For which of the following aqueous ion, spin only magnetic moment is 3.87 BM?

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

To determine the correct ion with a spin-only magnetic moment of 3.87 BM, we need to use the formula for the spin-only magnetic moment:

$\mu = \sqrt{n(n+2)}$

where $n$ is the number of unpaired electrons.

Given that the spin-only magnetic moment is 3.87 BM, we can set up the equation:

$3.87 = \sqrt{n(n+2)}$

Let's solve for $n$ by squaring both sides:

$3.87^2 = n(n+2)$

$14.9769 = n(n+2)$

Testing integer values for $n$, we find:

For $n = 3$, $n(n+2) = 3 \times 5 = 15$, which is approximately 14.9769.

This means the number of unpaired electrons is 3.

Now, we need to identify the transition metal ion with 3 unpaired electrons. Analyzing the given options:

$Ti^{2+}$: Titanium has electronic configuration: [Ar] 3d\)
$V^{2+}$: Vanadium has electronic configuration: [Ar] 3d\)
$Cr^{2+}$: Chromium has electronic configuration: [Ar] 3d\)
$Mn^{2+}$: Manganese has electronic configuration: [Ar] 3d\)

Based on this analysis, the correct answer is $V^{2+}$, which has 3 unpaired electrons.

Therefore, the ion with a spin-only magnetic moment of 3.87 BM is $V^{2+}$.

For which of the following aqueous ion, spin only magnetic moment is 3.87 BM?

The Correct Option is B

Solution and Explanation

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