Home › UP Board Class 12 › Inorganic Chemistry › The 'spin only' magnetic moment $\mu$ (in Bohr m…
The 'spin only' magnetic moment $\mu$ (in Bohr magnetons) of a transition metal ion with $n$ unpaired d-electrons is given by:
A$\mu = n$
B$\mu = \sqrt{n}$
C$\mu = \sqrt{n(n+2)}$
D$\mu = n(n+1)$
Answer & Solution
Correct answer: C. $\mu = \sqrt{n(n+2)}$
The spin-only formula is $\mu = \sqrt{n(n + 2)}\,\text{BM}$, where $n$ is the number of unpaired electrons. It comes from the spin angular momentum quantum number $S = n/2$ and the relation $\mu = 2\sqrt{S(S+1)}\,\text{BM} = \sqrt{n(n+2)}$. The actual magnetic moment can differ slightly due to orbital contribution; this is small for first-row transition ions in octahedral fields.
Related questions
Solid $PCl_5$ exist as:Which process of purification is represented by the following scheme $\mathrm{Ti}_{\text{(i) $\mathrm{A} + \mathrm{Na}_{2}\mathrm{CO}_{3} \longrightarrow \mathrm{B} + \mathrm{C}$Which of the following is a three-dimensional silicateAmong $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$, the stronger reducing agent in aqueous soHighest oxidation states of transition metals are typically observed only in their:Zinc has the lowest enthalpy of atomisation in the first transition series because:The actinoid contraction is greater than the lanthanoid contraction because: