Coordination Compound — NEET UG practice

1. Identify the type of isomerism each complex can show by looking at its ligand arrangement and composition.

2. Given $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ is a square planar $MA_2B_2$ type complex, it can arrange its two $\mathrm{NH}_3$ and two $\mathrm{Cl}^-$ ligands either adjacent or opposite to each other.

Therefore, it shows geometrical (cis-trans) isomerism → matches III.

3. Given $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$ has three identical symmetric bidentate en ligands wrapped around the metal in a propeller-like fashion with no plane of symmetry.

Hence, it shows optical isomerism → matches I.

4. Given $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2$ contains the ambidentate ligand $\mathrm{NO}_2^-$, which can bind to the metal through either its nitrogen atom (nitro) or its oxygen atom (nitrito, written as $\mathrm{ONO}^-$).

So, this pair of complexes are linkage isomers → matches IV.

5. Given $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ can also exist as $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2\mathrm{O}$, where a water molecule swaps places with a chloride ion inside vs. outside the coordination sphere.

This is called solvate (hydrate) isomerism → matches II.

6. Combining all matches: A-III, B-I, C-IV, D-II.

Hence, the answer is D.

1. Recall the definition: an ambidentate ligand has two different donor atoms, but only one of them binds to the metal at a time, depending on conditions.

2. Check each option against this definition.

Given ethane-1,2-diamine (en) has two nitrogen atoms as donor sites, both of the same type, so it is a bidentate ligand, not ambidentate.

Given the ethylenediaminetetraacetate ion (EDTA⁴⁻) has two nitrogen and four oxygen donor atoms, all of which bind simultaneously, so it is a hexadentate ligand.

Given oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$) has two oxygen donor atoms of the same type, both bonding together, so it is a bidentate ligand.

3. Given thiocyanate ($\mathrm{SCN}^-$) has two different donor atoms — sulfur and nitrogen — and can bond to the metal through either one, but not both at once.

This matches exactly the definition of an ambidentate ligand.

Hence, the answer is C.

1. For each complex, find the metal's oxidation state and d-electron count, then decide the hybridisation using valence bond theory, which fixes the shape.

2. Given $\left[\mathrm{Pt}\left(\mathrm{Cl}_2\right)\left(\mathrm{NH}_3\right)_2\right]$ is neutral with two $\mathrm{Cl}^-$ (charge $-1$ each) and two neutral $\mathrm{NH}_3$ ligands:

$$x + 2(-1) + 2(0) = 0 \implies x = +2$$

So $\mathrm{Pt}^{2+}$ has configuration $[\mathrm{Xe}]4f^{14}5d^8$; being a heavy 5d metal, it always pairs its d-electrons, using $dsp^2$ hybridisation.

Therefore, its shape is square planar → matches III.

3. Given $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ has $\mathrm{Co}^{3+}$ ($3d^6$), and the strong-field $\mathrm{NH}_3$ ligand pairs up the electrons, freeing two inner d-orbitals for $d^2sp^3$ hybridisation.

So, its shape is octahedral → matches I.

4. Given $\left[\mathrm{NiCl}_4\right]^{2-}$ has $\mathrm{Ni}^{2+}$ ($3d^8$), and the weak-field $\mathrm{Cl}^-$ ligand cannot force pairing, so the outer $sp^3$ orbitals are used instead of inner d-orbitals.

So, its shape is tetrahedral → matches IV.

5. Given $\left[\mathrm{Fe}(\mathrm{CO})_5\right]$ has iron in the $0$ oxidation state (since $\mathrm{CO}$ is neutral), giving 8 valence electrons that pair into the $3d$ orbitals under the strong-field $\mathrm{CO}$ ligand, leaving one $3d$, one $4s$, and three $4p$ orbitals for $dsp^3$ hybridisation.

So, its shape is trigonal bipyramidal → matches II.

6. Combining all matches: A-III, B-I, C-IV, D-II.

Hence, the answer is B.