Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

In the coordination complex $\mathrm{[Co(H_2O)_6]^{3+}}$, the central cobalt ion is in the $+3$ oxidation state, which corresponds to a $3d^6$ electronic configuration. In this specific octahedral environment, the crystal field splitting energy is sufficient to force the six $d$-electrons to pair up within the lower-energy $t_{2g}$ orbitals. Consequently, all six electrons are spin-paired, resulting in a total of zero unpaired electrons.

In octahedral complexes containing weak field ligands like $\mathrm{H}_2\mathrm{O}$, the $d$-electrons distribute across $t_{2g}$ and $e_g$ orbitals to form high spin configurations according to Hund's rule. The electron configurations for the given complexes are $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($t_{2g}^4 e_g^2$), $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($t_{2g}^5 e_g^2$), $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ($t_{2g}^4 e_g^2$), $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($t_{2g}^6 e_g^3$), and $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($t_{2g}^3 e_g^1$). Among these, the complexes with an even number of $t_{2g}$ electrons are $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, and $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, totaling three.

The number of moles of $\mathrm{AgCl}$ precipitated upon the addition of silver nitrate corresponds directly to the number of ionizable chloride ions present outside the coordination sphere of the complex. Since the reaction produces 2 moles of $\mathrm{AgCl}$ from the formula $\mathrm{CoCl}_3 \cdot n\mathrm{NH}_3$, exactly 2 chloride ions are located outside the coordination brackets, leaving 1 chloride ion inside to act as a ligand. For an octahedral complex, the coordination number must be 6; therefore, the sum of ammonia ligands and the single chloride ligand inside the sphere must equal 6, implying that $n = 5$. By analyzing the charge balance of the complex $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$, the 2 external chloride ions necessitate a charge of 2 on the coordination entity. Letting $x$ represent the oxidation state of cobalt, the presence of 1 chloride ligand contributes -1 while the 5 ammonia molecules are neutral, leading to the equation $x - 1 = 2$. Solving this yields $x = 3$, and combining these findings results in $x + n = 8$.

Molecular hybridization is determined by the steric number, which is the sum of bonding atoms and lone pairs around a central atom, or by the electronic configuration and ligand field effects in the case of coordination complexes. In Statement I, while $\mathrm{PF}_5$ exhibits $\mathrm{sp}^3\mathrm{d}$ hybridization, $\mathrm{BrF}_5$ has a steric number of six-comprising five bonding pairs and one lone pair-which corresponds to $\mathrm{sp}^3\mathrm{d}^2$ hybridization. Similarly, in Statement II, although $\mathrm{SF}_6$ exhibits $\mathrm{sp}^3\mathrm{d}^2$ hybridization, the $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ complex involves $d^2\mathrm{sp}^3$ hybridization because the strong-field ligands cause $d$-electron pairing to utilize the inner $3d$ orbitals, rendering both statements false.

Transition metal complexes frequently exhibit color due to d-d transitions, a process that occurs when an unpaired electron within a partially filled d-orbital absorbs visible light and excites to a higher energy level. In $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2\mathrm{O}$, the metal center exists as $\mathrm{Cu}^{2+}$ with a $3d^9 4s^0$ electronic configuration, providing the necessary unpaired electron to facilitate this specific transition. In contrast, the vibrant colors observed in species like $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, $\mathrm{K}_2\mathrm{CrO}_4$, and $\mathrm{KMnO}_4$ arise from charge transfer mechanisms rather than d-d transitions.

The presence of bridging carbonyl ligands depends on the electronic configuration and geometric requirements of the metal center to satisfy the 18-electron rule. In the solid-state structure of $[\mathrm{Co}_2(\mathrm{CO})_8]$, the two cobalt atoms are linked by a metal-metal bond and two bridging $\mathrm{CO}$ ligands, whereas the other complexes, $[\mathrm{Mn}_2(\mathrm{CO})_{10}]$, $[\mathrm{Os}_3(\mathrm{CO})_{12}]$, and $[\mathrm{Ru}_3(\mathrm{CO})_{12}]$, feature structures containing only terminal carbonyl ligands.

In $\mathrm{KMnO}_4$, manganese is in the $+7$ oxidation state, which corresponds to a $3d^0$ electronic configuration. Because there are no unpaired electrons, the spin-only magnetic moment is 0 BM. During the titration of oxalic acid in an acidic medium, the permanganate ion acts as an oxidizing agent and is reduced to the $\mathrm{Mn}^{2+}$ ion. This product features a $3d^5$ electronic configuration, which results in five unpaired electrons. The spin-only magnetic moment for this product is 6 BM. Therefore, the difference between the magnetic moment values of the product and $\mathrm{KMnO}_4$ is 6 BM.

Coordination complexes following the general formula $\mathrm{M(AA)}_2\mathrm{X}_2$ fundamentally adopt an octahedral geometry, which serves as the basis for understanding the spatial orientation of ligands around the central metal cation. In the specific case of the $\mathrm{Co}(\mathrm{en})_2\mathrm{Cl}_2^+$ complex, the two chloride ligands can occupy only two distinct relative positions within this octahedral framework: the trans position where they are 180 degrees apart, or the cis position where they are 90 degrees apart. Because this specific complex yields only two geometrical isomers, the assertion claiming a total count of three is incorrect, while the statement accurately identifying the complex's octahedral geometry is correct.

The reaction of nickel ions with dimethylglyoxime in the presence of ammonia yields a stable, red square planar complex characterized by two five-membered chelate rings, which invalidates the claim that six-membered rings are formed. Conversely, Prussian blue is a well-known mixed-valence coordination compound with the chemical structure $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$, confirming that it indeed contains iron atoms in both the +2 and +3 oxidation states.

The spectrochemical series categorizes ligands based on their ability to cause crystal field splitting, ranging from strong pi-acceptors to weak pi-donors. Because carbon monoxide acts as an exceptionally strong-field ligand and water maintains a moderate position, while fluoride and sulfide ions occupy the weaker end of the spectrum, their relative standing is determined by this specific ordering. Consequently, the sequence $\mathrm{CO} > \mathrm{H}_2\mathrm{O} > \mathrm{F}^- > \mathrm{S}^{2-}$ accurately reflects the decreasing order of ligand field strength.

Crystal field theory dictates how electrons occupy orbitals based on the specific geometry and ligand strength surrounding the central metal ion. For the octahedral complex $\left[ \mathrm{Co} \left( \mathrm{NH}_3 \right)_6 \right]^{3+}$, the $\mathrm{Co}^{+3}$ ion has a $3d^6$ configuration where the strong field ligand ensures all electrons pair up in the $t_{2g}$ orbitals ($t_{2g}^{2,2,2}$, $e_g^{0,0}$), resulting in zero unpaired electrons. In contrast, the tetrahedral $\left[ \mathrm{NiCl}_4 \right]^{2-}$ complex features a $\mathrm{Ni}^{+2}$ ion with a $3d^8$ configuration, where electrons populate the $e$ orbitals fully before splitting into the $t_2$ orbitals ($e^{2,2}$, $t_2^{2,1,1}$), which leaves two unpaired electrons behind. Combining these values results in a total of two unpaired electrons.

The spin-only magnetic moment of a transition metal complex is governed by the number of unpaired electrons ($n$) in the central metal ion, determined by the formula $\mu = \sqrt{n(n+2)}$ B.M. By examining the electronic configuration and ligand strength, we can assign the correct magnetic moment to each complex. The complex $[\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}$ contains $\mathrm{Cr}^{3+}$ ($3d^3$) with three unpaired electrons, resulting in a value of $3.87$ B.M. Similarly, $[\mathrm{NiCl}_4]^{2-}$ features $\mathrm{Ni}^{2+}$ ($3d^8$) in a weak field environment, possessing two unpaired electrons and a calculated magnetic moment of $2.83$ B.M.

The $[\mathrm{CoF}_6]^{3-}$ complex involves $\mathrm{Co}^{3+}$ ($3d^6$) with weak field ligands, promoting a high-spin state with four unpaired electrons that corresponds to $4.90$ B.M. Finally, in the $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ complex, strong field cyanide ligands cause all electrons in the $3d^8$ configuration of $\mathrm{Ni}^{2+}$ to pair, resulting in zero unpaired electrons and a magnetic moment of $0$ B.M.

In qualitative inorganic analysis, identifying unique colors associated with specific complexes allows for the detection of various ions. The coordination compound $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 \cdot x\mathrm{H}_2\mathrm{O}$ manifests as the deep, insoluble pigment known as Prussian Blue, while the complex $[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}]^{4-}$ is visually identified by an intense violet color. Similarly, the reaction of ferric ions with thiocyanate yields the soluble $[\mathrm{Fe}(\mathrm{SCN})]^{2+}$ complex, producing a striking blood-red solution. Finally, the formation of $(\mathrm{NH}_4)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$, commonly referred to as ammonium phosphomolybdate, is recognized during phosphate testing by the appearance of a characteristic yellow precipitate.

The electronic configuration of a metal ion, determined by its oxidation state, defines the number of unpaired electrons in the $d$-orbitals of a coordination complex. Vanadium in $[\mathrm{V}(\mathrm{H_2O})_6]^{3+}$ exists as $V^{3+}$ with a $3d^{2}$ configuration, which results in 2 unpaired electrons, an even number. Chromium in $[\mathrm{Cr}(\mathrm{H_2O})_6]^{2+}$ acts as $Cr^{2+}$ with a $3d^{4}$ configuration, yielding 4 unpaired electrons, which is also even. Conversely, nickel in $[\mathrm{Ni}(\mathrm{H_2O})_6]^{3+}$ has a $3d^{7}$ configuration with 3 unpaired electrons, and copper in $[\mathrm{Cu}(\mathrm{H_2O})_6]^{2+}$ possesses a $3d^{9}$ configuration with 1 unpaired electron, both representing odd numbers. Consequently, the total number of complexes with an even number of unpaired electrons is 2.

Hybridization describes the mixing of atomic orbitals to produce new, equivalent hybrid orbitals that dictate the specific geometric shape of a molecule based on the number and type of orbitals involved. An $sp^3$ hybridization mixes one s and three p orbitals to establish a tetrahedral structure, whereas $dsp^2$ hybridization is characterized by a square planar arrangement. In systems with five or six electron pairs, $sp^3d$ hybridization results in a trigonal bipyramidal geometry, and $sp^3d^2$ hybridization organizes orbitals into an octahedral shape. Linking these electronic configurations to their respective spatial orientations confirms that A corresponds to III, B to IV, C to I, and D to II.

A coordination complex exhibits optical isomerism only if it lacks both a plane of symmetry (POS) and a center of symmetry (COS), rendering it chiral. The $\text{cis} - [\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2]^{3-}$, $[\mathrm{Co}(\mathrm{en})_3]^{3+}$, $\text{cis}- [\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2]^{2+}$, and $\text{cis}- [\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$ isomers all satisfy this condition, as their asymmetrical arrangements prevent superimposition on their mirror images. In contrast, the trans configurations of $[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2]^{2+}$ and $[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2]^{3-}$ contain both a plane and a center of symmetry due to the opposing positions of their ligands, making them achiral and optically inactive. Thus, out of the six listed species, 4 complexes show optical isomerism.

The spin-only magnetic moment ($\mu$) of a transition metal complex depends directly on the number of unpaired electrons ($n$) residing in the metal's $d$-orbitals, governed by the relationship $\mu = \sqrt{n(n+2)}$ B.M. Because the magnetic moment increases with the count of unpaired electrons, determining this value for each metal center allows us to establish their relative order.

For the complex $Q = [\mathrm{V}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, the Vanadium ion is in the $+2$ oxidation state with a $3d^3$ configuration, resulting in $n=3$ unpaired electrons and a magnetic moment of $\mu = \sqrt{3(3+2)} = \sqrt{15}$ B.M. In $R = [\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, Iron is in the $+2$ oxidation state with a $3d^6$ configuration, and since water acts as a weak-field ligand, it does not force electron pairing, leaving $n=4$ unpaired electrons to produce a magnetic moment of $\mu = \sqrt{4(4+2)} = \sqrt{24}$ B.M. Finally, for $P = [\mathrm{FeF}_6]^{3-}$, the Iron is in the $+3$ oxidation state with a $3d^5$ configuration, where the weak-field fluoride ligand ensures all five electrons remain unpaired, yielding $\mu = \sqrt{5(5+2)} = \sqrt{35}$ B.M. Comparing these magnitudes, the values $\sqrt{15}$, $\sqrt{24}$, and $\sqrt{35}$ follow the increasing order $Q < R < P$.

Valence Bond Theory explains the geometry and hybridization of coordination compounds by evaluating the central metal ion's oxidation state and the field strength of the ligands involved. In $\mathrm{K}_2[\mathrm{Ni}(\mathrm{CN})_4]$, the nickel ion is in the +2 oxidation state with a $3d^8$ electronic configuration. Because cyanide is a strong field ligand, it induces electron pairing in the $3d$ orbitals, leaving one vacant for the $\mathrm{dsp}^2$ hybridization required for a square planar geometry. In the case of $[\mathrm{Ni}(\mathrm{CO})_4]$, the nickel atom is in a zero oxidation state with a $3d^8 4s^2$ configuration. Here, the carbonyl ligands act as a strong field influence, promoting the pairing of electrons and facilitating $\mathrm{sp}^3$ hybridization, resulting in a tetrahedral structure.

For the cobalt complexes, $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$ involves a $\mathrm{Co}^{3+}$ ion with a $3d^6$ configuration. Ammonia behaves as a strong field ligand in this environment, forcing the $3d$ electrons to pair up, which clears two inner $3d$ orbitals for $\mathrm{d}^2\mathrm{sp}^3$ hybridization. Finally, in $\mathrm{Na}_3[\mathrm{CoF}_6]$, the $\mathrm{Co}^{3+}$ ion remains in a $3d^6$ configuration, but fluoride acts as a weak field ligand. It does not cause electron pairing, forcing the metal to engage outer $4d$ orbitals for $\mathrm{sp}^3\mathrm{d}^2$ hybridization to accommodate the six ligands in an octahedral arrangement.

The nomenclature for coordination complexes dictates that ligands are listed alphabetically, with prefixes such as "di" or "bis" indicating the quantity of each type. In the complex $[\mathrm{PtBr}_2 (\mathrm{PMe}_3)_2]$, the two anionic bromide ligands are named "dibromo" and the two neutral trimethylphosphine ligands are named "bis(trimethylphosphine)," using "bis" because the name of the ligand itself implies a complexity that requires a specific multiplicative prefix. Given that the molecule is neutral and the two bromide ions account for a total charge of -2, the platinum metal must possess an oxidation state of II, which is appended in parentheses at the end of the name to complete the designation.

Ethane-1,2-diamine acts as a bidentate ligand by coordinating through two nitrogen donor atoms to form a ring structure, which defines it as a classic chelating ligand. In the industrial Hall-Heroult process for obtaining metallic aluminium, aluminium oxide is dissolved in molten cryolite to decrease the electrolytic cell's operating temperature and increase ionic conductivity. Silver leaching relies on cyanide ions to chemically isolate the metal from its ore, resulting in the formation of the soluble complex $\mathrm{Na}[\mathrm{Ag}(\mathrm{CN})_2]$. Other claims are chemically inaccurate, as Wilkinson's catalyst contains triphenylphosphine rather than simple phosphine, and $\mathrm{Ca}^{2+}$ forms a more stable complex with EDTA than $\mathrm{Mg}^{2+}$ does, meaning their stability constants are not similar.