Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct order of dipole moments for the given species is

The concept of dipole moment is one of the most fascinating intersections of chemistry and vector physics. It tells us not just what atoms are in a molecule, but how they share their electrons in three-dimensional space. In this problem, we are asked to compare the dipole moments of four very common, yet structurally distinct species: ${\text{BF}}_3$, ${\text{NH}}_4^{+}$, ${\text{NF}}_3$, and ${\text{NH}}_3$.

To solve this, we need to visualize each molecule, draw its individual bond dipoles, and then perform a mental vector addition. Let's dive in!

Let's start with Boron trifluoride (${\text{BF}}_3$). Boron is in group 13 and has three valence electrons, all of which form single bonds with highly electronegative fluorine atoms. According to VSEPR theory, these three electron domains will spread out as far as possible, resulting in a trigonal planar geometry with bond angles of exactly $120^{\circ }$.

Now, the B-F bond is definitely polar; fluorine is pulling electron density away from boron. However, dipole moment is a vector quantity. Because the molecule is perfectly flat and symmetrical, these three identical vectors pull in exactly opposite directions and completely cancel each other out. Imagine three equally strong people pulling a ring from three equally spaced directions—the ring won't move. Therefore, the net dipole moment of ${\text{BF}}_3$ is exactly zero ($\mu =0$).

Next, consider the ammonium ion (${\text{NH}}_4^{+}$). Nitrogen forms four single bonds with hydrogen atoms. This gives it a perfect tetrahedral geometry with bond angles of $109.5^{\circ }$. Just like in ${\text{BF}}_3$, the high degree of symmetry means that the four individual N-H bond dipoles perfectly balance each other out in 3D space. Thus, its net dipole moment is also zero ($\mu =0$).

So, right off the bat, we know that ${\text{BF}}_3={\text{NH}}_4^{+}$.

This is where the problem gets really interesting. Both Nitrogen trifluoride (${\text{NF}}_3$) and Ammonia (${\text{NH}}_3$) have a central nitrogen atom bonded to three other atoms, plus one lone pair of electrons. This gives both of them a trigonal pyramidal shape. Since they have the same shape, you might think their dipole moments would be similar. But they are drastically different! Why?

It all comes down to the direction of the vectors. In any molecule with a lone pair, the lone pair itself contributes to the overall dipole moment (called the orbital dipole), and it always points away from the central atom.

In ${\text{NF}}_3$, fluorine is much more electronegative than nitrogen. This means the three N-F bond dipoles point downwards, away from the nitrogen and towards the fluorines. However, the lone pair dipole points upwards. Because the bond dipoles and the lone pair dipole are pointing in opposite directions, they partially cancel each other out. This results in a very small net dipole moment ($\mu \approx 0.8\times 10^{-30}\text{ Cm}$).

In ${\text{NH}}_3$, the situation is reversed. Nitrogen is more electronegative than hydrogen. Therefore, the three N-H bond dipoles point upwards, towards the nitrogen. The lone pair dipole also points upwards. Since all the vectors are pointing in the same general direction, they reinforce each other! They add up to create a much larger net dipole moment ($\mu \approx 4.9\times 10^{-30}\text{ Cm}$).

By analyzing the geometry and vector addition for each species, we have our answer: 1. ${\text{BF}}_3$ and ${\text{NH}}_4^{+}$ have perfectly symmetrical geometries, so their dipole moments are zero. 2. ${\text{NF}}_3$ has opposing vectors, leading to a small dipole moment. 3. ${\text{NH}}_3$ has reinforcing vectors, leading to a large dipole moment.

Putting it all together in increasing order, we get: ${\text{BF}}_3={\text{NH}}_4^{+}<{\text{NF}}_3<{\text{NH}}_3$

This perfectly matches option (A). This problem is a beautiful reminder that in chemistry, shape and direction matter just as much as the atoms themselves!