1. Analyze each complex for oxidation state, magnetic nature, and type of isomerism.

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P. $[Cr(NH_3)_3Cl_2]Cl$

(i) Oxidation state of Cr

Let oxidation state of Cr be $x$. Inside the coordination sphere:

• $3NH_3$ are neutral
• $2Cl^-$ contribute $-2$

Since the complex ion is balanced by one outer $Cl^-$, the complex ion has charge $+1$.

So, $x-2=+1 \Rightarrow x=+3$

Thus, chromium is $Cr^{3+}$.

(ii) Electronic configuration and magnetism

Cr: $[Ar]3d^54s^1$

$Cr^{3+}$: remove 3 electrons $\Rightarrow 3d^3$

A $d^3$ configuration has 3 unpaired electrons, so it is paramagnetic.

(iii) Isomerism

The complex is of the type $[MA_3B_2]$ in a coordination number 5 situation; however in standard JEE treatment this compound is considered to show geometrical isomerism (cis-trans) for the two chloride ligands in the coordination sphere.

So, P is:

• Paramagnetic
• Exhibits cis-trans isomerism

Hence, $P \to 3$

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Q. $[Ti(H_2O)_5Cl](NO_3)_2$

(i) Oxidation state of Ti

Let oxidation state of Ti be $x$. Inside the coordination sphere:

• $5H_2O$ are neutral
• $1Cl^-$ gives $-1$

There are two outer $NO_3^-$ ions, so the complex ion has charge $+2$.

Thus, $x-1=+2 \Rightarrow x=+3$

So titanium is $Ti^{3+}$.

(ii) Electronic configuration and magnetism

Ti: $[Ar]3d^24s^2$

$Ti^{3+}$: remove 3 electrons $\Rightarrow 3d^1$

A $d^1$ system has 1 unpaired electron, hence paramagnetic.

(iii) Isomerism

The ligand inside is coordinated $Cl^-$ and outside there are $NO_3^-$ counter ions. Since $Cl^-$ and $NO_3^-$ can interchange between coordination sphere and ionisation sphere, it shows ionisation isomerism.

So, Q is:

• Paramagnetic
• Exhibits ionisation isomerism

Hence, $Q \to 1$

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R. $[Pt(en)(NH_3)Cl]NO_3$

(i) Oxidation state of Pt

Let oxidation state of Pt be $x$. Inside the coordination sphere:

• $en$ is neutral bidentate
• $NH_3$ is neutral
• $Cl^-$ gives $-1$

There is one outer $NO_3^-$, so complex ion charge is $+1$.

Thus, $x-1=+1 \Rightarrow x=+2$

So platinum is $Pt^{2+}$.

(ii) Electronic configuration and magnetism

Pt: $[Xe]4f^{14}5d^96s^1$

$Pt^{2+}$ is effectively $5d^8$.

For $Pt^{2+}$, especially in 4-coordinate complexes, it forms square planar complexes which are diamagnetic.

(iii) Isomerism

$[Pt(en)(NH_3)Cl]^+$ is square planar with one bidentate ligand $en$ occupying two adjacent positions, and $NH_3$ and $Cl$ occupy the remaining two positions. This does not give cis-trans isomerism here, but it can form an ionisation isomer by exchange with outer $NO_3^-$. For example: $[Pt(en)(NH_3)Cl]NO_3 \leftrightarrow [Pt(en)(NH_3)(NO_3)]Cl$

So, R is:

• Diamagnetic
• Exhibits ionisation isomerism

Hence, $R \to 4$

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S. $[Co(NH_3)_4(NO_3)_2]NO_3$

(i) Oxidation state of Co

Let oxidation state of Co be $x$. Inside the coordination sphere:

• $4NH_3$ are neutral
• $2NO_3^-$ contribute $-2$

One outer $NO_3^-$ means complex ion charge is $+1$.

Thus, $x-2=+1 \Rightarrow x=+3$

So cobalt is $Co^{3+}$.

(ii) Electronic configuration and magnetism

Co: $[Ar]3d^74s^2$

$Co^{3+}$: remove 3 electrons $\Rightarrow 3d^6$

In cobalt(III) ammine complexes, pairing occurs due to strong ligand field; thus it is low spin $d^6$, hence diamagnetic.

(iii) Isomerism

The complex is octahedral of type $[MA_4B_2]$, which shows cis-trans isomerism.

So, S is:

• Diamagnetic
• Exhibits cis-trans isomerism

Hence, $S \to 2$

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Final matching

$P-3,\quad Q-1,\quad R-4,\quad S-2$

This corresponds to Option B.

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2. Compare with stored correct answer

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored correct answer.