Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

To identify the species among $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Co}^{3+}$, and $\mathrm{Cr}^{2+}$ that functions as a strong oxidizing agent in aqueous media, we identify the ion with a high reduction potential, which is $\mathrm{Co}^{3+}$. Because $\mathrm{Co}^{3+}$ readily gains an electron to reduce to the more stable $\mathrm{Co}^{2+}$ oxidation state, it behaves as a strong oxidant. The electronic configuration of $\mathrm{Co}^{3+}$ (atomic number 27) is $[Ar] 3d^6$, which contains 4 unpaired electrons. Applying the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ with $n=4$ yields $\mu = \sqrt{4(4+2)} = \sqrt{24} \, \mathrm{BM}$, which is approximately 5 when rounded to the nearest integer.

The $\mathrm{EDTA}^{4-}$ ion acts as a hexadentate ligand, utilizing its six donor atoms to bind simultaneously to the metal center. Because it occupies six distinct coordination sites around the $\mathrm{Ca}^{2+}$ ion, the resulting complex adopts an octahedral coordination environment. This geometry is the standard spatial arrangement for any complex exhibiting a coordination number of six.

Paramagnetism in coordination complexes is dictated by the number of unpaired electrons present in the metal ion's d-orbitals, which determines the magnetic moment as calculated by the formula $\mu = \sqrt{n(n+2)}$ B.M. Among the given complexes, $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ possesses the highest number of unpaired electrons with $n=5$ and $\mu=5.92$ B.M., followed by $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ and $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, which both contain $n=4$ unpaired electrons resulting in a magnetic moment of 4.89 B.M. The complex $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ contains the fewest unpaired electrons, with $n=3$ resulting in a magnetic moment of 3.87 B.M. Because the degree of paramagnetism scales directly with the number of unpaired electrons, this cobalt complex exhibits the weakest paramagnetic character.

The spectrochemical series arranges ligands based on their capacity to cause $d$-orbital splitting in coordination complexes, with the relative field strength dictated by the electronic nature of the donor atom. Ligands like halides typically act as weak field donors, while neutral ligands like water and ammonia exhibit stronger interactions that result in greater crystal field splitting. Consequently, the experimentally established sequence proceeds from the weakest bromide ion to fluoride, increasing further with the inclusion of water and reaching a higher field strength with ammonia.

In tetrahedral coordination complexes, the $d$-orbitals of the metal ion split into a lower-energy $e$ set and a higher-energy $t_2$ set. Electrons fill the $e$ orbitals first, meaning the $t_2$ orbitals remain vacant unless the metal ion has more than four $d$-electrons. Applying this to the given species, $\mathrm{TiCl}_4$ contains the $\mathrm{Ti}^{+4}$ ion with a $d^0$ configuration ($e^0 t_2^0$), while $[\mathrm{MnO}_4]^{-}$ features the $\mathrm{Mn}^{+7}$ ion, which is also a $d^0$ system ($e^0 t_2^0$). Furthermore, $[\mathrm{FeO}_4]^{2-}$ contains the $\mathrm{Fe}^{+6}$ ion, resulting in a $d^2$ configuration ($e^2 t_2^0$). These three complexes contain no electrons in the $t_2$ orbital because their electron counts are sufficiently low to be accommodated entirely within the $e$ set. In contrast, the remaining species in the list possess $d$-electron configurations of $e^3 t_2^3$ and $e^4 t_2^3$, indicating that their $t_2$ orbitals are populated. Therefore, there are three complexes from the list that meet the specified condition.

Octahedral metal complexes exhibit crystal field splitting, where the five degenerate $d$-orbitals divide into a triplet of lower-energy $t_{2g}$ orbitals and a doublet of higher-energy $e_g$ orbitals. Since water acts as a weak-field ligand in these cases, electrons fill the available orbitals sequentially according to Hund's rule, guided by the specific number of $d$-electrons present in the central metal cation.

For the complex $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the chromium cation possesses a $3d^3$ configuration, placing all three electrons within the $t_{2g}$ set to yield $t_{2g}^3 e_{g}^0$. In $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the central $\mathrm{Fe}^{3+}$ ion contains five $d$-electrons, which populate the orbitals as $t_{2g}^3 e_{g}^2$. The $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ complex contains an $\mathrm{Ni}^{2+}$ ion with eight $d$-electrons, completely filling the $t_{2g}$ level before adding the remaining two to the $e_g$ orbitals, resulting in $t_{2g}^6 e_{g}^2$. Finally, for $[\mathrm{V}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the vanadium ion has two $d$-electrons that occupy the $t_{2g}$ orbitals, represented as $t_{2g}^2 e_{g}^0$.

The magnetic behavior of these complexes is dictated by the oxidation state of the nickel atom and the field strength of the attached ligands. In $[\mathrm{Ni}(\mathrm{CO})_4]$, nickel is in its zero oxidation state with a $3d^8 4s^2$ configuration, where the strong-field carbon monoxide ligands force all electrons to pair up, leaving zero unpaired electrons and resulting in a diamagnetic complex. In contrast, the $[\mathrm{NiCl}_4]^{2-}$ ion features nickel in a $+2$ oxidation state with a $3d^8$ configuration, and since chloride is a weak-field ligand, it cannot force pairing, which leaves two unpaired electrons and causes the complex to be paramagnetic.

The spectrochemical series ranks ligands according to their ability to induce splitting in metal $d$-orbitals, which directly determines their relative field strength. By arranging these specific ligands based on that experimental order, we observe a progression where $\mathrm{S}^{2-}$ acts as the weakest donor, followed by $\mathrm{C_2O_4^{2-}}$, $\mathrm{NH_3}$, and $\mathrm{en}$, culminating in $\mathrm{CO}$ as the strongest field ligand.

Meridional isomerism occurs in octahedral coordination complexes that follow the general formula $[ \mathrm{MA}_3 \mathrm{B}_3 ]$, where three identical ligands align along a meridian plane of the structure. This arrangement differs from the facial isomer, in which the three ligands occupy a single triangular face of the octahedron. Among the choices provided, the complex $[ \mathrm{Co} ( \mathrm{NH}_3)_3 ( \mathrm{NO}_2)_3 ]$ conforms to this required pattern, making it the one capable of existing as a meridional isomer.

Magnetic attraction in transition metal complexes is determined by the number of unpaired electrons in the metal ion's $d$-orbitals, with a higher count leading to stronger attraction. In the provided complexes, $[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ acts as a $d^{10}$ system with 0 unpaired electrons, while $[\mathrm{Co}(\mathrm{en})_3]^{3+}$ operates as a $d^6$ system that similarly contains 0 unpaired electrons. Furthermore, $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ features a $d^8$ configuration with 2 unpaired electrons, and $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ involves a $d^7$ configuration that results in 3 unpaired electrons. Because it possesses the highest number of unpaired electrons among these species, $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ exhibits the maximum attraction to the applied magnetic field.

The magnetic moment of a transition metal complex relates to the number of unpaired electrons through the spin-only formula $\mu = \sqrt{n(n+2)}$ Bohr magnetons. Given the observed magnetic moment of 6.06 BM, the complex possesses 5 unpaired electrons, which corresponds to the manganese ion being in the +2 oxidation state. Since each thiocyanate ligand carries a -1 charge, the charge balance of the complex is defined by the equation $2 + 6(-1) = -x$. Evaluating this expression as $2 - 6 = -x$ results in $-4 = -x$, which determines that the numerical value of $x$ is 4.

The magnetic properties of a metal ion are directly related to its electronic configuration, specifically the number of unpaired electrons, which is expressed by the formula $\mu = \sqrt{n(n+2)}$ where $\mu$ represents the magnetic moment in Bohr magnetons and $n$ represents the number of unpaired electrons. By substituting the given value of 4.90 into this formula, we get $\sqrt{n(n+2)} = 4.90$. Squaring both sides of the equation leads to $n(n+2) = 24.01$, which simplifies effectively to $n^2 + 2n - 24 = 0$. Factoring this quadratic equation yields $n = 4$, revealing that the ion contains 4 unpaired electrons.

The magnetic properties and stability of coordination complexes are determined by the electronic configuration of the metal ion and the field strength of the attached ligands. In $\mathrm{K}_3[\mathrm{Co}(\mathrm{CN})_6]$, the central cobalt ion is in the +3 oxidation state, yielding a $d^6$ electronic configuration. Because $\mathrm{CN}^{-}$ is a strong field ligand, it forces the pairing of electrons within the $d$-orbitals, resulting in zero unpaired electrons and confirming that the complex is diamagnetic and the most stable.

The spin-only magnetic moment depends on the number of unpaired electrons, $n$, according to the formula $\mu = \sqrt{n(n+2)}$ BM. Chromium in both $[\mathrm{Cr}(\mathrm{CN})_6]^{3-}$ and $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is in the $+3$ oxidation state, which corresponds to a $3d^3$ electronic configuration. Regardless of the ligand type, this configuration occupies the $t_{2g}$ orbitals with three unpaired electrons, yielding a magnetic moment of:

$\mu = \sqrt{3(3+2)} = \sqrt{15}$

Since both complexes possess the same spin-only magnetic moment, their ratio is 1.

The Crystal Field Splitting Energy (CFSE) in octahedral complexes is determined by the electronic distribution in the split $d$-orbitals, where the stabilization energy is calculated using the equation $\text{CFSE} = (-0.4 \times n(t_{2g}) + 0.6 \times n(e_g)) \Delta_0$. For the $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($d^2$) complex, both electrons occupy the lower $t_{2g}$ orbital, giving a CFSE of $-0.4 \times 2 = -0.8 \Delta_0$. The $[\mathrm{V}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($d^3$) complex fills the $t_{2g}$ subshell completely, resulting in a CFSE of $-0.4 \times 3 = -1.2 \Delta_0$. In the case of the $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ($d^4$) complex, the electron distribution corresponds to a value of $-0.8 \Delta_0$. Finally, the $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ ($d^5$) ion possesses a half-filled $d$-shell arrangement with three electrons in $t_{2g}$ and two in $e_g$, leading to a net CFSE calculation of $-0.4 \times 3 + 0.6 \times 2 = 0 \Delta_0$.

Chirality in octahedral coordination complexes arises when the molecule lacks both a plane of symmetry and a center of inversion, a condition satisfied by the cis isomer of complexes with the general formula $[M(AA)_2X_2]$. The complex cis-$[PtCl_2(en)_2]^{2+}$ possesses this specific geometry and is therefore optically active, whereas the trans isomer and other structures such as square planar $[PtCl_2(NH_3)_2]$ or octahedral trans-$[Co(NH_3)_4Cl_2]^{+}$ contain planes of symmetry that render them achiral.

The crystal field splitting energy ($\Delta_0$) for octahedral complexes depends primarily on the oxidation state of the metal ion and its $d$-electron configuration. Among the given $M^{3+}$ aqua complexes, the $[\mathrm{Cr}(\mathrm{OH}_2)_6]^{3+}$ ion possesses a $d^3$ configuration, where all three electrons reside in the lower-energy $t_{2g}$ orbitals, providing a uniquely stable electronic arrangement that results in a higher observed splitting energy. In contrast, the $d^4$ and $d^5$ configurations of the manganese and iron complexes are high-spin systems with electrons occupying the antibonding $e_g$ orbitals, which leads to greater interelectronic repulsions and smaller observed $\Delta_0$ values.

Solubility in ionic coordination compounds depends on the interplay between lattice energy and the ability of the ions to be stabilized by water solvation, with most large complex salts exhibiting negligible solubility. Among the compounds listed, $[\mathrm{Fe}_3(\mathrm{OH})_2(\mathrm{OAc})_6] \mathrm{Cl}$ is distinct for its ability to dissolve in water. In contrast, the Prussian blue complex $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ is insoluble, $\mathrm{K}_3[\mathrm{Co}(\mathrm{NO}_2)_6]$ displays very poor water solubility, and the heteropoly ammonium salt $(\mathrm{NH}_4)_3[\mathrm{As}(\mathrm{Mo}_3\mathrm{O}_{10})_4]$ is also insoluble.

The number of unpaired electrons in a coordination complex is determined by Crystal Field Theory, which considers the central metal ion's electronic configuration, the ligand field strength, and the resulting spin state of the complex.

In $[\mathrm{Ni}(\mathrm{CO})_4]$, the nickel atom is in the $0$ oxidation state with a $3d^8 4s^2$ configuration; because $\mathrm{CO}$ is a strong field ligand, the $4s$ electrons are forced into the $3d$ orbitals to form a diamagnetic $[\mathrm{Ar}] 3d^{10}$ configuration with $0$ unpaired electrons. For $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, the iron ion is $\mathrm{Fe}^{3+}$ with a $3d^5$ configuration, and the strong field ligand $\mathrm{CN}^-$ forces pairing in the $t_{2g}$ orbitals ($t_{2g}^5 e_g^0$), resulting in $1$ unpaired electron. The complex $[\mathrm{Cr}(\text{oxalate})_3]^{3-}$ contains $\mathrm{Cr}^{3+}$ with a $3d^3$ configuration, which inherently places one electron in each of the three $t_{2g}$ orbitals, yielding $3$ unpaired electrons regardless of ligand field strength.

In $[\mathrm{CoF}_6]^{3-}$, the cobalt ion is $\mathrm{Co}^{3+}$ with a $3d^6$ configuration, and the weak field ligand $\mathrm{F}^-$ results in a high-spin configuration $t_{2g}^4 e_g^2$ with $4$ unpaired electrons. Finally, $[\mathrm{FeF}_6]^{3-}$ contains $\mathrm{Fe}^{3+}$ with a $3d^5$ configuration and the weak field ligand $\mathrm{F}^-$, leading to a high-spin configuration $t_{2g}^3 e_g^2$ with $5$ unpaired electrons. Ordering these complexes by their number of unpaired electrons ($0 < 1 < 3 < 4 < 5$) yields the sequence $[\mathrm{Ni}(\mathrm{CO})_4] < [\mathrm{Fe}(\mathrm{CN})_6]^{3-} < [\mathrm{Cr}(\text{oxalate})_3]^{3-} < [\mathrm{CoF}_6]^{3-} < [\mathrm{FeF}_6]^{3-}$.

In the coordination complex $[\mathrm{CoCl}_4]^{2-}$, the central cobalt ion is in the $+2$ oxidation state. Given that neutral cobalt has an electronic configuration of $3d^7 4s^2$, the $\mathrm{Co}^{2+}$ ion retains seven electrons in its $3d$ subshell. In a tetrahedral crystal field, these $d$-orbitals split into a lower-energy $e$ set containing two orbitals and a higher-energy $t_2$ set containing three orbitals. Because chloride ions act as weak-field ligands, the seven electrons occupy these orbitals according to Hund's rule, filling the two $e$ orbitals completely with four electrons before distributing the remaining three electrons into the $t_2$ set, resulting in the configuration $e^4 t_2^3$. In this arrangement, the value of $m$ is 4, and the three electrons populating the $t_2$ orbitals are unpaired. Summing the value of $m$ with the three unpaired electrons yields a total of 7.