Given below are two statements:
Statement I: Among [mathrmCu(mathrmNH_3)_4]^2+$[\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+}$, [mathrmNi(mathrmen)_3]^2+$[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}$, [mathrmNi(mathrmNH_3)_6]^2+$[\mathrm{Ni}(\mathrm{NH}_{3})_{6}]^{2+}$ and [mathrmMn(mathrmH_2mathrmO)_6]^2+$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$, [mathrmMn(mathrmH_2mathrmO)_6]^2+$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$ has the maximum number of unpaired electrons.
Statement II: The number of pairs among \[NiCl_4]^2-, [Ni(CO)_4]\$\{[NiCl_{4}]^{2-}, [Ni(CO)_{4}]\}$, \[NiCl_4]^2-, [Ni(CN)_4]^2-\$\{[NiCl_{4}]^{2-}, [Ni(CN)_{4}]^{2-}\}$ and \[Ni(CO)_4], [Ni(CN)_4]^2-\$\{[Ni(CO)_{4}], [Ni(CN)_{4}]^{2-}\}$ that contain only diamagnetic species is two.
In the light of the above statements, choose the correct answer from the options given below:
A.textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
B.textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
C.textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
D.textStatement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
Solution & Explanation
### Core Logic
Evaluating Statement I:
- [Cu(NH_3)_4]^2+$[Cu(NH_3)_4]^{2+}$: Cu^2+$Cu^{2+}$ is 3d^9$3d^9$, 1 unpaired electron.
- [Ni(en)_3]^2+$[Ni(en)_3]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, in octahedral field, 2 unpaired electrons.
- [Ni(NH_3)_6]^2+$[Ni(NH_3)_6]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, 2 unpaired electrons.
- [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$: Mn^2+$Mn^{2+}$ is 3d^5$3d^5$, weak field ligand H_2O$H_2O$ leads to high spin, 5 unpaired electrons.
So [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true.
Evaluating Statement II:
- [Ni(CO)_4]$[Ni(CO)_4]$: Ni(0)$Ni(0)$ is 3d^8 4s^2$3d^8 4s^2$, strong field CO pairs electrons to 3d^10$3d^{10}$, diamagnetic (0 unpaired).
- [Ni(CN)_4]^2-$[Ni(CN)_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, strong field CN^-$CN^-$ forces pairing rightarrow dsp^2$\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired).
- [NiCl_4]^2-$[NiCl_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, weak field Cl^-$Cl^-$ does not pair rightarrow sp^3$\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired).
The pairs containing ONLY diamagnetic species:
- \[NiCl_4]^2-, [Ni(CO)_4]\$\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$rightarrow$\rightarrow$ 1 para, 1 dia (No)
- \[NiCl_4]^2-, [Ni(CN)_4]^2-\$\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$rightarrow$\rightarrow$ 1 para, 1 dia (No)
- \[Ni(CO)_4], [Ni(CN)_4]^2-\$\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$rightarrow$\rightarrow$ Both dia (Yes)
The number of such pairs is exactly ONE. Statement II says two, so it is false.
### Step 1: Final Conclusion
Statement I is true, Statement II is false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Keywords:#Magnetic Properties#unpaired electrons#JEE Main 2026 Morning Q65#Coordination Compounds JEE Main 2026
More Coordination Compounds Previous-Year Questions — Page 5
Q48jee_main_2025_04_april_eveningIsomerism in Coordination Compounds
A metal complex with a formula mathrmMCell_4cdot3mathrmNH_3$\mathrm{MC}\ell_{4}\cdot3\mathrm{NH}_{3}$ is involved in mathfraksp^3mathfrakd^2$\mathfrak{sp}^3\mathfrak{d}^2$ hybridisation. It upon reaction with excess of mathrmAgNO_3$\mathrm{AgNO}_3$ solution gives 'x' moles of AgCl. Consider 'x' is equal to the number of lone pairs of electron present in central atom of mathrmBrF_5$\mathrm{BrF}_5$ . Then the number of geometrical isomers exhibited by the complex is
Numerical Answer.Answer: 1.9 to 2.1
Solution
### Core Logic
1. Determine the value of x$x$:
- The central Bromine atom in BrF_5$BrF_5$ has 7 valence electrons. It forms 5 single bonds with fluorine, leaving 2 remaining electrons.
- Therefore, the number of lone pairs on Br in BrF_5$BrF_5$ is exactly 1 implies x = 1$\implies x = 1$.
2. Formulate the coordination sphere formula:
- Since x = 1$x = 1$, the complex yields 1 mole of AgCl$AgCl$ precipitate upon reaction with excess AgNO_3$AgNO_3$, meaning exactly 1 chloride ion sits outside the coordination sphere as an counter-ion.
- Rearranging the formula components around an octahedral coordination number of 6 gives the complex configuration:
[M(NH_3)_3Cl_3]Cl$$[M(NH_3)_3Cl_3]Cl$$
### Step 1: Isomer Analysis
Facial and meridional isomers representation for Q48
An octahedral complex of the type [Ma_3b_3]$[Ma_3b_3]$ exhibits exactly **2 geometrical isomers**:
- **Facial (fac)** isomer
- **Meridional (mer)** isomer
### Pattern Recognition
For [Ma_3b_3]$[Ma_3b_3]$ octahedral coordination types, don't waste time looking for optical active configurations. It splits cleanly into exactly two classical geometric forms: facial (all three identical ligands adjacent on a face) and meridional (ligands trace a meridian plane).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q31jee_main_2025_04_april_morningCrystal 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
Q34jee_main_2025_04_april_morningIsomerism 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
Q35jee_main_2025_07_april_eveningValency 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
More Coordination Compounds Questions — jee_main_2026_21_jan_morning
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