The quantum numbers of four electrons are given below:
I. n=4; l=2; \(m_l\)=-2; s=-\(\frac1{2}\)
II. n=3; l=2; \(m_l\)=1; s=+\(\frac1{2}\)
III. n=4; l=1; \(m_l\)=0; s=+\(\frac1{2}\)
IV. n=3; l=1; \(m_l\)=-1; s=+\(\frac1{2}\)
The correct decreasing order of energy of these electrons is:

1. IV>II>III>I
2. I>III>II>IV
3. III>I>II>IV
4. I>II>III>IV
Subtopic:  Pauli's Exclusion Principle & Hund's Rule | AUFBAU Principle |
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Which one of the following electrons in the ground state will have 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 |
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4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is:

1. 5f > 6p > 4d > 5p 2. 5f > 6p > 5p > 4d
3. 6p > 5f > 5p > 4d 4. 6p > 5f > 4d > 5p
Subtopic:  AUFBAU Principle |
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