Match Cobalt Platinum and Nickel Complexes with Properties

To match each complex in Column I with its corresponding properties in Column II, we analyze each complex individually:

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1. Complex (A): $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_2$

• Oxidation State:
  Let $x$ be the oxidation state of Cobalt ($\text{Co}$). $\text{NH}_3$ and $\text{H}_2\text{O}$ are neutral ligands, and $\text{Cl}^-$ has a $-1$ charge:
  $x + 4(0) + 2(0) + 2(-1) = 0 \implies x = +2$
  Hence, Cobalt is in the $+2$ oxidation state $\rightarrow$ (s).

• Magnetic Behaviour:
  $\text{Co}^{2+}$ has a $3d^7$ valence electronic configuration. In an octahedral ligand field, whether high-spin ($t_{2g}^5 e_g^2$, with $3$ unpaired electrons) or low-spin ($t_{2g}^6 e_g^1$, with $1$ unpaired electron), the complex contains unpaired electrons and is paramagnetic $\rightarrow$ (q).

• Isomerism:
  An octahedral complex of the type $[\text{MA}_4\text{B}_2]^{2+}$ (where $\text{A} = \text{NH}_3$ and $\text{B} = \text{H}_2\text{O}$) exhibits geometrical isomerism (cis and trans forms) $\rightarrow$ (p).

Matching for (A): $\text{A} \rightarrow \mathbf{p, q, s}$

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2. Complex (B): $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$

• Oxidation State:
  Let $x$ be the oxidation state of Platinum ($\text{Pt}$):
  $x + 2(0) + 2(-1) = 0 \implies x = +2$
  Hence, Platinum is in the $+2$ oxidation state $\rightarrow$ (s).

• Magnetic Behaviour:
  $\text{Pt}^{2+}$ is a $5d^8$ metal ion. All $4$-coordinate $5d^8$ complexes undergo $dsp^2$ hybridization to adopt a square planar geometry due to the high crystal field splitting energy ($\Delta_o$). In this square planar arrangement, all $8$ $d$-electrons pair up, leaving no unpaired electrons, making it diamagnetic $\rightarrow$ (r).

• Isomerism:
  Square planar complexes of the type $[\text{MA}_2\text{B}_2]$ exhibit geometrical isomerism (cis-platin and trans-platin) $\rightarrow$ (p).

Matching for (B): $\text{B} \rightarrow \mathbf{p, r, s}$

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3. Complex (C): $[\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}$

• Oxidation State:
  Let $x$ be the oxidation state of Cobalt ($\text{Co}$):
  $x + 5(0) + 1(-1) + 1(-1) = 0 \implies x = +2$
  Hence, Cobalt is in the $+2$ oxidation state $\rightarrow$ (s).

• Magnetic Behaviour:
  $\text{Co}^{2+}$ is a $3d^7$ metal ion. Surrounded by weak field ligands ($\text{H}_2\text{O}$ and $\text{Cl}^-$), it forms a high-spin octahedral complex with the configuration $t_{2g}^5 e_g^2$. It possesses $3$ unpaired electrons and is paramagnetic $\rightarrow$ (q).

• Isomerism:
  Complexes of the type $[\text{MA}_5\text{B}]$ have all positions equivalent and cannot form geometrical isomers $\rightarrow$ (p) is not applicable.

Matching for (C): $\text{C} \rightarrow \mathbf{q, s}$

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4. Complex (D): $[\text{Ni}(\text{H}_2\text{O})_6]\text{Cl}_2$

• Oxidation State:
  Let $x$ be the oxidation state of Nickel ($\text{Ni}$):
  $x + 6(0) + 2(-1) = 0 \implies x = +2$
  Hence, Nickel is in the $+2$ oxidation state $\rightarrow$ (s).

• Magnetic Behaviour:
  $\text{Ni}^{2+}$ is a $3d^8$ metal ion. In an octahedral field, its configuration is $t_{2g}^6 e_g^2$. With $2$ unpaired electrons, it is paramagnetic $\rightarrow$ (q).

• Isomerism:
  An octahedral complex of the type $[\text{MA}_6]$ has all six ligands identical and thus cannot exhibit geometrical isomerism $\rightarrow$ (p) is not applicable.

Matching for (D): $\text{D} \rightarrow \mathbf{q, s}$

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Final Correct Matches:

• (A) $\rightarrow$ p, q, s
• (B) $\rightarrow$ p, r, s
• (C) $\rightarrow$ q, s
• (D) $\rightarrow$ q, s