Question

Which of the following elements would lose an electron easily?

• A. $\mathrm{Mg}$
• B. $\mathrm{Na}$
• C. $\mathrm{K}$
• D. $\mathrm{Ca}$

Answer 1

The correct option is C

$\mathrm{K}$

Explanation for the correct option:

(c ) $\mathrm{K}$

• Ionization energy increases during the period and drops down the group.
• Among the elements stated above, $\mathrm{Na}$ and $\mathrm{Mg}$ correspond to the third period. $\mathrm{K}$ and $\mathrm{Ca}$, on the other hand, are from the fourth period.
• Because the fourth period is below the third period, it will have less ionization energy than the third period. Furthermore, $\mathrm{K}$ is to the left of $\mathrm{Ca}$.
• $\mathrm{K}$ has one electron in the outermost shell and the lowest ionization energy, therefore it can readily lose one electron.
• As a result, $\mathrm{K}$ possesses the ionization energy least amongst the elements specified above.
• Hence this option is correct

Explanation for the incorrect options:

(a) $\mathrm{Mg}$

• $\mathrm{Na}$ and $\mathrm{Mg}$ correspond to the third period.
• Thus they will have more ionization energy than fourth-period elements as they will face more nuclear charge
• Thus, this option is incorrect

(b) $\mathrm{Na}$

• $\mathrm{Na}$ and $\mathrm{Mg}$ correspond to the third period.
• Thus they will have more ionization energy than fourth-period elements as they will face more nuclear charge
• Thus, this option is incorrect

(d) $\mathrm{Ca}$

• $\mathrm{Ca}$ and $\mathrm{K}$ correspond to the fourth period.
• Thus $\mathrm{K}$ is to the left of $\mathrm{Ca}$ ,$\mathrm{Ca}$ will have more ionization energy than $\mathrm{K}$ as it have smaller size and more nuclear chargehas
• Thus, this option is incorrect

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