Question

Which of the following elements will have the largest atomic radii?

• A. $\mathrm{Cl}$
• B. $\mathrm{Br}$
• C. $\mathrm{K}$
• D. $\mathrm{Na}$

Answer 1

The correct option is C

$\mathrm{K}$

Explanation for the correct option:

(c ) $\mathrm{K}$

• Across the periodic table, atomic radii fluctuate predictably. The atomic radius grows from top to bottom in a group and diminishes from left to right across a period. As the atomic number increases over time, so does the nuclear charge, and therefore the nuclear force of attraction. As a result, atomic radius reduces over time.
• Among the elements stated above, $\mathrm{Na}$ and $\mathrm{Cl}$ correspond to the third phase. $\mathrm{K}$ and $\mathrm{Br}$, on the other hand, are from the fourth period.
• Because the third period is below the second period, it will have a bigger atomic radius than the second period.
• Furthermore, $\mathrm{K}$ is on the left side of the periodic table, while $\mathrm{Br}$ is on the right.
• Hence $\mathrm{K}$ has the largest atomic radius of the above-mentioned elements.
• Hence this option is correct

Explanation for the incorrect options:

(a) $\mathrm{Cl}$

• Among the elements stated above, $\mathrm{Na}$ and $\mathrm{Cl}$ correspond to the third phase. $\mathrm{K}$ and $\mathrm{Br}$, on the other hand, are from the fourth period.
• Fourth-period elements will have larger radii than third-period elements.
• Thus, this option is incorrect.

(b) $\mathrm{Br}$

• $\mathrm{K}$ and $\mathrm{Br}$, on the other hand, are from the fourth period.
• $\mathrm{K}$ is on the left side of the periodic table, while $\mathrm{Br}$ is on the right
• $\mathrm{Br}$ will have lesser radii than $\mathrm{K}$
• Thus, this option is incorrect.

(d) $\mathrm{Na}$

• Among the elements stated above, $\mathrm{Na}$ and $\mathrm{Cl}$ correspond to the third phase. $\mathrm{K}$ and $\mathrm{Br}$, on the other hand, are from the fourth period.
• Fourth-period elements will have larger radii than third-period elements.
• Thus, this option is incorrect.

Hence the correct option is (c ) $\mathrm{K}$