Problem 67 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Match the complexes in Column I with their properties listed in Column II. Column I (A) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}\) (B) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (C) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}\) (D) \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\) Column II (p) geometrical isomers (q) paramagnetic (r) diamagnetic (s) metal ion with \(+2\) oxidation state

Step-by-Step Solution

Answer

(A)-q,s; (B)-p,r,s; (C)-q; (D)-q,s.

Step 1: Determine Oxidation States

In each complex, identify the oxidation state of the metal ion. For (A) \([\text{Co(NH}_3\text{)}_4(\text{H}_2\text{O)}_2]\text{Cl}_2\), Co is in the \(+2\) oxidation state. For (B) \([\text{Pt(NH}_3\text{)}_2\text{Cl}_2]\), Pt is in the \(+2\) oxidation state. For (C) \([\text{Co(\text{H}_2O)}_5\text{Cl]}\text{Cl}\), Co is in the \(+3\) oxidation state. For (D) \([\text{Ni(\text{H}_2O)}_6]\text{Cl}_2\), Ni is in the \(+2\) oxidation state.

Step 2: Identify Magnetic Properties

Determine whether each complex is paramagnetic or diamagnetic by considering the electron configuration of the metal ions. Complex (A) with Co\(^2^+\) is paramagnetic. Complex (B) with Pt\(^2^+\) is diamagnetic. Complex (C) with Co\(^3^+\) has unpaired electrons, so it is paramagnetic. Complex (D) with Ni\(^2^+\) also has unpaired electrons, making it paramagnetic.

Step 3: Check for Geometrical Isomers

Identify the complexes that can exhibit geometrical isomerism. Complex (B) \([\text{Pt(NH}_3\text{)}_2\text{Cl}_2]\) is a square planar complex which can show geometrical isomerism. None of the other complexes in the list have the geometry that allows for geometric isomers.

Key Concepts

Oxidation States

In coordination chemistry, oxidation state is crucial for determining the electron count of a metal complex. It helps in predicting the chemical reactivity and stability of the compound. Oxidation state of a metal is the charge left on the metal atom after all the ligands are removed along with their associated electrons.
For complex (A) \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2}\mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}\), cobalt has an oxidation state of \(+2\). We know this because the complex has a neutral charge. Each chloride ion contributes a \(-1\) charge, totaling \(-2\) for two chlorides. The ammonia and water ligands are neutral. Thus, cobalt must be \(+2\) to balance the charges.

• (A) \[\text{Oxidation State: Co}^{2+}\]

Similarly, in (B) \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}]\), platinum is in the \(+2\) oxidation state. Ammonia is neutral and each chloride contributes \(-1\), so the net charge on Pt is \(+2\) to balance with the two chlorides.

• (B) \[\text{Oxidation State: Pt}^{2+}\]

For complex (C) \([\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\mathrm{Cl}]\mathrm{Cl}\), cobalt is in the \(+3\) oxidation state. The external chloride is \(-1\), leaving cobalt as \(+3\) to maintain the compound as neutral.

• (C) \[\text{Oxidation State: Co}^{3+}\]

Finally, for (D) \([\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]\mathrm{Cl}_{2}\), nickel is in the \(+2\) oxidation state. The two chlorides provide \(-2\) total charge, so nickel must be \(+2\) to neutral the overall charge.

• (D) \[\text{Oxidation State: Ni}^{2+}\]

Magnetic Properties

Knowing the magnetic properties of a complex can tell us about the arrangement of electrons and the nature of the bonding. Complexes can be paramagnetic, indicating unpaired electrons, or diamagnetic, indicating all electrons are paired.
Complexes with unpaired electrons exhibit magnetism when placed in a magnetic field. This is because each unpaired electron possesses a magnetic moment. In comparison, diamagnetic complexes are those with paired electrons and do not exhibit magnetism.

• Complex (A) \([\text{Co(NH}_3\text{)}_4(\text{H}_2\text{O)}_2]\text{Cl}_2\) is paramagnetic. Cobalt in the \(+2\) oxidation state has unpaired electrons.


• Complex (B) \([\text{Pt(NH}_3\text{)}_2\text{Cl}_2]\) is diamagnetic. Pt in the \(+2\) oxidation state often forms a square planar geometry, usually leading to paired electrons.


• Complex (C) \([\text{Co(H}_2\text{O)}_{5}\text{Cl}]\text{Cl}\) is paramagnetic due to the \(+3\) oxidation state of Co, which introduces unpaired electrons in the d-orbitals.


• Complex (D) \([\text{Ni(H}_2\text{O)}_6]\text{Cl}_2\) is paramagnetic. Ni \(+2\) also encounters unpaired electrons in its d-orbitals.

Geometrical Isomerism

Geometrical isomerism is an important concept in coordination chemistry and occurs in complexes where the arrangement of ligands around the metal center allows different spatial arrangements. It happens commonly in square planar and some octahedral complexes.
In complex (B) \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}]\), geometrical isomerism is possible due to its square planar shape. Here, the ammonia and chloride ligands can be arranged differently around the platinum center, leading to different isomers:

• Cis isomer: Where similar ligands, say \(\mathrm{NH}_3\), are adjacent to each other.


• Trans isomer: Where similar ligands are opposite each other across the metal center.

The remaining complexes [(A), (C), and (D)] do not exhibit geometrical isomerism due to their respective geometries (octahedral for A, C, and D) where such isomerism isn’t possible given the type and arrangement of ligands.