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The oxidation number of iron in the compound formed during brown ring test for mathrmNO_3^- ion is ________.

Numerical Answer Type:
Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

### Related Formula textFormula of Brown Ring Complex: [mathrmFe(mathrmH_2mathrmO)_5(mathrmNO)]^2+ ### Core Logic In this specific coordination complex, charge transfer occurs where nitric oxide transfers an electron to the iron center. As a result, textNO exists as a positive textNO^+ ligand, and iron drops to an unusual oxidation state: x + 5(0) + 1(+1) = +2 x + 1 = 2 implies x = +1 ### Step 1: Final Value Assignment Solving this charge balance confirms that the oxidation number of iron in the brown ring complex is +1. ### Pattern Recognition The brown ring test features a rare +1 oxidation state for iron because textNO coordinates as the positive nitrosonium cation (textNO^+). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Reference Study Guides

More Coordination Compounds Previous-Year Questions — Page 8

Q71 jee_main_2024_01_february_morning Definitions of Some Important Terms Pertaining to Coordination Compounds
Which of the following complex is homoleptic?
  • A. [Ni(CN)_4]^2-
  • B. [Ni(NH_3)_2Cl_2]
  • C. [Fe(NH_3)_4Cl_2]^+
  • D. [Co(NH_3)_4Cl_2]^+

Solution

### Core Logic A homoleptic complex is one in which the central metal atom/ion is bound to only one kind of donor group (ligand). A heteroleptic complex is one in which the central metal atom/ion is bound to more than one kind of donor group. ### Step 1: Analyze Options (1) [Ni(CN)_4]^2-: Only one type of ligand (CN^-). Homoleptic. (2) [Ni(NH_3)_2Cl_2]: Two types of ligands (NH_3 and Cl^-). Heteroleptic. (3) [Fe(NH_3)_4Cl_2]^+: Two types of ligands. Heteroleptic. (4) [Co(NH_3)_4Cl_2]^+: Two types of ligands. Heteroleptic. ### Pattern Recognition Homo = same, leptic = ligands. Look for the formula bracket containing only one symbol type after the central metal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q66 jee_main_2024_29_january_evening IUPAC Nomenclature of Coordination Complexes
The correct IUPAC name of mathrmK_2mathrmMnO_4 is
  • A. Potassium tetraoxopermanganate (VI)
  • B. Potassium tetraoxidomanganate (VI)
  • C. Dipotassium tetraoxidomanganate (VII)
  • D. Potassium tetraoxidomanganese (VI)

Solution

### Related Formula textOxidation State of Mn Evaluation: 2(+1) + x + 4(-2) = 0 ### Core Logic Solving for x: 2 + x - 8 = 0 implies x = +6 Since the complex is anionic, the metal name ends with the suffix '-ate', making it 'manganate(VI)'. The ligands are oxygen atoms, designated systematically as 'tetraoxido' or 'tetraoxo' according to newer recommendations. ### Step 1: Assembly Combining parts systematically yields the correct IUPAC string: Potassium tetraoxidomanganate(VI). ### Pattern Recognition Anionic metal centers must include the trailing '-ate' modifier followed immediately by their absolute Roman oxidation state indicators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q68 jee_main_2024_27_jan_morning Magnetic Properties of Coordination Complexes
Consider the following complex ions: P=[FeF_6]^3- Q=[V(H_2O)_6]^2+ R=[Fe(H_2O)_6]^2+ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
  • A. R < Q < P
  • B. R < P < Q
  • C. Q < R < P
  • D. Q < P < R

Solution

### Step 1: Evaluation of P=[FeF_6]^3- textFe^3+ rightarrow 3textd^5 Since textF^- is a weak field ligand, no pairing occurs. Number of unpaired electrons (n) = 5. mu = sqrt5(5+2) = sqrt35text BM ### Step 2: Evaluation of Q=[V(H_2O)_6]^2+ textV^2+ rightarrow 3textd^3 Number of unpaired electrons (n) = 3. mu = sqrt3(3+2) = sqrt15text BM ### Step 3: Evaluation of R=[Fe(H_2O)_6]^2+ textFe^2+ rightarrow 3textd^6 Since textH_2textO is a weak field ligand, configuration is textt_2textg^4 texte_textg^2. Number of unpaired electrons (n) = 4. mu = sqrt4(4+2) = sqrt24text BM ### Step 4: Comparison Comparing values: mu(Q) < mu(R) < mu(P) implies Q < R < P ### Pattern Recognition Count unpaired electrons strictly accounting for weak field vs strong field rules. Order scales monotonically with n. ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q62 jee_main_2024_29_jan_morning Applications of Coordination Compounds
Match List I with List II
LIST-I (Substances)LIST-II (Element Present)
A. Ziegler catalystI. Rhodium
B. Blood PigmentII. Cobalt
C. Wilkinson catalystIII. Iron
D. Vitamin B_12IV. Titanium
Choose the correct answer from the options given below:
  • A. textA-II, B-IV, C-I, D-III
  • B. textA-II, B-III, C-IV, D-I
  • C. textA-III, B-II, C-IV, D-I
  • D. textA-IV, B-III, C-I, D-II

Solution

### Core Logic Let us identify the central metal atom/element present in each of the given coordination complexes or biological molecules: 1. **Ziegler-Natta catalyst**: Used for polymerization of alkenes. Its chemical composition involves TiCl_4 and (C_2H_5)_3Al. Hence, the transition metal present is Titanium (IV). 2. **Blood pigment (Haemoglobin)**: The oxygen-carrying metalloprotein in red blood cells. It contains an Iron (III) central atom coordinated to a porphyrin ring. 3. **Wilkinson's catalyst**: Used for the hydrogenation of alkenes. Its formula is [RhCl(PPh_3)_3], meaning it contains Rhodium (I). 4. **Vitamin B_12 (Cyanocobalamin)**: A biologically important coordination compound containing Cobalt (II) at the center of a corrin ring. ### Step 1: Final Mapping Matching the elements: (A) Ziegler catalyst rightarrow (IV) Titanium (B) Blood Pigment rightarrow (III) Iron (C) Wilkinson catalyst rightarrow (I) Rhodium (D) Vitamin B_12 rightarrow (II) Cobalt This strictly maps to sequence A-IV, B-III, C-I, D-II. ### Pattern Recognition Always memorize the central metal for famous catalysts and biomolecules: Chlorophyll (Mg), Haemoglobin (Fe), Vitamin B12 (Co), Wilkinson (Rh), Ziegler-Natta (Ti). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d and f Block Elements

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