CBSE Class 12 Chemistry exam was conducted on February 27, 2025, from 10:30 AM to 1:30 PM. An estimated 17.5 lakh students are expected to appear for the exam across 7,800 centers in India and 26 other countries
The Chemistry theory paper has 70 marks, while 30 marks are allocated for the practical assessment. The paper is divided into Physical, Organic, and Inorganic Chemistry, and it includes numerical, conceptual, and application-based problems. The question paper includes multiple-choice questions (1 mark each), short-answer questions (3 marks each), and long-answer questions (5 marks each).
CBSE Class 12 Chemistry Question Paper 2025 (Set 3 – 56/1/3) with Answer Key
| CBSE Class 12 2025 Chemistry Question Paper with Answer Key |  Download |
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Standard electrode potential for Sn\(^{4+}\)/Sn\(^{2+}\) couple is +0.15 V and that for the Cr\(^{3+}\)/Cr couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be.
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The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr\(^{3+}\) ion (Atomic no. : Cr = 24) is ____________.
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In case of association, abnormal molar mass of solute will.
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Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve.
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Arrange the following compounds in increasing order of their boiling points :
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- (A) (ii) < (i) < (iii)
- (B) (i) < (ii) < (iii)
- (C) (iii) < (i) < (ii)
- (D) (iii) < (ii) < (i)
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The correct IUPAC name of [Pt(NH\(_3\))\(_2\)Cl\(_2\)]\(^{2+}\) is.
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The acid formed when propyl magnesium bromide is treated with CO\(_2\) followed by acid hydrolysis is :
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Acidified KMnO\(_4\) oxidises sulphite to.
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Which is the correct order of acid strength from the following ?.
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An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because \hspace{2cm}..
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The best reagent for converting propanamide into propanamine is.
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Which of the following statements is not true about glucose ?.
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Assertion (A) : All naturally occurring \(\alpha\)-amino acids except glycine are optically active..
Reason (R) : Most naturally occurring amino acids have L-configuration..
(A) Both Assertion (A) and Reason (R) are true and Reason R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
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Assertion (A) : The boiling point of ethanol is higher than that of methoxymethane..
Reason (R) : There is intramolecular hydrogen bonding in ethanol..
(A) Both Assertion (A) and Reason (R) are true and Reason R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
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Assertion (A) : The boiling points of alkyl halides decrease in the order RI \(>\) RBr \(>\) RCl \(>\) RF..
Reason (R) : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass..
(A) Both Assertion (A) and Reason (R) are true and Reason R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
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Assertion (A) : [Cr(H\(_2\)O)\(_6\)]Cl\(_2\) and [Fe(H\(_2\)O)\(_6\)]Cl\(_2\) are examples of homoleptic complexes..
Reason (R) : All the ligands attached to the metal are the same..
(A) Both Assertion (A) and Reason (R) are true and Reason R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
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Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Justify your answer..
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Complete and balance the following chemical equations :
(a) MnO\(_4^-\) + 3S\(_2\)O\(_3^{2-}\) + H\(_2\)O \(\rightarrow\).
(b) Cr\(_2\)O\(_7^{2-}\) + 3Sn\(^{2+}\) + 14H\(^+\) \(\rightarrow\).
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(A) Give reasons :
(a) Cooking is faster in pressure cooker than in an open pan..
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution ? What change in temperature would you observe after mixing liquids X and Y ?.
OR.
(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law ? Give an example..
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Identify A and B in each of the following reaction sequence :
(a) CH\(_3\)CH\(_2\)Cl \(\xrightarrow{NaCN}\) A \(\xrightarrow{H_2/Ni}\) B.
(b) C\(_6\)H\(_5\)NH\(_2\) \(\xrightarrow{NaNO_2/HCl, 0-5^\circ C}\) A \(\xrightarrow{H^+/H_2O, \Delta}\) B.
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What are the hydrolysis products of :
(a) Sucrose.
(b) Lactose.
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Hydrolysis involves breaking a bond (in this case, a glycosidic bond) by adding water, usually catalyzed by acid or enzymes.
(a) Sucrose: Sucrose (common table sugar) is a disaccharide composed of one unit of \(\alpha\)-glucose and one unit of \(\beta\)-fructose linked by an \(\alpha, \beta\)-1,2-glycosidic bond. Hydrolysis of sucrose yields one molecule of glucose and one molecule of fructose. \(\) Sucrose + \text{H_2\text{O \xrightarrow{\text{H^+ \text{ or Sucrase \text{Glucose + \text{Fructose \(\)
Hydrolysis Products of Sucrose: Glucose and Fructose..
(b) Lactose: Lactose (milk sugar) is a disaccharide composed of one unit of \(\beta\)-galactose and one unit of glucose (either \(\alpha\) or \(\beta\)) linked by a \(\beta\)-1,4-glycosidic bond. Hydrolysis of lactose yields one molecule of galactose and one molecule of glucose. \(\) \text{Lactose + \text{H_2\text{O \xrightarrow{\text{H^+ \text{ or Lactase \text{Galactose + \text{Glucose \(\)
Hydrolysis Products of Lactose: Galactose and Glucose..
Quick Tip: \textbf{Disaccharide Hydrolysis. Sucrose \(\rightarrow\) Glucose + Fructose. Lactose \(\rightarrow\) Galactose + Glucose. Maltose \(\rightarrow\) Glucose + Glucose. Hydrolysis breaks the glycosidic bond.
Henry's law constant for CO\(_2\) in water is \(1.67 \times 10^8\) Pa at 298 K. Calculate the number of moles of CO\(_2\) in 500 ml of soda water when packed under 2.53 \(\times\) 10\(^5\) Pa at the same temperature..
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According to Henry's Law, the partial pressure (\(p\)) of a gas above a liquid is directly proportional to the mole fraction (\(x\)) of the gas dissolved in the liquid: \(\) p = K_H x \(\)
where \(K_H\) is Henry's law constant.
Given:
Henry's law constant for CO\(_2\), \(K_H = 1.67 \times 10^8\) Pa.
Partial pressure of CO\(_2\) under which the soda water is packed, \(p_{CO_2} = 2.53 \times 10^5\) Pa.
Volume of water = 500 ml.
Temperature T = 298 K (though not directly needed for this calculation if K\(_H\) is given at this T).
First, calculate the mole fraction (\(x_{CO_2}\)) of CO\(_2\) dissolved in water: \(\) x_{CO_2 = \frac{p_{CO_2{K_H = \frac{2.53 \times 10^5 \, Pa{1.67 \times 10^8 \, \text{Pa \(\) \(\) x_{CO_2 \approx 1.515 \times 10^{-3 \(\)
Mole fraction is defined as: \(\) x_{CO_2 = \frac{\text{moles of CO_2{\text{moles of CO_2 + \text{moles of H_2O \(\)
Let \(n_{CO_2\) be the moles of CO\(_2\) and \(n_{H_2O}\) be the moles of water.
Since the solubility of CO\(_2\) is low (mole fraction is small), we can approximate \(n_{CO_2} + n_{H_2O} \approx n_{H_2O}\). \(\) x_{CO_2 \approx \frac{n_{CO_2{n_{H_2O \(\)
Now, calculate the moles of water in 500 ml. Assume the density of water is approximately 1 g/ml.
Mass of water = Volume \(\times\) Density = 500 ml \(\times\) 1 g/ml = 500 g.
Molar mass of water (H\(_2\)O) = 2(1.01) + 16.00 = 18.02 g/mol \(\approx\) 18 g/mol.
Moles of water \(n_{H_2O} = \frac{Mass}{Molar Mass} = \frac{500 \, g}{18 \, g/mol} \approx 27.78 \, mol\).
Now, calculate the moles of CO\(_2\): \(\) n_{CO_2 \approx x_{CO_2 \times n_{H_2O \(\) \(\) n_{CO_2 \approx (1.515 \times 10^{-3) \times (27.78 \, mol) \(\) \(\) n_{CO_2 \approx 0.04209 \, \text{mol \(\)
The number of moles of CO\(_2\) dissolved is approximately 0.0421 mol.
Quick Tip: \textbf{Henry's Law. \(p = K_H x\). Relates partial pressure of gas (\(p\)) to its mole fraction (\(x\)) in solution via Henry's constant (\(K_H\)). For dilute solutions, \(x_{gas} \approx n_{gas} / n_{solvent}\).
Calculate \(\Delta_r G^\circ\) and log K\(_C\) of the reaction..
2Cr(s) + 3Cd\(^{2+}\)(aq) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 3Cd(s).
Given \(E^\circ_{Cr^{3+}/Cr}\) = -0.74 V.
\(E^\circ_{Cd^{2+}/Cd}\) = -0.40 V.
[R = 8.314 J K\(^{-1}\) mol\(^{-1}\), F = 96500 C mol\(^{-1}\)].
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First, identify the half-reactions and calculate the standard cell potential (\(E^\circ_{cell}\)).
The reaction shows Cr(s) being oxidized to Cr\(^{3+}\) and Cd\(^{2+}\) being reduced to Cd(s).
Anode (Oxidation): Cr(s) \(\rightarrow\) Cr\(^{3+}\)(aq) + 3e\(^-\). \(E^\circ_{anode} = E^\circ_{Cr^{3+}/Cr} = -0.74\) V.
Cathode (Reduction): Cd\(^{2+}\)(aq) + 2e\(^-\) \(\rightarrow\) Cd(s). \(E^\circ_{cathode} = E^\circ_{Cd^{2+}/Cd} = -0.40\) V.
Calculate standard cell potential: \(\) E^\circ_{cell = E^\circ_{cathode - E^\circ_{anode = (-0.40 \, V) - (-0.74 \, \text{V) \(\) \(\) E^\circ_{cell = -0.40 + 0.74 = +0.34 \, \text{V \(\)
Next, determine the number of moles of electrons (\(n\)) transferred in the balanced overall reaction.
Oxidation: 2Cr \(\rightarrow\) 2Cr\(^{3+\) + 6e\(^-\)
Reduction: 3Cd\(^{2+}\) + 6e\(^-\) \(\rightarrow\) 3Cd
The number of electrons transferred is \(n = 6\).
Now calculate the standard Gibbs free energy change (\(\Delta_r G^\circ\)): \(\) \Delta_r G^\circ = -nFE^\circ_{cell \(\) \(\) \Delta_r G^\circ = -(6 \, mol) \times (96500 \, \text{C mol^{-1) \times (0.34 \, \text{V) \(\)
Note: 1 C \(\times\) 1 V = 1 J. \(\) \Delta_r G^\circ = - (6 \times 96500 \times 0.34) \, \text{J \(\) \(\) \Delta_r G^\circ = - 196860 \, \text{J = -196.86 \, \text{kJ \(\)
Finally, calculate the logarithm of the equilibrium constant (log K\(_C\)). The relationship is: \(\) \Delta_r G^\circ = -RT \ln K_c = -RT (2.303 \log_{10 K_c) \(\) \(\) \log_{10 K_c = -\frac{\Delta_r G^\circ{2.303 RT \(\)
Assuming standard temperature T = 298 K: \(\) \log_{10 K_c = -\frac{-196860 \, \text{J mol^{-1{2.303 \times (8.314 \, \text{J K^{-1 \, \text{mol^{-1) \times (298 \, \text{K) \(\) \(\) \log_{10 K_c = \frac{196860{2.303 \times 8.314 \times 298 \approx \frac{196860{5705.8 \approx 34.50 \(\)
Alternatively, using the Nernst equation relationship at standard conditions: \(\) E^\circ_{cell = \frac{RT{nF \ln K_c = \frac{2.303 RT{nF \log_{10 K_c \(\)
At T=298 K, \( \frac{2.303 RT{F} \approx 0.0591 \) V. \(\) E^\circ_{cell = \frac{0.0591{n \log_{10 K_c \(\) \(\) 0.34 = \frac{0.0591{6 \log_{10 K_c \(\) \(\) \log_{10 K_c = \frac{0.34 \times 6{0.0591 = \frac{2.04{0.0591 \approx 34.517 \(\)
Results: \(\) \Delta_r G^\circ \approx -196.9 \, kJ/mol \(\) \(\) \log_{10 K_c \approx 34.5 \(\) Quick Tip: \textbf{Electrochemistry Relations. \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\). \(\Delta G^\circ = -nFE^\circ_{cell}\). \(\Delta G^\circ = -RT \ln K_c\) or \(E^\circ_{cell} = (RT/nF) \ln K_c\). Determine \(n\) from balanced half-reactions.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature..
[Given : log 4 = 0.602, log 2 = 0.301, R = 8.314 J K\(^{-1}\) mol\(^{-1}\)].
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(A) Draw the structure of the major monohalo product for each of the following reaction :
(a) Ethylbenzene + Br\(_2\)/Heat.
(b) Methylcyclohexane + HBr.
(c) p-Cresol + HCl/Heat.
OR.
(B) How do you convert :
(a) Chlorobenzene to biphenyl.
(b) Propene to 1-Iodopropane.
(c) 2-bromobutane to but-2-ene..
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The elements of 3d transition series are given as : Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn.
Answer the following :
(a) Copper has exceptionally positive \(E^\circ_{M^{2+}/M}\) value, why ?.
(b) Which element is a strong reducing agent in +2 oxidation state and why ?.
(c) Zn\(^{2+}\) salts are colourless. Why ?.
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(a) Arrange the following compounds in increasing order of their boiling point :
(CH\(_3\))\(_2\)NH, CH\(_3\)CH\(_2\)NH\(_2\), CH\(_3\)CH\(_2\)OH..
(b) Give plausible explanation for each of the following :
(i) Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis..
(ii) Amides are less basic than amines..
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Define the following terms :
(a) Native Protein.
(b) Nucleotide.
(c) Essential amino acid.
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(a) Native Protein:
The term "native protein" refers to a protein in its functional, fully folded three-dimensional conformation, as it exists naturally within a biological system. This specific spatial arrangement (including secondary, tertiary, and potentially quaternary structure) is crucial for the protein's biological activity. When a protein loses this specific structure (e.g., due to heat, pH changes, chemicals), it undergoes denaturation and loses its function; the denatured state is non-native.
(b) Nucleotide:
A nucleotide is the basic structural unit (monomer) of nucleic acids like DNA and RNA. It consists of three components covalently linked together:
1. A nitrogenous base (either a purine: Adenine (A), Guanine (G); or a pyrimidine: Cytosine (C), Thymine (T) in DNA, Uracil (U) in RNA).
2. A pentose sugar (Deoxyribose in DNA, Ribose in RNA).
3. One or more phosphate groups (typically attached to the 5' carbon of the sugar).
Nucleotides also serve as energy carriers (e.g., ATP, GTP) and signaling molecules (e.g., cAMP).
(c) Essential Amino Acid:
Essential amino acids are amino acids that cannot be synthesized by an organism (specifically, humans in this context) in sufficient quantities to meet its physiological needs and must therefore be obtained from the diet. There are typically considered 9 essential amino acids for adult humans: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. Lack of any essential amino acid in the diet can lead to protein deficiency and health problems.
Quick Tip: \textbf{Biomolecule Definitions.} Native Protein: Functional, folded 3D structure. Nucleotide: Base + Sugar + Phosphate (monomer of nucleic acids). Essential Amino Acid: Cannot be synthesized by the body, must be obtained from diet.
Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of the OH group attached to the benzene ring. Since, the OH group increases the electron density more to the \(ortho\)- and \(para\)- positions, therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols..
Answer the following questions:
(a) What happens when phenol reacts with:
(b) Why phenol does not undergo protonation readily ?
(c) Which is a stronger acid -phenol or cresol? Give reason.
OR
(c) Write the 1UPAC name) of the product formed in the Reimer Tiemann reaction.
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The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction..
Answer the following questions:
(i) What is a rate determining step?.
(ii) Define complex reaction.
(iii) What is the effect of temperature on the rate constant of a reaction?
OR.
(a) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions?.
(b) The conversion of molecule X to Y follows second-order kinetics. If the concentration of X is increased 3 times, how will it affect the rate of formation of Y?.
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(A).
(a) Carry out the following conversions :
(i) Ethanal to But-2-enal.
(ii) Propanoic acid to ethane.
(b) An alkene A with molecular formula C\(_5\)H\(_{10}\) on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C..
OR.
(B) An organic compound (A) (molecular formula C\(_8\)H\(_{16}\)O\(_2\)) was hydrolyzed with dilute sulphuric acid to get a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives But-1-ene. Identify (A), (B) and (C) and write chemical equations for the reactions involved..
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(A) In the following complex ions, explain the type of hybridisation, shape and magnetic property :
[Fe(H2o)6]2+
[NiCI4]2-
(At. Nos. : Fe= 26. Ni = 28)
(B). For the complex \([Fe(en)_2Cl_2]^+ \), identify:
(B) (a) Write IUPAC names of the following:
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(a) Oxidation number of iron:
In the complex \([Fe(en)_2Cl_2]^+\), \(en\) (ethylenediamine) is a neutral ligand, and \(Cl\) is a monodentate anionic ligand. Let the oxidation number of Fe be \(x\). For two chloride ions, each with a charge of -1, the charge balance equation is: \[ x + 2(-1) + 0 = +1 \quad (since the overall charge of the complex is +1) \]
Thus, \(x = +3\). Therefore, the oxidation number of iron is +3.
(b) Hybridization and Shape of the Complex:
Since the complex involves 2 bidentate ethylenediamine ligands and 2 chloride ions, the coordination number of iron is 6. In this case, iron will be in an octahedral arrangement with respect to the ligands. Hence, the hybridization of Fe is \(d^2sp^3\), and the shape of the complex is octahedral.
(c) Magnetic Behaviour of the Complex:
The electronic configuration of Fe\(^{3+}\) is \(3d^5\). Since there are 5 unpaired electrons in the \(d\)-orbitals, the complex will be paramagnetic.
(d) Optical Isomerism:
Since the complex contains two bidentate ligands (ethylenediamine), it may exhibit optical isomerism. This is because the complex does not have a plane of symmetry, making it capable of existing as non-superimposable mirror images. Therefore, the complex exhibits optical isomerism.
(e) IUPAC Name of the Complex:
The IUPAC name of the complex is bis(ethylenediamine)chloridoiron(III) chloride.
% Section (B)
OR.
(B) (a) Using IUPAC norms, write the names of the following:
% Option
(i) \([Co(NH_3)_4Cl(NO_3)]Cl\)
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(ii) \(K_3[Fe(CN)_6]\)
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(iii) \([Cr(C_2O_4)_3]^{3-}\)
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Solution:
(i) IUPAC Name:
For \([Co(NH_3)_4Cl(NO_3)]Cl\), the central metal is cobalt, and the ligands are four ammine (NH\(_3\)), one chloride (Cl), and one nitrate (NO\(_3\)). The IUPAC name is tetraamminechloridonitratocobalt(III) chloride.
(ii) IUPAC Name:
For \(K_3[Fe(CN)_6]\), the central metal is iron, and the ligand is cyanide (CN). The complex is in the +3 oxidation state of iron. The IUPAC name is potassium hexacyanoferrate(III).
(iii) IUPAC Name:
For \([Cr(C_2O_4)_3]^{3-}\), the central metal is chromium, and the ligand is oxalate (\(C_2O_4^{2-}\)). The IUPAC name is tris(oxalato)chromium(III).
(b) What is crystal field splitting energy? Why low spin tetrahedral complexes are not formed?.
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Solution:
Crystal Field Splitting Energy (CFSE):
Crystal Field Splitting Energy is the energy difference between two sets of degenerate orbitals in a metal complex, caused by the interaction between the central metal ion and the ligands. In octahedral complexes, the d-orbitals split into two sets, \(e_g\) (higher energy) and \(t_{2g}\) (lower energy), and the difference between these energy levels is called the crystal field splitting energy (\(\Delta\)). The value of \(\Delta\) determines the stability and magnetic properties of the complex.
Low Spin Tetrahedral Complexes:
Tetrahedral complexes are generally high spin because the crystal field splitting energy (\(\Delta_t\)) in tetrahedral geometry is much smaller than in octahedral geometry. In tetrahedral complexes, the splitting energy between the \(e\) and \(t_2\) orbitals is relatively small, and it does not favor the pairing of electrons. Hence, low spin tetrahedral complexes are rare. The low spin state is usually favored in octahedral complexes with strong field ligands.
Quick Tip: \textbf{Magnetic Properties and Field Strength.} In octahedral complexes, stronger field ligands lead to larger crystal field splitting energy (\(\Delta\)), which may favor the low spin state (fewer unpaired electrons). In tetrahedral complexes, the splitting energy is smaller, making them typically high spin with more unpaired electrons.
(A).
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
Sn(s) | Sn\(^{2+}\) (0.004 M) || H\(^+\) (0.02 M) | H\(_2\)(g) (1 Bar) | Pt(s).
(Given: \(E^\circ_{Sn^{2+}/Sn}\) = -0.14 V, \(E^\circ_{H^+/H_2(g), Pt}\) = 0.00V).
(b) Account for the following ; % Note: Sub-question (b) from image 10, page 23 appears here.
(i) On the basis of E\(^\circ\) values, O\(_2\) gas should be liberated at anode but it is Cl\(_2\) gas which is liberated in the electrolysis of aqueous NaCl..
(ii) Conductivity of CH\(_3\)COOH decreases on dilution..
OR.
(B).
(a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use..
(b) Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01M Cr\(^{3+}\) (aq) and 1.0 x 10\(^{-4}\) M H\(^+\)(aq)..
The half cell reaction is.
Cr\(_2\)O\(_7^{2-}\)(aq) + 14H\(^+\)(aq) + 6e\(^-\) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 7H\(_2\)O(l).
and the standard electrode potential is given as E\(^\circ\) = 1.33 V..
[Given : log 10 = 1].
View Solution
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