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Step-by-Step Solution
Step 1: Write the Electron Configuration of Fe
The atomic number of iron (Fe) is 26. In its ground state, the electron configuration can be written as:
[Ar]\; 3d^6\,4s^2
This shows there are 6 electrons in the 3d orbital and 2 in the 4s orbital.
Step 2: Determine the Number of Unpaired Electrons
In the 3d6 configuration of iron, there are 4 unpaired electrons. This can be ascertained by filling the five 3d orbitals according to Hund’s rule, which maximizes the number of unpaired electrons.
Step 3: Apply the Spin-Only Magnetic Moment Formula
The spin-only magnetic moment ( \mu ) for an ion or atom with n unpaired electrons is given by:
\mu \;=\;\sqrt{n \bigl(n + 2\bigr)}\;\text{BM}
Here, n = 4 , so:
\mu \;=\;\sqrt{4\times(4 + 2)}\;=\;\sqrt{24}\;\text{BM}
Step 4: Substitute the Values of the Square Roots
Given: \sqrt{3} = 1.73 and \sqrt{2} = 1.41 , we can express \sqrt{24} as:
\sqrt{24} = \sqrt{4 \times 6} = 2\,\sqrt{6} = 2\,\sqrt{3\times 2} = 2 \times 1.73 \times 1.41 \approx 4.8786\;\text{BM}
Step 5: Express the Result in the Required Format
The numerical value is approximately 4.88\;\text{BM} . In terms of 10^{-1} :
4.88 \,\text{BM} \;=\;48.8\times 10^{-1}\,\text{BM}
Rounding off to the nearest integer (within the context of the factor in front) gives:
49\times 10^{-1}\,\text{BM}
Therefore, the spin-only magnetic moment of atomic iron (Fe) in its ground state is:
\boxed{49\times 10^{-1}\,\text{BM}}
