Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

Crystal Field Theory determines electron pairing based on the oxidation state of the central metal and the field strength of the surrounding ligands. The complex $[\mathrm{FeF}_6]^{3-}$ contains $\mathrm{Fe}^{3+}$ ($3d^5$) with weak-field fluoride ligands, resulting in 5 unpaired electrons, while $[\mathrm{CoF}_6]^{3-}$ has $\mathrm{Co}^{3+}$ ($3d^6$) and weak-field fluoride ligands, resulting in 4 unpaired electrons. In contrast, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ features $\mathrm{Ni}^{2+}$ ($3d^8$) forced into a diamagnetic state by strong-field cyanide ligands, and $[\mathrm{Ni}(\mathrm{CO})_4]$ involves neutral nickel ($3d^8 4s^2$) where strong-field carbonyl ligands induce full pairing in the $3d$ orbitals, meaning both contain 0 unpaired electrons.

Homoleptic complexes are defined by the presence of only a single type of ligand coordinated to the central metal ion, while the electronic spin state of the complex is dictated by the field strength of those ligands relative to the metal's pairing energy. Among the provided species, $[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]^{2-}$ is heteroleptic because it contains two distinct types of ligands, whereas $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$, $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, and $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ are all homoleptic. Within this group, the strong field nature of the cyanide and ammonia ligands induces electron pairing, resulting in low spin configurations for $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ and $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, in contrast to the high spin states observed in the remaining homoleptic complexes containing weak field ligands.

Valence Bond Theory predicts the geometry and hybridization of metal complexes based on the interplay between the coordination number and the ligand field strength. Strong field ligands induce electron pairing within the central metal ion's $d$-orbitals, enabling inner $d$-orbital participation, whereas weak field ligands lack this ability, forcing the use of higher-energy outer orbitals. For a coordination number of 6, the metal center typically adopts either an inner orbital $d^2 sp^3$ or an outer orbital $sp^3 d^2$ configuration, while a coordination number of 4 generally leads to tetrahedral $sp^3$ or square planar $dsp^2$ geometries.

In the complex $[\mathrm{CoF}_6]^{3-}$, the $\mathrm{Co}^{3+}$ ion interacts with fluoride, which is a weak field ligand, necessitating an outer orbital hybridization of $sp^3 d^2$. The complex $[\mathrm{NiCl}_4]^{2-}$ contains $\mathrm{Ni}^{2+}$ with chloride ligands, which are also weak field, resulting in a tetrahedral $sp^3$ hybridization. When considering $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, the strong field nature of the ammonia ligands forces the electrons of the $\mathrm{Co}^{3+}$ ion to pair up, which opens the inner $d$-orbitals for bonding and yields $d^2 sp^3$ hybridization. Finally, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ involves a strong field cyanide ligand bound to $\mathrm{Ni}^{2+}$, which facilitates a square planar geometry characterized by $dsp^2$ hybridization.

The $d$-electronic configuration of an octahedral complex is determined by the number of unpaired electrons, which relates to the magnetic moment through the spin-only formula $\mu = \sqrt{n(n+2)}$. For a magnetic moment of 3.95 BM, solving this expression shows that $n = 3$ unpaired electrons are present. Since a $Co^{2+}$ ion has a $3d^7$ configuration, distributing these seven electrons into the $t_{2g}$ and $e_g$ orbitals to accommodate three unpaired electrons leads to the $t_{2g}^5 e_g^2$ configuration.

The spin state of a coordination complex is determined by the competition between the crystal field splitting energy and the energy required to pair electrons. In octahedral complexes, when the splitting energy $\Delta_0$ is less than the pairing energy $P$, the electrons occupy higher orbitals rather than pairing up, resulting in a high spin complex. Similarly, since the crystal field splitting in tetrahedral complexes $\Delta_t$ is always significantly less than the pairing energy $P$, electrons avoid pairing whenever possible, which makes the formation of low spin complexes in these geometries exceptionally rare.

The spin-only magnetic moment is determined by the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ denotes the number of unpaired electrons. The central ion in $[\mathrm{Mn}(\mathrm{SCN})_6]^{4-}$ is $\mathrm{Mn}^{2+}$, a $3d^5$ system that forms a high-spin complex with a weak field ligand, leading to negligible crystal field splitting ($\Delta_o \approx 0$). With five unpaired electrons in this configuration, the magnetic moment calculates to $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ B.M., which aligns with the given experimental value of 5.96 B.M.

The magnitude of the standard electrode potential for the $M^{3+}/M^{2+}$ redox couple is highest for Cobalt, reflecting the specific stability associated with its oxidation states. In the complex $[Co(CN)_6]^{4-}$, the central Cobalt atom exists in the $+2$ oxidation state, which corresponds to an electronic configuration of $3d^7$. Because the cyanide ligand creates a strong crystal field, the d-orbitals undergo significant splitting that forces a low-spin electron distribution in an octahedral environment. The seven valence electrons completely fill the lower energy $t_{2g}$ orbitals with six electrons, and the single remaining electron resides in the higher energy $e_g$ orbital, resulting in 1 electron in the $e_g$ set.

The color of these coordination complexes is governed by the electronic environment of the metal ions, where only $\mathrm{K}_3[\mathrm{Co}(\mathrm{NO}_2)_6]$ and $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$ exhibit a yellow color. Both complexes feature metal ions with a $3d^6$ configuration-$\mathrm{Co}^{3+}$ and $\mathrm{Fe}^{2+}$, respectively-that interact with strong field ligands to force a low-spin pairing of electrons into the $t_{2g}$ orbitals, resulting in a $t_{2g}^6 e_g^0$ arrangement. Because all electrons are paired, the number of unpaired electrons is zero, leading to a spin-only magnetic moment of $\mu = \sqrt{0(0+2)} = 0$ B.M. for each complex. Consequently, the sum of these magnetic moment values is $0$.

The electronic configuration of the central metal ion and the field strength of the ligands determine the hybridization and magnetic properties of coordination complexes. For $[\mathrm{MnBr}_4]^{2-}$, the $\mathrm{Mn}^{2+}$ ion in a $d^5$ configuration binds to weak-field bromide ligands, resulting in $sp^3$ hybridization and paramagnetism, while $[\mathrm{FeF}_6]^{3-}$ contains $\mathrm{Fe}^{3+}$ in a $d^5$ state that interacts with weak-field fluoride ligands to form an octahedral complex with $sp^3 d^2$ hybridization and paramagnetic character.

In contrast, the complex $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ features $\mathrm{Co}^{3+}$ in a $d^6$ configuration with strong-field oxalate ligands, forcing electron pairing to achieve $d^2 sp^3$ hybridization and a diamagnetic state. Finally, $[\mathrm{Ni}(\mathrm{CO})_4]$ consists of a nickel center with a $d^{10}$ configuration and strong-field carbonyl ligands, leading to $sp^3$ hybridization and a diamagnetic nature due to the absence of unpaired electrons.

Crystal Field Stabilization Energy reflects the energy gained by an electron configuration relative to the average energy of the split $d$-orbitals, where strong field ligands like ethylenediamine exert a greater influence by forcing electron pairing. In the $[\mathrm{Co}(\mathrm{en})_3]^{3+}$ complex, the $\mathrm{Co}^{3+}$ center adopts a $d^6$ electronic state and, due to the presence of the strong ethylenediamine ligands, fills the $t_{2g}$ orbitals completely to reach a $t_{2g}^6 e_g^0$ configuration.

This configuration maximizes the number of electrons in the lower-energy $t_{2g}$ set, resulting in the calculation:
$CFSE = (6 \times (-0.4) + 0 \times 0.6) \Delta_0 = -2.4 \Delta_0$
This value represents the greatest stabilization among the given complexes because the full $t_{2g}$ subshell minimizes the overall energy of the metal-ligand system.

A homoleptic octahedral complex composed of six identical monodentate ligands follows the $[\mathrm{Ma}_6]$ type, which creates a highly symmetrical geometry that prevents the formation of any stereoisomers. In contrast, while cis-platin and trans-platin are indeed heteroleptic, they are complexes of $\mathrm{Pt}$ with the chemical formula $[\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}_2]$, rather than complexes of $\mathrm{Pd}$. Therefore, the initial statement correctly describes the geometric constraints of such homoleptic systems, whereas the second statement is erroneous due to the incorrect identification of the central metal atom.

Hybridization in coordination complexes depends on whether the ligands utilize inner $d$-orbitals or outer $d$-orbitals for bonding, with $sp^3d^2$ hybridization specifically indicating the use of outer $d$-orbitals. For $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, $[\mathrm{CrF}_6]^{3-}$, and $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$, the central metal ions $\mathrm{Co}^{3+}$ ($3d^6$), $\mathrm{Cr}^{3+}$ ($3d^3$), and $\mathrm{Mn}^{3+}$ ($3d^4$) form inner orbital complexes, resulting in $d^2sp^3$ hybridization. Conversely, $\mathrm{SF}_6$ forms an $sp^3d^2$ hybridized structure, and the complexes $[\mathrm{CoF}_6]^{3-}$ and $[\mathrm{MnCl}_6]^{3-}$ also utilize the outer $d$-orbitals, leading to $sp^3d^2$ hybridization. These three species are the ones involved in $sp^3d^2$ hybridization.

Transition metal oxides generally shift from basic to acidic character as the oxidation state of the metal increases, a principle used to classify the compounds $\mathrm{VO}_2$, $\mathrm{V}_2 \mathrm{O}_3$, $\mathrm{V}_2 \mathrm{O}_5$, $\mathrm{CrO}_3$, and $\mathrm{Mn}_2 \mathrm{O}_7$. Among these specific oxides, only $\mathrm{CrO}_3$ and $\mathrm{Mn}_2 \mathrm{O}_7$ function as acidic oxides, establishing the value of $X$ as 2. The primary valency $Y$ of cobalt in the coordination compound $[\mathrm{Co}(\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{NH}_2)_3]_2(\mathrm{SO}_4)_3$ represents the metal's oxidation state, which is determined by satisfying charge neutrality for the complex. Given that ethylenediamine is a neutral ligand and each sulfate ion carries a $2-$ charge, the balanced equation $2(x) + 3(-2) = 0$ yields $x = 3$, identifying $Y$ as 3 and making the final sum of $X+Y$ equal to 5.

The electronic configuration of a central metal ion within a coordination complex is defined by its oxidation state, which dictates the number of valence electrons remaining after ionization. By determining the charge on each metal center, one can verify which complexes possess exactly four electrons in their $3d$ subshell.

In complex (A), the iron ion exists in the $\mathrm{Fe}^{6+}$ state, resulting in a $[\mathrm{Ar}] 3d^2$ configuration. Complex (B) contains manganese as $\mathrm{Mn}^{3+}$, which possesses the required $[\mathrm{Ar}] 3d^4$ configuration. While complex (C) features iron as $\mathrm{Fe}^{3+}$ with a $[\mathrm{Ar}] 3d^5$ arrangement, complex (D) involves chromium in the $\mathrm{Cr}^{2+}$ state, which also corresponds to the $[\mathrm{Ar}] 3d^4$ configuration. Finally, the nickel ion in complex (E) is $\mathrm{Ni}^{4+}$, resulting in a $[\mathrm{Ar}] 3d^6$ configuration, which does not meet the criteria.

Crystal Field Theory dictates the electronic distribution in transition metal complexes based on the metal's oxidation state and the ligand field strength. Strong-field ligands create a large energy gap between the $t_{2g}$ and $e_g$ orbitals, promoting electron pairing in lower energy levels, while weak-field ligands produce a smaller gap, favoring the occupation of higher-energy orbitals to maximize unpaired electrons.

For the $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ complex, $\mathrm{Fe}^{3+}$ possesses a $d^5$ configuration, and the strong-field cyanide ligands force the electrons into a $t_{2g}^5 e_{g}^0$ arrangement, resulting in 1 unpaired electron. The $[\mathrm{FeF}_6]^{3-}$ complex also involves $\mathrm{Fe}^{3+}$ with $d^5$, but the weak-field fluoride ligands permit a $t_{2g}^3 e_{g}^2$ distribution, yielding 5 unpaired electrons. With the $[\mathrm{CoF}_6]^{3-}$ complex, $\mathrm{Co}^{3+}$ has a $d^6$ configuration; the weak-field fluoride ligands lead to a $t_{2g}^4 e_{g}^2$ arrangement with 4 unpaired electrons. Finally, for $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$, $\mathrm{Mn}^{3+}$ presents a $d^4$ configuration, and the strong-field cyanide ligands constrain the electrons to a $t_{2g}^4 e_{g}^0$ state, leaving 2 unpaired electrons.

The stability of transition metal complexes is dictated by the magnitude of their Crystal Field Stabilization Energy (CFSE), where a larger negative value signifies greater stability. By calculating the CFSE for each complex, we find that $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$ ($\mathrm{Mn}^{3+}$, $t_{2g}^4$) yields $-1.6 \Delta_0$, $[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ ($\mathrm{Co}^{2+}$, $t_{2g}^6 e_g^1$) gives $-1.8 \Delta_0$, $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ ($\mathrm{Fe}^{3+}$, $t_{2g}^5$) results in $-2.0 \Delta_0$, and $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ ($\mathrm{Fe}^{2+}$, $t_{2g}^6$) provides $-2.4 \Delta_0$. Comparing these absolute values shows an increasing stability trend corresponding to the sequence $[\mathrm{Mn}(\mathrm{CN})_6]^{3-} < [\mathrm{Co}(\mathrm{CN})_6]^{4-} < [\mathrm{Fe}(\mathrm{CN})_6]^{3-} < [\mathrm{Fe}(\mathrm{CN})_6]^{4-}$.

Octahedral complexes coordinated by three bidentate ligands, such as the $[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ complex produced in this reaction, lack both a plane of symmetry and a center of inversion. This inherent structural asymmetry renders the molecule chiral, meaning it cannot be superimposed on its own mirror image. Consequently, the complex exists as a pair of distinct enantiomers, specifically the d and l forms, which corresponds to a total of 2 optical isomers.

A homoleptic complex consists of a central metal atom bonded exclusively to one type of ligand, a classification requiring verification of both the uniformity of the ligands and the parity of the central metal's $d$ electrons. Regarding the candidates, $[\mathrm{FeO}_4]^{2-}$ is homoleptic but contains $\mathrm{Fe}^{6+}$ with a $3d^2$ configuration, which is an even number, while $[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]^{2-}$ and $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$ are heteroleptic due to their mixed ligand environments. In contrast, $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ is a homoleptic complex where the $\mathrm{Fe}^{3+}$ ion possesses a $3d^5$ configuration, and $[\mathrm{CoCl}_4]^{2-}$ is also homoleptic with a $\mathrm{Co}^{2+}$ ion having a $3d^7$ configuration, both of which provide the necessary odd number of $d$ electrons.

The central Bromine atom in $\mathrm{BrF}_5$ possesses 7 valence electrons and forms 5 covalent bonds with fluorine, which results in $(7-5)/2 = 1$ lone pair on the central atom, establishing $x = 1$. This value indicates that only one chloride ion resides outside the coordination sphere, defining the complex formula as $[\mathrm{M}(\mathrm{NH}_3)_3\mathrm{Cl}_3]\mathrm{Cl}$. Because the complex exhibits $sp^3d^2$ hybridisation with a coordination number of 6, it corresponds to the octahedral $\mathrm{[Ma_3b_3]}$ type, which exhibits two geometrical isomers: the facial (fac) isomer, where the three identical ligands occupy the vertices of a triangular face, and the meridional (mer) isomer, where the three identical ligands are positioned along a meridian.

The phenomenon of freezing point depression depends on the number of solute particles in a solution, as described by the formula $\Delta T_{\mathrm{f}} = i K_{\mathrm{f}} m$. By substituting the given experimental data into this relationship, we get the equation $0.558 = i \times 1.86 \times 0.1$. Solving this for the Van't Hoff factor yields $i = \frac{0.558}{0.186} = 3$, which indicates that the complex dissociates into 3 ions upon dissolving.

Since the coordination number of Chromium must be 6, we examine the molecular structure to find a configuration that fits this dissociation pattern. The compound $[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$ fulfills the criteria because it ionizes in water to form the complex ion $[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}]^{2+}$ along with 2 chloride ions, providing the required total of 3 ions.