Question
The calculated 'spin-only' magnetic moment of $Ti^{2+}$ ($3d^2$) is :
This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$2.84 \text{ BM}$
Calculate Spin-Only Magnetic Moment for $Ti^{2+}$
The spin-only magnetic moment ($\mu_s$) is calculated using the formula:
$ \mu_s = \sqrt{n(n+2)} \text{ BM} $
Where '$n$' represents the number of unpaired electrons.
Steps for Calculation:
- Identify the ion and configuration: The question concerns the $Ti^{2+}$ ion, which has a $3d^2$ electron configuration.
- Determine unpaired electrons: In a $3d^2$ configuration, there are 2 unpaired electrons in the d-orbitals. Therefore, $n = 2$.
- Apply the formula: Substitute $n=2$ into the spin-only magnetic moment formula: $ \mu_s = \sqrt{2(2+2)} $ $ \mu_s = \sqrt{2 \times 4} $ $ \mu_s = \sqrt{8} $
- Calculate the value: The square root of 8 is approximately 2.828. $ \mu_s \approx 2.828 \text{ BM} $
- Compare with options: The calculated value of approximately $2.828$ BM is closest to $2.84$ BM.
Thus, the calculated 'spin-only' magnetic moment of $Ti^{2+}$ is $2.84$ BM.
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