JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Atomic Structure

Correct statements for an element with atomic number 9 are
A. There can be 5 electrons for which

m_{s} = + \frac{1}{2}

and 4 electrons for which

m_{s} = - \frac{1}{2}

B. There is only one electron in

p_{z}

orbital
C. The last electron goes to orbital with

n = 2

and

l = 1

Choose the correct answer from the options given below:

Answer: (B)

Solution

The element of atomic number 9 is fluorine, with configuration

1 s^{2} 2 s^{2} 2 p^{5}

. Across its nine electrons, five can be assigned spin

+ \frac{1}{2}

(one each in

1 s

2 s

and the three

2 p

orbitals) and the remaining four spin

- \frac{1}{2}

, so that statement is correct. The last electron enters the

2 p

subshell, for which

n = 2

and

l = 1

, so that statement is also correct. The single unpaired

2 p

electron need not reside specifically in the

p_{z}

orbital, so that claim is wrong, and the angular nodes of the occupied orbitals sum to

0 + 0 + 1 + 1 + 1 = 3

rather than one, so that claim is wrong too. Only the statements about the spin distribution and about

n = 2, l = 1

are correct.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option