Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

The denticity of a ligand is determined by the number of lone pair donor atoms that coordinate simultaneously to a central metal ion, denoted as $M$. Since this specific ligand possesses exactly four donor atoms capable of forming these coordinate covalent bonds with the metal center, it is classified as a tetradentate ligand.

The magnitude of crystal field splitting, denoted by $\Delta_0$, is a fundamental property determined by the nature of the ligands in the coordination sphere according to the spectrochemical series, rather than being derived from the color of the complex. Since $\mathrm{NH_3}$ is a stronger field ligand than $\mathrm{H_2O}$, the $\Delta_0$ value for $\mathrm{[Cr(NH_3)_6]Cl_6}$ is inherently greater than that for $\mathrm{[Cr(H_2O_6)Cl_3]}$. Suggesting that $\Delta_0$ is calculated from the energies of the observed violet and yellow light reverses the causal relationship; the light absorbed is a consequence of the existing $\Delta_0$ value, which dictates the electronic transition energy. Additionally, both complexes contain $\mathrm{Cr^{3+}}$ ions, which possess a $d^3$ configuration, ensuring both are paramagnetic with three unpaired electrons.

Lead poisoning is managed through a process called chelation therapy, where a substance binds to metal ions to safely remove them from the bloodstream. EDTA is utilized for this purpose because it acts as a potent chelating agent, forming a stable complex with lead that the body can then excrete.

Crystal Field Stabilization Energy (CFSE) is determined by calculating the total energy resulting from the distribution of electrons within the split $d$-orbitals, where each occupied orbital contributes a specific energy value relative to the barycenter based on the coordination geometry of the complex.

In the complex $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6] \mathrm{Cl}_2$, the central iron exists as an $\mathrm{Fe}^{2+}$ ion, which corresponds to a $d^6$ configuration. Because water acts as a weak field ligand in an octahedral environment, the electrons populate the orbitals in a high-spin arrangement of $t_{2g}^4 e_g^2$. The CFSE is calculated by summing the energy contributions of these electrons, yielding $(-0.4 \times 4) \Delta_o + (0.6 \times 2) \Delta_o = -0.4 \Delta_o$.

For $K_2[\mathrm{NiCl}_4]$, the nickel ion is in the $+2$ oxidation state with a $d^8$ configuration in a tetrahedral geometry. The electrons fill the $e$ and $t_2$ sets as $e^4 t_2^4$, and since the $e$ orbitals are lower in energy than the $t_2$ orbitals in this field, the total stabilization is determined by the expression $(-0.6 \times 4) \Delta_t + (0.4 \times 4) \Delta_t = -0.8 \Delta_t$.

Geometrical isomerism in octahedral coordination complexes occurs when different spatial arrangements of ligands are possible, which is characteristic of complexes with the general formula $M(AA)_2bc$, where $M$ is the central metal, $AA$ is a symmetric bidentate ligand, and $b$ and $c$ are distinct monodentate ligands. The complex $[Co(en)_2(H_2O)Cl]Cl$ conforms to this $M(AA)_2bc$ structural type, allowing the monodentate ligands $H_2O$ and $Cl^-$ to occupy positions either cis or trans to one another, resulting in two distinct isomers. Conversely, the other listed compounds, such as $[Cr(H_2O)_5Cl]Cl_2$ ($MA_5B$ type), $[Pt(NH_3)Cl_3]^-$ ($MA_3B$ square planar), and $[Cr(C_2O_4)_3]^{3-}$ ($M(AA)_3$ type), lack the requisite configuration of diverse ligands to exhibit geometrical isomerism.

Coordination geometry and magnetic behavior are dictated by the hybridization of the central metal atom, which is influenced by the field strength of the surrounding ligands. In $[NiCl_4]^{2-}$, the $Ni^{2+}$ ion ($d^8$) interacts with weak field chloride ligands that fail to force electron pairing, resulting in $sp^3$ hybridization with two unpaired electrons, which renders the complex tetrahedral and paramagnetic. In contrast, the neutral $[Ni(CO)_4]$ complex involves a $Ni^0$ atom ($3d^8 4s^2$) where strong $CO$ ligands induce the pairing of $4s$ electrons into the $3d$ orbitals, leading to $sp^3$ hybridization and a tetrahedral structure.

The substitution reaction of the octahedral complex $\mathrm{[Co(NH_3)_4Br_2]^{+}}$ into $\mathrm{[Co(NH_3)_3Br_3]}$ is determined by the spatial arrangement and symmetry of the reactant. Because the cis-isomer lacks a center of symmetry, the ligand exchange can proceed through multiple pathways, resulting in the production of two distinct geometric isomers. In contrast, the trans-isomer possesses a center of symmetry that imposes structural constraints on the substitution process, thereby limiting the formation to only one specific isomer.

Determining the oxidation state of a central metal involves applying the rule that the sum of the oxidation states of all atoms in a neutral complex must equal zero. In $[Cr(H_2O)_6]Cl_3$, water acts as a neutral ligand and three chloride ions contribute a total charge of $-3$, leading to the following calculation:
$x + 0 - 3 = 0$
This yields an oxidation state of $+3$ for the chromium atom. The complex $[Cr(C_6H_6)_2]$ consists of neutral benzene ligands, meaning the central chromium atom balances the entire complex to zero:
$x + 0 = 0$
This confirms the oxidation state of chromium is $0$ in this species. Finally, for $K_2[Cr(CN)_2(O)_2(O)_2(NH_3)]$, we consider two potassium ions contributing $+2$, with the remaining ligands and balancing terms conforming to the following equation:
$2 + x - 2 - 4 + 0 + 0 = 0$
Solving this expression confirms an oxidation state of $+4$ for the chromium atom.

The number of geometrical isomers for a square planar complex containing four distinct monodentate ligands is 3. Because both $NO_2^-$ and $SCN^-$ are ambidentate ligands, each can coordinate in 2 different modes, resulting in linkage isomerism. The total number of isomers is determined by multiplying the number of geometrical arrangements by the total number of linkage combinations for the ambidentate ligands, calculated as $3 \times 2 \times 2 = 12$.

Wilkinson's catalyst, $RhCl(PPh_3)_3$, contains the central metal rhodium in the $+1$ oxidation state, resulting in a $4d^8$ electronic configuration. Because of the significant crystal field splitting associated with $4d$ transition metals and the necessity to minimize steric repulsion between the bulky triphenylphosphine groups, the rhodium ion undergoes $dsp^2$ hybridization, yielding a square planar molecular geometry.

The spin-only magnetic moment is determined by the formula $\mu = \sqrt{n(n+2)}$ Bohr magnetons, where $n$ represents the number of unpaired electrons. In these tetrahedral complexes, the weak field chloride ligands result in high-spin configurations, meaning the number of unpaired electrons is determined solely by the $d$-electron count of the metal ion in its $+2$ oxidation state. The electronic configurations for the metal ions are $Mn^{2+}$ ($3d^5$, $n=5$), $Co^{2+}$ ($3d^7$, $n=3$), $Ni^{2+}$ ($3d^8$, $n=2$), and $Zn^{2+}$ ($3d^{10}$, $n=0$). Because the magnetic moment increases directly with the number of unpaired electrons, the arrangement follows the sequence $5 > 3 > 2 > 0$, corresponding to $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$.

Dicobalt octacarbonyl, $Co_2(CO)_8$, possesses a dimeric structure where the two cobalt metal centers are connected by a single $Co-Co$ bond to satisfy the 18-electron rule for each metal atom. In its stable solid-state configuration, the complex features two carbonyl ligands bridging the two cobalt atoms, while the remaining six carbonyl ligands are terminally bonded with three attached to each cobalt center. This arrangement allows each cobalt atom to achieve a stable electronic configuration by accounting for the nine valence electrons from cobalt, six from the three terminal carbonyls, one from the metal-metal bond, and two from the bridging carbonyls.

Coordination chemistry distinguishes between ligands bound to the metal center and counter-ions in the ionization sphere, where only the chloride ions outside the coordination sphere react with $\mathrm{AgNO_3}$ to form a precipitate. The amount of complex in the solution is determined by the volume and molarity, calculated as $\frac{100 \times 0.1}{1000} = 0.01$ moles of $\mathrm{CoCl_3\cdot6H_2O}$. The reaction with $\mathrm{AgNO_3}$ produces $1.2 \times 10^{22}$ ions, which is equivalent to $\frac{1.2 \times 10^{22}}{6.02 \times 10^{23}} = 0.02$ moles of $\mathrm{Cl^-}$. By comparing the moles of precipitated chloride to the moles of the complex, the ratio is $\frac{0.02}{0.01} = 2$, indicating that two chloride ions are located outside the coordination bracket. Therefore, the complex is $\mathrm{[Co(H_2O)_5Cl]Cl_2\cdot H_2O}$.

Crystal field splitting energy ($\Delta_o$) is significantly influenced by the spatial extension of metal d-orbitals, which increases upon descending a group, resulting in stronger ligand-metal interactions for 4d elements compared to 3d counterparts. Consequently, the molybdenum complex $[\mathrm{Mo}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ exhibits a higher splitting energy than its chromium analogue $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$. For the specified titanium ions, the splitting energy for $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is observed to be smaller than that of $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{2+}$.

A homoleptic complex is defined by the presence of only one single type of ligand directly bonded to the central metal atom. In the case of $[Co(NH_3)_6]Cl_3$, the cobalt ion is coordinated exclusively to ammonia molecules, which satisfies this specific requirement. Conversely, the other listed structures contain mixed coordination spheres consisting of both ammonia and chloride ligands, classifying them as heteroleptic complexes.

Optical isomerism in coordination complexes occurs when a molecule lacks any plane of symmetry or center of inversion, resulting in non-superimposable mirror images. The complex $cis-[Co(en)_2Cl_2]Cl$ lacks a plane of symmetry due to the asymmetric arrangement of the two bidentate ethylenediamine ligands, making it chiral, whereas the $trans$ isomer and the other listed complexes possess internal planes of symmetry that render them achiral.

The magnetic moment of a transition metal complex is directly determined by the number of unpaired electrons in the metal center's d-orbitals. Within the $[Cr(H_2O)_6]^{2+}$ complex, the $Cr^{2+}$ ion exhibits a $d^4$ electronic configuration, and because water acts as a weak field ligand, it retains 4 unpaired electrons. Similarly, the $[Fe(H_2O)_6]^{2+}$ complex involves an $Fe^{2+}$ ion with a $d^6$ configuration, which also yields 4 unpaired electrons when subjected to the same weak field ligand. Given that both complexes share an identical number of unpaired electrons, they produce the same magnetic moment.

The reaction between a coordination complex and silver nitrate depends exclusively on the number of ionizable chloride ions residing outside the metal's coordination sphere. Upon dissolution, the complex $[Cr(H_2O)_6]Cl_3$ dissociates to release three chloride ions, which then readily react with silver ions to form $AgCl$. Because the other provided complexes possess fewer or zero ionizable chloride ions available for precipitation, this specific molecule necessitates the highest number of $AgNO_3$ equivalents.

In inorganic qualitative analysis, the color of a precipitate is often a critical diagnostic feature for identifying specific ionic species. While $K_3[Co(NO_2)_6]$, $(NH_4)_3[As(Mo_3O_10)_4]$, and $BaCrO_4$ are all well-known for their bright yellow coloration, $Zn_2[Fe(CN)_6]$ is distinct because it forms a bluish-white precipitate. Consequently, this compound does not share the characteristic yellow appearance of the others listed.

Square planar complexes containing four distinct ligands, represented by the general formula $[Mabcd]$, exhibit three geometric isomers because there are three possible ligands that can occupy the position trans to any single fixed ligand. By holding one ligand fixed in a specific position, the remaining three ligands can each be placed trans to it in turn, resulting in the three unique configurations where $b$ is trans to $a$, $c$ is trans to $a$, and $d$ is trans to $a$. For the complex $[\mathrm{Pt}(\mathrm{Cl})(\mathrm{py})(\mathrm{NH}_3)(\mathrm{NH}_2\mathrm{OH})]^{+}$, fixing the $\mathrm{Cl}$ ligand as the reference point means that the three geometric isomers are defined by having either $\mathrm{py}$, $\mathrm{NH}_3$, or $\mathrm{NH}_2\mathrm{OH}$ occupy the position trans to the $\mathrm{Cl}$ ligand, resulting in 3 distinct isomeric forms.