Chemistry · Structure of Atom

JEE Main 2025 — 8 April, Evening Shift — Question 1

Correct statements for an element with atomic number 9 are

A. There can be 5 electrons for which $\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}$ 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 $I=1$

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

Choose the correct answer from the options given below :

Option A:
A and B only
Option B:
C and D only
Option C:
A and C only
Correct
Option D:
A, C and D only

Answer: C

Step-by-step solution

$\mathrm{Z}=9$

Electronic configuration $\Rightarrow$

There can be $5 \mathrm{e}^{-}$ for which $\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}$ and $4 \mathrm{e}^{-}$ for which $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$

There can be $2 \mathrm{e}^{-}$ in $\mathrm{p}_{\mathrm{z}}$ orbital

Last $\mathrm{e}^{-}$ enters $2 \mathrm{p}(\mathrm{n}=2, \mathrm{I}=1)$

Angular nodes = 1

$\therefore$ Sum of angular nodes $=0+0+3=3$

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Electronic Configuration: Rules, Aufbau Principle and Exceptions

The energy of an electron in first Bohr orbit of H atom is -13.6 eV . The magnitude of energy value of electron in the first excited state…

Answer: C

Step-by-step solution

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