\([Cr(H_2O)_6]Cl_3\) (atomic number of \(Cr \) = 24) has a magnetic moment of 3.83 B.M. the correct distribution of \(3d\) electrons in the chromium atom is:

\(3d^1_{xy}\, \ 3d^1_{yz}\, \ 3d^1_{xz}\)

The correct answer is option 4. \(3d^1_{xy}\, \ 3d^1_{yz}\, \ 3d^1_{xz}\).

Let us break down the problem step by step to understand how to determine the correct distribution of the \(3d\) electrons in the chromium ion in the complex \([Cr(H_2O)_6]Cl_3\).

The complex is given as \([Cr(H_2O)_6]Cl_3\).

\(H_2O\) is a neutral ligand (it does not contribute to the charge).

\(Cl_3\) is composed of three chloride ions, each carrying a \(-1\) charge.

The complex itself is neutral, so the sum of the charges of all components must equal zero:

\(\text{Oxidation state of } Cr + 3(-1) = 0\)

\(\text{Oxidation state of } Cr = +3\)

Therefore, the chromium ion in this complex is in the \(+3\) oxidation state, denoted as \(Cr^{3+}\).

Next, we need to determine the electronic configuration of chromium in its \(+3\) oxidation state.

Atomic number of chromium (Cr): 24

Ground-state electronic configuration of chromium (Cr): \([Ar] 3d^5 4s^1\)

When chromium loses three electrons to form \(Cr^{3+}\), it loses the two \(4s\) electrons and one of the \(3d\) electrons:

Removing two electrons from the \(4s\) orbital and one from the \(3d\) orbital leaves:

\([Ar] 3d^3\)

Thus, the electronic configuration of \(Cr^{3+}\) is \([Ar] 3d^3\), meaning it has three electrons in the \(3d\) orbitals.

The magnetic moment (\(\mu\)) of a metal ion is associated with the number of unpaired electrons it has and is given by the formula:

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

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

We are told that the magnetic moment is 3.83 Bohr Magnetons (B.M.). To find the number of unpaired electrons (\(n\)), we can equate and solve for \(n\):

\(3.83 = \sqrt{n(n+2)}\)

Squaring both sides:

\(n(n+2) = (3.83)^2 = 14.67 \approx 15\)

The value of \(n(n+2) = 15\) is achieved when \(n = 3\).

This tells us that the \(Cr^{3+}\) ion has three unpaired electrons.

The next step is to determine how these three unpaired electrons are distributed among the \(3d\) orbitals.

Crystal Field Theory and Orbital Splitting

In an octahedral field, such as the one formed by the six water ligands around the chromium ion, the five \(3d\) orbitals split into two sets:

\(t_{2g}\) orbitals (lower energy): \(d_{xy}\), \(d_{yz}\), \(d_{xz}\)

\(e_g\) orbitals (higher energy): \(d_{x^2 - y^2}\), \(d_{z^2}\)

The \(t_{2g}\) orbitals lie at a lower energy level because they point between the ligands in the octahedral structure, experiencing less repulsion. In contrast, the \(e_g\) orbitals are aligned along the axes, directly facing the ligands, and thus are at a higher energy level due to greater repulsion.

For \(Cr^{3+}\) with a \(3d^3\) configuration:

The three unpaired electrons will occupy the lower-energy \(t_{2g}\) orbitals first, one in each orbital, to minimize repulsion.

So, the electron configuration in the \(3d\) orbitals is:

\(3d^1_{xy}, \ 3d^1_{yz}, \ 3d^1_{xz}\)

Each of these \(t_{2g}\) orbitals (\(d_{xy}\), \(d_{yz}\), \(d_{xz}\)) will contain one unpaired electron.

Conclusion

Given the magnetic moment of 3.83 B.M., which corresponds to 3 unpaired electrons, and the nature of orbital splitting in an octahedral field, the correct distribution of the \(3d\) electrons in the \(Cr^{3+}\) ion is: option 4: \(3d^1_{xy}, \ 3d^1_{yz}, \ 3d^1_{xz}\)