Problem 167 - A Complete Resource Book in Chemistry for JEE Main

Question

The 'spin-only' magnetic moment [in units of Bohr magneton \(\left.\left(\mu_{\mathrm{B}}\right)\right]\) of \(\mathrm{Ni}^{2+}\) in aqueous solution would be (Atomic number of \(\mathrm{Ni}=28\) ) [2006] (a) \(2.84\) (b) \(4.90\) (c) 0 (d) \(1.73\)

Step-by-Step Solution

Answer

The 'spin-only' magnetic moment of \(\mathrm{Ni}^{2+}\) is approximately \(2.84\), corresponding to option (a).

Step 1: Determine the Electron Configuration

First, we need to find the electron configuration of the Ni atom, which is 28 electrons: \([\text{Ar}] 4s^2 3d^8\). Now consider Ni\(^{2+}\), which means it has lost 2 electrons, typically from the 4s orbital. So, the configuration becomes \([\text{Ar}] 3d^8\).

Step 2: Calculate the Number of Unpaired Electrons

For Ni\(^{2+}\), the electron configuration of \(3d^8\) implies the presence of 8 electrons in the \(3d\) subshell. As \(d\) orbitals can hold a total of 10 electrons, arrange these 8 electrons to minimize pairing, resulting in 2 unpaired electrons.

Step 3: Calculate the Spin-Only Magnetic Moment

The formula for the spin-only magnetic moment \(\mu_{\text{spin}}\) is given by: \[\mu_{\text{spin}} = \sqrt{n(n+2)}\] where \(n\) is the number of unpaired electrons. Substitute \(n = 2\) into the formula: \[\mu_{\text{spin}} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\]

Step 4: Compare with Available Options

The calculated spin-only magnetic moment is approximately \(2.83\). Compare this result against the provided options. The closest match is option (a), which is \(2.84\).

Key Concepts

Electron Configuration

Understanding the electron configuration of an atom is essential for figuring out its magnetic properties. Electron configuration describes how electrons are distributed in an atom's orbital shells. For an atom like nickel (\(\text{Ni}\), with an atomic number of 28), its neutral state electron configuration is \([\text{Ar}] 4s^2 3d^8\). This means that after filling the inner shells (similar to the noble gas argon), nickel has two electrons in the 4s orbital and eight electrons in the 3d orbital.
However, when nickel becomes a Ni\(^{2+}\) ion, it loses two electrons. Typically, electrons are removed from the outermost shell first, which in this case is the 4s orbital. Consequently, the electron configuration for Ni\(^{2+}\) is \([\text{Ar}] 3d^8\). This focuses our analysis on the 3d orbital, where we need to determine electron pairing.

Unpaired Electrons

Once we have the electron configuration for Ni\(^{2+}\), the next step is to consider how electrons are arranged in the 3d subshell. This subshell can hold a maximum of 10 electrons but in Ni\(^{2+}\), it occupies 8. According to Hund's Rule, electrons will fill a set of degenerate orbitals one at a time, with parallel spins, to minimize electron repulsion. This results in maximizing the number of unpaired electrons.

For the Ni\(^{2+}\) ion with the configuration \([\text{Ar}] 3d^8\), the electrons are distributed to fill five d-orbitals. Generally, this leads to two unpaired electrons because the other six electrons will pair up. This is an essential step as the unpaired electrons determine the net magnetic moment.

Magnetic Moment Calculation

The magnetic moment of a transition metal complex can often be assessed using the spin-only formula. This formula relates directly to the number of unpaired electrons (\(n\)) in the d-block element's configuration. The spin-only magnetic moment (\(\mu_{\text{spin}}\)) is calculated using the equation:
\[\mu_{\text{spin}} = \sqrt{n(n+2)}\]
This equation relies solely on the number of unpaired electrons without considering orbital contribution, making it a common approach for first-row transition metals like nickel. Substituting \(n = 2\) in the formula, we get:
\[\mu_{\text{spin}} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\]
Hence, the calculated magnetic moment for the Ni\(^{2+}\) ion closely matches the experimentally observed value, aligning with option (a), \(2.84\). This confirms that using the number of unpaired electrons provides a good estimation of the magnetic properties in transition metals.