Question

Magnetic moment 2.83 BM is given by which of the following ions?
(At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28)

• A. Ni²⁺
• B. Cr³⁺
• C. Mn²⁺
• D. Ti³⁺

Answer 1

The Correct Option is A

To determine which ion has a magnetic moment of 2.83 Bohr Magneton (BM), we need to calculate the magnetic moment using the formula for each of the given ions. The magnetic moment can be calculated using the formula:

\(\mu = \sqrt{n(n+2)}\) BM

where \(n\) is the number of unpaired electrons in the ion.

Let's analyze each of the ions:

1. Ni²⁺: The atomic number of Ni is 28. The electronic configuration of Ni is [Ar] 3d⁸ 4s². When it is in the Ni²⁺ state, it loses two electrons from the 4s orbital. So, the electronic configuration becomes [Ar] 3d⁸, having 2 unpaired electrons. Therefore, the magnetic moment is calculated as:
   \(\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\) BM.
2. Cr³⁺: The atomic number of Cr is 24. The electronic configuration of Cr is [Ar] 3d⁵ 4s¹. In the Cr³⁺ state, it loses three electrons leading to the configuration [Ar] 3d³, having 3 unpaired electrons. Therefore, the magnetic moment is:
   \(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) BM.
3. Mn²⁺: The atomic number of Mn is 25. The electronic configuration of Mn is [Ar] 3d⁵ 4s². In Mn²⁺ state, it loses two electrons from the 4s orbital, having the configuration [Ar] 3d⁵ with 5 unpaired electrons. Therefore, the magnetic moment is:
   \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) BM.
4. Ti³⁺: The atomic number of Ti is 22. The electronic configuration of Ti is [Ar] 3d² 4s². In the Ti³⁺ state, it loses three electrons leading to the configuration [Ar] 3d¹, having 1 unpaired electron. Therefore, the magnetic moment is:
   \(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\) BM.

Therefore, based on the calculations, Ni²⁺ has a magnetic moment of approximately 2.83 BM, which matches the given condition. Thus, the correct answer is Ni²⁺.