JEE Main 2025 Apr 8 Shift 2, Chemistry Q15: Quantum Numbers

JEE Main2025Apr 8, Shift 2Chemistry

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 the $p_{z}$ orbital

C. The last electron goes to an orbital with $n = 2$ and $l = 1$

D. The sum of angular nodes of all the atomic orbitals is 1

Choose the correct answer from the options given below:

A
C and D Only
B
A and C Only
C
A, C and D Only
D
A and B Only

Solution

Fluorine (Z = 9) has the ground-state configuration $1 s^{2} 2 s^{2} 2 p^{5}$ , filled according to Hund's rule.

Counting electron spins across all filled orbitals gives 5 electrons with $m_{s} = + \frac{1}{2}$ and 4 electrons with $m_{s} = - \frac{1}{2}$ , so statement A is correct.

The single unpaired 2p electron may occupy any one of $p_{x}$ , $p_{y}$ or $p_{z}$ depending on orbital-filling convention, so it is not necessarily confined to $p_{z}$ ; statement B is not a reliably correct claim.

The last (differentiating) electron of fluorine enters a 2p orbital, for which $n = 2$ and $l = 1$ , so statement C is correct.

The number of angular nodes for any orbital equals its azimuthal quantum number $l$ : both 1s and 2s orbitals have 0 angular nodes each, while each of the three 2p orbitals has 1 angular node, so the total angular nodes summed over the five occupied orbitals is $0 + 0 + 1 + 1 + 1 = 3$ , not 1, making statement D incorrect.

Only statements A and C are correct, matching option (2).

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q15: Quantum Numbers

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