Question

Using the VSEPR theory, predict the molecular structure of each of the following molecules. a. \(\mathrm{NCl}_{3}\) b. \(\mathrm{H}_{2} \mathrm{Se}\) c. \(\mathrm{SiCl}_{4}\)

Short answer

The molecular structures of the given molecules are: a. \(\mathrm{NCl}_{3}\): Trigonal pyramidal b. \(\mathrm{H}_{2} \mathrm{Se}\): Bent c. \(\mathrm{SiCl}_{4}\): Tetrahedral

Step-by-step solution

Determine the central atom

For each molecule, we need to identify the central atom, which is the atom with the lowest electronegativity (excluding hydrogen). The central atom is surrounded by all other atoms in the molecule. a. In \(\mathrm{NCl}_{3}\), N is the central atom. b. In \(\mathrm{H}_{2} \mathrm{Se}\), Se is the central atom. c. In \(\mathrm{SiCl}_{4}\), Si is the central atom.

Draw the Lewis structure

For each molecule, draw the Lewis structure to determine the arrangement of electron pairs around the central atom. a. \(\mathrm{NCl}_{3}\): Nitrogen has 5 valence electrons, and each chlorine has 7 valence electrons. The Lewis structure looks like this: N is connected to each Cl by a single bond, with one lone pair on N and three lone pairs on each Cl. b. \(\mathrm{H}_{2} \mathrm{Se}\): Selenium has 6 valence electrons and each hydrogen has 1 valence electron. The Lewis structure consists of Se connected to each hydrogen atom by a single bond, with two lone pairs on Se. c. \(\mathrm{SiCl}_{4}\): Silicon has 4 valence electrons, and each chlorine has 7 valence electrons. The Lewis structure consists of Si connected to each Cl by a single bond, with no lone pairs on Si and three lone pairs on each Cl.

Determine the number of bonding and lone pairs around the central atom

a. \(\mathrm{NCl}_{3}\): Nitrogen has 3 bonding pairs (with chlorine atoms) and 1 lone pair. b. \(\mathrm{H}_{2} \mathrm{Se}\): Selenium has 2 bonding pairs (with hydrogen atoms) and 2 lone pairs. c. \(\mathrm{SiCl}_{4}\): Silicon has 4 bonding pairs (with chlorine atoms) and no lone pairs.

Predict the molecular geometry using VSEPR theory

a. \(\mathrm{NCl}_{3}\): 3 bonding pairs and 1 lone pair around nitrogen result in a trigonal pyramidal molecular geometry. b. \(\mathrm{H}_{2} \mathrm{Se}\): 2 bonding pairs and 2 lone pairs around selenium result in a bent molecular geometry. c. \(\mathrm{SiCl}_{4}\): 4 bonding pairs and no lone pairs around silicon result in a tetrahedral molecular geometry. In conclusion, the molecular structures of the given molecules are: a. \(\mathrm{NCl}_{3}\): Trigonal pyramidal b. \(\mathrm{H}_{2} \mathrm{Se}\): Bent c. \(\mathrm{SiCl}_{4}\): Tetrahedral