Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

Crystal Field Stabilization Energy (CFSE) is determined by the electron configuration in an octahedral ligand field, where $[Cr(H_2O)_6]^{2+}$ ($d^4$, high-spin) has a configuration of $t_{2g}^3 e_g^1$ yielding a CFSE of $-0.6 \Delta_o$, whereas $[Mn(H_2O)_6]^{2+}$ ($d^5$, high-spin) has a configuration of $t_{2g}^3 e_g^2$ resulting in a CFSE of $0 \Delta_o$. Given that a CFSE of $-0.6 \Delta_o$ represents a higher stabilization energy than $0 \Delta_o$, the first statement is true. Regarding magnetic moments, potassium ferricyanide contains $Fe^{3+}$ ($d^5$) in a low-spin octahedral environment with a $t_{2g}^5$ configuration having one unpaired electron and a spin-only magnetic moment of $\sqrt{1(1+2)} = 1.73$ BM, while sodium ferrocyanide contains $Fe^{2+}$ ($d^6$) in a low-spin environment with a $t_{2g}^6$ configuration having zero unpaired electrons and a magnetic moment of $0$ BM, which confirms the second statement is true.

The geometry, hybridization, and magnetic behavior of coordination complexes are determined by the oxidation state of the central metal ion and the field strength of the surrounding ligands, which dictate how electrons occupy the available orbitals. For the ion $[\mathrm{NiCl}_4]^{2-}$, nickel is in the $+2$ oxidation state with a $3d^8$ configuration; because chloride is a weak-field ligand, it does not induce electron pairing, resulting in $\mathrm{sp}^3$ hybridization, a tetrahedral geometry, and two unpaired electrons that render the complex paramagnetic. In the case of $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$, nickel is also in the $+2$ state, and for a coordination number of six, the complex adopts an $\mathrm{sp}^3\mathrm{d}^2$ outer-orbital octahedral geometry that maintains unpaired electrons, making it paramagnetic. Conversely, $[\mathrm{Ni}(\mathrm{CO})_4]$ involves nickel in the $0$ oxidation state, where the strong-field carbonyl ligands force electron pairing to achieve $\mathrm{sp}^3$ hybridization, resulting in a tetrahedral geometry that is diamagnetic, which contradicts the paramagnetic classification provided in the premise. Finally, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ contains nickel in the $+2$ oxidation state, where the strong-field cyanide ligands promote the pairing of $d$-electrons to allow for $\mathrm{dsp}^2$ hybridization, creating a square planar geometry that is diamagnetic.

The molar mass of $\mathrm{CrCl}_3 \cdot 6\mathrm{H}_2\mathrm{O}$ is calculated as $52 + 3(35.5) + 6(18) = 266.5$ g/mol. Dividing the given mass of $5.33$ g by this molar mass results in $0.02$ mol of the complex. Similarly, the moles of $\mathrm{AgCl}$ are determined by dividing $8.61$ g by the molar mass of $\mathrm{AgCl}$, which is $108 + 35.5 = 143.5$ g/mol, yielding $0.06$ mol of $\mathrm{AgCl}$.

Because the substance is a 1:3 electrolyte, it dissociates according to $\left[\mathrm{Cr}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]\mathrm{Cl}_3 \rightarrow \left[\mathrm{Cr}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{3+} + 3\mathrm{Cl}^-$, meaning 1 mole of the complex releases 3 moles of chloride ions that react with silver nitrate. The ratio of the moles of complex reacted to the moles of $\mathrm{AgCl}$ formed is $\frac{0.02}{0.06} = \frac{1}{3}$. Multiplying this ratio by $10^2$ gives $33.33$, which is $33$ when rounded to the nearest integer.

A coordination complex exhibits paramagnetism when it possesses one or more unpaired electrons in its central metal ion orbitals, a state dictated by the specific electronic configuration and the influence of ligand field splitting. Among the listed compounds, $\left[\mathrm{MnBr}_4\right]^{2-}$ contains a $d^5$ metal ion with weak-field bromide ligands, resulting in five unpaired electrons, while $\left[\mathrm{NiCl}_4\right]^{2-}$ has a $d^8$ configuration in a tetrahedral geometry, leaving two unpaired electrons. The $\left[\mathrm{CoF}_6\right]^{3-}$ complex features a $d^6$ cobalt ion under the influence of weak fluoride ligands, leading to a high-spin state with four unpaired electrons, and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ involves a $d^4$ manganese ion that retains two unpaired electrons in the $t_{2g}$ orbitals despite the presence of cyanide ligands. Furthermore, $\left[\mathrm{Ti}(\mathrm{CN})_6\right]^{3-}$ possesses a $d^1$ configuration with a single unpaired electron, and $\left[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$ contains a $d^9$ copper ion, which leaves one unpaired electron in the $e_g$ set. Conversely, $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, and $\left[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3\right]^{3-}$ are diamagnetic because their specific ligand environments and metal configurations force complete electron pairing within the orbitals. Consequently, there are a total of 6 paramagnetic complexes in the provided group.

The spin-only magnetic moment is determined by the number of unpaired electrons, $n$, in the central metal ion using the relationship $\mu = \sqrt{n(n+2)}$ BM. Because water functions as a weak field ligand in all these complexes, we determine the unpaired electrons directly from the metal ion's configuration without pairing. For $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($3d^4$ configuration), the four unpaired electrons result in a magnetic moment of 4.90 BM, while $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($3d^7$ configuration) has three unpaired electrons, yielding 3.87 BM. Similarly, $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($3d^9$ configuration) possesses one unpaired electron with a value of 1.73 BM, and $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($3d^5$ configuration) contains five unpaired electrons, giving 5.92 BM.

The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \, \text{BM}$, where $n$ represents the number of unpaired electrons; a value greater than $3.0 \, \text{BM}$ is only achieved when $n \ge 3$, since $\sqrt{3(3+2)} \approx 3.87$ and $\sqrt{2(2+2)} \approx 2.83$. The high-spin octahedral electronic configurations and their corresponding unpaired electron counts are: $\mathrm{Ti}^{2+}$ ($d^2$, $n=2$), $\mathrm{V}^{3+}$ ($d^2$, $n=2$), $\mathrm{Fe}^{2+}$ ($d^6$, $t_{2g}^4 e_g^2$, $n=4$), $\mathrm{Co}^{2+}$ ($d^7$, $t_{2g}^5 e_g^2$, $n=3$), and $\mathrm{Ni}^{2+}$ ($d^8$, $t_{2g}^6 e_g^2$, $n=2$). Comparing these values, only $\mathrm{Fe}^{2+}$ and $\mathrm{Co}^{2+}$ possess at least three unpaired electrons to satisfy $\mu > 3.0 \, \text{BM}$. Summing the unpaired electrons for these ions, $4 + 3 = 7$.

In low spin octahedral complexes, the strong crystal field induces electron pairing within the $t_{2g}$ orbitals before any electrons occupy the $e_g$ orbitals, thereby minimizing the total number of unpaired electrons. Applying this to the given ions results in the configurations $Cr^{3+}$ ($3d^3$: 3 unpaired), $Mn^{3+}$ ($3d^4$: $t_{2g}^4$, 2 unpaired), $Fe^{3+}$ ($3d^5$: $t_{2g}^5$, 1 unpaired), and $Co^{3+}$ ($3d^6$: $t_{2g}^6$, 0 unpaired), which establishes the decreasing order of $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$.

The splitting energy $\Delta_o$ is directly proportional to the field strength of the ligands and inversely proportional to the wavelength of absorbed light, denoted by the relationship $\lambda = \frac{hc}{\Delta_o}$. Based on the spectrochemical series, ligand strength decreases in the order $CN^- > NH_3 > H_2O > F^-$, meaning $[Co(CN)_6]^{3-}$ has the largest splitting energy and shortest wavelength, while $[CoF_6]^{3-}$ has the smallest splitting energy and longest wavelength. Applying this logic to the given complexes leads to the splitting order $III > I > IV > II > V$, which translates to the final wavelength absorption order $III < I < IV < II < V$.

The nickel(II) dimethylglyoxime complex is a neutral, square planar coordination compound that is insoluble in water, which makes statement B incorrect. Regarding electronic configuration, the nickel ion in this $dsp^2$ hybridized complex is diamagnetic, meaning it contains zero unpaired d-electrons, rendering statement C false. Finally, statement E is incorrect because the structure consists of two bidentate dimethylglyoxime ligands that form exactly two five-membered chelate rings with the central nickel ion, rather than four.

The first-row transition metal that does not liberate $\mathrm{H}_2$ gas from dilute $\mathrm{HCl}$ is copper because it possesses a positive standard reduction potential. When $\mathrm{CuSO}_4$ reacts with excess $\mathrm{KCN}$, the $\mathrm{Cu}^{2+}$ ions are reduced to $\mathrm{Cu}^+$ and sequestered into the highly stable complex ion $[\mathrm{Cu}(\mathrm{CN})_4]^{3-}$. Because the dissociation constant of this complex is extremely low, the concentration of free copper ions remains insufficient to exceed the solubility product constant of copper sulfide, resulting in zero moles of precipitate formed upon the addition of $\mathrm{H}_2\mathrm{S}(g)$.

The magnetic properties of coordination complexes are determined by the oxidation state of the central metal atom and the field strength of the ligands, which dictate the orbital hybridization and electron pairing. In $Ni(CO)_4$, the neutral nickel atom adopts $sp^3$ hybridization with all electrons paired due to the strong field of carbonyl ligands, and in $[Ni(CN)_4]^{2-}$, the $Ni^{2+}$ ion forms a square planar $dsp^2$ complex with paired electrons, making both diamagnetic. Conversely, the weak field $Cl^-$ ligands in $[NiCl_4]^{2-}$ induce $sp^3$ hybridization with two unpaired electrons in the $d$-orbitals, resulting in a paramagnetic complex.

Valence Bond Theory classifies coordination complexes based on ligand field strength, where strong field ligands force electron pairing to utilize inner $d$-orbitals ($d^2sp^3$ hybridization for coordination number 6), while weak field ligands prevent pairing, necessitating the use of higher energy $nd$-orbitals ($sp^3d^2$ hybridization). In $[Co(NH_3)_6]^{3+}$, the strong field $NH_3$ ligand induces pairing of the $Co^{3+}$ ($3d^6$) electrons, resulting in an inner orbital complex with $d^2sp^3$ hybridization. Similarly, for $[MnCl_6]^{3-}$, the weak field chloride ligand fails to pair the $Mn^{3+}$ ($3d^4$) electrons, resulting in an outer orbital complex with $sp^3d^2$ hybridization.

The statement concerning $[CoF_6]^{3-}$ is incorrect because the weak field fluoride ligand necessitates an outer orbital complex with $sp^3d^2$ hybridization, not $d^2sp^3$. Conversely, $[FeF_6]^{3-}$ is correctly identified as an outer orbital complex with $sp^3d^2$ hybridization, as the weak field fluoride ligand cannot pair the $Fe^{3+}$ ($3d^5$) electrons. Finally, $[Ni(CN)_4]^{2-}$ utilizes $dsp^2$ hybridization to form a square planar geometry due to the strong field cyanide ligand, rendering the stated $sp^3$ tetrahedral hybridization incorrect.

In a tetrahedral complex, the five degenerate d-orbitals of the central metal ion split into two distinct groups due to the differing electrostatic repulsion experienced from the ligands. Because the ligands approach the central atom from paths that lie between the coordinate axes, the $d_{xy}$, $d_{yz}$, and $d_{zx}$ orbitals-which are also oriented between the axes-experience significantly less repulsion and exist at a lower energy level. Conversely, the $d_{x^{2-}y^2}$ and $d_{z^2}$ orbitals point along the axes, where they encounter greater repulsion and are consequently pushed to a higher energy level. This creates a definitive energy relationship characterized by $(d_{xy} = d_{yz} = d_{zx}) < (d_{x^{2-}y^2} = d_{z^2})$, where the set of three orbitals directed between the axes is consistently lower in energy than the degenerate set of two orbitals directed along the axes.

Coordination isomerism occurs in salts featuring both cationic and anionic coordination spheres where the metal ions or ligands allow for the exchange of components between the two spheres to form distinct structural isomers. This condition is successfully met in $[\mathrm{Co}(\mathrm{NH}_3)_6] [\mathrm{Cr}(\mathrm{CN})_6]$, $[\mathrm{Fe}(\mathrm{NH}_3)_6] [\mathrm{Co}(\mathrm{CN})_6]$, and $[\mathrm{Co}(\mathrm{NH}_3)_6] [\mathrm{Fe}(\mathrm{CN})_6]$, as these combinations involve different metal centers that permit ligand migration to create unique chemical arrangements. Conversely, complexes like $[\mathrm{Ag}(\mathrm{NH}_3)_2] [\mathrm{Ag}(\mathrm{CN})_2]$ and $[\mathrm{Co}(\mathrm{NH}_3)_6] [\mathrm{Co}(\mathrm{CN})_6]$ do not exhibit this isomerism, as swapping the ligands between two identical metal centers fails to produce a chemically distinct compound.

The conductivity of the chromium complex corresponding to a $1:2$ electrolyte indicates that two chloride ions are outside the coordination sphere, while one is coordinated to the chromium ion. Passing the complex solution through a cation exchanger replaces the complex cation $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]^{2+}$ with hydrogen ions, leaving only the two ionizable chloride ions in the solution available to react with silver nitrate.

The molar mass of $\text{CrCl}_3 \cdot 6\text{H}_2\text{O}$ is calculated as $52 + 3(35.5) + 6(18) = 266.5 \text{ g mol}^{-1}$. Given 2.75 g of the complex, the number of moles is $\frac{2.75}{266.5} \approx 0.01032 \text{ mol}$. Each mole of the complex yields two moles of chloride ions, so the moles of $\text{AgCl}$ formed are $2 \times 0.01032 = 0.02064 \text{ mol}$. The mass of $\text{AgCl}$ formed is $0.02064 \times 143.5 \approx 2.96 \text{ g}$, which rounds to 3 g.

The complex $[\mathrm{Ni}(\mathrm{PPh}_3)_2\mathrm{Cl}_2]$ features $\mathrm{Ni}^{2+}$ in a $d^8$ configuration, adopting a paramagnetic tetrahedral geometry that precludes geometrical isomerism because all four ligand positions in a tetrahedron are equivalent, rendering statement A incorrect. The presence of partially filled $d$-orbitals allows for $d-d$ electronic transitions that impart color to the substance, contradicting the assertion that the complex is white, which makes statement B incorrect. Furthermore, the CFSE for a tetrahedral $d^8$ configuration is calculated as $4(-0.6\Delta_t) + 4(0.4\Delta_t) = -0.8\Delta_t$, which converts to approximately $-0.355\Delta_o$ since $\Delta_t = \frac{4}{9}\Delta_o$, making statement D incorrect as it falsely equates this energy to $-0.8\Delta_o$.

Strong field ligands induce electron pairing in octahedral complexes to form inner-orbital $d^2sp^3$ hybrids, while weak field ligands result in outer-orbital $sp^3d^2$ hybridization. Complexes $[\mathrm{FeF}_6]^{3-}$, $[\mathrm{MnCl}_6]^{3-}$, and $[\mathrm{CoF}_6]^{3-}$ feature weak field ligands and therefore undergo $sp^3d^2$ hybridization. Among the $d^2sp^3$ hybridized complexes, $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ has a $d^6$ configuration with all electrons paired in the $t_{2g}$ orbitals, rendering it diamagnetic. In contrast, $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ ($d^5$) and $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$ ($d^4$) utilize $d^2sp^3$ hybridization while retaining one and two unpaired electrons, respectively, confirming their paramagnetism. Consequently, exactly two of the listed complexes satisfy both the hybridization and paramagnetism criteria.

The spin-only magnetic moment of a metal complex is governed by the number of unpaired electrons in the $d$-orbitals, which is determined by the metal's oxidation state and the field strength of the ligands. Using the relation $\mu = \sqrt{n(n+2)}$ B.M., where $n$ denotes the number of unpaired electrons, we can quantify the magnetic character of these iron complexes.

In $\mathrm{K}_3[\mathrm{Fe}(\mathrm{OH})_6]$, the central iron atom is in the $+3$ oxidation state, corresponding to a $3d^5$ electronic configuration. Since the hydroxide ion is a weak field ligand, it does not induce electron pairing, resulting in five unpaired electrons in the $t_{2g}^3 e_g^2$ configuration and a magnetic moment of 5.92 B.M. In $\mathrm{K}_4[\mathrm{Fe}(\mathrm{OH})_6]$, the iron is in the $+2$ oxidation state with a $3d^6$ configuration. Given the weak field nature of the ligands, the electrons distribute as $t_{2g}^4 e_g^2$, leaving four unpaired electrons and yielding a magnetic moment of 4.90 B.M.

The $[\mathrm{CrCl}_3(\mathrm{py})_3]$ complex follows the $MA_3B_3$ type octahedral geometry, which exhibits geometrical isomerism resulting in two distinct arrangements: the facial (fac) isomer, where the three identical ligands occupy the vertices of one triangular face, and the meridional (mer) isomer, where the three identical ligands are arranged meridian to the central atom, totaling 2 stereoisomers. For the $[\mathrm{CrCl}_2(\mathrm{ox})_2]^{3-}$ complex, categorized as the $M(A)_2(L-L)_2$ type where $(L-L)$ represents the bidentate oxalate ligand, the chlorine ligands can be arranged in either a cis or trans configuration. The trans isomer is achiral and represents 1 stereoisomer, whereas the cis isomer lacks a center or plane of symmetry, existing as a pair of non-superimposable mirror images (enantiomers), which adds 2 stereoisomers for a total of 3.

The Crystal Field Stabilization Energy of a metal complex reaches zero when its $d$-orbitals are symmetrically occupied, effectively canceling out the stabilization energy. For the $\mathrm{K}_3[\mathrm{Fe}(\mathrm{SCN})_6]$ complex, the central iron atom is in the $+3$ oxidation state with a $3d^5$ electronic configuration. As $\mathrm{SCN}^-$ is a weak field ligand, these five electrons arrange singly across the $t_{2g}$ and $e_g$ energy levels to yield a $t_{2g}^3 e_g^2$ configuration, which results in a net CFSE of zero.