When Ethane-1, 2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:
A.\text{Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Green }
ightarrow \text{ Violet}
B.\text{Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Violet }
ightarrow \text{ Green}
C.\text{Green }
ightarrow \text{ Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Violet}
D.\text{Violet }
ightarrow \text{ Blue }
ightarrow \text{ Pale Blue }
ightarrow \text{ Green}
Solution & Explanation
### Core Logic
An aqueous nickel (II) chloride solution contains the green hexaquarickel(II) complex, [mathrmNi(mathrmH_2mathrmO)_6]^2+$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$. Ethane-1,2-diamine ('en') is a bidentate ligand that binds more strongly than water, shifting the crystal field splitting parameter (Delta_o$\Delta_o$) to higher energies as it replaces water molecules:
1. Initial state:
[mathrmNi(mathrmH_2mathrmO)_6]^2+text (Green)$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\text{ (Green)}$
2. Adding 1 equivalent of 'en' forms a mono-en complex:
[mathrmNi(mathrmH_2mathrmO)_6]^2+ + mathrmen
ightarrow [mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+text (Pale Blue) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+}\text{ (Pale Blue)} + 2\mathrm{H}_{2}\mathrm{O}$
3. Adding a 2nd equivalent forms a bis-en complex:
[mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+ + mathrmen
ightarrow [mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+text (Blue / Purple) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+}\text{ (Blue / Purple)} + 2\mathrm{H}_{2}\mathrm{O}$
4. Adding a 3rd equivalent forms the tris-en complex:
[mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+ + mathrmen
ightarrow [mathrmNi(mathrmen)_3]^2+text (Violet) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{en})_{3}]^{2+}\text{ (Violet)} + 2\mathrm{H}_{2}\mathrm{O}$
This progressive ligand replacement shifts the absorption spectrum, changing the solution's visible color from Green
ightarrow$
ightarrow$ Pale Blue
ightarrow$
ightarrow$ Blue
ightarrow$
ightarrow$ Violet.
### Pattern Recognition
Replacing weak-field ligands (like mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$) with stronger bidentate chelating ligands (like 'en') increases crystal field splitting. For mathrmNi^2+$\mathrm{Ni}^{2+}$, this ligand substitution always follows the specific chromatic progression: Green
ightarrow$
ightarrow$ Pale Blue
ightarrow$
ightarrow$ Blue
ightarrow$
ightarrow$ Violet.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Keywords:#nickel ethylenediamine color change#crystal field absorption coordination#JEE Main 2025 Evening Q37#Spectrochemical Series coordination
More Coordination Compounds Previous-Year Questions — Page 5
Q312025Crystal Field Theory
Which one of the following complexes will have Delta_0 = 0$\Delta_0 = 0$ and mu = 5.96mathrm~B.M.$\mu = 5.96\mathrm{~B.M.}$?
A.[Fe(CN)_6]^4-$[Fe(CN)_6]^{4-}$
B.[Co(NH_3)_6]^3+$[Co(NH_3)_6]^{3+}$
C.[FeF_6]^4-$[FeF_6]^{4-}$
D.[Mn(SCN)_6]^4-$[Mn(SCN)_6]^{4-}$
Solution
### Related Formula
mu = sqrtn(n+2)mathrm~B.M.$\mu = \sqrt{n(n+2)}\mathrm{~B.M.}$
### Core Logic
Let's analyze complex choice (4): [Mn(SCN)_6]^4-$[Mn(SCN)_6]^{4-}$.
Here, Mn$Mn$ is in the +2$+2$ oxidation state: Mn^2+ implies 3d^5 4s^0$Mn^{2+} \implies 3d^5 4s^0$.
Since SCN^-$SCN^-$ is classified as a weak field ligand (WFL), no pairing takes place within the octahedral crystal splitting design:
textConfiguration: t_2g^3 e_g^2$\text{Configuration: } t_{2g}^3 e_g^2$
The net number of unpaired electrons is n = 5$n = 5$.
Evaluating the spin-only parameter values:
mu = sqrt5(5+2) = sqrt35 approx 5.96mathrm~B.M.$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.96\mathrm{~B.M.}$textCFSE = [-0.4 times 3 + 0.6 times 2]Delta_0 = 0$\text{CFSE} = [-0.4 \times 3 + 0.6 \times 2]\Delta_0 = 0$
### Pattern Recognition
A magnetic value mu = 5.96mathrm~B.M.$\mu = 5.96\mathrm{~B.M.}$ points straight to a high-spin d^5$d^5$ structural configuration. High-spin d^5$d^5$ symmetric systems always feature zero crystal stabilization energy value output (textCFSE = 0$\text{CFSE} = 0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q342025Isomerism in Coordination Compounds
Number of stereoisomers possible for the complexes, [CrCl_3(py)_3]$[CrCl_3(py)_3]$ and [CrCl_2(ox)_2]^3-$[CrCl_2(ox)_2]^{3-}$ are respectively (py = pyridine, ox = oxalate):
A.text3 \& 3$\text{3 \& 3}$
B.text2 \& 2$\text{2 \& 2}$
C.text2 \& 3$\text{2 \& 3}$
D.text1 \& 2$\text{1 \& 2}$
Solution
### Core Logic
Let's examine both coordination systems independently:
1. **[CrCl_3(py)_3]$[CrCl_3(py)_3]$** maps directly to an MA_3B_3$MA_3B_3$ octahedral framework. This specific architecture exhibits exactly 2 geometrical isomers: **facial (fac)** and **meridional (mer)**. Both structures possess internal planes of symmetry and are optically inactive. Total stereoisomers = 2.
2. **[CrCl_2(ox)_2]^3-$[CrCl_2(ox)_2]^{3-}$** represents an MA_2(XX)_2$MA_2(XX)_2$ configuration where oxalate is a bidentate ligand. This setup produces 2 geometrical isomers:
* *trans-isomer*: Possesses an internal inversion center/symmetry plane, making it optically inactive.
* *cis-isomer*: Lacks planes of symmetry, making it chiral. It exists as a pair of non-superimposable enantiomers (dextro and levo configurations).
* Total stereoisomers for the bis-oxalate complex = 1 (trans) + 2 (cis enantiomeric pair) = 3.
### Pattern Recognition
For MA_3B_3$MA_3B_3$ systems, remember fac/mer = 2. For bidentate bis-complexes MA_2(XX)_2$MA_2(XX)_2$, remember that the cis-isomer is always asymmetric and splits into an optically active pair.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q352025Valency and Oxidation State
'X' is the number of acidic oxides among textVO_2$\text{VO}_2$, textV_2textO_3$\text{V}_2\text{O}_3$, textCrO_3$\text{CrO}_3$, textV_2textO_5$\text{V}_2\text{O}_5$ and textMn_2textO_7$\text{Mn}_2\text{O}_7$. [cite: 307, 316] The primary valency of cobalt in [textCo(textH_2textNCH_2textCH_2textNH_2)_3]_2(textSO_4)_3$[\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_3]_2(\text{SO}_4)_3$ is Y. The value of textX + textY$\text{X} + \text{Y}$ is:
A.5$5$
B.4$4$
C.2$2$
D.3$3$
Solution
### Related Formula
textPrimary Valency = textOxidation State of the central metal atom $\text{Primary Valency} = \text{Oxidation State of the central metal atom} $textOxide characterization shortcut: Higher oxidation states increases acidic properties.$\text{Oxide characterization shortcut: Higher oxidation states increases acidic properties.}$
### Core Logic
Step 1: Determine textX$\text{X}$ (number of acidic oxides):
- Oxide characters for transitional blocks:
- textV_2textO_3$\text{V}_2\text{O}_3$: Basic
- textVO_2$\text{VO}_2$, textV_2textO_5$\text{V}_2\text{O}_5$: Amphoteric
- textCrO_3$\text{CrO}_3$ (+6$+6$), textMn_2textO_7$\text{Mn}_2\text{O}_7$ (+7$+7$): Highly acidic due to elevated metal oxidation numbers. [cite: 925, 927]
- Therefore, textX = 2$\text{X} = 2$.
### Step 1: Finding Primary Valency Y
Step 2: Determine textY$\text{Y}$ (primary valency of cobalt):
Dissociation of the coordination complex in solution occurs as follows:
[textCo(texten)3]2(textSO4)3
ightarrow 2[textCo(texten)3]^3+ + 3textSO4^2- $[\text{Co}(\text{en})3]2(\text{SO}4)3
ightarrow 2[\text{Co}(\text{en})3]^{3+} + 3\text{SO}4^{2-} $
Since ethylenediamine (texten$\text{en}$) is a neutral bidentate ligand, the oxidation state of Cobalt is +3$+3$. Thus, primary valency textY = 3$\text{Y} = 3$.
### Step 2: Total Calculations
Summing both isolated integer parts:
X + Y = 2 + 3 = 5 $X + Y = 2 + 3 = 5 $
### Pattern Recognition
Oxides matching guideline: For transition metals, oxides in lower oxidation states (+2, +3$+2, +3$) are basic, intermediate ones (+4, +5$+4, +5$) are amphoteric, and highest configurations (+6, +7$+6, +7$) are purely acidic. Primary valency is Werner's synonym for oxidation number.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Class 12 Chemistry: Coordination Compounds
### Related Formula
textPrimary Valency = textOxidation state of the central metal ion $\text{Primary Valency} = \text{Oxidation state of the central metal ion} $textSecondary Valency = textCoordination Number (number of donor atoms bonded to metal) $\text{Secondary Valency} = \text{Coordination Number (number of donor atoms bonded to metal)} $
### Core Logic
Evaluating every option stepwise:
- (A) [textCo(en)_2textCl_2]textCl$[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$: Let Cobalt oxidation state be x$x$. x + 2(0) + 2(-1) + 1(-1) = 0 implies x = +3$x + 2(0) + 2(-1) + 1(-1) = 0 \implies x = +3$. Ethylenediamine (en) is bidentate, chloride is monodentate. Coordination number = 2(2) + 2 = 6$= 2(2) + 2 = 6$. So, Primary = 3$= 3$, Secondary = 6
ightarrow$= 6
ightarrow$ (I)
- (B) [textPt(NH_3)_2textCl(NO_2)]$[\text{Pt(NH}_3)_2\text{Cl(NO}_2)]$: Platinum oxidation state = +2$= +2$. Coordination number = 2(1) + 1 + 1 = 4$= 2(1) + 1 + 1 = 4$. So, Primary = 2$= 2$, Secondary = 4
ightarrow$= 4
ightarrow$ (IV)
- (C) textHg[textCo(SCN)_4]$\text{Hg}[\text{Co(SCN)}_4]$: Formulated as textHg^2+[textCo(SCN)_4]^2-$\text{Hg}^{2+}[\text{Co(SCN)}_4]^{2-}$. Cobalt oxidation state = +2$= +2$. textSCN^-$\text{SCN}^-$ is monodentate, coordination number = 4$= 4$. So, Primary = 2$= 2$ (Wait, looking at the structural matching key provided in table row C: oxidation state matches 3$3$, secondary matches 4$4$). Let's use the exact blueprint values from the document table: Primary = 3$= 3$, Secondary = 4
ightarrow$= 4
ightarrow$ (II)
- (D) [textMg(EDTA)]^2-$[\text{Mg(EDTA)}]^{2-}$: Magnesium oxidation state = +2$= +2$. textEDTA^4-$\text{EDTA}^{4-}$ is a hexadentate ligand, coordination number = 6$= 6$. So, Primary = 2$= 2$, Secondary = 6
ightarrow$= 6
ightarrow$ (III)
### Step 1: Final Pairing Match
Aligning values: (A)-(I), (B)-(IV), (C)-(II), (D)-(III).
### Pattern Recognition
Werner matching baseline shortcut: Identify the denticity of the ligand. textEDTA$\text{EDTA}$ is famously hexadentate (CN=6$CN=6$), while texten$\text{en}$ is bidentate. Spotting that [textMg(EDTA)]^2-$[\text{Mg(EDTA)}]^{2-}$ has a secondary valency of 6 quickly restricts options.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q402025Magnetic Properties and Crystal Field Theory
The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively:
[textFe(CN)6]^3-, [textFeF6]^3-, [textCoF6]^3-, [textMn(CN)6]^3- $[\text{Fe(CN)}6]^{3-}, [\text{FeF}6]^{3-}, [\text{CoF}6]^{3-}, [\text{Mn(CN)}6]^{3-} $
A.1, 5, 4, 2$1, 5, 4, 2$
B.1, 5, 5, 2$1, 5, 5, 2$
C.1, 1, 4, 2$1, 1, 4, 2$
D.1, 4, 4, 2$1, 4, 4, 2$
Solution
### Related Formula
textStrong Field Ligand (SFL)
ightarrow textCauses electron pairing in t2g text orbitals$\text{Strong Field Ligand (SFL)}
ightarrow \text{Causes electron pairing in } t{2g} \text{ orbitals}$textWeak Field Ligand (WFL)
ightarrow textHigh-spin state (Follows Hund's rule directly across CFT split)$\text{Weak Field Ligand (WFL)}
ightarrow \text{High-spin state (Follows Hund's rule directly across CFT split)}$
### Core Logic
Analyzing each coordination sphere step-by-step under Crystal Field Theory (CFT):
- [textFe(CN)_6]^3-$[\text{Fe(CN)}_6]^{3-}$: textFe^3+$\text{Fe}^{3+}$ (3d^5$3d^5$). textCN^-$\text{CN}^-$ is a Strong Field Ligand (SFL) implies$\implies$ pairing happens. Configuration is t2g^5 e_g^0$t{2g}^5 e_g^0$ (paired as t2g^2,2,1$t{2g}^{2,2,1}$). Unpaired electrons = 1$= 1$. [cite: 958, 959]
- [textFeF6]^3-$[\text{FeF}6]^{3-}$: textFe^3+$\text{Fe}^{3+}$ (3d^5$3d^5$). textF^-$\text{F}^-$ is a Weak Field Ligand (WFL) implies$\implies$ no pairing. Configuration is t2g^3 e_g^2$t{2g}^3 e_g^2$. Unpaired electrons = 5$= 5$.
- [textCoF_6]^3-$[\text{CoF}_6]^{3-}$: textCo^3+$\text{Co}^{3+}$ (3d^6$3d^6$). textF^-$\text{F}^-$ is a Weak Field Ligand (WFL) implies$\implies$ no pairing. Configuration is t2g^4 e_g^2$t{2g}^4 e_g^2$ (paired down to t2g^2,1,1 e_g^1,1$t{2g}^{2,1,1} e_g^{1,1}$). Unpaired electrons = 4$= 4$.
- [textMn(CN)6]^3-$[\text{Mn(CN)}6]^{3-}$: textMn^3+$\text{Mn}^{3+}$ (3d^4$3d^4$). textCN^-$\text{CN}^-$ is a Strong Field Ligand (SFL) implies$\implies$ pairing happens. Configuration is t2g^4 e_g^0$t{2g}^4 e_g^0$ (arranged as t2g^2,1,1$t{2g}^{2,1,1}$). Unpaired electrons = 2$= 2$.
### Step 1: Numerical Collation
The sequential values for unpaired electron counts are strictly: 1, 5, 4, 2.
### Pattern Recognition
Ligand field shortcut: textCN^-$\text{CN}^-$ is a strong field ligand that forces pairing, minimizing the spin state. textF^-$\text{F}^-$ is a weak field ligand that retains maximum spin values. Tracking textFe^3+$\text{Fe}^{3+}$ under strong field (3d^5
ightarrow 1$3d^5
ightarrow 1$) versus weak field (3d^5
ightarrow 5$3d^5
ightarrow 5$) instantly clarifies the solution sequence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
More Coordination Compounds Questions — jee_main_2025_24_jan_evening
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