Question

1. predict the molecular shape of ch3br (1 point) *

linear
bent
trigonal planar
trigonal pyramidal
tetrahedral

1. predict the molecular shape of ncl3 (1 point) *

linear
bent
trigonal planar
trigonal pyramidal
tetrahedral

Explanation:

Question 18 (Predict the molecular shape of \( \text{CH}_3\text{Br} \))

Step 1: Determine the central atom and bonding

The central atom in \( \text{CH}_3\text{Br} \) is carbon (\( \text{C} \)). Carbon has 4 valence electrons. It forms 4 single bonds: 3 with hydrogen (\( \text{H} \)) atoms and 1 with a bromine (\( \text{Br} \)) atom. So, the steric number (number of bonding pairs + lone pairs) for \( \text{C} \) is 4 (all bonding pairs, no lone pairs on \( \text{C} \)).

Step 2: Predict the molecular shape

For a central atom with a steric number of 4 and no lone pairs, the electron - pair geometry and molecular shape are tetrahedral. This is because the four bonding pairs of electrons around the carbon atom will arrange themselves in a way that maximizes the distance between them, which corresponds to a tetrahedral geometry.

Step 1: Determine the central atom and bonding

The central atom in \( \text{NCl}_3 \) is nitrogen (\( \text{N} \)). Nitrogen has 5 valence electrons. It forms 3 single bonds with chlorine (\( \text{Cl} \)) atoms. So, the number of bonding pairs is 3, and the number of lone pairs on \( \text{N} \) is \( 5 - 3= 1 \) (since each bond uses 1 electron from \( \text{N} \)). The steric number (bonding pairs + lone pairs) is \( 3 + 1=4 \).

Step 2: Predict the molecular shape

The electron - pair geometry for a steric number of 4 is tetrahedral. However, when there is 1 lone pair and 3 bonding pairs, the molecular shape is trigonal pyramidal. The lone pair of electrons on the nitrogen atom repels the bonding pairs, causing the three \( \text{N - Cl} \) bonds to form a trigonal pyramidal shape.

Answer:

E. Tetrahedral

Question 19 (Predict the molecular shape of \( \text{NCl}_3 \))