Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans-platin are heteroleptic complexes of Pd. In the light of the above statements, choose the correct answer from the options given below:

Let us evaluate both statements individually: * **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is [Ma6]. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.** * **Statement II**: Cis-platin and trans-platin have the chemical formula [Pt(NH3)2Cl2]. While they are indeed heteroleptic complexes, they are coordination coordinates of **Platinum (Pt)**, not Palladium (Pd). **Statement II is false.** Always read element symbols with immense focus in coordination chemistry. Changing a single letter from Pt to Pd creates a false assertion trap designed to test parsing alertness rather than chemical difficulty. Class 12 Chemistry: Coordination Compounds

Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans-platin are heteroleptic comp

Let us evaluate both statements individually: * **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is [Ma6]. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.** * **Statement II**: Cis-platin and trans-platin have the chemic

Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans-platin are heteroleptic complexes of Pd. In the light of the above statements, choose the correct answer from the options given below:

Let us evaluate both statements individually: * **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is [Ma6]. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.** * **Statement II**: Cis-platin and trans-platin have the chemical formula [Pt(NH3)2Cl2]. While they are indeed heteroleptic complexes, they are coordination coordinates of **Platinum (Pt)**, not Palladium (Pd). **Statement II is false.** Always read element symbols with immense focus in coordination chemistry. Changing a single letter from Pt to Pd creates a false assertion trap designed to test parsing alertness rather than chemical difficulty. Class 12 Chemistry: Coordination Compounds

Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans-platin are heteroleptic comp

Let us evaluate both statements individually: * **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is [Ma6]. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.** * **Statement II**: Cis-platin and trans-platin have the chemic

Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will…

Year	2026	2025	2024	Total
Questions	19	34	15	68

Q26 jee_main_2025_29_jan_evening Magnetic Properties of Coordination Compounds

The calculated spin-only magnetic moments of

K₃[Fe(OH)₆]

and

K₄[Fe(OH)₆]

respectively are: (1) 4.90 and 4.90 B.M. (2) 5.92 and 4.90 B.M. (3) 3.87 and 4.90 B.M. (4) 4.90 and 5.92 B.M.

A. 4.90 and 4.90 B.M.
B. 5.92 and 4.90 B.M.
C. 3.87 and 4.90 B.M.
D. 4.90 and 5.92 B.M.

Solution

Related Formula

mu = sqrtn(n+2) B.M.

Core Logic

In $K₃[Fe(OH)₆]$ $K_{3}[Fe(OH)_{6}]$, iron is in the $+3$ $+3$ oxidation state ( $Fe³⁺ = 3d⁵$ $Fe^{3+} = 3d^{5}$). Since $OH⁻$ $OH^{-}$ is a weak field ligand, no pairing of electrons takes place. The number of unpaired electrons ( $n$ $n$) is $5$ $5$.

mu = sqrt5(5+2) = sqrt35 approx 5.92 B.M.

In $K₄[Fe(OH)₆]$ $K_{4}[Fe(OH)_{6}]$, iron is in the $+2$ $+2$ oxidation state ( $Fe²⁺ = 3d⁶$ $Fe^{2+} = 3d^{6}$). Since $OH⁻$ $OH^{-}$ is a weak field ligand, no pairing occurs. The number of unpaired electrons ( $n$ $n$) is $4$ $4$.

mu = sqrt4(4+2) = sqrt24 approx 4.90 B.M.

Pattern Recognition

Identify the ligand field strength first. $OH⁻$ $OH^{-}$ is a weak field ligand in the spectrochemical series, so it does not cause pairing in either $Fe²⁺$ $Fe^{2+}$ or $Fe³⁺$ $Fe^{3+}$ configurations.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q37 jee_main_2025_29_jan_evening Homoleptic Complexes and Electronic Configurations

Identify the homoleptic complexes with odd number of d electrons in the central metal. (A)

[FeO₄]²⁻

(B)

[Fe(CN)₆]³⁻

(C)

[Fe(CN)₅NO]²⁻

(D)

[CoCl₄]²⁻

(E)

[Co(H₂O)₃F₃]

Choose the correct answer from the options given below:

A. (B) and (D) only
B. (C) and (E) only
C. (A), (B) and (D) only
D. (A), (C) and (E) only

Solution

Core Logic

A complex is homoleptic if the metal is bound to only one kind of donor ligand group.

(A) $[FeO₄]²⁻$ $[FeO_4]^{2-}$ is homoleptic, but $Fe⁺⁶$ $Fe^{+6}$ corresponds to a $3d²$ $3d^2$ (even) electronic configuration.
(B) $[Fe(CN)₆]³⁻$ $[Fe(CN)_6]^{3-}$ is homoleptic. $Fe⁺³$ $Fe^{+3}$ corresponds to a $3d⁵$ $3d^5$ (odd) configuration.
(C) $[Fe(CN)₅NO]²⁻$ $[Fe(CN)_5NO]^{2-}$ is heteroleptic (contains two types of ligands).
(D) $[CoCl₄]²⁻$ $[CoCl_4]^{2-}$ is homoleptic. $Co⁺²$ $Co^{+2}$ corresponds to a $3d⁷$ $3d^7$ (odd) configuration.
(E) $[Co(H₂O)₃F₃]$ $[Co(H_2O)_3F_3]$ is heteroleptic.

Pattern Recognition

Filter by 'homoleptic' first to instantly eliminate multi-ligand mixed structures like options (C) and (E).

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q46 jee_main_2025_29_jan_evening Crystal Field Theory and Colors of Complexes

Consider the following low-spin complexes

K₃[Co(NO₂)₆]

,

K₄[Fe(CN)₆]

,

K₃[Fe(CN)₆]

,

Cu₂[Fe(CN)₆]

and

Zn₂[Fe(CN)₆]

. The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer is nearest integer)

Numerical Answer. Answer: 0 to 0

Solution

Core Logic

From the given list, the complexes exhibiting a distinct yellow color are $K₃[Co(NO₂)₆]$ $K_{3}[Co(NO_{2})_{6}]$ and $K₄[Fe(CN)₆]$ $K_{4}[Fe(CN)_{6}]$.

Let's calculate the spin-only magnetic moments for these low-spin configurations:

For $K₃[Co(NO₂)₆]$ $K_{3}[Co(NO_{2})_{6}]$, cobalt is in $+3$ $+3$ oxidation state ( $Co³⁺ = 3d⁶$ $Co^{3+} = 3d^{6}$).

In the presence of the strong ligand field ( $NO₂^-$ $NO_2^-$), all six electrons pair up completely in the $t2g$ $t_{2g}$ orbitals:

t2g⁶ eg⁰ implies n = 0 unpaired electrons implies mu = 0 BM

Crystal Field Theory and Colors of Complexes diagram for Q46 - JEE Main 2025 Evening

For $K₄[Fe(CN)₆]$ $K_{4}[Fe(CN)_{6}]$, iron is in $+2$ $+2$ oxidation state ( $Fe²⁺ = 3d⁶$ $Fe^{2+} = 3d^{6}$).

In the strong field of cyanide ligands ( $CN^-$ $CN^-$), pairing is complete:

t2g⁶ eg⁰ implies n = 0 unpaired electrons implies mu = 0 BM

Therefore, the sum of their spin-only magnetic moments is $0 + 0 = 0$ $0 + 0 = 0$.

Pattern Recognition

Low-spin $d⁶$ $d^6$ octahedral complexes always yield a fully closed-shell $t2g⁶$ $t_{2g}^6$ arrangement with zero unpaired electrons, leading deterministically to a magnetic moment of 0 BM.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q36 jee_main_2025_28_jan_morning Borax Bead Test and Crystal Field Split

The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is

A. $Ti³⁺$ $\mathrm{Ti}^{3+}$
B. $Ni²⁺$ $\mathrm{Ni}^{2+}$
C. $Mn²⁺$ $\mathrm{Mn}^{2+}$
D. $Cr³⁺$ $\mathrm{Cr}^{3+}$

Solution

Core Logic

Nickel ( $Ni²⁺$ $\mathrm{Ni}^{2+}$) exhibits a $d⁸$ $d^8$ electronic profile. In regular octahedral complex splits: $t2g⁶ eg²$ $t_{2g}^6 e_g^2$ Because the lower $t2g$ $t_{2g}$ subshell is fully paired and the higher $eg$ $e_g$ contains exactly 2 electrons matching Hund's rules, this orbital distribution remains configurationally identical under both strong-field and weak-field environments. Additionally, $Ni²⁺$ $\mathrm{Ni}^{2+}$ compounds produce a characteristic violet bead during hot cycles in a non-luminous flame within the qualitative borax matrix.

Pattern Recognition

Sees: Configuration invariant to ligand strength + qualitative test combination. Shortcut: A $d⁸$ $d^8$ structure in octahedral splitting always stays high-spin/low-spin identical, pointing strictly to $Ni²⁺$ $\mathrm{Ni}^{2+}$.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: The d-and f-Block Elements

Q jee_main_2025_03_april_morning Crystal Field Theory - Color and Spectrochemical Series

The correct order of the complexes

[Co(NH₃)₅(H₂O)]³⁺

(A),

[Co(NH₃)₆]³⁺

(B),

[Co(CN)₆]³⁻(C)

and

[CoCl(NH₃)₅]²⁺

(D) in terms wavelength of light absorbed is :

A. $D>A>B>C$ $D>A>B>C$
B. $C>B>D>A$ $C>B>D>A$
C. $D>C>B>A$ $D>C>B>A$
D. $C>B>A>D$ $C>B>A>D$

Solution

Related Formula

The energy of light absorbed is inversely proportional to the wavelength absorbed:

E = Δₒ = (hc)/(λ) λ \propto (1)/(Δₒ)

Core Logic

All complexes share the same central metal ion, $Co³⁺$ $\text{Co}^{3+}$. The magnitude of the crystal field splitting energy ( $Δₒ$ $\Delta_o$) depends exclusively on the relative ligand field strength listed in the spectrochemical series:

Cl^- < H₂O < NH₃ < CN^-

Step 1: Ordering Energies and Wavelengths

The splitting energy order is:

CFSE: C (highest) > B > A > D (lowest)

Inverting this sequence to match absorption wavelength values yields:

λabsorbed: D > A > B > C

Pattern Recognition

Shortcut: Stronger field ligand $\implies$ larger splitting gap $\implies$ high photon energy $\implies$ shorter absorbed wavelength. Since $CN^-$ $\text{CN}^-$ is a strong field ligand, complex C must absorb the shortest wavelength, putting it at the very end.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Coordination Compounds Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ordering Energies and Wavelengths

Pattern Recognition

Chapter Mix

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