You are given the following two complex X and Y which are isomer of each other , X is Hg `[Co(SCN)_(4)]`. It is further given that 'spin only' magnetic moment of X is found to be 3.78 B.M. and that of Y is 1.73 B.M. Then which of the following is correct ?

Options

A. Anion of X will be tetrahedral and that of Y will be square planar
B. Anion of X will be square planar but that of Y will be tetrahedral
C. Both the anions will be tetrahedral
D. Both the anions will be square planar

Explanation

To solve the problem, we need to analyze the two complexes X and Y, which are isomers of each other, and compare their magnetic properties and geometries.

### Step-by-Step Solution:

1. **Identify the Complexes**:
- Complex X is given as \( \text{Hg[Co(SCN)_4]} \).
- Complex Y is given as \( \text{Hg[Co(NCS)_4]} \).

2. **Determine the Spin-Only Magnetic Moment for X**:
- The spin-only magnetic moment formula is given by:
\[ \mu = \sqrt{n(n + 2)} \text{ B.M.} \]
- For complex X, the magnetic moment is 3.78 B.M.
- Setting up the equation:
\[ \sqrt{n(n + 2)} = 3.78 \]
- Squaring both sides:
\[ n(n + 2) = (3.78)^2 = 14.2884 \]
- Solving the quadratic equation \( n^2 + 2n - 14.2884 = 0 \):
- Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = 2, c = -14.2884 \).
- Calculate the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-14.2884) = 4 + 57.1536 = 61.1536 \]
- Thus,
\[ n = \frac{-2 \pm \sqrt{61.1536}}{2} \approx 3 \text{ (only the positive root is valid)} \]
- Therefore, complex X has **3 unpaired electrons**.

3. **Determine the Geometry and Hybridization of X**:
- Since \( \text{SCN}^- \) is a weak field ligand, it does not cause pairing of electrons in cobalt.
- Cobalt in \( \text{Co}^{2+} \) has the electronic configuration of \( [Ar] 3d^7 \).
- With 3 unpaired electrons, the hybridization is \( sp^3 \) leading to a **tetrahedral geometry**.

4. **Determine the Spin-Only Magnetic Moment for Y**:
- For complex Y, the magnetic moment is 1.73 B.M.
- Setting up the equation:
\[ \sqrt{n(n + 2)} = 1.73 \]
- Squaring both sides:
\[ n(n + 2) = (1.73)^2 = 2.9929 \]
- Solving the quadratic equation \( n^2 + 2n - 2.9929 = 0 \):
- Using the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2.9929)}}{2 \cdot 1} \]
- Calculate the discriminant:
\[ 4 + 11.9716 = 15.9716 \]
- Thus,
\[ n = \frac{-2 \pm \sqrt{15.9716}}{2} \approx 1 \text{ (only the positive root is valid)} \]
- Therefore, complex Y has **1 unpaired electron**.

5. **Determine the Geometry and Hybridization of Y**:
- Since \( \text{NCS}^- \) is a strong field ligand, it causes pairing of electrons in cobalt.
- Cobalt in \( \text{Co}^{2+} \) has the electronic configuration of \( [Ar] 3d^7 \).
- With 1 unpaired electron, the hybridization is \( dsp^2 \) leading to a **square planar geometry**.

6. **Conclusion**:
- Complex X is tetrahedral with 3 unpaired electrons.
- Complex Y is square planar with 1 unpaired electron.
- The correct statement regarding the complexes is that X is tetrahedral and Y is square planar.

### Final Answer:
The correct option is that complex X is tetrahedral and complex Y is square planar.