Question

An element of d-block (Z) of 4th period has spin only magnetic moment of its $Z^{3+}$ form is 3.9 BM, then find minimum atomic number of element (Z).

Answer 1

Correct Answer: 24

To find the atomic number (Z) of the d-block element in the 4th period whose $Z^{3+}$ ion has a spin-only magnetic moment of 3.9 Bohr Magneton (BM), we need to determine the number of unpaired electrons in the ion.

The formula for spin-only magnetic moment (µ) is given by:
$$\mu = \sqrt{n(n+2)} \text{ BM}$$
where n = number of unpaired electrons.

Given that µ = 3.9 BM, we plug it into the equation to find n:
$$3.9 = \sqrt{n(n+2)}$$

Squaring both sides, we get:
$$15.21 = n(n+2)$$

Solving this quadratic equation:
$$n^2 + 2n - 15.21 = 0$$

Using the quadratic formula, $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where a = 1, b = 2, and c = -15.21:

Calculating the discriminant:
$$b^2 - 4ac = 2^2 - 4 \times 1 \times (-15.21) = 4 + 60.84 = 64.84$$

Solving for n:
$$n = \frac{-2 \pm \sqrt{64.84}}{2} = \frac{-2 \pm 8.05}{2}$$

Taking the positive root, we find:
$$n = \frac{6.05}{2} \approx 3.025$$
Since n must be an integer (number of electrons), round it to n = 3.

Now, to find Z, remember that the element loses three electrons to form $Z^{3+}$. Thus, in the $Z^{3+}$ state, the remaining electron configuration within the d-orbital has 3 unpaired electrons.

Considering the d-block of the 4th period, the losing of 3 electrons initially involves elements like Ti (22), V (23), Cr (24), etc.
Cr in its $+3$ oxidation state will have the configuration [$Ar$] 3d³, giving 3 unpaired electrons, fitting the required condition.

Therefore, the atomic number Z of the element is 24, which fits within the specified range.

Atomic number: 24