Answer the question on the basis of passage given below:

All transition elements display typical metallic properties like high melting point, boiling point, tensile strength, high malleability and ductility, luster, etc. Their metallic character increases down the group and elements of 4d and 5d series exhibit higher metallic characters. They show variable oxidation states and forms a number of coloured compounds due to the presence of unpaired electrons which shows d-d transitions and makes them paramagnetic.

A group-IV element X shows magnetic moment of 1.73 BM, its oxidation state is?

$X^{3+}$

The correct answer is Option (3) → $X^{3+}$.

To determine the oxidation state of the group-IV element \(X\) that shows a magnetic moment of 1.73 BM (Bohr Magneton), we can use the following reasoning:

Understanding Magnetic Moments

Magnetic Moment and Unpaired Electrons:

The magnetic moment (\(\mu\)) of a transition metal complex is related to the number of unpaired electrons (\(n\)) in the d-orbitals according to the formula:

\(\mu = \sqrt{n(n + 2)} \text{ BM}\)

The value of 1.73 BM indicates the presence of one unpaired electron.

Possible Oxidation States of Group IV Elements

Group IV elements (like carbon, silicon, germanium, tin, and lead) can exhibit various oxidation states, including \(+4\), \(+2\), and even \(0\) or \(+3\) under specific conditions.

Analyzing the Given Oxidation States

1. \(X^{+}\):

This would likely have an oxidation state of \(0\) or very low, which does not correlate with a magnetic moment.

2. \(X^{2+}\):

In the \(+2\) state, the element would likely have two unpaired electrons if it's d-block, leading to a higher magnetic moment.

3. \(X^{3+}\):

A \(+3\) oxidation state would typically have one unpaired electron (assuming a d^5 configuration) which correlates with a magnetic moment around 1.73 BM.

4. \(X^{4+}\):

A \(+4\) oxidation state typically would lead to all d-orbitals being paired, resulting in no unpaired electrons and therefore a magnetic moment of \(0\).

Conclusion

Given that the magnetic moment of 1.73 BM corresponds to one unpaired electron, the oxidation state that is consistent with this is: \(X^{3+}\)