Question

X, Y are the complexes of \(M^{n+}\) ion. The spin only magnetic moment values of X, Y respectively are 3.87 BM, 1.73 BM. \(M^{n+}\) is:

• A. \(Mn^{2+}\)
• B. \(Co^{2+}\)
• C. \(Fe^{2+}\)
• D. \(Cr^{3+}\)

Hint:

Remember: \[ 1.73 \rightarrow 1 \text{ unpaired electron} \] \[ 2.83 \rightarrow 2 \text{ unpaired electrons} \] \[ 3.87 \rightarrow 3 \text{ unpaired electrons} \]

Answer 1

The Correct Option is C

Concept: The spin-only magnetic moment is \[ \mu = \sqrt{n(n+2)} \] where \(n\) is the number of unpaired electrons.

Step 1: Determine unpaired electrons in X.
Given \[ \mu = 3.87 \] This corresponds to \[ n=3 \] unpaired electrons.

Step 2: Determine unpaired electrons in Y.
Given \[ \mu = 1.73 \] This corresponds to \[ n=1 \] unpaired electron.

Step 3: Identify the metal ion.
A \(d^6\) ion such as \[ Fe^{2+} \] can show both high-spin and low-spin complexes. Thus it can have 4, 2, 1 or other possible unpaired electron arrangements depending on ligand field strength. Hence the given magnetic moments are consistent with \[ \boxed{Fe^{2+}} \]