The ‘spin only’ magnetic moment of Ni²⁺ in aqueous solution would be:

• A. √6 BM
• B. √15 BM
• C. √2 BM
• D. √8 BM

To determine the ‘spin only’ magnetic moment of Ni²⁺ in an aqueous solution, we first need to understand the electronic configuration of nickel and how it behaves in its +2 oxidation state. Nickel has an atomic number of 28, and its ground state electron configuration is [Ar] 3d⁸ 4s². When it loses two electrons to form Ni²⁺, it loses the two 4s electrons, resulting in a configuration of 3d⁸.

Understanding the Magnetic Moment

The magnetic moment can be calculated using the formula:

μ = √(n(n + 2))

where μ is the magnetic moment in Bohr magnetons (BM) and n is the number of unpaired electrons. The presence of unpaired electrons is what contributes to the magnetic properties of a substance.

Determining Unpaired Electrons

In the case of Ni²⁺ with a 3d⁸ configuration, we need to visualize how the electrons are arranged in the d-orbitals. The 3d orbitals can hold a total of 10 electrons, and in the case of Ni²⁺, we have 8 electrons. The filling of the d-orbitals follows Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up.

• 3d¹ (1 electron in one orbital)
• 3d² (2 electrons in two orbitals)
• 3d³ (3 electrons in three orbitals)
• 3d⁴ (4 electrons in four orbitals)
• 3d⁵ (5 electrons in five orbitals)
• 3d⁶ (6 electrons, 3 paired and 3 unpaired)
• 3d⁷ (7 electrons, 4 paired and 3 unpaired)
• 3d⁸ (8 electrons, 4 paired and 2 unpaired)

For Ni²⁺, the configuration results in 2 unpaired electrons. This is because, in the 3d⁸ configuration, the electrons will pair up in the lower energy orbitals first, leaving two electrons unpaired in the higher energy orbitals.

Calculating the Magnetic Moment

Now that we know there are 2 unpaired electrons (n = 2), we can substitute this value into the magnetic moment formula:

μ = √(2(2 + 2)) = √(2 × 4) = √8

Final Answer

Thus, the ‘spin only’ magnetic moment of Ni²⁺ in aqueous solution is √8 BM. Therefore, the correct answer is:

D. √8 BM