Question

1. which of the following compounds exhibits only dispersion and dipole-dipole intermolecular forces?

a. nh₃
b. n₂
c. co₂
d. ccl₄
e. pf₃

Explanation:

Step1: Analyze each compound's polarity and intermolecular forces

• Option A: $\ce{NH3}$

$\ce{NH3}$ has a trigonal pyramidal shape, is polar, and has hydrogen bonding (due to $\ce{N - H}$ bonds) in addition to dispersion and dipole - dipole forces. So it's not just dispersion and dipole - dipole.

• Option B: $\ce{N2}$

$\ce{N2}$ is a diatomic molecule with a non - polar covalent bond. It only has dispersion forces (London dispersion forces), no dipole - dipole forces since the molecule is non - polar.

• Option C: $\ce{CO2}$

$\ce{CO2}$ has a linear shape with a symmetric distribution of electron density (the two $\ce{C = O}$ bonds are polar but cancel each other out). So $\ce{CO2}$ is non - polar and only has dispersion forces, no dipole - dipole forces.

• Option D: $\ce{CCl4}$

$\ce{CCl4}$ has a tetrahedral shape. The $\ce{C - Cl}$ bonds are polar, but the symmetric tetrahedral geometry makes the molecule non - polar. So it only has dispersion forces, no dipole - dipole forces.

• Option E: $\ce{PF3}$

$\ce{PF3}$ has a trigonal pyramidal shape (similar to $\ce{NH3}$ but with $\ce{P}$ instead of $\ce{N}$). The $\ce{P - F}$ bonds are polar, and the molecular geometry is asymmetric, so the molecule is polar. It does not have hydrogen bonding (since there is no $\ce{H}$ bonded to a highly electronegative atom like $\ce{N}$, $\ce{O}$, or $\ce{F}$). So the intermolecular forces in $\ce{PF3}$ are dispersion forces and dipole - dipole forces.

Answer:

E. $\ce{PF3}$