Question

Which of the following elements has the electronic configuration $3{\mathrm{s}}^1$?

• A. Sodium
• B. Chlorine
• C. Oxygen
• D. Helium

Answer 1

The correct option is A

Sodium

The explanation of the correct option:

A) Sodium

• Sodium atom ($\mathrm{Na}$) belongs to the first group also known as Alkali metals. Its atomic number is $11$
• It has a single electron in its outermost shell and exists as${\mathrm{Na}}^{+}$. All alkali metals have a single electron in their outermost shell.
• The electronic configuration of $\mathrm{Na}$is $1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^6,3{\mathrm{s}}^1\mathrm{or}\left[\mathrm{Ne}\right]3{\mathrm{s}}^1$Therefore, it has $3{\mathrm{s}}^1$in its configuration.

The explanation for incorrect options:

B) Chlorine

• Chlorine belongs to the group $17$also known as halogens. Its atomic number is $17$
• Chlorine contains $5$electrons in its outermost shell with just one required to complete its octet.
• The electronic configuration of chlorine is : $1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^6,3{\mathrm{s}}^{2}3{\mathrm{p}}^5\mathrm{or}\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^5$

C) Oxygen

• Oxygen belongs to group number $16$which is chalcogens. the atomic number of oxygen is $8$
• Oxygen requires only two electrons to complete its octet.
• The electronic configuration goes as: $1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^4\mathrm{or}\left[\mathrm{He}\right]2{\mathrm{s}}^{2}2{\mathrm{p}}^4$

D) Helium

• Helium belongs to the noble gas group. Noble gases have complete an octet and are stable and inert as well.
• The atomic number of helium is $10$
• The electronic configuration of Helium is :$1{\mathrm{s}}^2,2{\mathrm{s}}^{2}2{\mathrm{p}}^6$

Hence, A) Sodium is the correct option.