Animated Solution for Chemistry - Coordination Compounds: Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists. $\{\text{en}={\text{H}}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2\text{ ; atomic numbers ; Ti}=22\text{ ; Cr}=24\text{ ; Co}=27\text{ ; Pt}=78\}$

The world of coordination chemistry is a fascinating interplay of geometry, electron counting, and subtle structural shifts. In this problem, we are tasked with decoding the magnetic properties and isomerism of four distinct coordination complexes. Let's embark on this journey and break down each complex step by step.

The Chromium Complex

Our first candidate is $[Cr(N{H}_3{)}_{4}C{l}_2]Cl$.

To understand its magnetism, we first need to find the oxidation state of the central metal. Ammonia ($N{H}_3$) is a neutral ligand, and each chloride ($C{l}^{-}$) carries a $-1$ charge. Balancing the charges, we find that Chromium is in the $+3$ oxidation state.

Chromium's neutral electron configuration is $[Ar]3d^{5}4s^1$. Removing three electrons leaves us with a $3d^3$ system. In an octahedral crystal field, these three electrons will singly occupy the lower energy $t_{2g}$ orbitals. Because there are three unpaired electrons, this complex is strongly paramagnetic.

Now, what about its isomerism? The complex has the general formula $M{a}_4{b}_2$ (where 'a' is $N{H}_3$ and 'b' is $Cl$). This specific geometry allows for cis-trans isomerism. The two chloride ligands can either be adjacent to each other (cis) or opposite each other (trans). It cannot show ionisation isomerism because exchanging the inner chloride with the outer chloride yields the exact same molecule. Thus, P matches with (3).

The Titanium Complex

Next up is $[Ti(H_{2}O{)}_{5}Cl](N{O}_3{)}_2$.

Titanium here is also in the $+3$ oxidation state. Its neutral configuration is $[Ar]3d^{2}4s^2$, so $T{i}^{3+}$ is a $3d^1$ system. With exactly one unpaired electron, this complex is paramagnetic.

Looking at its structure, it follows the $M{a}_{5}b$ pattern. Because five ligands are identical, it is impossible to create distinct geometric arrangements; hence, no cis-trans isomerism. However, notice the anions! We have an inner chloride and outer nitrates. If we swap them, we get $[Ti(H_{2}O{)}_5(N{O}_3)]Cl(N{O}_3)$, a completely different compound that yields different ions in solution. This is the hallmark of ionisation isomerism. Therefore, Q matches with (1).

The Platinum Complex

Our third complex is $[Pt(en)(N{H}_3)Cl]N{O}_3$.

Platinum is in the $+2$ oxidation state, making it a $5d^8$ system. Here is a golden rule of coordination chemistry: $P{t}^{2+}$ almost exclusively forms square planar complexes, and due to the large crystal field splitting of the 5d orbitals, these complexes are always diamagnetic (all electrons are paired).

For isomerism, we have a bidentate ligand, ethylenediamine ('en'). Because 'en' is a relatively short chain, it can only connect to adjacent (cis) positions on the square plane. It cannot stretch across the metal to form a trans isomer. So, no geometrical isomerism. But just like compound Q, we can swap the inner chloride with the outer nitrate to form $[Pt(en)(N{H}_3)(N{O}_3)]Cl$. This means it exhibits ionisation isomerism. Thus, R matches with (4).

The Cobalt Complex

Finally, we have $[Co(N{H}_3{)}_4(N{O}_3{)}_2]N{O}_3$.

Cobalt is in the $+3$ oxidation state, which is a $3d^6$ system. When paired with $N{H}_3$, $C{o}^{3+}$ typically forms a low-spin complex. The crystal field splitting is large enough that all six electrons pair up in the lower $t_{2g}$ orbitals. With zero unpaired electrons, this complex is diamagnetic.

Structurally, it is an $M{a}_4{b}_2$ complex (four ammonias and two nitrates inside the coordination sphere). Just like compound P, this allows for cis-trans isomerism. It cannot show ionisation isomerism because swapping an inner nitrate with an outer nitrate doesn't change the complex. Therefore, S matches with (2).

By systematically breaking down the oxidation states, electron configurations, and structural formulas, we have successfully decoded the matrix: - P matches with 3 - Q matches with 1 - R matches with 4 - S matches with 2

This problem beautifully illustrates how a solid grasp of Crystal Field Theory and structural geometry can unlock the secrets of even the most intimidating coordination compounds!