Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

The $[Co(NH_3)_3Cl_3]$ complex adopts an octahedral geometry where the three chlorine atoms occupy the corners of one triangular face, resulting in a facial isomer. In this specific orientation, each of the three chlorine atoms is adjacent to the other two, and since adjacent positions in an octahedral coordination sphere are separated by 90 degrees, the $Cl-Co-Cl$ bond angles are all 90 degrees.

The magnetic moment of a transition metal complex is determined by the number of unpaired electrons in the central metal ion's $d$-orbitals, which can be calculated using the spin-only formula $\mu = \sqrt{n(n+2)}$. In the complex $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, manganese exists as a $\mathrm{Mn}^{2+}$ ion, which has an electronic configuration of $3d^5$ given that its atomic number is 25. Because water acts as a weak field ligand in this octahedral environment, all five of these electrons remain unpaired. Substituting $n=5$ into the formula yields the following result:
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \text{ B.M.}$
Rounding this calculated value to the nearest integer provides an answer of 6.

Ambidentate ligands are defined by their possession of multiple potential donor atoms, which allows them to coordinate with a central metal ion through different atoms depending on the specific reaction conditions. Familiar examples include the nitrite ion ($\mathrm{NO}_2^{-}$) and the thiocyanate ion ($\mathrm{NCS}^{-}$), both of which are chemically recognized for this dual-binding capability. In contrast, ligands such as oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$), ethylene diamine, and water ($\mathrm{H}_2\mathrm{O}$) do not exhibit this property, as they function strictly as polydentate or monodentate ligands without the structural flexibility to vary their donor atom. Since every other provided group contains at least one of the known ambidentate species, only the set consisting of $\mathrm{C}_2\mathrm{O}_4^{2-}$, ethylene diamine, and $\mathrm{H}_2\mathrm{O}$ is entirely free of ambidentate ligands.

Naming coordination complexes requires identifying the counter ion first, followed by the ligands, and finally the central metal atom with its oxidation state included in parentheses. For $\mathrm{K}_3 [ \mathrm{Co} ( \mathrm{C}_2 \mathrm{O}_4 )_3 ]$, the presence of the potassium counter ion signifies that the complex $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$ is anionic, which dictates that the central metal cobalt must end in the suffix "-ate." Because there are three oxalate ligands, each carrying a charge of -2, the oxidation state of the central cobalt atom is calculated as +3, resulting in the final name potassium trioxalatocobaltate(III).

The Crystal Field Stabilization Energy (CFSE) for octahedral complexes is determined by the formula $CFSE = (-0.4 \times n_{t2g} + 0.6 \times n_{eg}) \Delta_o$, where $n_{t2g}$ and $n_{eg}$ denote the number of electrons in the $t_{2g}$ and $e_g$ orbitals, respectively.

For $[\mathrm{Cu}(\mathrm{NH}_3)_6]^{2+}$, the $\mathrm{Cu}^{2+}$ ion has a $d^9$ configuration, which in an octahedral field splits as $t_{2g}^6 e_g^3$, resulting in $CFSE = (-0.4 \times 6 + 0.6 \times 3) \Delta_o = -0.6 \Delta_o$, matching entry I.

For $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the $\mathrm{Ti}^{3+}$ ion possesses a $d^1$ configuration ($t_{2g}^1 e_g^0$), yielding $CFSE = (-0.4 \times 1 + 0.6 \times 0) \Delta_o = -0.4 \Delta_o$, matching entry IV.

For $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, the $\mathrm{Fe}^{3+}$ ion has a $d^5$ configuration; because $\mathrm{CN}^-$ is a strong-field ligand, it adopts a low-spin arrangement of $t_{2g}^5 e_g^0$, providing $CFSE = (-0.4 \times 5 + 0.6 \times 0) \Delta_o = -2.0 \Delta_o$, matching entry II.

For $[\mathrm{NiF}_6]^{4-}$, the $\mathrm{Ni}^{2+}$ ion has a $d^8$ configuration ($t_{2g}^6 e_g^2$), resulting in $CFSE = (-0.4 \times 6 + 0.6 \times 2) \Delta_o = -1.2 \Delta_o$, which matches entry III. This leads to the correct mapping of A-I, B-IV, C-II, D-III.

The occurrence of a low spin and diamagnetic octahedral complex depends on the presence of strong field ligands that induce sufficient crystal field splitting to force the pairing of all electrons within the lower energy orbitals. In the case of $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, the $\mathrm{Co}^{3+}$ central metal ion possesses a $3d^6$ configuration, and the ammonia molecules function as strong field ligands. These ligands force all six $d$-electrons to pair within the $t_{2g}$ energy level, leaving no unpaired electrons and resulting in a stable, diamagnetic low spin configuration.

The magnetic moment of a coordination complex depends primarily on the number of unpaired electrons, which is determined by whether the ligands are strong field or weak field. In both the complexes $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ and $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the iron is in the +3 oxidation state, meaning the reason statement is factual. However, the strong field ligand $\mathrm{CN}^-$ causes electron pairing, resulting in only one unpaired electron and a magnetic moment of $\sqrt{1(1+2)} = 1.74$ BM. In contrast, $\mathrm{H}_2\mathrm{O}$ acts as a weak field ligand that leaves five unpaired electrons, producing a higher magnetic moment of $\sqrt{5(5+2)} = 5.92$ BM. Because the difference in magnetic moment is driven by the nature of the ligands rather than the shared oxidation state, the reason does not explain the assertion despite being true.

Octahedral complexes with the formula $[\mathrm{Cr}(\mathrm{ox})_2\mathrm{ClBr}]^{3-}$ possess stereoisomers defined by the geometric orientation of the monodentate ligands and their inherent symmetry. Placing the $\mathrm{Cl}$ and $\mathrm{Br}$ atoms at trans-positions yields one geometric isomer that contains a plane of symmetry and is optically inactive. In the cis-orientation, the ligands are positioned at 90 degrees to each other, creating an asymmetric structure that lacks a plane of symmetry and consequently exists as two distinct enantiomers. This combination of one trans-isomer and two cis-enantiomers results in a total of 3 stereoisomers.

The classification of carbonyl ligands as either terminal or bridging depends on whether they coordinate to a single metal atom or span across two metal centers. In the complex $\mathrm{W}(\mathrm{CO})_6$, the coordination sphere consists solely of six terminal carbonyl groups arranged octahedrally around the tungsten atom, resulting in zero bridging carbonyls. Similarly, the molecular structure of $\mathrm{Mn}_2(\mathrm{CO})_{10}$ features a direct metal-metal bond with five terminal carbonyls attached to each manganese center, leaving no ligands in a bridging position. Consequently, the sum of bridging carbonyls in both species is zero.

Primary valency refers to the oxidation state of the central metal ion, whereas secondary valency reflects the coordination number, indicating the total number of ligand donor atoms directly attached to that metal. In the complex $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$, the cobalt atom must have an oxidation state of $+3$ to balance the two outer chloride ions and the one inner chloride ligand, establishing the primary valency as 3. Given that the metal is bonded to five ammonia molecules and one inner chloride ion within the coordination sphere, it forms a total of 6 coordinate bonds, identifying the secondary valency as 6.

Double salts are formed by the stoichiometric combination of two distinct salts that maintain their individual identities within a single crystal lattice and dissociate into their component ions when dissolved in water. Applying this principle, $\mathrm{FeSO}_4 \cdot (\mathrm{NH}_4)_2\mathrm{SO}_4 \cdot 6\mathrm{H}_2\mathrm{O}$ and $\mathrm{K}_2\mathrm{SO}_4 \cdot \mathrm{Al}_2(\mathrm{SO}_4)_3 \cdot 24\mathrm{H}_2\mathrm{O}$ qualify as double salts. Conversely, compounds such as $\mathrm{CuSO}_4 \cdot 4\mathrm{NH}_3 \cdot \mathrm{H}_2\mathrm{O}$ and $\mathrm{Fe}(\mathrm{CN})_2 \cdot 4\mathrm{KCN}$ function as complex compounds rather than additive combinations, because they involve coordinate covalent bonding that remains intact in solution.

The number of unpaired electrons in a coordination complex is determined by the electronic configuration of the metal ion and the strength of the coordinating ligands, which influence the splitting of d-orbitals. Strong field ligands cause electrons to pair in lower-energy orbitals, while weak field ligands allow for a high-spin distribution.

For $[\mathrm{Cr}(\mathrm{CN})_6]^{3-}$, chromium is in the $+3$ oxidation state with a $3d^3$ configuration. These electrons occupy the lower energy $t_{2g}$ orbitals singly, resulting in 3 unpaired electrons. In $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, iron is in the $+2$ oxidation state with a $3d^6$ configuration. Since water is a weak field ligand, the electrons remain high-spin, giving 4 unpaired electrons.

The complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ contains cobalt in the $+3$ oxidation state with a $3d^6$ configuration. Here, ammonia acts as a strong field ligand, forcing the six electrons to pair in the $t_{2g}$ orbitals, which leaves 0 unpaired electrons. Finally, for $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$, nickel is in the $+2$ oxidation state with a $3d^8$ configuration, where the electrons occupy the orbitals such that 2 unpaired electrons remain.

When ferric chloride reacts with potassium ferrocyanide, the free ferric ions and hexacyanoferrate(II) ions combine to form an insoluble, deep blue pigment historically called Prussian blue. To ensure the electrical neutrality of the resulting coordination complex, four iron(III) cations must balance three hexacyanoferrate(II) anions, leading to the specific chemical formula $\mathrm{Fe_4[Fe(CN)_6]_3}$. This precipitation reaction serves as a definitive qualitative test for the presence of iron ions in an aqueous solution.

The hybridization of a coordination complex depends on the coordination number, the metal's electronic configuration, and the field strength of the ligands. For the tetrahedral complex $\mathrm{[Ni(CO)_4]}$, the nickel center utilizes $\mathrm{sp^3}$ hybridization, whereas the square planar $\mathrm{[Cu(NH_3)_4]^{2+}}$ complex involves $\mathrm{dsp^2}$ hybridization. Regarding the octahedral iron complexes, the distinct ligand field strengths determine the orbital usage; $\mathrm{[Fe(NH_3)_6]^{2+}}$ acts as a strong-field complex, inducing electron pairing to form an inner-orbital $d^2\mathrm{sp^3}$ configuration. Conversely, $\mathrm{[Fe(H_2O)_6]^{2+}}$ functions as a weak-field complex where the electrons do not pair, necessitating the use of higher energy orbitals for outer-orbital $\mathrm{sp^3 d^2}$ hybridization.

The cobalt ion in $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is in the +3 oxidation state with a $3d^6$ electronic configuration, and because ammonia serves as a strong field ligand, it forces the electrons in the d-orbitals to pair up. This arrangement vacates two inner d-orbitals, which hybridize with one 4s and three 4p orbitals to produce $d^2 \mathrm{sp}^3$ hybridization. Because all electrons are paired as a result of this ligand interaction, the complex exhibits diamagnetic behavior.

In medicinal chemistry, specific platinum-based complexes serve as potent agents for cancer treatment. Cis-platin, represented by the chemical formula $\text{cis}-[\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}_2]$, is a widely used chemotherapy drug that works by interfering with the replication of cancerous cells to inhibit tumour growth. Because this drug falls under the chemical category of coordination compounds of platinum, both this specific agent and the general class of platinum coordination compounds are recognized for their anti-tumour properties. In contrast, compounds like EDTA and D-Penicillamine are typically utilized as chelating agents for therapeutic detoxification and heavy metal poisoning rather than for the treatment of malignancies.

The energy of the light absorbed by a coordination complex is inversely proportional to its wavelength $\lambda$, following the relation $E = \frac{hc}{\lambda}$. Since the crystal field splitting energy $\Delta$ increases with the strength of the ligand field, complexes with stronger field ligands exhibit larger splitting and thus absorb light at shorter wavelengths, while those with weaker ligands show smaller splitting and absorb at longer wavelengths.

By arranging the coordination entities according to their ligand field strength, we can identify their respective absorption values. The $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ complex, featuring the strong-field cyanide ligand, undergoes the largest splitting and absorbs at the shortest wavelength of 310 nm. Between the cobalt ammine complexes, $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ absorbs at 475 nm, whereas replacing an amine with a weaker chloride ligand in $[\mathrm{CoCl}(\mathrm{NH}_3)_5]^{2+}$ decreases the crystal field splitting, shifting the absorption to 535 nm. Finally, the $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_4]^{2+}$ complex involves the weakest field ligand environment among these, corresponding to the longest absorption wavelength of 600 nm.

Paramagnetism in coordination complexes arises from the presence of one or more unpaired electrons in the d-orbitals of the central metal ion, a state dictated by the crystal field splitting energy relative to the electron pairing energy. In the case of $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, the $\mathrm{Ni}^{2+}$ ion is $d^8$ and $\mathrm{CN}^-$ is a strong field ligand, forcing all electrons to pair in a $dsp^2$ square planar geometry, rendering it diamagnetic, whereas $[\mathrm{Ni}(\mathrm{CO})_4]$ contains $\mathrm{Ni}^{0}$ with a $d^{10}$ configuration where all electrons are naturally paired, also making it diamagnetic. Conversely, $[\mathrm{NiCl}_4]^{2-}$ features $\mathrm{Ni}^{2+}$ in a $d^8$ configuration with weak field ligands in a tetrahedral geometry, resulting in two unpaired electrons and thus paramagnetism. The complex $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ contains $\mathrm{Fe}^{2+}$ ($d^6$) with a strong field ligand, leading to a low-spin $t_{2g}^6 e_g^0$ diamagnetic configuration.

For the remaining species, $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ features $\mathrm{Cu}^{2+}$ in a $d^9$ configuration, which necessitates one unpaired electron, making it paramagnetic. The complex $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ contains $\mathrm{Fe}^{3+}$ ($d^5$) paired with a strong field ligand, creating a low-spin $t_{2g}^5$ configuration with one unpaired electron, which is paramagnetic. Finally, $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ involves $\mathrm{Fe}^{2+}$ ($d^6$) and a weak field ligand, resulting in a high-spin $t_{2g}^4 e_g^2$ configuration containing four unpaired electrons, which is also paramagnetic. Based on these electronic configurations, the paramagnetic species are $[\mathrm{NiCl}_4]^{2-}$, $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$, $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, and $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, identifying a total of four.

The $\mathrm{Co}^{2+}$ ion features a $3d^7$ electronic configuration, and when it binds with the weak field ligand $\mathrm{H}_2\mathrm{O}$ to form an octahedral complex, the $d$-orbitals split into a lower energy $t_{2g}$ set and a higher energy $e_g$ set. Electrons fill these orbitals in a high-spin arrangement, resulting in an occupancy of $t_{2g}^5 e_g^2$. Since the three $t_{2g}$ orbitals hold five electrons, two orbitals are completely paired while the third holds a single electron. This configuration leaves 1 unpaired electron in the $t_{2g}$ set.

The spin-only magnetic moment of a transition metal complex is determined by the number of unpaired electrons in its d-orbitals, which is influenced by the field strength of the surrounding ligands. In the case of $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, iron exists as $\mathrm{Fe}^{3+}$ with a 3d5 configuration, and since water is a weak-field ligand, no pairing occurs, leaving five unpaired electrons. Substituting the number of unpaired electrons into the equation $\mu = \sqrt{n(n+2)}$ yields the square root of 35, which corresponds to 5.92 B.M. In $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$, the iron ion also holds a 3d5 configuration, but cyanide acts as a strong-field ligand that forces the electrons to pair up, resulting in only one unpaired electron. Consequently, applying the equation $\mu = \sqrt{n(n+2)}$ with this single unpaired electron gives the square root of 3, which results in a magnetic moment of 1.732 B.M.