Question

A neutral molecule \(XF_{3}\) has a zero dipole moment. The element X is most likely:

• A. chlorine
• B. boron
• C. nitrogen
• D. carbon
• E. bromine

Hint:

Zero dipole moment in \(AX_n\) molecules usually implies the central atom has no lone pairs (e.g., \(BF_3, CH_4, PCl_5, SF_6\)). If there are lone pairs, the symmetry is broken (unless it's a specific case like \(XeF_4\)), resulting in a net dipole.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The dipole moment of a molecule is the vector sum of its individual bond dipoles.
For a molecule with polar bonds (like $X-F$) to have a net dipole moment of exactly zero, its geometry must be perfectly symmetrical so that the individual bond dipole vectors cancel each other out completely.
According to VSEPR theory, an $AB_3$ type molecule is symmetrical and non-polar only if it has a trigonal planar geometry, meaning there are no lone electron pairs on the central atom.
Step 2: Key Formula or Approach:
1. Determine the number of valence electrons for the proposed central atom $X$.
2. Determine the steric number (Number of bonding pairs + Number of lone pairs) for the $XF_3$ molecule.
3. Predict the molecular geometry using VSEPR theory.
4. Assess if that geometry allows for the cancellation of dipole moments.
Step 3: Detailed Explanation:
Let's evaluate each option to see which central atom leads to a trigonal planar geometry with zero lone pairs.
(a) Chlorine (Group 17): Has 7 valence electrons. In $ClF_3$, there are 3 bonding pairs and $(7-3)/2 = 2$ lone pairs. Steric number = 5 ($sp^3d$). The geometry is T-shaped. This is an asymmetrical shape, so $ClF_3$ has a non-zero dipole moment.
(b) Boron (Group 13): Has 3 valence electrons. In $BF_3$, it shares all 3 electrons with fluorine, resulting in 3 bonding pairs and 0 lone pairs. Steric number = 3 ($sp^2$). The geometry is perfectly trigonal planar. Because all three $B-F$ bonds are identical and arranged symmetrically at 120-degree angles, their dipole moments cancel out exactly. Net dipole moment is zero.
(c) Nitrogen (Group 15): Has 5 valence electrons. In $NF_3$, there are 3 bonding pairs and $(5-3)/2 = 1$ lone pair. Steric number = 4 ($sp^3$). The geometry is trigonal pyramidal. Because of the lone pair at the apex, the bond dipoles do not cancel out. Net dipole moment is non-zero.
(d) Carbon (Group 14): Has 4 valence electrons. A neutral $CF_3$ would be a radical species with an unpaired electron, not a stable, standard neutral molecule in this context. It generally forms stable molecules with 4 bonds like $CF_4$.
(e) Bromine (Group 17): Similar to chlorine, $BrF_3$ has 3 bonding pairs and 2 lone pairs, resulting in a T-shaped geometry and a non-zero dipole moment.
Step 4: Final Answer:
The element X is most likely Boron.