Question

Which of the following elements contains maximum number of unpaired electrons?

• A. Fluorine
• B. Sodium
• C. Nitrogen
• D. Oxygen

Hint:

Always apply Hund's Rule of Maximum Multiplicity. It states that for a given electron subshell (like $p$, $d$, or $f$), every orbital is singly occupied with one electron before any one orbital is doubly occupied, maximizing unpaired spins.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
To solve this, we must write down the ground-state electron configurations for each given element and use Hund's Rule to count the number of unpaired electrons in their outermost orbitals.

Step 2: Detailed Explanation:
Let's evaluate each option sequentially:
(a) Fluorine ($Z=9$): Configuration is $1s^2 \ 2s^2 \ 2p^5$. In the $2p$ subshell, electrons are arranged as $\uparrow\downarrow, \uparrow\downarrow, \uparrow$. It has 1 unpaired electron.
(b) Sodium ($Z=11$): Configuration is $1s^2 \ 2s^2 \ 2p^6 \ 3s^1$. The $3s$ subshell contains 1 unpaired electron.
(c) Nitrogen ($Z=7$): Configuration is $1s^2 \ 2s^2 \ 2p^3$. According to Hund's Rule, the three electrons in the $2p$ subshell will singly occupy the $p_x$, $p_y$, and $p_z$ orbitals ($\uparrow, \uparrow, \uparrow$). It has 3 unpaired electrons.
(d) Oxygen ($Z=8$): Configuration is $1s^2 \ 2s^2 \ 2p^4$. In the $2p$ subshell, electrons are arranged as $\uparrow\downarrow, \uparrow, \uparrow$. It has 2 unpaired electrons.

Step 3: Final Answer:
Nitrogen contains the maximum number (three) of unpaired electrons. Thus, option (c) is correct.