Question

The correct order of decreasing polarisability of the ions is

• A. $Cl^-$, $Br^-$, $I^-$, $F^-$
• B. $F^-$, $I^-$, $Br^-$, $Cl^-$
• C. $F^-$, $Cl$, $Br^-$, $I^-$
• D. $I^-$, $Br^-$, $Cl^-$, $F^-$.

Answer 1

The Correct Option is D

To determine the order of decreasing polarizability of the given ions, we must first understand the concept of polarizability. Polarizability is the tendency of an electron cloud of an ion or atom to become distorted, which depends largely on the size of the ion. Larger ions have more diffuse electron clouds that are more easily distorted.

In this case, we are comparing the polarizability of halide ions: $Cl^-$, $Br^-$, $I^-$, and $F^-$. Let's analyze their properties:

1. $I^-$: Iodide ion has the largest size among these ions. Its electron cloud is the most easily distorted, making it highly polarizable.
2. $Br^-$: Bromide ion is smaller than iodide but larger than chloride and fluoride, which makes it less polarizable than $I^-$ but more than the others.
3. $Cl^-$: Chloride ion is smaller than iodide and bromide ions, leading to less polarizability compared to them.
4. $F^-$: Fluoride ion is the smallest ion among them, resulting in a highly compact electron cloud that is not easily distorted, hence it has the least polarizability.

Therefore, the correct order of decreasing polarizability is: $I^-$, $Br^-$, $Cl^-$, $F^-$.

Thus, the correct answer is the option: $I^-$, $Br^-$, $Cl^-$, $F^-$.

This understanding is in line with the fact that polarizability increases with an increase in ionic size and vice versa.