Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

According to the spectrochemical series, the magnitude of crystal field splitting ($\Delta$) for halide ligands follows the order $I^- < Br^- < Cl^- < F^-$, which directly correlates to the energy of absorbed light ($E = h\nu$). Since $Br^-$ is a weaker field ligand than $Cl^-$, the $[CoBr_4]^{2-}$ complex experiences smaller splitting and thus absorbs light of lower energy than $[CoCl_4]^{2-}$, confirming that Statement I is true. Conversely, because $I^-$ is a weaker field ligand than $Cl^-$, the energy separation between the d-orbitals in $[CoI_4]^{2-}$ is smaller than in $[CoCl_4]^{2-}$, which makes Statement II false.

The strength of metallic bonding in transition metals is directly proportional to the number of unpaired electrons available for delocalization, which means that a greater number of unpaired electrons requires significantly more energy to break the atomic lattice, resulting in higher enthalpies of atomization. Crystal field theory demonstrates that the geometric arrangement of ligands dictates the splitting pattern of d-orbitals, where the octahedral field in $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ lowers the energy of the $d_{xy}, d_{xz}, d_{yz}$ set relative to the $d_{x^{2}-y^{2}}, d_{z^{2}}$ set ($d_{xy}=d_{xz}=d_{yz} < d_{x^{2}-y^{2}}=d_{z^{2}}$), whereas the tetrahedral field in $[\mathrm{Ni}(\mathrm{Cl})_4]^{2-}$ forces an inverted energy hierarchy such that the $d_{x^{2}-y^{2}}, d_{z^{2}}$ set is lower than the $d_{xy}, d_{xz}, d_{yz}$ set ($d_{x^{2}-y^{2}}=d_{z^{2}} < d_{xy}=d_{xz}=d_{yz}$).

Determining the number of unpaired electrons in octahedral complexes relies on Crystal Field Theory, where the distribution of electrons between $t_{2g}$ and $e_g$ orbitals depends on the strength of the ligand field. A $d^3$ configuration possesses three unpaired electrons, whereas a low-spin $d^4$ system forces pairing within the lower energy orbitals to yield two unpaired electrons. High-spin complexes maximize unpaired electrons by populating the $d$-orbitals singly before pairing, resulting in five unpaired electrons for $d^5$ and four for $d^6$. A low-spin $d^7$ system accommodates six electrons in the $t_{2g}$ set and one in the $e_g$ set, leaving one unpaired electron. Summing these individual counts as $3 + 2 + 5 + 4 + 1 = 15$ provides the final result.

Metal carbonyls are stabilized through a synergic bonding mechanism that imparts both $\sigma$ and $\pi$ character to the metal-carbon interaction. The $\sigma$-component forms by the donation of a lone pair from the carbonyl carbon into a vacant metal orbital, while the $\pi$-component is established when electrons are donated from a filled metal d-orbital into the vacant $\pi^*$ antibonding orbital of the CO ligand. Because this synergic effect actually strengthens the metal-carbon bond, the assertion that such interactions weaken the bond is inaccurate.

Square planar complexes of the type $[\mathrm{Mabcd}]$ exhibit three distinct geometrical isomers because fixing the position of one ligand allows the remaining three ligands to be arranged in three unique spatial configurations, so $X = 3$. The octahedral complex $[\mathrm{CrCl}_2(\mathrm{ox})_2]^{3-}$ exists in cis and trans geometrical forms; the cis-isomer is chiral and therefore optically active, whereas the trans-isomer possesses a plane of symmetry, rendering it optically inactive, so $Y = 1$. The octahedral complex $[\mathrm{Co}(\mathrm{NH}_3)_3(\mathrm{NO}_2)_3]$ follows the $[\mathrm{Ma}_3\mathrm{b}_3]$ configuration, which exhibits geometrical isomerism in the facial (fac) and meridional (mer) forms, so $Z = 2$. Calculating the sum yields $X + Y + Z = 3 + 1 + 2 = 6$.

The paramagnetism of transition metal complexes is determined by the number of unpaired electrons, which is dictated by the $d$-electron configuration of the metal ion and the field strength of the ligands. In $[\mathrm{CoF}_6]^{3-}$ ($d^6$ high-spin), $[\mathrm{TiF}_6]^{3-}$ ($d^1$), and $[\mathrm{Fe(CN)}_6]^{3-}$ ($d^5$ low-spin), the central metal ions possess 4, 1, and 1 unpaired electrons respectively, making them paramagnetic, while $\mathrm{V}_2\mathrm{O}_5$ ($\mathrm{V}^{5+}$, $d^0$) is diamagnetic, verifying that Statement I is correct. For Statement II, the respective numbers of unpaired electrons for $\mathrm{K}_4[\mathrm{Fe(CN)}_6]$ ($d^6$ low-spin), $\mathrm{K}_3[\mathrm{Fe(CN)}_6]$ ($d^5$ low-spin), $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]\mathrm{SO}_4 \cdot \mathrm{H}_2\mathrm{O}$ ($d^6$ high-spin), and $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_3$ ($d^5$ high-spin) are 0, 1, 4, and 5, which validates the inequality $0 < 1 < 4 < 5$ and confirms Statement II is also correct.

Crystal field theory dictates the arrangement of electrons in transition metal complexes, where the coordination geometry and ligand field strength determine if electrons remain unpaired. For $[\mathrm{Ni}(\mathrm{CO})_4]$, nickel exists in the 0 oxidation state with a $3d^8 4s^2$ configuration, and the strong field ligand $\mathrm{CO}$ forces pairing, leading to 0 unpaired electrons. The complex $[\mathrm{NiCl}_4]^{2-}$ contains $\mathrm{Ni}^{2+}$ with a $3d^8$ configuration coordinated with the weak field ligand $\mathrm{Cl}^-$ in a tetrahedral geometry, resulting in 2 unpaired electrons due to the $e^4 t_2^4$ occupancy. In the square planar complexes $[\mathrm{PtCl}_2(\mathrm{NH}_3)_2]$, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, and $[\mathrm{Pt}(\mathrm{CN})_4]^{2-}$, the metal centers ($\mathrm{Ni}^{2+}$ or $\mathrm{Pt}^{2+}$) have $d^8$ configurations, and the strong field ligands ensure all $d$ electrons are paired, yielding 0 unpaired electrons for each species. The total number of unpaired electrons is calculated as $0 + 2 + 0 + 0 + 0 = 2$.

The magnitude of crystal field splitting, $\Delta_0$, is governed by the field strength of the ligands, as defined by the spectrochemical series where stronger ligands induce a larger splitting of the $d$-orbitals. Ordering the ligands-cyanide, ethylenediamine, water, and fluoride-by their donor strength determines the sequence of their corresponding energy splitting values.

The cyanide ligand is the strongest in this set and induces the highest splitting of 26,600 cm⁻¹ for $[\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}$, which corresponds to IV. Ethylenediamine acts as the next strongest ligand, producing a splitting of 22,300 cm⁻¹ for $[\mathrm{Cr}(\mathrm{en})_{3}]^{3+}$, which aligns with III. The water ligand is weaker, leading to a splitting of 17,400 cm⁻¹ for $[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}$, matching II. Finally, the fluoride ion is the weakest ligand, resulting in the smallest splitting value of 15,060 cm⁻¹ for $[\mathrm{CrF}_{6}]^{3-}$, which corresponds to I.

Coordination compounds dissociate in aqueous solution such that only the ions outside the coordination sphere become available for precipitation reactions. A 2 L portion of the mixture contains 0.2 mol of $[Co(NH_3)_5SO_4]Br$ and 0.2 mol of $[Co(NH_3)_5Br]SO_4$. Upon addition of excess $AgNO_3$, the bromide ion dissociates from $[Co(NH_3)_5SO_4]Br$ to form 0.2 mol of $AgBr$ precipitate 'Y' according to the reaction $Br^- + Ag^+ \rightarrow AgBr(s)$. When the other 2 L sample reacts with excess $BaCl_2$, the sulfate ion dissociates from $[Co(NH_3)_5Br]SO_4$ to form precipitate 'Z' via the reaction $SO_4^{2-} + Ba^{2+} \rightarrow BaSO_4(s)$, which yields 0.2 mol of $BaSO_4$.

Valence bond theory and crystal field theory dictate the hybridization, geometry, and magnetic properties of coordination complexes based on the oxidation state of the central metal ion and the field strength of the ligands. In $K_3[Co(CO_3)_3]$, the oxidation state of cobalt is $+3$, corresponding to a $3d^6$ electronic configuration. Because the carbonate ($CO_3^{2-}$) ion acts as a weak field ligand, it does not cause electron pairing, leading to $sp^3d^2$ hybridization, an octahedral geometry, and four unpaired electrons ($t_{2g}^4 e_g^2$) which results in a spin-only magnetic moment of $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM.

For $[Ni(CN)_4]^{2-}$, the strong field ligand $CN^-$ causes pairing of the $d$-electrons in the $Ni^{2+}$ ion ($3d^8$), resulting in $dsp^2$ hybridization, square planar geometry, and diamagnetic behavior where $\mu = 0$ BM. The complex $[MnBr_4]^{2-}$ consists of $Mn^{2+}$ ($3d^5$) coordinated with weak field $Br^-$ ligands, resulting in $sp^3$ hybridization, tetrahedral geometry, and five unpaired electrons, which corresponds to a magnetic moment of $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM. Finally, $[CoF_6]^{3-}$ involves $Co^{3+}$ ($3d^6$) with weak field $F^-$ ligands, yielding an outer orbital $sp^3d^2$ octahedral complex with four unpaired electrons and a magnetic moment of $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM. Since both Statement I and Statement II correctly describe these properties, the assertions are valid.

In a tetrahedral crystal field, the $d$-orbitals split into a lower-energy $e$ set and a higher-energy $t_2$ set, where occupying the $e$ orbitals stabilizes the system by $-\frac{3}{5} \Delta_{t}$ per electron, while occupying $t_2$ orbitals raises energy by $+\frac{2}{5} \Delta_{t}$. The total $\mathrm{CFSE}$ is determined by the formula $\mathrm{CFSE}= \left[ -\frac{3}{5} n_{1} + \frac{2}{5} n_{2} \right] \Delta_{t}$, where $n_1$ and $n_2$ are the number of electrons in the $e$ and $t_2$ orbitals, respectively.

For a $d^2$ metal ion, both electrons enter the lower energy $e$ set, resulting in $2 \times (-\frac{3}{5}) = -1.2 \Delta_{t}$, which matches III. A $d^4$ configuration distributes two electrons into each set, giving $2 \times (-\frac{3}{5}) + 2 \times (\frac{2}{5}) = -0.4 \Delta_{t}$, corresponding to IV. A $d^6$ ion places three electrons in each level, calculating as $3 \times (-\frac{3}{5}) + 3 \times (\frac{2}{5}) = -0.6 \Delta_{t}$, matching I. Finally, the $d^8$ configuration places four electrons in each set, yielding $4 \times (-\frac{3}{5}) + 4 \times (\frac{2}{5}) = -0.8 \Delta_{t}$, which aligns with II.

The stability of coordination complexes is heavily influenced by the chelation effect, where multidentate ligands form stable rings around the central metal atom. Because the oxalate ion in $[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ acts as a bidentate chelating agent, this complex demonstrates superior stability compared to the others mentioned which involve monodentate ligands. In contrast, when comparing $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ and $[\mathrm{Cu}(\mathrm{en})_2]^{2+}$, the complex containing ethylenediamine is actually more stable due to this same chelation phenomenon, rendering the assertion that $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ is more stable incorrect.

When considering the bonding in $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$, the iron exists in the $\mathrm{Fe}^{2+}$ state with a $3d^6$ electronic configuration. The presence of a strong field ligand like cyanide induces electron pairing, which empties the necessary orbitals to facilitate $d^{2}sp^{3}$ hybridization. Finally, linkage isomerism is a distinct property displayed by complexes containing ambidentate ligands, such as $\mathrm{NO}_2^{-}$, which can coordinate to a metal center through either its nitrogen or oxygen atom. Since both $\mathrm{NO}_2^{-}$ and $\mathrm{SCN}^{-}$ are inherently ambidentate, the statement denying this characteristic is false, whereas the complex $[\mathrm{Fe}(\mathrm{NO}_2)_3\mathrm{Cl}_3]^{3-}$ is correctly identified as one capable of exhibiting linkage isomerism.

The spin-only magnetic moment $\mu$ is determined by the number of unpaired electrons $n$ through the relation $\mu = \sqrt{n(n+2)}$ BM. For the $[MnBr_4]^{2-}$ complex, the $Mn^{2+}$ ion ($3d^5$) occupies a tetrahedral geometry with weak field ligands, resulting in $n=5$. For $[Cu(H_2O)_6]^{2+}$, the $Cu^{2+}$ ion ($3d^9$) has $n=1$ unpaired electron. The $[Ni(CN)_4]^{2-}$ complex contains $Ni^{2+}$ ($3d^8$), where the strong field $CN^-$ ligand induces pairing to form a square planar geometry, resulting in $n=0$. In $[Ni(H_2O)_6]^{2+}$, the $Ni^{2+}$ ($3d^8$) ion in an octahedral field with weak field ligands retains two unpaired electrons, so $n=2$. Arranging these complexes according to their respective $n$ values of $5$ (A), $1$ (B), $0$ (C), and $2$ (D) in increasing order results in $0 < 1 < 2 < 5$, which corresponds to the sequence C $<$ B $<$ D $<$ A.

The third ionization energy ($\mathrm{IE}_3$) represents the energy required to remove an electron from the $M^{2+}$ ion, which generally increases across the transition series as the effective nuclear charge rises. For scandium, the removal of the third electron is relatively easier, corresponding to a value of 2393 kJ/mol, whereas for zinc, removing an electron from the stable, fully filled $3d^{10}$ subshell is exceptionally difficult, requiring 3837 kJ/mol. Since the values for manganese and copper fall between these extremes, the increasing order of $\mathrm{Sc} < \mathrm{Mn} < \mathrm{Cu} < \mathrm{Zn}$ confirms that the first statement is true.

The Crystal Field Stabilization Energy (CFSE) depends on both the oxidation state of the central metal ion and the field strength of the ligands. Because $\mathrm{Co}^{3+}$ ions induce greater splitting than $\mathrm{Co}^{2+}$ ions, and ethylenediamine functions as a stronger field ligand compared to water, the stabilization energies follow the order:
$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+} < [\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+} < [\mathrm{Co}(\mathrm{en})_3]^{3+}$
This hierarchy validates that the second statement is also true.

In all three complexes, the central nickel ion is in the $+2$ oxidation state with a $3d^8$ electronic configuration. Regardless of whether the geometry is octahedral or tetrahedral, the distribution of these electrons consistently leaves two unpaired electrons in each case. The frequency of light absorbed is directly proportional to the magnitude of the crystal field splitting energy, which is significantly larger in octahedral complexes like $[\mathrm{Ni}(\mathrm{en})_3]^{2+}$ and $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$ compared to the tetrahedral complex $[\mathrm{NiCl}_4]^{2-}$. Because the ligand field strength of ethylenediamine exceeds that of ammonia, the octahedral complex with ethylenediamine experiences the greatest splitting, leading to the overall trend where the absorption frequency of (A) is greater than (C), which is greater than (B).

The behavior of coordination compounds in aqueous solution depends on the distinction between ligands bound inside the coordination sphere and counter-ions residing outside. Only the counter-ions that dissociate completely when dissolved are available to participate in precipitation reactions with reagents like silver nitrate. In the complex $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2]\mathrm{Cl}$, the chlorine atom outside the square bracket acts as the ionizable counter-ion, while the two chlorine atoms inside are firmly bonded to the chromium center and do not ionize.

Because the ratio of this specific complex to its dissociable chloride ion is $1:1$, every mole of the complex produces one mole of $AgCl$ upon reaction with excess silver nitrate. We determine the initial amount of the complex by multiplying its molar concentration by its volume in liters, calculating $0.05 \times 0.1 = 0.005$ moles. Given the $1:1$ stoichiometry, this leads to the formation of exactly $5 \times 10^{-3}$ moles of $AgCl$, identifying the required integer as 5.

Statement I is true because calculating the number of unpaired electrons based on electron configuration and ligand field splitting reveals that $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ ($d^9$) has one, $[\mathrm{Ni}(\mathrm{en})_3]^{2+}$ and $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$ ($d^8$ octahedral) have two, and $[\mathrm{Mn}(\mathrm{H_2O})_6]^{2+}$ ($d^5$ high-spin) possesses five unpaired electrons, which is the maximum. Statement II is false because evaluating the magnetic properties shows that only the pair $\{ [\mathrm{Ni}(\mathrm{CO})_4], [\mathrm{Ni}(\mathrm{CN})_4]^{2-} \}$ consists entirely of diamagnetic species, whereas the pairs containing $[\mathrm{NiCl}_4]^{2-}$ (which is paramagnetic) do not satisfy this condition, resulting in a total count of only one such pair rather than two.

Crystal Field Theory explains that transition metal d-orbitals are degenerate when in a free atomic state, but this symmetry is broken upon the formation of an octahedral complex due to electrostatic repulsions between metal electrons and incoming ligands. This interaction causes the d-orbitals to split, where the $e_g$ orbitals are destabilized by an energy shift of $+0.6 \Delta_{o}$ and the $t_{2g}$ orbitals are stabilized by an energy shift of $-0.4 \Delta_{o}$ to maintain the system's overall average energy.

The number of electrons in each species is determined by the formula $Z - \text{charge}$, yielding 18 for $\mathrm{Sc}^{3+}$, 22 for $\mathrm{Cr}^{2+}$, 22 for $\mathrm{Mn}^{3+}$, 24 for $\mathrm{Co}^{3+}$, and 23 for $\mathrm{Fe}^{3+}$. Comparing these, only $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ share an identical number of electrons, which establishes that $n=2$.

The formation of $n=2$ moles of $\mathrm{AgCl}$ indicates the presence of two ionizable chloride ions in the complex $[Co(en)_2(NH_3)Cl]Cl_2$. In this octahedral complex, the central metal ion is $\mathrm{Co}^{3+}$, which possesses a $3d^6$ electronic configuration. Under the influence of the strong field ligand ethylenediamine, the electrons pair within the $t_{2g}$ orbitals to form a low-spin $t_{2g}^6 e_g^0$ configuration, resulting in 6 electrons in the $t_{2g}$ orbital.

The crystal field stabilization energy (CFSE) for octahedral complexes is calculated using the formula $CFSE = (-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g}) \Delta_o$. For the $[\text{Cr}(\text{oxalate})_3]^{3-}$ complex, the $Cr^{3+}$ ($d^3$) ion exhibits an electronic configuration of $t_{2g}^3 e_g^0$, resulting in a CFSE magnitude of $1.2 \Delta_o$. Conversely, for the $[\text{Co}(\text{oxalate})_3]^{3-}$ complex, the $Co^{3+}$ ($d^6$) ion under the condition $\Delta_o \gg P$ adopts a low-spin configuration of $t_{2g}^6 e_g^0$, which results in a CFSE magnitude of $2.4 \Delta_o$. The value of $n$ is determined by the ratio of these stabilization energies, $\frac{2.4 \Delta_o}{1.2 \Delta_o} = 2$.