The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

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 metal ion with a magnetic moment of 3.87 Bohr Magnetons (B.M.), we can use the formula for calculating the magnetic moment:

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

Where $n$ is the number of unpaired electrons.

Let's calculate the number of unpaired electrons:

Rearranging the formula to solve for $n$:

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

Squaring both sides:

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

Calculating the square of 3.87:

$15.00 = n(n+2)$

Solving the quadratic equation $n^2 + 2n - 15 = 0$ gives the number of unpaired electrons:

Factorizing the equation:

$(n + 5)(n - 3) = 0$

Thus, $n = 3$ (since electron count cannot be negative).

Now, let's examine the provided options to identify which metal ion corresponds to $n = 3$:

$T_1^{2+}$: Incorrect, as thallium generally exists in a +1 or +3 oxidation state, not a transition metal.
$Mn^{2+}$: Incorrect, manganese in this state has 5 unpaired electrons.
$V^{2+}$: Correct, vanadium in the +2 state has 3 unpaired electrons (due to electron configuration [Ar] 3d³).
$Cr^{2+}$: Incorrect, chromium in this state has 4 unpaired electrons.

Thus, the metal ion $V^{2+}$ has a magnetic moment of 3.87 B.M. due to its 3 unpaired electrons, matching our calculation.

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

The Correct Option is C

Solution and Explanation

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