atom, Bond Pairs, and Lone Pairs for \(\mathrm{OCl}_{2}\)

In \(\mathrm{OCl}_{2}\) (oxygen dichloride), the central atom is Oxygen (O) with two chlorine atoms bonded to it. Since Oxygen typically forms two bonds and has two lone pairs, the central Oxygen atom has two bond pairs (with Cl) and two lone pairs.

(b) Predict the geometry for \(\mathrm{OCl}_{2}\)

With two bond pairs and two lone pairs, the central Oxygen atom has four electron groups repelling each other. The molecular geometry will be bent or V-shaped, with an electron-pair geometry of tetrahedral.

(c) Determine the Central Atom, Bond Pairs, and Lone Pairs for \(\mathrm{SnCl}_{3}^{-}\)

In \(\mathrm{SnCl}_3^-\) (tin trichloride ion), the central atom is tin (Sn) with three chlorine atoms bonded to it. Tin, being part of Group 14, typically has 4 electrons in its valence shell. However, in this ion, only three electrons are bonded to chlorine atoms, leaving one lone pair and resulting in three bond pairs and one lone pair.

(c) Predict the geometry for \(\mathrm{SnCl}_{3}^{-}\)

With three bond pairs and one lone pair, the central tin atom has four electron groups repelling each other. The molecular geometry will be trigonal pyramidal, with an electron-pair geometry of tetrahedral.

(d) Determine the Central Atom, Bond Pairs, and Lone Pairs for \(\mathrm{CS}_{2}\)

In \(\mathrm{CS}_{2}\) (carbon disulfide), the central atom is carbon (C) with two sulfur atoms bonded to it. Carbon, being part of group 14, typically forms four bonds and has no lone pairs. In this molecule, carbon forms two double bonds with the two sulfur atoms, resulting in two bond pairs and no lone pairs on the central atom.

(d) Predict the geometry for \(\mathrm{CS}_{2}\)

With two bond pairs and no lone pairs, the central carbon atom has two electron groups repelling each other. The molecular geometry will be linear, with an electron-pair geometry of linear as well.