Coordination Compounds – JEE Main ChemistryPractice Questions & PYQs

Geometrical isomerism in coordination complexes depends on the ability of ligands to occupy different spatial positions relative to each other within the coordination sphere. Square planar complexes of the form $[Pt(NH_3)_2Cl(NO_2)]$ accommodate distinct cis and trans arrangements, whereas $[Pt(NH_3)_3Cl]^+$ does not possess enough ligand variety to support such configuration changes. Likewise, octahedral complexes of the type $[Pt(NH_3)_4ClBr]^{2+}$ allow for isomeric forms because the two different halogen ligands can be positioned adjacent or opposite one another, while the symmetric $[Pt(NH_3)Cl_5]^-$ structure remains fixed in a single geometric orientation.

The electronic behavior of transition metal complexes is dictated by the strength of the ligand field relative to pairing energy and the geometry of the complex. For a Co(III) center, strong field ligands force electrons into a low-spin, diamagnetic state, which results in zero magnetic moment rather than a high one, making that statement incorrect. Conversely, in the high-spin scenario defined by $\Delta_0 < P$, electrons populate the orbitals individually before pairing, leading to the $t_{2g}^4 e_g^2$ configuration, which is a correct assessment.

The absorption of light by a complex is inversely proportional to the crystal field splitting energy as described by the relationship $\Delta = \frac{hc}{\lambda}$. Because ethylenediamine is a significantly stronger field ligand than the fluoride ion, the cobalt-en complex possesses a larger splitting energy and consequently absorbs light at a shorter wavelength than the cobalt-fluoride complex. Finally, the tetrahedral splitting energy is defined by the relationship $\Delta_t = \frac{4}{9} \Delta_0$. Applying this to an octahedral splitting of 18,000 cm$^{-1}$ yields $\Delta_t = 8,000 \text{ cm}^{-1}$, which contradicts the value provided in the statement.

In the complex $[\mathrm{Ru}(\mathrm{en})_3]\mathrm{Cl}_2$, the central ruthenium ion exists in the $+2$ oxidation state, resulting in a $d^6$ electronic configuration. Because ruthenium is a $4d$ series transition metal, it experiences a significantly large crystal field splitting energy that inherently forces the electrons to pair up within the lower-energy orbitals, leading to a low-spin configuration of $t_{2g}^{6} e_{g}^{0}$. Conversely, the iron ion in $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$ is also in the $+2$ oxidation state with a $d^6$ configuration, but it is interacting with water, which functions as a weak field ligand. For $3d$ metals in an octahedral environment with weak field ligands, the crystal field splitting energy is insufficient to cause pairing, which forces the electrons to populate both the $t_{2g}$ and $e_{g}$ sets according to Hund's rule, resulting in a high-spin $t_{2g}^{4} e_{g}^{2}$ configuration.

Octahedral complexes with the general formula $[Ma_3b_3]$ exhibit facial (fac) and meridional (mer) isomerism based on the spatial arrangement of the three identical ligands. In the facial isomer, these three ligands occupy the corners of one triangular face of the octahedron, while the meridional isomer places them along an equatorial plane or meridian. The complex $[Co(NH_3)_3(NO_2)_3]$ adheres to this $[Ma_3b_3]$ structure, enabling the existence of these distinct geometric forms, unlike the other provided coordination compounds which do not satisfy this specific requirement.

The central concept determining isomerism in this coordination compound is the presence of the ambidentate ligand $\mathrm{NO}_2^-$. Because this ligand can bind to the platinum center through either its nitrogen atom or its oxygen atom, it allows for different modes of connectivity, giving rise to linkage isomerism. In this specific complex, the two $\mathrm{NO}_2^-$ groups can arrange themselves in three distinct ways: both ligands binding through nitrogen, one binding through nitrogen and one through oxygen, or both binding through oxygen. Since the compound does not exhibit geometrical or optical isomerism, these three possibilities represent the distinct structural forms of the complex, yielding isomers $[\mathrm{Pt}(\mathrm{en})(\mathrm{NO}_2)_2]$, $[\mathrm{Pt}(\mathrm{en})(\mathrm{NO}_2)(\mathrm{ONO})]$, and $[\mathrm{Pt}(\mathrm{en})(\mathrm{ONO})_2]$.

The existence of optical isomers depends on molecular chirality, which requires that a structure lacks any internal plane of symmetry. In complexes with $sp^3$ hybridisation, the metal adopts a tetrahedral geometry, which inherently possesses planes of symmetry in the $MA_2B_2$ arrangement, meaning zero optical isomers are possible. Similarly, a $dsp^2$ hybridised complex results in a square planar geometry, where the flat, two-dimensional nature of the molecule provides an automatic plane of symmetry that makes it optically inactive, also resulting in zero optical isomers.

Coordination complexes exhibit isomerism primarily based on their specific spatial geometry, which dictates the arrangement of ligands around the central metal ion. Because $[\mathrm{Ni}(\mathrm{NH}_3)_2\mathrm{Cl}_2]$ adopts a tetrahedral structure, all ligand positions are equivalent, effectively preventing the formation of distinct geometrical or optical isomers. In contrast, the remaining complexes-specifically the octahedral $[\mathrm{Ni}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+}$, the $[\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}_2]$ complex, and the octahedral $[\mathrm{Ni}(\mathrm{en})_3]^{2+}$-all possess geometries that accommodate various forms of isomerism.

In a square pyramidal $ML_5$ complex, the bond angles consist of eight 90° angles (four among the base ligands and four between the axial and equatorial ligands) and two 180° angles (formed across the diagonals of the square base), resulting in a total of 10 angles. In a trigonal bipyramidal $ML_5$ complex, there are six 90° angles (between the axial and equatorial ligands), three 120° angles (between the equatorial ligands), and one 180° angle (between the two axial ligands), which also contributes a sum of 10 angles. Combining these, the total number of 90°, 120°, and 180° L-M-L angles for the two complexes is $10 + 10 = 20$.

The Crystal Field Stabilization Energy of a transition metal complex is determined by the net energy contribution of electrons occupying the split $d$-orbitals according to the specific geometry. In an octahedral ligand field, the orbitals split into a lower $t_{2g}$ set and a higher $e_{g}$ set, causing a high-spin $d^{6}$ configuration to arrange as $(t_{2g})^{4}(e_{g})^{2}$. The net stabilization is calculated as:

$(4)(-0.4\Delta_{0}) + 2(0.6\Delta_{0}) = -0.4\Delta_{0}$

Conversely, in a tetrahedral field, the splitting energy is inverted, with the $e$ orbitals at a lower energy level than the $t_{2}$ orbitals. Distributing six electrons for a high-spin configuration results in a $(e)^{3}(t_{2})^{3}$ arrangement, where the stabilization energy is determined by:

$3(-0.6\Delta_{t}) + 3(0.4\Delta_{t}) = -0.6\Delta_{t}$

Ruthenium, with an atomic number of 44, exists in the $+2$ oxidation state in the $[\mathrm{Ru}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ complex, resulting in a $4d^6$ electronic configuration. When these six electrons occupy the $d$-orbitals of an octahedral field, the condition $\Delta_0 > P$ forces the electrons to pair up within the lower-energy $t_{2g}$ orbitals rather than occupying the higher-energy $e_{g}$ set. This arrangement leads to a configuration of $(t_{2g})^6 (e_g)^0$, meaning all electrons are paired and the number of unpaired electrons is $n = 0$. Consequently, applying the magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM yields a value of 0 BM.

Paramagnetic behavior is directly proportional to the number of unpaired electrons in a coordination complex, where a higher count of unpaired $d$-electrons leads to stronger paramagnetism. The complex $[\mathrm{Pd}(\mathrm{gly})_2]$ involves a $d^8$ metal center in a square planar geometry, which is characteristically diamagnetic because all electrons are paired. Similarly, the iron complex $[\mathrm{Fe}(\mathrm{en})(\mathrm{bpy})(\mathrm{NH}_3)_2]^{2+}$ is essentially diamagnetic, and the $[\mathrm{Ti}(\mathrm{NH}_3)_6]^{3+}$ ion contains only one unpaired electron in its $d^1$ configuration. In contrast, $[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2]^{-}$ acts as the most paramagnetic species among these options because, despite the $\Delta_0 > P$ condition, it possesses the highest number of unpaired electrons.

The geometry of square planar complexes containing four distinct ligands allows for exactly three unique arrangements, establishing that $n=3$. Applying this value to the formula $[\mathrm{Fe(CN)_6}]^{n-6}$ identifies the complex as $[\mathrm{Fe(CN)_6}]^{3-}$, where the iron center is in the $+3$ oxidation state with a $3d^5$ electronic configuration. Since cyanide acts as a strong field ligand, it causes the five electrons to occupy the lower $t_{2g}$ set, resulting in one unpaired electron. The spin-only magnetic moment for this single unpaired electron is $\sqrt{1(1+2)} = 1.73$ B.M. The crystal field stabilization energy is then calculated by placing these five electrons in the $t_{2g}$ orbitals:
$CFSE = [(-0.4 \times 5) + (0.6 \times 0)] \Delta_0 = -2.0 \Delta_0$

The splitting of d-orbitals in a coordination complex is primarily determined by the relationship between the crystal field splitting energy $\Delta_0$ and the pairing energy $P$. In the $[\mathrm{CoF}_3(\mathrm{H_2O})_3]$ complex, the cobalt center exists in the $+3$ oxidation state, yielding a $d^6$ electronic configuration that adopts a high-spin distribution of $t_{2g}^4 e_g^2$ because $\Delta_0 < P$. We find the stabilization energy by calculating the total energy deviation from the barycenter, represented by the expression $(-0.4 \times 4 + 0.6 \times 2) \Delta_{0}$. This arithmetic results in a final value of $-0.4 \Delta_{0}$.

According to IUPAC nomenclature rules for coordination compounds, ligands are ordered alphabetically, followed by the central metal atom with its oxidation state in parentheses, and finally the counter-ion. For the complex $[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$, the ligands ammine ($NH_3$), chlorido ($Cl^-$), and methanamine ($NH_2CH_3$) are arranged alphabetically as diammine, chlorido, and methanamine, while the oxidation state of platinum ($Pt$) is determined to be +2 by solving the equation $x + 2(0) + (-1) + 0 = +1$. This results in the complete name diamminechlorido(methanamine)platinum(II) chloride.

Concentrated $\mathrm{H}_{2}\mathrm{SO}_{4}$ acts as a dehydrating agent that selectively removes water molecules held within the crystal lattice rather than those coordinated to the metal center. To determine the molecular structure, calculate the total molar mass of $\mathrm{H}_{12}\mathrm{O}_{6}\mathrm{Cl}_{3}\mathrm{Cr}$ by summing the atomic masses, which gives:
$52 + 3(35) + 6(16) + 12(1) = 265$
The 13.5% mass loss is due to the evaporation of these lattice water molecules. By setting up the ratio of the mass of the water to the total mass, the equation becomes:
$\frac{x \times 18}{265} \times 100 = 13.5$
Solving for $x$ yields $x = 2$, indicating that two water molecules are non-coordinated and exist outside the coordination sphere. Comparing this result with the provided options identifies $[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{4}\mathrm{Cl}_{2}]\mathrm{Cl} \cdot 2\mathrm{H}_{2}\mathrm{O}$ as the complex containing exactly two molecules of lattice water.

The hybridization of a central atom defines the specific set of atomic orbitals that combine to form equivalent hybrid orbitals, which directly determines the number and type of d-orbitals involved in bonding. For the complex $[Ni(CN)_4]^{2-}$, the central nickel ion adopts $dsp^2$ hybridization, which explicitly utilizes exactly one d-orbital in the mixing process to facilitate its square planar geometry. In contrast, the other structures provided involve hybridization schemes such as $d^2sp^3$ or $sp^3d^2$, which necessitate the participation of two d-orbitals to accommodate their respective ligand arrangements.

Optical activity in coordination complexes requires the molecule to be chiral, which fundamentally means the structure must lack a plane of symmetry. Many metal complexes with ligands like ethane-1,2-diamine ($en$) are achiral due to the presence of internal symmetry planes, but the configuration in option D creates a non-superimposable mirror image by effectively breaking that symmetry. This absence of a plane of symmetry is the specific requirement for demonstrating optical activity, distinguishing the compound in option D from the more symmetric arrangements found in the other choices.

Square planar complexes featuring four distinct monodentate ligands, represented as $[M(abcd)]$, possess three unique geometric arrangements where each ligand occupies a specific relative position. Within this coordination sphere, the ligands SCN and $NO_2$ are ambidentate, meaning each can coordinate through two different donor atoms, thereby creating two distinct linkage isomers for each site. Combining these factors requires multiplying the 3 geometric isomers by the 2 possibilities for the SCN group and the 2 possibilities for the $NO_2$ group, yielding a grand total of 12 isomers for the complex $[M(F)(Cl)(SCN)(NO_2)]$.

In an octahedral complex like $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the five $d$-orbitals of the central metal ion split into two distinct energy sets, $t_{2g}$ and $e_g$, due to the electrostatic field created by the surrounding ligands. The $t_{2g}$ orbitals, which comprise $d_{xy}$, $d_{yz}$, and $d_{zx}$, are degenerate because they exist at the same energy level within this specific crystal field geometry. Consequently, any pairing of these orbitals, such as $d_{zx}$ and $d_{yz}$, constitutes a degenerate set.

The coordination complex formed from cobalt(III) chloride and ethylenediamine in a 1:2 mole ratio is $[Co(en)_2Cl_2]Cl$, which exhibits geometrical isomerism owing to the different spatial orientations of the chloride ligands relative to the bidentate ethylenediamine ligands. The cis-isomer possesses a structure where the two chloride ligands are adjacent, lacking a center of symmetry and thereby enabling optical activity, which corresponds to the violet-colored product A. In contrast, the trans-isomer features the chloride ligands at a $180^\circ$ angle, resulting in a centrosymmetric structure that is optically inactive, identifying it as the green-colored product B.