A metal ion (M) of group-4 is showing magnetic moment of 1.73 BM, its oxidation state is :

\(M^{3+}\)

The correct answer is option 3. \(M^{3+}\).

The magnetic moment of a metal ion can be used to determine its oxidation state and electronic configuration. The magnetic moment (\(\mu\)) in Bohr Magneton (BM) is related to the number of unpaired electrons in the ion by the formula:

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

where \(n\) is the number of unpaired electrons.

In this case, the metal ion has a magnetic moment of 1.73 BM.

Calculate the number of unpaired electrons:

We can use the formula to estimate the number of unpaired electrons:

\(\mu = \sqrt{n(n+2)} = 1.73 \, \text{BM}\)

Solving for \(n\):

For \(n = 1\):

\(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}\)

Thus, the ion has 1 unpaired electron.

Group 4 elements are Titanium (Ti), Zirconium (Zr), and Hafnium (Hf).

Let us consider Titanium (Ti) first, as it is a typical group 4 metal.

The electron configuration of Titanium (Ti) in its neutral state is:

\([Ar] 4s^2 3d^2\)

Now, we'll look at the oxidation states of Titanium (Ti):

In the \(Ti^{+}\) state: \( [Ar] 4s^2 3d^1 \) → 1 unpaired electron (this matches the observed magnetic moment of 1.73 BM).

In the \(Ti^{2+}\) state: \( [Ar] 3d^2 \) → 2 unpaired electrons.

In the \(Ti^{3+}\) state: \( [Ar] 3d^1 \) → 1 unpaired electron (this also matches the magnetic moment of 1.73 BM).

In the \(Ti^{4+}\) state: \( [Ar] 3d^0 \) → 0 unpaired electrons (diamagnetic, no magnetic moment).

Both \(Ti^{+}\) and \(Ti^{3+}\) states would give a magnetic moment of 1.73 BM due to 1 unpaired electron. However, \(Ti^{3+}\) is a more common oxidation state for titanium compared to \(Ti^+\).

Thus, the correct answer is option 3: \(M^{3+}\)