Question

A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86 BM. The atomic number of the metal is

• A. 25
• B. 26
• C. 22
• D. 23

Answer 1

The Correct Option is D

\[ 22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \quad 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \quad 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \quad 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \]
The spin-only magnetic moment is given by:
\[ \mu_s = \sqrt{n(n+2)} \, \text{BM}, \]
where \(n\) is the number of unpaired electrons.
For \(\mu_s = 3.86 \, \text{BM}\):
\[ 3.86 = \sqrt{n(n+2)}. \]
Squaring both sides:
\[ 3.86^2 = n(n+2) \implies 14.9 \approx n(n+2). \]
Solving for \(n\),
we find: \[ n = 3. \]
The element with \(n = 3\) unpaired electrons in its \(+2\) oxidation state can be identified as follows: Configuration in \(+2\) state:
\[ 22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \, (n=2), \quad 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \, (n=3), \quad 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \, (n=5), \quad 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \, (n=4). \]
Thus, the element is \(V\) (Vanadium) with atomic number 23.

Answer 2

Step 1: Given data and concept
We are told that a first-row transition metal ion in its +2 oxidation state has a spin-only magnetic moment of 3.86 BM (Bohr Magneton). We need to find the atomic number of the metal. 

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

Step 2: Calculate number of unpaired electrons
Given that \(\mu = 3.86\,\text{BM}\), we can write:
\[ 3.86 = \sqrt{n(n+2)} \] Squaring both sides:
\[ 3.86^2 = n(n+2) \] \[ 14.9 = n^2 + 2n \] Now, solving for \(n\):
\[ n^2 + 2n - 14.9 = 0 \] Approximating, \(n \approx 4\) (since \(4(4+2) = 24\), which gives \(\sqrt{24} = 4.90\) too high, and for \(n=3\), \(\sqrt{15} = 3.87 \approx 3.86\)).

Therefore, \(n = 3\) unpaired electrons.

Step 3: Identify the metal ion configuration
We are dealing with a first-row transition metal in the +2 oxidation state (3d series). Let’s recall the general pattern:
 

Metal	Configuration (M²⁺)	Unpaired Electrons
Sc²⁺	3d¹	1
Ti²⁺	3d²	2
V²⁺	3d³	3
Cr²⁺	3d⁴	4
Mn²⁺	3d⁵	5
Fe²⁺	3d⁶	4
Co²⁺	3d⁷	3
Ni²⁺	3d⁸	2
Cu²⁺	3d⁹	1

The ion with 3 unpaired electrons and located early in the 3d series is \( \text{V}^{2+} \).

Step 4: Verify with magnetic moment
For \( \text{V}^{2+} \) (3d³):
\[ \mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\,\text{BM} \] which matches perfectly with the given value \(3.86\,\text{BM}\).

Step 5: Atomic number
Vanadium (V) has atomic number \(23\).

Final answer
 

23