Exams › NEET › Chemistry

Electronic configuration of Cr2+: [Ar] 3d4

1. μn = √(n(n+2))
2. μn = √(4(4+2)) = √24 BM = 4.9 BM
3. μn = √(n(n+2)) = √24 BM
4. μn = √(4(4+2)) = √24 BM = 4.9 BM

Correct answer: μn = √(4(4+2)) = √24 BM = 4.9 BM

Solution

The electronic configuration of Cr2+ is [Ar] 3d4. The number of unpaired electrons (n) is 4. Using the formula μn = √(n(n+2)), the magnetic moment is √(4(4+2)) = √24 ≈ 4.9 BM.

Related NEET Chemistry questions

• Protons and neutrons are also called
• From the \( \alpha \) -particle scattering experiment, Rutherford concluded that: This question has multiple correct options
• Which of following was not explained by Rutherford?
• The energy of an electron in the ground state of \( \boldsymbol{H} \) atom is \( -\mathbf{1 3 . 6 e V} \) The negative sign indicates that:
• Outer electronic configuration of Cl⁻ is:
• The sub-shell with lowest value of (n + l) is filled up first. When two or more sub-shells have the same (n + l) value, the sub-shell with lowest value of 'n' is filled up first. Therefore, the correct order is:

⚔️ Practice NEET Chemistry free + battle 1v1 →