The quantum numbers of four electrons are given below:

A. \(n\)=3; \(l\)=2; \(m_l\)=1; and s=+\(\frac1{2}\)

B. \(n\)=4; \(l\)=1; \(m_l\)=0; and s=+\(\frac1{2}\)

C. \(n\)=4; \(l\)=2; \(m_l\)=-2; and s=-\(\frac1{2}\)

D. \(n\)=3; \(l\)=1; \(m_l\)=-1; and s=+\(\frac1{2}\)

The correct decreasing order of energy of these electrons is:
1. C>A>B>D
2. C>B>A>D
3. C>D>A>B
4. A>C>D>B
Subtopic:  Pauli's Exclusion Principle & Hund's Rule | AUFBAU Principle |
 78%
Level 2: 60%+
NEET - 2024
Hints

Which one of the following electrons in the ground state will have the least amount of energy?

1. An electron in hydrogen atom.
2. An electron in 2p orbital of carbon atom.
3. The electron of copper atom present in 4s orbital.
4. The outermost electron in sodium atom.
Subtopic:  AUFBAU Principle |
 68%
Level 2: 60%+
NEET - 2022
Hints

Which of the following is the correct arrangement of 4d, 5p, 5f, and 6p orbitals in the order of decreasing energy?

1. 5f > 6p > 4d > 5p 2. 5f > 6p > 5p > 4d
3. 6p > 5f > 5p > 4d 4. 6p > 5f > 4d > 5p
Subtopic:  AUFBAU Principle |
 69%
Level 2: 60%+
NEET - 2019
Hints

advertisementadvertisement