[Cr (H_{2}*O) 6 ]Cl 3 (atomic number of r = 24 has a mag-netic moment of 3.83 B.M. The correct distribution of 3d-electrons in the chromium present in the complex is: al n (a) 3d xy ^ 1 ,3d yz ^ 1 ,3d zx ^ 1 (b) 3d xy ^ 1 ,3d yz ^ 1 ,3d z^ 2 ^ 1 (c) 3d (x ^ 2 - y ^ 2) ^ 1 3d z ^ 2 ^ 1 (d) 3d xy ^ 1 ,3d (x^ 2 -y^ 2 ) ^ 1 ,3d xx ^ 1

1. Oxidation State of Chromium

In $[Cr(H_2O)_6]Cl_3$ water is a neutral ligand and chloride is the counter-ion outside the bracket. Therefore the cation is $[Cr(H_2O)_6]^{3+}$, giving oxidation state $+3$ to chromium.

2. Ground-State Electron Configuration of $Cr^{3+}$

Atomic $Cr$ is $[Ar]3d^54s^1$. Remove first the $4s$ and then two $3d$ electrons to obtain

So, only three $3d$ electrons remain.

3. Crystal-Field Splitting in an Octahedral Field

Water is a weak-field ligand, so the complex is high-spin. The octahedral crystal field splits $3d$ into a lower-energy $t_{2g}$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$) and a higher-energy $e_g$ set ($d_{z^2}$, $d_{x^2-y^2}$).

Because $$ \Delta_o < P

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

\mu \approx \sqrt{3\times5} = \sqrt{15} \approx 3.87\text{ B.M.}

(d_{xy})^1,(d_{yz})^1,(d_{xz})^1,(d_{z^2})^0,(d_{x^2-y^2})^0

🧒 Imagine Chromium as a set of 3 empty boxes and 3 chocolates

1. The boxes are called $d_{xy}$, $d_{yz}$, and $d_{xz}$.
2. The chocolates are the 3 electrons left in chromium after it loses 3 of them (becoming $Cr^{3+}$).
3. Mom (the rule) says: "Put one chocolate in each box before adding a second one anywhere."
4. So each box gets one chocolate and they all stay "unpaired" (alone).
5. Because 3 chocolates are unpaired, the toy magnet in your hand shows a strength of about 3.8 B.M.—exactly what the question tells us!

Therefore, the electrons sit one‐by‐one in $d_{xy}$, $d_{yz}$ and $d_{xz}$.

Step 1 – Oxidation State

$[Cr(H_2O)_6]Cl_3 \;\to\; [Cr(H_2O)_6]^{3+} + 3Cl^-$ $\therefore \; Oxidation\;state\;of\;Cr = +3$

Step 2 – Electronic Configuration of $Cr^{3+}$

Atomic chromium: $[Ar]3d^54s^1$ Remove three electrons (first from $4s$, then $3d$): $Cr^{3+} = [Ar]3d^3$

Step 3 – Crystal-Field Splitting

In an octahedral field: $t_{2g}: \; d_{xy},\, d_{yz},\, d_{xz}$ $e_g: \; d_{z^2},\, d_{x^2-y^2}$ Water is a weak-field ligand, so electrons remain high-spin and occupy $t_{2g}$ singly:

Step 4 – Magnetic Moment Verification

Number of unpaired electrons, $n = 3$.

Given value is 3.83 B.M., matching closely.

Step 5 – Choose Correct Option

The distribution corresponds to Option (a): $d_{xy}^1,d_{yz}^1,d_{xz}^1$.