# Which of these are expected to be the most polarizable?

## A) Which of the following atoms would you expect to be most polarizable: N,P,As,Sb? I wrote down what I thought was correct, but I am a bit unsure if I am actually correct. I said the most polarizable would be Sb. Next would be As, P, and N. Am I correct to say that polarizability decreases across a period and increases down a group? The way I understood was that the more electrons there are in an atom, the more polarizable the atom. B) Place the molecules in order of decreasing polarizability: $G e C {l}_{4}$, $C {h}_{4}$, $S i C {l}_{4}$, $S i {H}_{4}$, and $G e B {r}_{4}$ When writing the answer, wouldn't I have to find the molar mass and list them in order of smallest molar mass to greatest molar mass? I would highly appreciate it if someone could help to clarify this for me.

Jul 21, 2018

You may want to read here for another example.

The idea is that the atom that is the least electronegative AND the largest radius is the most polarizable. This happens to be the lower left of the periodic table.

• Smaller electronegativity means it does not want to pull electrons towards itself as easily.

• Larger radius means it has less pull on those electrons anyway purely due to their distance away from the nucleus.

For molecules, you just want to consider the one that consists of the largest atoms. Secondarily, try to pick the one where each atom is as small electronegativity as possible.

A) $\text{Sb}$ is the most polarizable.

Its electron cloud is most easily distorted by another incoming atom, because it is largest and because it is least electronegative among the choices.

B) ${\text{GeBr}}_{4}$ should be the most polarizable for three reasons:

1. Out of all these molecules (${\text{GeCl}}_{4}$, ${\text{CH}}_{4}$, ${\text{SiCl}}_{4}$, ${\text{SiH}}_{4}$, ${\text{GeBr}}_{4}$), I would choose the one with the largest atoms overall, because their electrons stick out the most and thus are easiest to polarize.

2. $\text{Br}$ is the least electronegative among all of the second atoms ($\text{H", "Cl", "Br}$), and $\text{Ge}$ is the least electronegative among all the first atoms ($\text{Ge", "C", "Si}$).

3. Because both atoms are easily polarizable in ${\text{GeBr}}_{4}$, neither atom concentrates the electron density towards itself, which makes the electron cloud more loose.