Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

The hybridization of central metal atoms in coordination complexes is dictated by the nature of the ligands and the specific geometry required to minimize electronic repulsions. Carbonyl ligands like $\mathrm{CO}$ are strong-field ligands that promote tetrahedral geometry in $\mathrm{Ni}(\mathrm{CO})_4$, resulting in $sp^3$ hybridization. For $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, the strong cyanide ligand facilitates electron pairing within the nickel ion, enabling a square planar geometry characterized by $dsp^2$ hybridization. Cobalt complexes further illustrate ligand field effects, where strong ligands in $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ force pairing in the inner d-orbitals to achieve $d^2 sp^3$ hybridization, whereas the weak-field fluoride ligands in $[\mathrm{CoF}_6]^{3-}$ cannot force such pairing and instead utilize outer d-orbitals, leading to $sp^3 d^2$ hybridization.

The magnetic behavior of a transition metal complex is determined by the presence or absence of unpaired electrons within the d-orbitals of the central metal ion. In the complex $\mathrm{K}_3[\mathrm{Cu}(\mathrm{CN})_4]$, the copper ion exists in the $+1$ oxidation state, which results in a completely filled $d^{10}$ electronic configuration. Because all d-orbitals are fully occupied, there are no unpaired electrons, rendering the complex diamagnetic, whereas the remaining complexes contain metal ions with unpaired electrons that make them paramagnetic.

The reaction of $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ with excess ammonia in the presence of oxygen leads to the oxidation of the metal center, producing the stable octahedral complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$. In this product, the central cobalt ion exists in the $+3$ oxidation state, which corresponds to a $3d^6$ electronic configuration. Because ammonia acts as a strong field ligand, it creates a large crystal field splitting that promotes a low-spin arrangement of these electrons. In an octahedral ligand field, the $d$-orbitals split into the lower-energy $t_{2g}$ set and the higher-energy $e_g$ set. With all six electrons forced to pair up in the lower energy level to minimize the total energy of the system, they occupy the $t_{2g}$ orbitals exclusively, resulting in the $t_{2g}^6 e_g^0$ configuration and placing 6 electrons within the $t_{2g}$ subshell.

According to Werner's theory, the primary valency of a central metal ion corresponds to its oxidation state, which is satisfied by the negative ions associated with the coordination complex. Since the complex with the formula $\mathrm{CoCl}_3(\mathrm{NH}_3)_4$ acts as a $1:1$ electrolyte, it must dissociate into a positively charged complex ion and a singly charged negative ion, necessitating a structural arrangement of $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]\mathrm{Cl}$. To determine the oxidation state, we assign $x$ to the charge of the cobalt ion and account for the neutral ammonia ligands and two chloride ligands that each carry a charge of $-1$ to balance the overall $+1$ charge of the complex cation, leading to the equation $x + 4(0) + 2(-1) = +1$. Solving this equation results in $x = 3$, which defines the primary valency of the central metal ion.

Hybridization is determined by the central atom's electronic environment and coordination geometry, where the square planar $[PtCl_4]^{2-}$ complex exhibits $dsp^2$ hybridization due to the $5d^8$ electronic configuration of $Pt^{2+}$. The molecule $BrF_5$ has a steric number of $6$ arising from $5$ bonding pairs and $1$ lone pair, leading to $sp^3d^2$ hybridization, whereas $PCl_5$ involves $5$ bonding pairs for a steric number of $5$, resulting in $sp^3d$ hybridization. The octahedral $[Co(NH_3)_6]^{3+}$ complex utilizes $d^2sp^3$ hybridization because the strong-field $NH_3$ ligands induce electron pairing in the $Co^{3+}$ ($3d^6$) center.

Crystal Field Stabilization Energy in octahedral complexes is determined by the specific distribution of electrons across the split $t_{2g}$ and $e_g$ energy levels. In $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, iron exists in the $+3$ oxidation state, resulting in a $d^5$ electronic configuration. Because cyanide acts as a strong field ligand, it forces the electrons to pair within the lower-energy $t_{2g}$ orbitals, creating an inner orbital complex with a $t_{2g}^5 e_g^0$ arrangement. Given that each electron in the $t_{2g}$ set is stabilized by $-\frac{2}{5} \Delta_0$, the total CFSE is calculated as $5 \times (-\frac{2}{5} \Delta_0)$, which yields a value of $-2 \Delta_0$.

Metal complexes featuring metals in low oxidation states are typically electron-rich, creating a need to redistribute excess electron density to maintain stability. Ligands that possess strong $\pi$-acceptor character facilitate this stabilization through a synergic bonding mechanism, where the metal transfers electron density from its filled d-orbitals into the empty antibonding orbitals of the ligand. This back bonding interaction becomes significantly more effective when the metal is in a lower oxidation state, as the increased availability of d-electrons strengthens the overall metal-ligand bond.

The stability of a coordination complex within a crystal field depends primarily on the strength of the ligands involved and the resulting electronic configuration of the central metal ion. Among the options provided, the cyanide ion acts as the strongest field ligand, which creates a large octahedral splitting energy, $\Delta_0$, and forces the electrons into a lower energy arrangement. The $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ complex possesses a $d^6$ configuration that, due to this strong ligand field, results in a substantial stabilization energy calculated as $-2.40 \Delta_0 + 2P$. Consequently, this specific configuration yields the highest overall stabilization among all the choices listed.

The absorption of light in coordination complexes is governed by the crystal field splitting energy, denoted as $\Delta_0$, where the wavelength of the absorbed light ($\lambda$) is inversely proportional to this energy. Stronger ligands exert a greater influence on the d-orbitals, resulting in a larger $\Delta_0$ and consequently a shift toward shorter absorbed wavelengths. Between the two inner-orbital complexes provided, $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$ and $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$, the cyanide ion is a significantly stronger field ligand than ammonia. Therefore, $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$ exhibits the larger crystal field splitting and absorbs light at the shortest wavelength.

For the central iron atom in $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$, the metal exists in the $+3$ oxidation state, which corresponds to a $d^5$ electronic configuration. Due to the strong-field nature of the cyanide ligands, the electrons are forced into a low-spin arrangement, leaving only one unpaired electron in the $t_{2g}$ orbitals. Using the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ represents the number of unpaired electrons, the calculation is $\sqrt{1(1+2)} = \sqrt{3} \approx 1.732$ B.M. Rounding this value to the nearest integer yields 2.

The detection of nitrate ions relies on the formation of a distinct brown interface where a nitrate-containing solution is layered over an acidified ferrous sulfate mixture. This visual marker results from the creation of the specific coordination complex $[Fe(H_2O)_5(NO)]SO_4$, which arises when nitric oxide binds to the hexa-aqua iron(II) species. Because the true identity of the product is a pentaaquanitrosyliron(II) sulfate complex, the term nitroferrous sulphate is chemically inaccurate and does not describe the species formed.

Coordination compounds consist of an inner coordination sphere and an outer ionization sphere, where only the ions located outside the brackets are free to undergo precipitation reactions. To obtain a white precipitate with $\mathrm{BaCl}_2$, the compound must contain a sulfate ion ($\mathrm{SO_4^{2-}}$) in the outer sphere, which reacts to form insoluble $\mathrm{BaSO_4}$. Since $\mathrm{[Co(NH_3)_5Br]SO_4}$ features a sulfate counter ion, it reacts with barium to form this precipitate, yet it lacks the chloride ions necessary to produce a white precipitate upon the addition of $\mathrm{AgNO}_3$. Other candidate compounds either contain chloride in the outer sphere, which would react with silver nitrate, or lack the required sulfate ion to participate in the barium reaction.

The formation of Prussian blue is fundamentally driven by the requirement for electrical neutrality when combining ferric cations with ferrocyanide anions. To stabilize the resulting precipitate, the total positive charge contribution from four $\mathrm{Fe}^{3+}$ ions, totaling $12+$, must exactly balance the total negative charge provided by three $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ units, which equals $12-$. This specific stoichiometric ratio leads to the insoluble coordination complex represented by the chemical equation $4\mathrm{Fe}^{3+}+3[\mathrm{Fe}(\mathrm{CN})_6]^{4-} \rightarrow \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$.

The structure of $[\mathrm{Co}_2(\mathrm{CO})_8]$ is defined by a direct metal-metal bond that links the two cobalt atoms, establishing $X=1$. The ligand arrangement in this complex includes two bridging carbonyl groups and six terminal carbonyl ligands attached to the metal centers, which defines the value $Y=6$. Calculating the required sum of these structural features results in $X+Y=7$.

For the tetrahedral $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_4]^{2+}$ complex, the crystal field splitting energy $\Delta_t$ is related to the octahedral splitting energy $\Delta_0$ of the $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ complex by the expression $\Delta_t = \frac{4}{9} \Delta_0$. The light absorbed during the $d-d$ transition provides the energy $\Delta_t$, which is calculated using the photon energy equation:

$\Delta_t = \frac{hc}{\lambda}$

By substituting the constants $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$, $c = 3.08 \times 10^8\ \mathrm{m\,s^{-1}}$, and the wavelength $\lambda = 600 \times 10^{-9}\ \mathrm{m}$, one can determine the value of $\Delta_t$. Rearranging the relationship to $\Delta_0 = \frac{9}{4} \Delta_t$ and applying these values leads to an octahedral crystal field splitting energy of $745 \times 10^{-21}\ \mathrm{J}.$

The structure of copper sulfate pentahydrate, represented as $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_4]\mathrm{SO}_4 \cdot \mathrm{H}_2\mathrm{O}$, reveals that the central $\mathrm{Cu}^{2+}$ cation is bonded to four water molecules. These coordination bonds form exclusively through the oxygen atoms of the water ligands, confirming the presence of $\mathrm{Cu}-\mathrm{O}$ bonds. Because the sulfate group remains outside the primary coordination sphere and does not bond directly to the metal, the claim that S-based ligands coordinate with the copper ion is inaccurate.

The complex $[\mathrm{Cu}(\mathrm{en})_2(\mathrm{SCN})_2]$ contains the ambidentate ligand thiocyanate, which can coordinate to the central metal atom through either the nitrogen or the sulfur atom, leading to linkage isomerism. By examining the possible donor atom combinations for these two ligands, we identify three stable coordination arrangements: both ligands binding through nitrogen, both ligands binding through sulfur, or one ligand binding through nitrogen while the other binds through sulfur. These combinations result in three distinct isomeric forms: $[\mathrm{Cu}(\mathrm{en})_2(\mathrm{NCS})_2]$, $[\mathrm{Cu}(\mathrm{en})_2(\mathrm{SCN})_2]$, and $[\mathrm{Cu}(\mathrm{en})_2(\mathrm{NCS})(\mathrm{SCN})]$.

The magnetic moment of a transition metal complex is governed by the spin-only formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ represents the number of unpaired electrons in the metal's $d$-orbitals. Complexes with fewer unpaired electrons display lower magnetic moments, while those with more unpaired electrons exhibit higher values, making the electron configuration and ligand strength the primary determinants of the result.

In the case of $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, the central iron ion has a $d^5$ configuration. Because cyanide is a strong field ligand, it induces significant pairing of electrons, leaving only $n=1$ unpaired electron and resulting in a magnetic moment of 1.73 B.M. For $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$, the manganese ion has a $d^4$ configuration; the strong field ligand forces pairing such that $n=2$ unpaired electrons remain, which produces a magnetic moment of 2.82 B.M. Transitioning to $[\mathrm{MnCl}_6]^{3-}$, the complex is high spin with a $d^4$ configuration, which maximizes the number of unpaired electrons at $n=4$, yielding a magnetic moment of 4.89 B.M. Finally, $[\mathrm{FeF}_6]^{3-}$ contains the iron ion in a $d^5$ configuration with weak field fluoride ligands that prevent pairing, leaving all five electrons unpaired with $n=5$, which creates the highest magnetic moment of 5.91 B.M. Comparing these values confirms the increasing order of magnetic moments as $B < D < C < A$.

The Jahn-Teller theorem states that any non-linear molecular system in a degenerate electronic state is inherently unstable and will undergo distortion to lift the degeneracy, thereby lowering its total energy and symmetry. For copper(II) complexes, which possess a $d^9$ electronic configuration, the single electron in the $e_g$ set of orbitals creates an unequal occupancy that drives this spontaneous structural deformation. In an ideal octahedral environment, such as in $[ \mathrm{Cu}( \mathrm{H}_2 \mathrm{O})_6] \mathrm{SO}_4$, all six coordination sites are occupied by identical water ligands, allowing the Jahn-Teller effect to operate fully to minimize the repulsion and stabilize the system. Conversely, in complexes containing different ligands, the structure is already geometrically constrained and tetragonally distorted due to the varying ligand field strengths, so the specific spontaneous distortion driven by the Jahn-Teller effect is most significant in the system that begins with the highest degree of octahedral symmetry.

In an acidic environment, potassium permanganate acts as a strong oxidizing agent, reducing manganese from the +7 oxidation state to the +2 oxidation state and producing $\mathrm{Mn}^{2+}$ ions, which contain 5 unpaired electrons in their 3d orbitals. We calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n + 2)}$, where $n$ is the number of unpaired electrons. Inserting $n = 5$ into the equation yields $\mu = \sqrt{5(5 + 2)} = \sqrt{35}$, which is approximately 5.92 B.M. Rounding this result to the nearest integer provides 6 B.M.

Geometrical isomerism in octahedral coordination complexes requires the arrangement of ligands to vary in space, a condition satisfied only by structures of the type $[\mathrm{MA}_4\mathrm{B}_2]$ or $[\mathrm{MA}_3\mathrm{B}_3]$ that allow for distinct cis and trans configurations. Converting the provided substances into their coordination forms reveals that $\mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3$ becomes $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]\mathrm{Cl}$, which matches the $[\mathrm{MA}_4\mathrm{B}_2]$ structural type and thus exhibits geometrical isomerism. The remaining substances, $\mathrm{CoCl}_3 \cdot 5 \mathrm{NH}_3$ which forms $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$, $\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3$ which forms $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$, and $\mathrm{CoCl}(\mathrm{NO}_3)_2 \cdot 5 \mathrm{NH}_3$ which forms $[\mathrm{Co}(\mathrm{NH}_3)_5(\mathrm{NO}_3)]\mathrm{Cl}(\mathrm{NO}_3)$, all possess either $[\mathrm{MA}_5\mathrm{B}]$ or $[\mathrm{MA}_6]$ type structures that do not support cis-trans forms. Consequently, there is 1 complex from the list that possesses the necessary structural requirements.