Question

The 'Spin only' Magnetic moment for \([Ni(NH_3​)_6​]^{2+}\) is _____ × \(10^{−1}\) BM. (given = Atomic number of Ni : 28)

Answer 1

Correct Answer: 28

Problem Statement: 
Ammonia (NH₃) acts as a WFL (Weak Field Ligand) with Ni²⁺, and the hybridization of the complex \([Ni(NH₃)_6]^{2+}\) is \( sp^3d^2 \).

Electronic Configuration of Ni²⁺:
The electron configuration of \( Ni^{2+} \) (with atomic number 28) is:
\[ Ni^{2+} = 3d^8 \] This implies that the 4s orbital is empty, and there are 8 electrons in the 3d orbital.

Electron Configuration Diagram:
The 3d orbitals are filled as follows, showing the electron arrangement for the 3d orbitals of Ni²⁺:

\[ \text{Ni}^{2+} = \text{3d}^8: \quad \uparrow\downarrow \ \uparrow\downarrow \ \uparrow\downarrow \ \uparrow \ \uparrow \]

Hybridization:
The hybridization of the complex \([Ni(NH₃)_6]^{2+}\) is \( sp^3d^2 \), as it involves six ligands (NH₃), suggesting the use of six orbitals (one from each of the 3d, 4s, and 4p orbitals) to form the bonds.

Number of Unpaired Electrons:
There are 2 unpaired electrons in the 3d orbitals, which will contribute to the magnetic properties of the complex.

Magnetic Moment Calculation:
The magnetic moment \( \mu \) is given by the formula: \[ \mu = \sqrt{n(n + 2)} \, \text{BM}, \] where \( n \) is the number of unpaired electrons. Here, \( n = 2 \) (since there are 2 unpaired electrons), so: \[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.82 \, \text{BM}. \] Thus, the magnetic moment is \( 2.82 \, \text{BM}. \)

Conclusion:
The magnetic moment \( \mu = 28.2 \times 10^{-1} \, \text{BM} \), which corresponds to \( x = 28 \).

Answer 2

NH$_3$ acts as a weak field ligand with Ni$^{2+}$.
\[ \text{Ni}^{2+} = 3d^8 \]

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\[ \text{No. of unpaired electrons} = 2 \]
\[ \mu = \sqrt{n(n+2)} = \sqrt{8} = 2.82 \, \text{BM} \]
\[ 28.2 \times 10^{-1} \, \text{BM} \]
\[ x = 28 \]