JEE Chemistry 2023 Question with Solution

Given: The calculated magnetic moment is $\mu = 4.90 \, \text{BM}$.

Find: The number of unpaired electrons, $n$.

Use the expression for magnetic moment:

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

Substitute the given value:

$$4.90 = \sqrt{n(n+2)}$$

Squaring both sides:

$$(4.90)^2 = n(n+2)$$ $$24.01 = n(n+2)$$

So,

$$24.01 = n^2 + 2n$$

Rearranging,

$$n^2 + 2n - 24.01 = 0$$

Solving gives:

$$n = 4$$

Therefore, the metal ion has $4$ unpaired electrons.

Given: Magnetic moment of the complex is $4.9 \, \text{B.M.}$.

Find: The number of unpaired electrons.

The magnetic moment of a complex is related to the number of unpaired electrons by:

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

Substitute the value:

$$4.9 = \sqrt{n(n+2)}$$

Square both sides:

$$24.01 = n(n+2)$$

Testing the integer value $n = 4$,

$$4(4+2) = 24$$

which matches $24.01$ approximately.

So, the correct answer is $4$.