Question

Which of the following elements does not lose an electron easily?

• A. $\mathrm{Na}$
• B. $\mathrm{F}$
• C. $\mathrm{Mg}$
• D. $\mathrm{Al}$

Answer 1

The correct option is B

$\mathrm{F}$

The explanation for the correct option:

B) $\mathbf{F}$

• The atomic number of fluorine is $9$ and its electronic configuration is: $\left[\mathrm{He}\right]2{\mathrm{s}}^{2}2{\mathrm{p}}^5$
• It requires only one electron to complete its octet and become stable.
• Hence, it is very hard to remove an electron and takes a huge amount of energy to remove. It has large ionization energy.

The explanation for the incorrect option is:

A)$\mathbf{Na}$

• The atomic number of sodium is $11$and its electronic configuration is: $\left[\mathrm{He}\right]3{\mathrm{s}}^1$
• To complete an octet state, sodium has to remove one electron from its shell and be nearest to a noble gas.
• It is very easy to remove an electron from sodium and it has small ionization energy.

C)$\mathbf{Mg}$

• The atomic number of magnesium is $12$ and its electronic configuration is:$\left[\mathrm{Ne}\right]3{\mathrm{s}}^2$
• To complete its octet state, magnesium has to remove two electrons from its outermost shell.
• It is very easy to remove those two electrons and magnesium has small ionization energy as well.

D)$\mathbf{Al}$

• The atomic number of aluminium is and its electronic configuration is: $\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^1$
• To obtain an octet state, aluminium has to remove three electrons from its outermost shell.
• It is easy to remove three electrons and they get removed easily as well. They have low ionization energy value as well.

Hence, option B) $\mathbf{F}$is the correct option.