Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

The magnetic and structural properties of nickel complexes are governed by the oxidation state of the metal and the field strength of the surrounding ligands. $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ uses strong field ligands to force a square planar arrangement with $dsp^2$ hybridization and a diamagnetic $d^8$ configuration, while $[\mathrm{Ni}(\mathrm{CO})_4]$ adopts a tetrahedral structure with $sp^3$ hybridization and a diamagnetic $d^{10}$ configuration, which contradicts the description of it as paramagnetic. Similarly, comparing $[\mathrm{NiCl}_4]^{2-}$ and $[\mathrm{Ni}(\mathrm{CO})_4]$ reveals distinct electronic structures; while both are tetrahedral, the former is a paramagnetic $d^8$ species with two unpaired electrons, whereas the latter is a diamagnetic $d^{10}$ species. These differences in magnetic behavior and electronic configuration demonstrate that both statements are inaccurate.

The spin only magnetic moment depends on the number of unpaired electrons in the metal center of a coordination complex. By accounting for the charge of the cyanide ligands, iron is identified to be in the +3 oxidation state, resulting in a 3d5 electronic configuration. Since cyanide is a strong field ligand, it forces the d-electrons to pair up, leaving just 1 unpaired electron. Applying the formula $\mu_s = \sqrt{n(n+2)}$ B.M. with $n=1$ leads to $\sqrt{3}$, which is approximately 1.73. Rounding this value to the nearest integer results in 2 B.M.

Paramagnetism in coordination complexes occurs whenever the central metal ion possesses unpaired electrons within its $d$-orbitals, a condition governed by the oxidation state of the metal and the splitting effect of the ligands. The complexes $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$, $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, and $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ are all diamagnetic because the strong field cyanide ligands force the electrons into paired configurations. Conversely, $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ contains one unpaired electron despite the strong ligand field, and $[\mathrm{Ti}(\mathrm{CN})_6]^{3-}$ also exhibits one unpaired electron due to its $d^1$ configuration. Counting these specific examples confirms that there are 2 paramagnetic complexes.

Identifying the complex that reacts with silver nitrate to precipitate two moles of silver chloride requires examining the coordination sphere of each compound. The complex $\mathrm{NiCl}_2 \cdot 6 \mathrm{H}_2\mathrm{O}$ acts as an ionic compound with the structure $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$, where the two chloride ions are positioned outside the coordination sphere and are readily available to react with excess $\mathrm{AgNO}_3$. This distinguishes it from the other complexes provided, which do not release two chloride ions in the same manner.

The central nickel ion in this octahedral complex is in the $+2$ oxidation state, leading to a $d^8$ electronic configuration. In an octahedral crystal field, these electrons occupy the $d$-orbitals such that the $t_{2g}$ orbitals are fully filled with six electrons, while the remaining two electrons reside in the $e_g$ set. This configuration results in $n=2$ unpaired electrons. The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$, which gives $\mu = \sqrt{2(2+2)} = \sqrt{8}$. This value is approximately 2.83, and rounding to the nearest integer provides 3.

Synergic bonding is the fundamental interaction responsible for the stability of metal carbonyl complexes, occurring through a dual process of $\sigma$-donation from the carbonyl carbon to the metal and $\pi$-back-donation from filled metal $d$-orbitals into the empty $\pi^*$ antibonding orbitals of the carbon monoxide ligand. Since $[\mathrm{Cr}(\mathrm{CO})_6]$, $[\mathrm{Mn}(\mathrm{CO})_5]$, and $[\mathrm{Mn}_2(\mathrm{CO})_{10}]$ all function as coordination compounds containing carbon monoxide directly bonded to a metal center, each of these three species possesses the necessary metal-ligand architecture to facilitate this electron back-donation. Therefore, all three listed complexes exhibit synergic bonding.

The magnetic moment ($\mu$) is calculated using the relation $\mu = \sqrt{n(n+2)}\,\mathrm{BM}$, where $n$ represents the number of unpaired electrons; a value of 1.73 BM corresponds to $n=1$. $\mathrm{O}_2^+$ contains 15 electrons, with a molecular orbital configuration that leaves one unpaired electron in the $\pi_{2p}^*$ orbital. The complex $[\mathrm{Cu}(\mathrm{NH}_3)_4] \mathrm{Cl}_2$ contains $\mathrm{Cu}^{2+}$ with a $3d^9$ configuration, which results in one unpaired electron. Similarly, $\mathrm{O}_2^-$ has 17 electrons and one unpaired electron in the $\pi_{2p}^*$ orbital. However, $\mathrm{CuI}$ consists of $\mathrm{Cu}^+$ and $\mathrm{I}^-$ ions, both of which have completely filled shells, specifically a $3d^{10}$ configuration for the copper ion and a full octet for the iodide ion, leaving zero unpaired electrons and resulting in a magnetic moment of zero.

Valence bond theory determines the hybridization of a central metal atom by considering the coordination number and the field strength of the attached ligands, which dictates whether electrons in the $d$-orbitals undergo pairing. In $\text{K}_2[\text{Ni}(\text{CN})_4]$, the nickel ion exists in the $+2$ oxidation state with a $3d^8$ electronic configuration, and the strong field $\text{CN}^-$ ligands force the two unpaired electrons to pair up, vacating one $3d$ orbital to participate in $dsp^2$ hybridization. The other compounds either feature weak field ligands that fail to induce pairing, resulting in $sp^3$ or $sp^3d^2$ geometries, or involve a nickel complex where the metal adopts $sp^3$ hybridization due to the tetrahedral geometry.

The hybridization and magnetic behavior of a coordination complex are determined by the electronic configuration of the central metal ion and the field strength of the surrounding ligands. Manganese in the complex $[{MnCl}_{6}]^{3-}$ is in the $+3$ oxidation state, yielding a $3d^{4}$ electronic configuration, and because chloride ions are weak-field ligands, they lack the strength to force pairing among these electrons. This results in the utilization of the outer $4d$ orbitals for bond formation, leading to $sp^{3}d^{2}$ hybridization, and the presence of four unpaired electrons ensures that the complex is paramagnetic.

The magnetic moment of a transition metal complex is fundamentally determined by the number of unpaired electrons present in the central metal ion. To determine this for the complex $[\mathrm{Co}(\mathrm{CN})_6]^{4-}$, we first identify the oxidation state of the cobalt center. Because the cyanide ligand carries a charge of $-1$, the metal must exist in the $+2$ oxidation state to balance the overall complex charge of $-4$. A neutral cobalt atom has an atomic number of 27, so the $\mathrm{Co}^{2+}$ ion possesses a $3d^7$ electronic configuration. Since cyanide acts as a strong field ligand, it forces the $d$-orbital electrons to pair up, leaving only 1 unpaired electron. Applying the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM, where $n$ represents the number of unpaired electrons, we substitute $n=1$ to calculate $\mu = \sqrt{3}$ BM. This results in a value of approximately 1.73 BM, which is typically rounded to 2 BM in this context.

In coordination chemistry, the secondary valency of a central metal ion is equivalent to its coordination number, which represents the total number of donor atoms directly bonded to the metal. The reaction of $\mathrm{FeCl}_3$ with oxalic acid in the presence of $\mathrm{KOH}$ yields the potassium trioxalatoferrate(III) complex, formulated as $\mathrm{K}_3[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]$. Within this complex, the central iron atom is surrounded by three oxalate ions. Since an oxalate ion acts as a bidentate ligand providing two donor sites, the total coordination number is calculated as $3 \times 2 = 6$. Thus, the secondary valency of iron in this product is 6.

Crystal field theory provides the basis for understanding how the interaction between metal ions and ligands dictates the hybridization and magnetic properties of coordination complexes. In the first group of compounds, the presence of strong field ligands such as cyanide, along with the chelating effect of the oxalate ion, creates a large energy separation between $d$-orbitals. This environment forces the $d$-electrons of the $\mathrm{Mn}^{3+}$, $\mathrm{Fe}^{3+}$, and $\mathrm{Co}^{3+}$ centers to pair up, which effectively clears the necessary inner $d$-orbitals for $d^2sp^3$ hybridization.

Conversely, the second group of complexes features chloride and fluoride ions, which are categorized as weak field ligands. Because these ligands cannot induce spin pairing, the $d$-electrons remain unpaired according to Hund's rule. Consequently, the $\mathrm{Mn}^{3+}$ ion in $[\mathrm{MnCl}_6]^{3-}$ retains a high-spin $d^4$ configuration with four unpaired electrons, while the $\mathrm{Fe}^{3+}$ ion in $[\mathrm{FeF}_6]^{3-}$ maintains a $d^5$ configuration with five unpaired electrons, rendering both species paramagnetic. Since the hybridization assignments in the first statement and the magnetic descriptions in the second statement align perfectly with these principles, both statements are factually accurate.

The magnetic moment value of $3.87\ \mathrm{BM}$ is determined by the spin-only formula $\mu = \sqrt{n(n+2)}$, which indicates that the metal ion must have three unpaired electrons, as $\sqrt{3(3+2)} = \sqrt{15} \approx 3.87$. Evaluating the crystal field stabilisation energy requires the distribution of these electrons into $t_{2g}$ and $e_g$ orbitals according to the relation $\text{CFSE} = [(n_{t_{2g}}) \times (-0.4) + (n_{e_{g}}) \times 0.6] \Delta_0$. For a $\mathrm{Co}^{2+}$ ion in an octahedral aqua complex, the $d^7$ electronic configuration $t_{2g}^5 e_g^2$ perfectly yields a CFSE of $[5 \times (-0.4) + 2 \times 0.6] \Delta_0 = -0.8 \Delta_0$. Since these parameters align precisely with the provided data, the metal ion is identified as $\mathrm{Co}^{2+}$.

An atomic number of 29 corresponds to copper, which in its divalent cationic state, $\text{Cu}^{2+}$, adopts an electronic configuration of [Ar] $3d^9$. Distributing these electrons across the five 3d orbitals reveals that four orbitals contain pairs, leaving exactly one unpaired electron. Given the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$, substituting $n = 1$ results in $\mu = \sqrt{3}$. This calculation yields approximately 1.73 BM, which corresponds to 2 BM when rounded.

The iron center in $[\mathrm{Fe}(\mathrm{CO})_{4}(\mathrm{C}_{2}\mathrm{O}_{4})]^{+}$ carries a $+3$ oxidation state because the neutral carbonyl ligands and the $-2$ oxalate ligand must balance the complex charge of $+1$, leading to a $[\mathrm{Ar}] 3d^5$ electronic configuration.

Because carbon monoxide acts as a strong field ligand, it forces the $d$-electrons to pair within the $t_{2g}$ orbitals, arranging them as $t_{2g}^{2,2,1}$ with the $e_g$ set remaining empty. This leaves exactly one unpaired electron, so the spin-only magnetic moment is calculated as $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM.

The coordination complex $\left[\mathrm{Co}(\mathrm{Ox})_2(\mathrm{Br})(\mathrm{NH}_3)\right]^{2-}$ follows the generic structural type $\left[\mathrm{M}(\mathrm{AA})_2 \mathrm{BC}\right]$, which is capable of exhibiting both optical and geometrical isomerism. Because the cis-isomer is chiral, it exists as a pair of non-superimposable mirror images, resulting in two distinct optical forms. The trans-isomer, where the monodentate ligands occupy opposite coordination sites, exhibits no optical activity and provides one additional unique form. Summing the two enantiomers from the cis-configuration with the single trans-isomer yields 3 possible stereoisomers.

Geometric isomerism in coordination complexes is defined by the specific spatial arrangement of ligands around the central metal atom. For the square planar $[NiBr_2(PPh_3)_2]$, the trans isomer is characterized by the two bromide ligands occupying positions directly opposite each other, resulting in a bond angle of 180 degrees across the nickel center. In contrast, the meridional isomer of the octahedral $[Co(NH_3)_3(NO_2)_3]$ complex exists when the three identical ligands are arranged in a single plane passing through the metal, distinct from the facial arrangement where they cluster on one side of the octahedron.

The magnitude of the crystal field stabilisation energy (CFSE) is fundamentally determined by the field strength of the coordinating ligands when the central metal ion remains constant. According to the spectrochemical series, these specific ligands can be ordered by their increasing splitting capability as $\mathrm{F}^- < \mathrm{H}_2\mathrm{O} < \mathrm{NH}_3 < \mathrm{en}$. Since each step in this series represents a more effective interaction with the d-orbitals of the $\mathrm{Co}^{3+}$ center, the resulting complexes increase in their CFSE value following this exact progression, yielding the sequence $[\mathrm{CoF}_6]^{3-} < [\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+} < [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} < [\mathrm{Co}(\mathrm{en})_3]^{3+}$, which corresponds to $A < B < C < D$.

The octahedral splitting energy, $\Delta_o$, of an octahedral complex represents the energy difference between the $t_{2g}$ and $e_g$ orbitals, which is equal to the energy of the photon absorbed during the electronic transition. Using the Planck-Einstein relation $\Delta_o = \frac{hc}{\lambda}$, where $h = 6.626 \times 10^{-34}\,\mathrm{Js}$ is Planck's constant, $c = 3 \times 10^{8}\,\mathrm{ms}^{-1}$ is the speed of light, and the wavelength $\lambda = 498 \times 10^{-9}\,\mathrm{m}$, the energy is calculated as $\Delta_o = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{498 \times 10^{-9}}$. Computing this expression results in approximately $3.99 \times 10^{-19}\,\mathrm{J}$, which rounds to 4.

Denticity represents the number of donor atoms through which a ligand coordinates to a central metal ion to form a complex. Biuret, which possesses the chemical structure $NH_2CONHCONH_2$, functions as a bidentate ligand because it attaches to the metal center using two donor atoms. By utilizing these two coordination sites simultaneously, the molecule establishes a denticity value of 2.

The magnetic moment of a transition metal complex, calculated as the spin-only value, is determined by the number of unpaired electrons ($n$) present in the $d$-orbitals, which is defined by the formula $\mu = \sqrt{n(n+2)}$. This value depends on both the oxidation state of the central metal and the nature of the ligands, which dictate whether electrons pair up or remain unpaired.

For the complex $[\mathrm{FeCl}_4]^{2-}$, the iron center is in the $+2$ oxidation state, resulting in a $3d^6$ electronic configuration. Because the chloride ion is a weak-field ligand, it does not induce electron pairing, leaving four electrons unpaired. Substituting $n=4$ into the formula $\sqrt{n(n+2)}$ yields $\sqrt{4(4+2)}$, which simplifies to $\sqrt{24}$ or 4.90 BM.

In the case of $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$, the cobalt ion exists in the $+3$ oxidation state, which also corresponds to a $3d^6$ configuration. However, the oxalate ligand is a strong-field ligand that forces all $d$-electrons to pair up within the lower energy orbitals. With zero unpaired electrons, the magnetic moment becomes 0 BM.

Finally, for the species $\mathrm{MnO}_4^{2-}$, manganese is in the $+6$ oxidation state, resulting in a $3d^1$ configuration. This single electron remains unpaired, so with $n=1$, the expression $\sqrt{1(1+2)}$ results in $\sqrt{3}$, which equals 1.73 BM.