Match the LIST-I with LIST-II.

choose the correct answer from the options given below

The spin-only magnetic moment ($$\mu$$) is calculated using the formula:

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

where $$n$$ is the number of unpaired electrons

A. $$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Cr}$$ is in the $$+2$$ state ($$\text{Cr}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Cr}$$ is $$[\text{Ar}]\,3d^5 4s^1$$.
  Removing 2 electrons gives $$\text{Cr}^{2+}: [\text{Ar}]\,3d^4$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, so it forms a high-spin complex ($$t_{2g}^3 \, e_g^1$$).
• Unpaired Electrons ($$n$$): $$4$$
• Magnetic Moment: $$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\text{ BM}$$
• Match: A $$\rightarrow$$ III

B. $$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Co}$$ is in the $$+2$$ state ($$\text{Co}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Co}$$ is $$[\text{Ar}]\,3d^7 4s^2$$.
  Removing 2 electrons gives $$\text{Co}^{2+}: [\text{Ar}]\,3d^7$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, leading to a high-spin complex ($$t_{2g}^5 \, e_g^2$$).
• Unpaired Electrons ($$n$$): $$3$$
• Magnetic Moment: $$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ BM}$$
• Match: B $$\rightarrow$$ I

C. $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Cu}$$ is in the $$+2$$ state ($$\text{Cu}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Cu}$$ is $$[\text{Ar}]\,3d^{10} 4s^1$$.
  Removing 2 electrons gives $$\text{Cu}^{2+}: [\text{Ar}]\,3d^9$$.
• Crystal Field Splitting: High-spin arrangement ($$t_{2g}^6 \, e_g^3$$).
• Unpaired Electrons ($$n$$): $$1$$
• Magnetic Moment: $$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ BM}$$
• Match: C $$\rightarrow$$ IV

D. $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Mn}$$ is in the $$+2$$ state ($$\text{Mn}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Mn}$$ is $$[\text{Ar}]\,3d^5 4s^2$$.
  Removing 2 electrons gives $$\text{Mn}^{2+}: [\text{Ar}]\,3d^5$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, leading to a high-spin complex ($$t_{2g}^3 \, e_g^2$$).
• Unpaired Electrons ($$n$$): $$5$$
• Magnetic Moment: $$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\text{ BM}$$
• Match: D $$\rightarrow$$ II
  

Hence, Option D is correct.