Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Coordination Compounds
From the quantum nuances of Crystal Field Splitting and Synergic Bonding to the 3D topology of Optical Isomerism. Master the chemistry of transition metal complexes.
Transition metal complexes defy standard covalent rules. To conquer JEE Advanced, you must perfectly predict electronic configurations using the Spectrochemical Series, navigate the spatial traps of fac/mer and cis/trans isomerism, and mathematically balance Werner's Primary and Secondary valencies.
View Solution
Case (a): Strong Field Ligand ($\Delta_o > P$)
Because splitting energy is massive, electrons prefer to pair up in the lower $t_{2g}$ level rather than jump to the $e_g$ level.
Configuration: $t_{2g}^6 e_g^0$.
CFSE = $[6 \times (-0.4\Delta_o)] + [0 \times (0.6\Delta_o)] + \text{Pairing Energy of newly paired electrons}$.
In a spherically symmetric $d^6$ field, there is 1 inherent pair. In $t_{2g}^6$, there are 3 pairs. The extra pairs generated by the crystal field = $3 - 1 = 2P$.
CFSE = $-2.4\Delta_o + 2P$.
Case (b): Weak Field Ligand ($\Delta_o < P$)
Because pairing requires more energy than jumping the gap, electrons fill all five orbitals singly before pairing.
Configuration: $t_{2g}^4 e_g^2$.
CFSE = $[4 \times (-0.4\Delta_o)] + [2 \times (0.6\Delta_o)]$
CFSE = $-1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$.
Since there is exactly 1 pair here (the same as the spherical average), no extra pairing energy term ($P$) is added to the CFSE change.
View Solution
Step 1: Analyze Metal Oxidation States
- In $[V(CO)_6]^-$, Vanadium is in the $-1$ oxidation state.
- In $Cr(CO)_6$, Chromium is in the $0$ oxidation state.
- In $[Mn(CO)_6]^+$, Manganese is in the $+1$ oxidation state.
Step 2: Relate Oxidation State to Back-Bonding
A negatively charged metal center ($V^{-1}$) is highly electron-rich. It aggressively pushes electron density backward into the empty $\pi^*$ antibonding orbitals of the $CO$ ligands ($\pi$-back bonding).
Conversely, a positively charged metal center ($Mn^{+1}$) is electron-deficient. It holds onto its d-electrons tightly, heavily restricting $\pi$-back donation to the $CO$ ligands.
Step 3: Bond Strength and Frequency
Maximal back-donation (in $V^{-1}$) severely populates the $\pi^*$ orbital of $CO$, drastically reducing the $C-O$ bond order (weakening the bond), which lowers the IR stretching frequency.
Minimal back-donation (in $Mn^{+1}$) leaves the $C-O$ triple bond relatively intact and strong, resulting in the highest IR stretching frequency.
View Solution
Step 1: Precipitation Stoichiometry
$0.1 \text{ mole}$ of complex yields $0.1 \text{ mole}$ of $AgCl$. This ratio (1:1) proves that exactly one chloride ion per formula unit is ionizable (located outside the square brackets).
Step 2: Construct the Coordination Sphere
The remaining two chloride ions and all four ammonia molecules must be firmly locked inside the coordination sphere to satisfy Cobalt's fixed secondary valency (coordination number), which is almost universally 6 for $Co^{3+}$.
Inside the brackets: $4 \times NH_3$ and $2 \times Cl^-$. Total ligands = 6.
Outside the brackets: $1 \times Cl^-$.
Step 3: The Formulation
Structure: $[Co(NH_3)_4Cl_2]Cl$.
Step 4: Assign Valencies
- Primary Valency: Corresponds to the oxidation state of the central metal. Cobalt is $+3$. It is satisfied by the total of 3 chloride ions (both inside and outside).
- Secondary Valency: Corresponds to the coordination number. Cobalt is bonded to 6 ligands inside the sphere. Thus, secondary valency = 6.
View Solution
Step 1: Oxidation State and Ligands
In metal carbonyls, the oxidation state of the metal is typically zero. $Mn = 0$.
The complex $Mn_2(CO)_{10}$ is symmetric. Each Manganese atom is coordinated to exactly 5 $CO$ ligands.
Step 2: The Metal-Metal Bond
For the overall molecule to be stable and diamagnetic, the two Manganese atoms must share an electron pair, forming a single $Mn-Mn$ covalent bond. This bond contributes exactly 1 electron to the EAN of each Manganese atom.
Step 3: Calculate EAN per Manganese atom
EAN = (Atomic Number of Mn) - (Oxidation State) + ($2 \times$ number of $CO$ ligands) + ($1 \times$ Metal-Metal bond)
EAN = $25 - 0 + (2 \times 5) + 1$
EAN = $25 + 10 + 1 = 36$.
Step 4: EAN Rule Verification
The noble gas at the end of Manganese's period (Period 4) is Krypton ($Kr$), which has an atomic number of 36. Since the calculated EAN matches Krypton exactly, the complex obeys the rule.
View Solution
Step 1: The Trans Isomer
In the trans isomer, the two Chloride ligands are placed $180^{\circ}$ opposite to each other (axial positions). The two bidentate 'en' ligands occupy the equatorial plane.
Symmetry check: The equatorial plane itself acts as a perfect internal plane of symmetry ($\sigma_h$), cutting the molecule in half. Therefore, the trans isomer is superimposable on its mirror image. It is optically inactive (achiral).
Step 2: The Cis Isomer
In the cis isomer, the two Chloride ligands are placed $90^{\circ}$ adjacent to each other. The bulky 'en' rings wrap around the remaining positions like a propeller.
Symmetry check: This arrangement completely destroys all planes of symmetry and centers of inversion. Because it lacks internal symmetry, the cis isomer is optically active (chiral). It exists as a non-superimposable pair of enantiomers (d- and l-forms).
Step 3: Total Count
- 1 trans isomer (achiral).
- 2 cis isomers (a pair of enantiomers).
Total stereoisomers = $1 + 2 = 3$.
View Solution
Step 1: Metal Oxidation State
Let oxidation state of Co be $x$. $x + 3(-2) = -3 \implies x = +3$.
$Co^{3+}$ has a $3d^6$ configuration.
Step 2: The Spectrochemical Exception
Normally, oxalate is a weak field ligand ($H_2O < ox < NH_3$). One would expect it to form a high-spin, outer-orbital complex.
However, $Co^{3+}$ has an exceptionally high charge density. This massive positive charge pulls the ligands in much closer than a $+2$ ion would. This drastic decrease in bond length exponentially increases the crystal field splitting energy ($\Delta_o$).
Step 3: The Resulting Field
Because of the $Co^{3+}$ ion, $\Delta_o$ becomes overwhelmingly larger than the pairing energy ($P$). Therefore, oxalate is forced to act as a Strong Field Ligand in this specific complex.
Step 4: Configuration and Hybridization
Since it acts as a strong field, the 6 d-electrons pair up entirely in the lower $t_{2g}$ orbitals ($t_{2g}^6 e_g^0$).
This leaves two inner $3d$ orbitals empty. The hybridization is $d^2sp^3$ (inner orbital complex).
Because all electrons are perfectly paired, the complex is diamagnetic ($\mu = 0$).
View Solution
Step 1: The Geometry of Approach
In an octahedral complex, 6 ligands approach perfectly along the Cartesian axes directly into the lobes of the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals, causing massive repulsion and large splitting.
In a tetrahedral complex, only 4 ligands approach, and they do so between the axes. They never point directly at any specific d-orbital lobe. This indirect approach results in much weaker electrostatic repulsion.
Step 2: The Mathematical Relationship
Due to having only 4 ligands (instead of 6) and the indirect angle of approach, the crystal field splitting energy for a tetrahedral field ($\Delta_t$) is mathematically derived to be significantly smaller than an octahedral field:
$$ \Delta_t = \frac{4}{9} \Delta_o $$
Step 3: Thermodynamic Consequence
Because $\Delta_t$ is less than half the magnitude of $\Delta_o$, the energy gap between the lower $e$ set and upper $t_2$ set is always very small. It is practically impossible for $\Delta_t$ to ever exceed the pairing energy ($P$) of electrons.
Since $\Delta_t \ll P$ in all standard cases, electrons will always jump the small gap and occupy the higher orbitals before pairing up. Therefore, tetrahedral complexes are always high-spin.
View Solution
Step 1: Determine Unpaired Electrons
$\mu = \sqrt{n(n+2)} = 3.87 \text{ B.M.}$
This corresponds mathematically to exactly $n = 3$ unpaired electrons ($\sqrt{15} \approx 3.87$).
Step 2: Evaluate Standard Assumptions (The Trap)
The overall complex cation is $[Fe(H_2O)_5(NO)]^{2+}$. If we assume $NO$ is a neutral ligand, then Iron must be $Fe^{2+}$.
$Fe^{2+}$ has a $3d^6$ configuration. With weak field water ligands, $d^6$ would have $4$ unpaired electrons ($t_{2g}^4 e_g^2$). This contradicts the experimental data of $n=3$.
Step 3: The True Nature of the NO Ligand
In this specific complex, the $NO$ molecule acts as a non-innocent ligand. It donates one of its own anti-bonding electrons directly to the Iron atom, oxidizing itself to the nitrosonium cation ($NO^+$).
Because the ligand brings a $+1$ charge, the oxidation state of Iron must drop to balance the overall $+2$ charge of the sphere.
$x + 0 + (+1) = +2 \implies x = +1$.
Step 4: Configuration of $Fe^{+1}$
Iron ($Z=26$) is $[Ar] 4s^2 3d^6$. $Fe^{+1}$ is $[Ar] 3d^7$.
In the presence of the ligands, it adopts a high-spin configuration ($t_{2g}^5 e_g^2$). This configuration possesses exactly 3 unpaired electrons, perfectly matching the $3.87 \text{ B.M.}$ experimental observation!
Pair A: $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$
Pair B: $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$
Pair C: $[Pd(C_6H_5)_2(SCN)_2]$ and $[Pd(C_6H_5)_2(NCS)_2]$
View Solution
Pair A: Ionization Isomerism
The sulfate ion and the bromide ion are swapping positions between the inner coordination sphere (acting as a ligand) and the outer ionization sphere (acting as a counter-ion). In aqueous solution, the first gives a precipitate with $AgNO_3$ (due to free $Br^-$), while the second gives a precipitate with $BaCl_2$ (due to free $SO_4^{2-}$).
Pair B: Coordination Isomerism
Both the cation and the anion are complex coordination spheres. The entire set of ligands ($NH_3$ and $CN^-$) are completely swapped between the two different metal centers (Cobalt and Chromium). This is exclusively seen in salts where both halves are complex ions.
Pair C: Linkage Isomerism
The thiocyanate ion ($SCN^-$) is an ambidentate ligand. It can coordinate to the metal either through the Sulfur atom (thiocyanato) or through the Nitrogen atom (isothiocyanato). The chemical formula remains identical, but the point of atomic linkage shifts.
View Solution
Step 1: The First Substitution
Start with $[PtCl_4]^{2-}$, a perfect square planar complex. All four positions are equivalent $Cl^-$ ligands.
Add 1 eq of $NH_3$. It replaces any one $Cl^-$ to form $[Pt(NH_3)Cl_3]^-$.
Step 2: Evaluating Trans-Directors
We must add the second $NH_3$. Where does it go? We must look at the ligands currently on the metal. We have one $NH_3$ and three $Cl^-$ ligands.
According to the Trans Effect series, $Cl^-$ is a significantly stronger trans-director than $NH_3$. This means the $Cl^-$ ligands possess a stronger ability to weaken the bond directly opposite to them.
Step 3: The Second Substitution
- The $Cl^-$ that is trans to the $NH_3$ is protected because $NH_3$ is a weak trans-director.
- However, the two $Cl^-$ ligands that are trans to each other mutually weaken each other's bonds via their strong trans-effect.
Therefore, the incoming second $NH_3$ nucleophile will effortlessly kick out one of these mutually weakened, heavily labilized $Cl^-$ ligands.
Step 4: Final Geometry
Because the substituted $Cl^-$ was trans to another $Cl^-$, the new $NH_3$ group is placed exactly $90^{\circ}$ adjacent to the first $NH_3$ group. This rigidly dictates a Cis geometry.
View Solution
Step 1: Electronic Configuration of Cu(II)
$Cu^{2+}$ has a $3d^9$ configuration. In an octahedral crystal field, this splits into $t_{2g}^6 e_g^3$.
Step 2: Analyze Degeneracy
The $t_{2g}$ level is perfectly, symmetrically filled (6 electrons).
The higher energy $e_g$ level contains 3 electrons distributed across two degenerate orbitals ($d_{x^2-y^2}$ and $d_{z^2}$). This creates an asymmetrical filling: one orbital has 2 electrons, and the other has only 1.
Step 3: The Electronic Repulsion
Nature places the pair of electrons into the $d_{z^2}$ orbital, leaving the single electron in the $d_{x^2-y^2}$ orbital. Because the $d_{z^2}$ orbital is now packed with more electron density, it exerts a massive electrostatic repulsion against the two ligands approaching along the z-axis (the axial ligands).
Step 4: The Distortion (Z-out)
To relieve this intense repulsion, the two axial ligands are physically pushed further away from the metal center. This stretches the axial bonds, breaking the perfect octahedral symmetry into a tetragonal distortion (z-out elongation), which subsequently lowers the energy of the $d_{z^2}$ orbital, stabilizing the entire complex.
View Solution
Step 1: Analyze Titanium Complex
In $[Ti(H_2O)_6]^{3+}$, Titanium is in the $+3$ oxidation state. It has a $3d^1$ configuration ($t_{2g}^1 e_g^0$).
When struck by visible light, this single electron absorbs a photon of exactly the right energy to jump from the lower $t_{2g}$ orbital across the $\Delta_o$ gap into the higher $e_g$ orbital. This is a classic $d-d$ transition. It absorbs yellow-green light, transmitting the complementary purple color. Because $d-d$ transitions are Laporte-forbidden (they are symmetrically restricted), the color is relatively faint/moderate in intensity.
Step 2: Analyze Permanganate Complex
In $MnO_4^-$, Manganese is in the maximum $+7$ oxidation state. It has a $3d^0$ configuration. There are exactly zero d-electrons available. Therefore, a $d-d$ transition is physically impossible.
The intense color arises because the $Mn^{7+}$ ion is extremely electron-deficient, while the surrounding Oxygen ligands are electron-rich. When struck by light, an electron from the Oxygen ligand's full p-orbital is temporarily excited directly into the empty d-orbital of the Manganese atom. This is a Ligand-to-Metal Charge Transfer (LMCT).
Step 3: Intensity Difference
Because LMCT transitions are fully symmetry-allowed (they follow Laporte selection rules), the probability of absorption is phenomenally high, resulting in an exceptionally intense, dark coloration compared to $d-d$ transitions.
View Solution
Step 1: Baseline Nickel Configuration
In both complexes, Nickel is in the $+2$ oxidation state. $Ni^{2+}$ has a $3d^8$ configuration.
Step 2: Analyze $[Ni(CN)_4]^{2-}$ (Strong Field)
Cyanide ($CN^-$) is a massive, exceptionally Strong Field Ligand at the very top of the spectrochemical series. Its immense ligand field forces the 8 d-electrons to pair up tightly, completely vacating one of the inner $3d$ orbitals (the $d_{x^2-y^2}$ orbital).
To accommodate 4 ligands, the metal uses this one empty $3d$ orbital, one $4s$ orbital, and two $4p$ orbitals.
Hybridization: $dsp^2$.
Geometry: Square Planar.
Magnetic Nature: Diamagnetic (all electrons paired).
Step 3: Analyze $[NiCl_4]^{2-}$ (Weak Field)
Chloride ($Cl^-$) is a very Weak Field Ligand (a $\pi$-donor). It lacks the thermodynamic power to force the electrons to pair up against their repulsion. The 8 d-electrons remain spread out, leaving two of them unpaired. The inner $3d$ orbitals remain occupied and unavailable.
To accommodate 4 ligands, the metal must use the outer empty orbitals: one $4s$ and three $4p$ orbitals.
Hybridization: $sp^3$.
Geometry: Tetrahedral.
Magnetic Nature: Paramagnetic (2 unpaired electrons).
View Solution
Step 1: The Reactions
Compare the formation reactions from the aqueous hexaaqua complex:
Reaction 1 (Monodentate): $[Cd(H_2O)_6]^{2+} + 4 CH_3NH_2 \rightleftharpoons [Cd(H_2O)_2(CH_3NH_2)_4]^{2+} + 4 H_2O$
Reaction 2 (Bidentate): $[Cd(H_2O)_6]^{2+} + 2 \text{ en} \rightleftharpoons [Cd(H_2O)_2(en)_2]^{2+} + 4 H_2O$
Step 2: Enthalpy ($\Delta H$) Check
In both cases, four $Cd-O$ bonds are broken, and four $Cd-N$ bonds are formed. The bond strengths are virtually identical. Therefore, $\Delta H$ is roughly the same for both reactions. Enthalpy cannot explain the difference.
Step 3: Entropy ($\Delta S$) Check
Look at the stoichiometry (number of independent particles).
In Reaction 1: 5 particles (1 complex + 4 amines) $\rightarrow$ 5 particles (1 complex + 4 waters). $\Delta S \approx 0$.
In Reaction 2: 3 particles (1 complex + 2 'en') $\rightarrow$ 5 particles (1 complex + 4 waters). Here, the number of independent, chaotic particles in the solution drastically increases! This results in a massive, highly positive increase in Entropy ($\Delta S \gg 0$).
Step 4: Gibbs Free Energy
$\Delta G = \Delta H - T\Delta S$. A massive positive $\Delta S$ makes the $-T\Delta S$ term highly negative, resulting in a much more negative $\Delta G$ for the chelate complex. Since $\Delta G = -RT \ln \beta$, a more negative $\Delta G$ exponentially skyrockets the stability constant ($\beta$).
View Solution
Step 1: 3D Visualization
The complex consists of a central Chromium atom surrounded octahedrally by three bidentate oxalate ligands. Because they must bridge adjacent $90^{\circ}$ positions, the three rings arrange themselves in a 3D structure that resembles the three blades of a propeller or a spiral staircase.
Step 2: Symmetry Check
If you slice the molecule through any plane, the bridging chelate rings break the symmetry. There is no plane of symmetry ($\sigma$) and no center of inversion ($i$). The entire $[M(AA)_3]$ geometry is inherently chiral due to its axial chirality (helicity).
Step 3: Enantiomer Designation
Because it is asymmetric, it perfectly forms non-superimposable mirror images. One enantiomer twists like a right-handed screw ($\Delta$, delta), and the other twists like a left-handed screw ($\Lambda$, lambda).
View Solution
Step 1: The Facial (fac) Isomer
If the three identical 'a' ligands (and the three 'b' ligands) are placed on mutually adjacent $90^{\circ}$ positions, they form the three corners of one triangular face of the octahedron. This is the fac-isomer.
Symmetry check: A plane can be sliced exactly diagonally through the metal and one 'a' and one 'b' ligand, mirroring the other two sets. It possesses a plane of symmetry and is optically inactive.
Step 2: The Meridional (mer) Isomer
If one set of three identical ligands is rearranged so that two of them are $180^{\circ}$ opposite to each other (trans) while the third is $90^{\circ}$ to both, they form an arc that cuts directly across the equator (meridian) of the octahedron. This is the mer-isomer.
Symmetry check: The plane containing all three 'a' ligands and the metal atom acts as a perfect plane of symmetry. It is also optically inactive.
View Solution
Step 1: The Resting Catalyst
Ligands: 3 neutral Triphenylphosphines ($PPh_3$) and 1 anionic Chloride ($Cl^-$). Total complex is neutral.
Oxidation state of Rh: $x + 0 + (-1) = 0 \implies x = +1$.
Coordination Number: 4 ligands = 4.
Rhodium(I) is $d^8$. A 4-coordinate $d^8$ complex with bulky/strong $\pi$-acceptor ligands generally adopts a Square Planar geometry.
Step 2: Oxidative Addition of $H_2$
The $H-H$ bond breaks, and both Hydrogen atoms attach directly to the Rhodium metal center as separate Hydride ($H^-$) ligands.
Because two new ligands are added, the Coordination Number jumps from $4 \rightarrow \mathbf{6}$ (changing geometry to Octahedral).
Because both hydrogens act as $-1$ hydrides, the metal must lose two electrons to them, increasing its oxidation state. Oxidation state jumps from $+1 \rightarrow \mathbf{+3}$ ($Rh^{III}$, a $d^6$ species).
View Solution
Step 1: Calculate $\beta_4$
The overall formation constant is the product of the stepwise constants:
$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$
$\beta_4 = 10^{4.0} \times 10^{3.2} \times 10^{2.7} \times 10^{2.0} = 10^{(4.0 + 3.2 + 2.7 + 2.0)} = 10^{11.9}$.
Step 2: Explain the Decrease
1. Statistical Factor: Initially, the $Cu^{2+}$ has 4 empty spots (coordinated waters) to accept an $NH_3$. The probability of an attack is high. For the last step ($K_4$), there is only 1 available spot left, drastically lowering the probability of successful coordination.
2. Electrostatic/Steric Factor: The incoming $NH_3$ ligand is bulky and carries a dipole. As more $NH_3$ ligands crowd around the metal, steric hindrance increases. Furthermore, replacing neutral water with neutral ammonia doesn't change the charge, but the sheer crowding of electron clouds repels the next incoming ligand, making each successive addition thermodynamically harder.
View Solution
Step 1: Metal Oxidation State
In both, Iron is $+3$ ($3d^5$ configuration).
Step 2: Analyze $[Fe(CN)_6]^{3-}$ (Strong Field)
Cyanide is a potent strong field ligand. It forces the 5 electrons to pair up against Hund's rule: $t_{2g}^5$ (1 unpaired electron).
This vacates two of the inner $3d$ orbitals. The metal hybridizes two $3d$, one $4s$, and three $4p$ orbitals.
Hybridization: $d^2sp^3$ (Because it uses inner $(n-1)d$ orbitals, it is an Inner Orbital Complex).
Magnetic Moment: $n=1 \implies \mu = \sqrt{1(3)} = 1.73 \text{ B.M.}$
Step 3: Analyze $[FeF_6]^{3-}$ (Weak Field)
Fluoride is a weak field ligand. It cannot force pairing. The 5 electrons remain singly occupied: $t_{2g}^3 e_g^2$ (5 unpaired electrons).
Because the inner $3d$ orbitals are full, the metal must reach out to the empty, higher-energy $4d$ orbitals to hybridize.
Hybridization: $sp^3d^2$ (Because it uses outer $nd$ orbitals, it is an Outer Orbital Complex).
Magnetic Moment: $n=5 \implies \mu = \sqrt{5(7)} = 5.92 \text{ B.M.}$
View Solution
Step 1: Order the Ligand Field Strengths
According to the Spectrochemical Series: $F^- \text{ (weakest)} < NH_3 \text{ (moderate)} < CN^- \text{ (strongest)}$.
Therefore, the splitting energy $\Delta_o$ order is: $[CoF_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}$.
Step 2: Correlate Energy to Wavelength
Energy is inversely proportional to wavelength ($E \propto 1/\lambda$).
- Largest $\Delta_o$ requires highest energy $\implies$ shortest wavelength.
- Smallest $\Delta_o$ requires lowest energy $\implies$ longest wavelength.
Step 3: Assign the Wavelengths
- Strongest field ($CN^-$) gets shortest $\lambda$: $[Co(CN)_6]^{3-}$ absorbs at $290 \text{ nm}$ (UV region).
- Moderate field ($NH_3$) gets middle $\lambda$: $[Co(NH_3)_6]^{3+}$ absorbs at $475 \text{ nm}$ (Blue region, transmits Yellow/Orange).
- Weakest field ($F^-$) gets longest $\lambda$: $[CoF_6]^{3-}$ absorbs at $700 \text{ nm}$ (Red region, transmits Green/Blue).
View Solution
Step 1: The Nitro Isomer
In the nitro complex, the Nitrogen atom donates its lone pair to the metal.
Formula: $[Co(NH_3)_5(NO_2)]^{2+}$. This complex is typically yellow.
Step 2: The Nitrito Isomer
In the nitrito complex, one of the Oxygen atoms donates its lone pair to the metal.
Formula: $[Co(NH_3)_5(ONO)]^{2+}$. This complex is typically red.
Step 3: IR Spectroscopy Differentiation
The internal bonds of the ligand vibrate at specific frequencies.
- In the Nitrito ($-O-N=O$) linkage, there is a distinct single $N-O$ bond and a double $N=O$ bond. IR will show two vastly different stretching frequencies (one high for double, one low for single).
- In the Nitro ($-NO_2$) linkage, resonance delocalizes the double bond across both $N-O$ bonds (giving them identical partial double-bond character). IR will show symmetric and asymmetric stretching frequencies that are very close together, characteristic of the $NO_2$ group.
View Solution
Step 1: The Active Structure
To be a hexadentate ligand, EDTA must use its 2 Nitrogen lone pairs and 4 Oxygen lone pairs. The oxygen lone pairs are only highly available when the four carboxylic acid groups are fully deprotonated into carboxylate anions ($-COO^-$).
Active form: The $EDTA^{4-}$ anion.
Step 2: The pH Barrier
Carboxylic acids are weak acids ($pKa \approx 2-3$). In a heavily acidic solution (massive excess of $H^+$ ions), Le Chatelier's principle drives the equilibrium backward. The $H^+$ ions protonate the carboxylate anions, converting them back into neutral $-COOH$ groups.
Step 3: Loss of Chelation
A neutral $-COOH$ group is a terrible ligand because the oxygen's lone pair is heavily delocalized into the carbonyl $\pi$-system via resonance. Without the dense negative charge of the carboxylate anions, EDTA loses four of its six "teeth." Furthermore, the Nitrogen lone pairs can also be protonated into $-NH^+$ groups, completely destroying its ability to bind the metal.
View Solution
Step 1: Configuration Analysis
$Cr^{3+}$ has a $3d^3$ configuration. The 6 ammonia ligands provide 12 electrons. Total = $3 + 12 = 15$ electrons. It is 3 electrons short of the 18-electron rule.
Step 2: Crystal Field Splitting ($t_{2g}$ and $e_g$)
In an octahedral field, the d-orbitals split into the lower energy $t_{2g}$ set (3 orbitals) and the higher energy $e_g$ set (2 orbitals). The $e_g$ orbitals are strongly anti-bonding (or non-bonding depending on the exact MO diagram) with respect to the metal-ligand sigma bonds.
Step 3: The $d^3$ Sweet Spot
With exactly 3 d-electrons, Chromium perfectly half-fills the lower $t_{2g}$ energy level ($t_{2g}^3$). This provides exceptional symmetrical stability and maximum Exchange Energy.
Crucially, the higher energy, destabilizing $e_g$ orbitals remain completely empty. To reach 18 electrons, the complex would have to push electrons into these high-energy anti-bonding orbitals, which would actually destabilize the complex and weaken the metal-ligand bonds.
View Solution
Step 1: Calculate the Van't Hoff factor ($i$)
$\Delta T_f = i \times K_f \times m$
$0.0558 = i \times 1.86 \times 0.01$
$0.0558 = i \times 0.0186$
$i = \frac{0.0558}{0.0186} = 3$.
Step 2: Interpret the Ions
Since $i = 3$ and it is $100\%$ dissociated, exactly 1 mole of the complex yields 3 moles of ions in solution. This must consist of one complex cation and two counter-anions. Therefore, exactly 2 Chloride ions must reside outside the coordination sphere.
Step 3: Construct the Sphere
Total formula is $PtCl_4 \cdot 4NH_3$. If 2 $Cl^-$ are outside, then 2 $Cl^-$ and all 4 $NH_3$ must be inside to satisfy Platinum's secondary valency of 6.
Sphere: $[Pt(NH_3)_4Cl_2]$.
Counter ions: $Cl_2$.
View Solution
Step 1: The Low Temperature State (Low Spin)
At $80 \text{ K}$, there is very little thermal energy available. The system settles into the absolute lowest potential energy state. Even if $\Delta_o$ is only slightly greater than $P$, the electrons will pair up in the lower $t_{2g}$ orbitals.
Configuration: $t_{2g}^6 e_g^0$. All electrons paired $\implies \mu = 0 \text{ B.M.}$ (Diamagnetic).
Step 2: The Entropy Factor
The Low Spin state has a lower Enthalpy ($H$), but the High Spin state ($t_{2g}^4 e_g^2$) has vastly more arrangements and higher spin multiplicity, meaning it possesses a significantly higher Entropy ($S$).
Step 3: The High Temperature State (High Spin)
As temperature rises to $300 \text{ K}$, the thermodynamic term $-T\Delta S$ begins to dominate the Gibbs Free Energy equation ($\Delta G = \Delta H - T\Delta S$). The massive entropic advantage of the High Spin state overcomes the slight enthalpy penalty of jumping the $\Delta_o$ gap.
The electrons unpair and populate the higher $e_g$ orbitals. Configuration: $t_{2g}^4 e_g^2$. This yields 4 unpaired electrons.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ B.M.}$
Mastering the Transition Metals
Congratulations on conquering these 25 ultra-challenging problems on Coordination Compounds! Transition metal chemistry is a beautiful intersection of quantum mechanics, 3D geometry, and classical thermodynamics. In JEE Advanced, rarely will a question ask just for a name; you must be prepared to link magnetic moments to hybridization, translate freezing point depressions into Werner structures, and decode the subtle IR spectra shifts caused by synergic back-bonding. Keep visualizing those octahedral axes, and visit Chemca.in for more elite masterclasses!
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