Question

Which of the following is least ionic?

• A. $\mathrm{KCl}$
• B. ${\mathrm{BaCl}}_2$
• C. $\mathrm{AgCl}$
• D. ${\mathrm{CaCl}}_2$

Answer 1

The correct option is C

$\mathrm{AgCl}$

The explanation for the correct option:

(c) $\mathrm{AgCl}$

1. The compound which are held together by the ionic bond is called ionic compounds.
2. They are formed by ions having opposite charges.
3. The are a crystalline solid which are usually formed when metal reacts with nonmetals.
4. Now, according to the fajan's rule, if the cation has a smaller size and the anion has a larger size, polarisation of the electron cloud increases which causes more overlap between anion and cation due to which covalent character of the ionic bond increases.
5. As we move from left to right across the periodic table, the size of the element decreases because of an increase in effective nuclear charge.
6. Since, $\mathrm{Ag}$ belongs to d block, and $\mathrm{K}$, $\mathrm{Ba}$, $\mathrm{Ca}$ belongs to s block, $\mathrm{Ag}$ will be the smallest metal among the others.
7. $\mathrm{Cl}$ when gains electron and makes anion, its size increases.
8. So, according to fajan's rule, $\mathrm{Ag}$ and $\mathrm{Cl}$ bond will be covalent in nature.
9. Therefore, $\mathrm{AgCl}$ will be least ionic.

The explanation for incorrect options:

(a) $\mathrm{KCl}$

1. Potassium ($\mathrm{K}$) belongs to group 1.
2. $\mathrm{K}$ is the largest among all.
3. So according to fajan's rule, $\mathrm{KCl}$ is the least covalent and most ionic compound.

(b) ${\mathrm{BaCl}}_2$

1. Barium ($\mathrm{Ba}$) belongs to group 2.
2. $\mathrm{Ba}$ is larger than $\mathrm{Ag}$ and $\mathrm{Ca}$
3. So, ${\mathrm{BaCl}}_2$ is more ionic than $\mathrm{AgCl}$ and ${\mathrm{CaCl}}_2$

(d) ${\mathrm{CaCl}}_2$

1. Calcium ($\mathrm{Ca}$) belongs to group 2.
2. $\mathrm{Ca}$is larger than $\mathrm{Ag}$
3. So, ${\mathrm{CaCl}}_2$is more ionic than $\mathrm{AgCl}$

Therefore, Option (c) is correct, $\mathrm{AgCl}$ will be least ionic.