Maharashtra Board HSC Chemistry Question Paper Solution March 2023
Complete Solutions with Detailed Explanations
SECTION − A
Q.1. Select and write the correct answer for the following multiple choice type of questions: [10]
The relation between radius of sphere and edge length in body centered cubic lattice is given by formula:
Explanation: In a BCC lattice, atoms touch along the body diagonal. The length of the body diagonal is \( \sqrt{3}a \), which equals \( 4r \). Therefore, \( 4r = \sqrt{3}a \) or \( r = \frac{\sqrt{3}}{4}a \).
The pH of weak monoacidic base is 11.2, its OH⁻ ion concentration is:
Explanation: \( \text{pH} + \text{pOH} = 14 \)
\( 11.2 + \text{pOH} = 14 \implies \text{pOH} = 2.8 \)
\( [\text{OH}^-] = \text{antilog}(-2.8) = 10^{-2.8} = 10^{-3} \times 10^{0.2} \)
From log tables, antilog(0.2) ≈ 1.585.
Therefore, \( [\text{OH}^-] = 1.585 \times 10^{-3} \) mol dm⁻³.
Which of the following correctly represents integrated rate law equation for a first order reaction in gas phase?
Explanation: For reaction \( A(g) \rightarrow B(g) + C(g) \), initial pressure is \( P_i \). At time t, total pressure \( P = P_i + x \), so \( x = P - P_i \). The pressure of reactant A remaining is \( P_A = P_i - x = P_i - (P - P_i) = 2P_i - P \). Substituting into the rate law gives the formula in option (B).
The spin only magnetic moment of Mn²⁺ ion is _______.
Explanation: Mn (Z=25) has configuration [Ar] 3d⁵ 4s². Mn²⁺ has configuration [Ar] 3d⁵. Number of unpaired electrons (n) = 5.
\( \mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.916 \) BM.
The correct formula of a complex having IUPAC name Tetraamminedibromoplatinum (IV) bromide is ________.
Explanation: Platinum(IV) means charge +4. Ligands: 4 NH₃ (neutral) and 2 Br⁻ (-1 each). Charge on sphere = +4 + 0 + 2(-1) = +2. To balance this +2 charge, two bromide ions (Br⁻) are needed outside the sphere. Formula: [PtBr₂(NH₃)₄]Br₂.
The allylic halide, among the following is _______.
Explanation: In allylic halides, the halogen atom is bonded to an sp³ hybridized carbon atom next to a carbon-carbon double bond (C=C). Option (D) fits this definition perfectly.
The product of following reaction is:
CH₃ – CH = CH – CH₂ – CHO \(\xrightarrow{i) LiAlH_4 \ ii) H_3O^+}\) ________?
Explanation: LiAlH₄ is a strong reducing agent that reduces aldehydes (-CHO) to primary alcohols (-CH₂OH). It typically does not reduce isolated carbon-carbon double bonds. Thus, the double bond remains intact.
Ozonolysis of 2, 3 dimethyl but-2-ene, followed by decomposition by Zn dust and water gives ________.
Explanation: Structure: \( (CH_3)_2C = C(CH_3)_2 \). Breaking the double bond and adding oxygen gives two molecules of \( (CH_3)_2C=O \), which is Acetone (Propan-2-one).
The glycosidic linkage present in maltose is ________.
The monomer of natural rubber is ________.
Explanation: Natural rubber is cis-1,4-polyisoprene. Its monomer is Isoprene (2-methylbuta-1,3-diene).
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Q.2. Answer the following questions: [8]
Write the name of the technique used to know geometry of nanoparticles.
Write the name of the product formed by the action of LiAlH₄/ ether on acetamide.
Reaction: \( CH_3CONH_2 + 4[H] \rightarrow CH_3CH_2NH_2 + H_2O \)
Write the structure of the product formed when chlorobenzene is treated with sodium metal in the presence of dry ether.
Structure: Ph – Ph (Two benzene rings connected directly).
Write the chemical composition of cryolite.
Write the name of platinum complex used in the treatment of cancer.
Write the SI unit of cryoscopic constant.
Write the correct condition for spontaneity in terms of Gibbs energy.
Calculate molar conductivity for 0.5 M BaCl₂ if its conductivity at 298K is 0.01 Ω⁻¹ cm⁻¹.
Formula: \( \Lambda_m = \frac{1000 \times k}{C} \)
Calculation: \( \Lambda_m = \frac{1000 \times 0.01}{0.5} = \frac{10}{0.5} = 20 \)
Result: \( 20 \, \Omega^{-1} cm^2 mol^{-1} \) (or \( S \cdot cm^2 \cdot mol^{-1} \)).
SECTION − B
Attempt any EIGHT of the following questions: [16]
Distinguish between lanthanides and actinides.
| Lanthanides | Actinides |
|---|---|
| 1. Last electron enters into 4f orbital. | 1. Last electron enters into 5f orbital. |
| 2. Binding energy of 4f electrons is higher. | 2. Binding energy of 5f electrons is lower. |
| 3. Besides +3, they show +2 and +4 oxidation states in few cases. | 3. Besides +3, they show higher oxidation states like +4, +5, +6, +7. |
| 4. They are non-radioactive (except Promethium). | 4. They are all radioactive. |
Calculate the mole fraction of solute, if the vapour pressure of pure benzene at certain temperature is 640 mmHg and vapour pressure of solution of a solute in benzene is 600 mmHg.
\( P^0_1 \) (Pure solvent) = 640 mmHg
\( P \) (Solution) = 600 mmHg
To find: Mole fraction of solute (\( x_2 \)).
Formula: Relative lowering of vapour pressure \( \frac{P^0_1 - P}{P^0_1} = x_2 \)
Calculation:
\( x_2 = \frac{640 - 600}{640} \)
\( x_2 = \frac{40}{640} \)
\( x_2 = \frac{1}{16} = 0.0625 \)
Result: Mole fraction of solute is 0.0625.
Define: Green chemistry. Write two advantages of nanoparticle and nanotechnology.
Advantages of Nanotechnology:
1. Revolutionized electronics with faster and smaller devices.
2. Development of smart drugs and targeted drug delivery systems in medicine (e.g., curing cancer).
3. Self-cleaning surfaces (e.g., lotus effect).
Explain the following terms:
i. Substitutional impurity defect
ii. Interstitial impurity defect
ii. Interstitial impurity defect: This defect arises when a foreign atom occupies the interstitial void (empty space) between the host atoms in the crystal lattice. Example: Steel (Carbon atoms in Iron lattice).
Write the chemical reactions for the following:
i. Chlorobenzene is heated with fuming H₂SO₄
ii. Ethyl bromide is heated with silver acetate
\( \text{C}_6\text{H}_5\text{Cl} + \text{H}_2\text{SO}_4 (\text{fuming}) \xrightarrow{\Delta} \text{p-Chlorobenzenesulphonic acid (Major)} + \text{o-Chlorobenzenesulphonic acid} + \text{H}_2\text{O} \)
ii. Action of Silver Acetate on Ethyl Bromide:
\( \text{C}_2\text{H}_5\text{Br} + \text{CH}_3\text{COOAg} \xrightarrow{\Delta} \text{CH}_3\text{COOC}_2\text{H}_5 (\text{Ethyl Acetate}) + \text{AgBr} \downarrow \)
Define: Acidic buffer solution. Write the relationship between solubility and solubility product for PbI₂.
Relationship for PbI₂:
Dissociation: \( \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \)
Let \( S \) be the solubility in mol/L.
\( [\text{Pb}^{2+}] = S \), \( [\text{I}^-] = 2S \)
\( K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = (S)(2S)^2 = 4S^3 \)
\( \therefore K_{sp} = 4S^3 \)
What is the action of the following reagents on ethyl amine:
i. Chloroform and caustic potash
ii. Nitrous acid
\( \text{C}_2\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc)} \xrightarrow{\Delta} \text{C}_2\text{H}_5\text{NC} (\text{Ethyl isocyanide}) + 3\text{KCl} + 3\text{H}_2\text{O} \)
(Offensive smelling substance formed).
ii. Action of Nitrous Acid (HNO₂):
\( \text{C}_2\text{H}_5\text{NH}_2 + \text{HNO}_2 \xrightarrow{NaNO_2/HCl, 273K} \text{C}_2\text{H}_5\text{OH} (\text{Ethyl alcohol}) + \text{N}_2 \uparrow + \text{H}_2\text{O} \)
Calculate standard Gibbs energy change at 25°C for the cell reaction
\( \text{Cd}_{(s)} + \text{Sn}^{2+}_{(aq)} \rightarrow \text{Cd}^{2+}_{(aq)} + \text{Sn}_{(s)} \)
Given: \( E^\circ_{Cd} = -0.403V \), \( E^\circ_{Sn} = -0.136V \)
Cathode (Reduction): Sn (accepts electrons)
Anode (Oxidation): Cd (loses electrons)
\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.136) - (-0.403) = +0.267 \, V \)
2. Calculate \( \Delta G^\circ \):
Formula: \( \Delta G^\circ = -nFE^\circ_{cell} \)
Here \( n = 2 \) electrons, \( F = 96500 \, C/mol \)
\( \Delta G^\circ = -2 \times 96500 \times 0.267 \)
\( \Delta G^\circ = -193000 \times 0.267 \)
\( \Delta G^\circ = -51531 \, J/mol \)
\( \Delta G^\circ = -51.531 \, kJ/mol \)
Write chemical reaction for the preparation of glucose from sucrose. Write structure of D-ribose.
\( \text{C}_{12}\text{H}_{22}\text{O}_{11} (\text{Sucrose}) + \text{H}_2\text{O} \xrightarrow{\text{dil. HCl} / \Delta} \text{C}_6\text{H}_{12}\text{O}_6 (\text{Glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{Fructose}) \)
Structure of D-Ribose:
It is an aldopentose.
CHO – CH(OH) – CH(OH) – CH(OH) – CH₂OH
(In Fischer projection, all three OH groups on chiral carbons are on the right side).
Define Extensive property. Calculate the work done during the expansion of 2 moles of an ideal gas from 10 dm³ to 20 dm³ at 298 K in vacuum.
Calculation:
Expansion is in vacuum.
External pressure \( P_{ext} = 0 \).
Work formula: \( W = -P_{ext} \Delta V \)
\( W = -0 \times (20 - 10) = 0 \)
Work done is Zero.
Write the reactions for the formation of nylon 6,6 polymer.
Reaction:
\( n \text{HOOC}-(\text{CH}_2)_4-\text{COOH} + n \text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2 \)
\( \xrightarrow{\Delta, \text{Polymerization}} \)
\( [-\text{OC}-(\text{CH}_2)_4-\text{CONH}-(\text{CH}_2)_6-\text{NH}-]_n + 2n\text{H}_2\text{O} \)
(Formation of Polyamide Nylon 6,6).
Draw structures of the following compounds:
i. chloric acid
ii. peroxy disulphuric acid
Cl is central atom. Double bond to two Oxygens (one O is OH?). No, structure is: H-O-Cl(=O)(=O). Cl has oxidation state +5.
ii. Peroxy disulphuric acid (\( H_2S_2O_8 \)):
Structure contains a peroxy linkage (-O-O-) between two Sulfur atoms.
HO-SO₂-O-O-SO₂-OH.
SECTION − C
Attempt any EIGHT of the following questions: [24]
Define Osmosis. How will you determine molar mass of non volatile solute by elevation of boiling point?
Molar Mass Determination:
1. The elevation in boiling point (\( \Delta T_b \)) is directly proportional to molality (\( m \)).
\( \Delta T_b = K_b \times m \)
2. Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 W_2}{M_2 W_1} \)
Where \( W_2 \) = mass of solute, \( M_2 \) = molar mass of solute, \( W_1 \) = mass of solvent.
3. Substituting m in the first equation:
\( \Delta T_b = K_b \times \frac{1000 W_2}{M_2 W_1} \)
4. Rearranging for \( M_2 \):
\( M_2 = \frac{1000 K_b W_2}{\Delta T_b W_1} \)
Convert the following:
i. Ethyl alcohol into ethyl acetate
ii. Phenol into benzene
iii. Diethyl ether into ethyl chloride
By Esterification with Acetic Acid.
\( \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
ii. Phenol to Benzene:
Reduction with Zinc dust.
\( \text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO} \)
iii. Diethyl ether to Ethyl chloride:
Action of \( PCl_5 \).
\( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{PCl}_5 \xrightarrow{\Delta} 2\text{C}_2\text{H}_5\text{Cl} + \text{POCl}_3 \)
A weak monobasic acid is 10% dissociated in 0.05 M solution. What is percent dissociation in 0.15 M solution?
Given: \( \alpha_1 = 10\% = 0.1 \), \( C_1 = 0.05 \, M \)
\( K_a = C_1 \alpha_1^2 \)
\( K_a = 0.05 \times (0.1)^2 = 0.05 \times 0.01 = 5 \times 10^{-4} \)
Step 2: Calculate \( \alpha_2 \) for \( C_2 = 0.15 \, M \)
\( \alpha_2 = \sqrt{\frac{K_a}{C_2}} \)
\( \alpha_2 = \sqrt{\frac{5 \times 10^{-4}}{0.15}} = \sqrt{\frac{5}{15} \times 10^{-3} \times \frac{10}{10}} \text{ (Adjusting powers)} \)
Actually: \( \frac{5 \times 10^{-4}}{0.15} = \frac{5}{1500} \times 100 \times 10^{-4} \) ... easier: \( \frac{5 \times 10^{-4}}{15 \times 10^{-2}} = \frac{1}{3} \times 10^{-2} = 0.333 \times 10^{-2} = 33.3 \times 10^{-4} \)
\( \alpha_2 = \sqrt{33.3 \times 10^{-4}} \approx 5.77 \times 10^{-2} \)
\( \alpha_2 = 0.0577 \)
Percent dissociation = \( 0.0577 \times 100 = 5.77\% \).
Explain dehydrohalogenation reaction of 2-chlorobutane. Write use and environmental effect of CFC.
When 2-chlorobutane is boiled with alcoholic KOH, hydrogen and halogen are removed to form an alkene. It follows Saytzeff rule (More substituted alkene is major product).
\( \text{CH}_3-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3 + \text{KOH (alc)} \rightarrow \)
Major Product: But-2-ene (\( \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \)) (80%)
Minor Product: But-1-ene (\( \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3 \)) (20%)
CFC (Chlorofluorocarbons):
Use: Used as refrigerants in ACs and refrigerators, and as propellants in aerosols.
Environmental Effect: They cause depletion of the Ozone layer in the stratosphere, allowing harmful UV radiation to reach earth.
2000 mmol of an ideal gas expanded isothermally and reversibly from 20 L to 30 L at 300 K, calculate the work done in the process (R = 8.314 JK⁻¹ mol⁻¹).
\( n = 2000 \, \text{mmol} = 2 \, \text{mol} \)
\( V_1 = 20 \, L, V_2 = 30 \, L \)
\( T = 300 \, K \)
\( R = 8.314 \, J K^{-1} mol^{-1} \)
Formula: \( W_{max} = -2.303 nRT \log_{10}(\frac{V_2}{V_1}) \)
Calculation:
\( W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}(\frac{30}{20}) \)
\( W = -11488.29 \times \log_{10}(1.5) \)
\( \log_{10}(1.5) \approx 0.1761 \)
\( W = -11488.29 \times 0.1761 \approx -2023 \, J \)
\( W = -2.023 \, kJ \)
Result: Work done is -2.023 kJ (Work done by the gas).
What are interstitial compounds? Give the classification of alloys with examples.
Classification of Alloys:
1. Ferrous Alloys: Contain Iron as main component. Ex: Stainless Steel, Nickel Steel.
2. Non-Ferrous Alloys: Do not contain Iron. Ex: Brass (Cu+Zn), Bronze (Cu+Sn).
(Alternatively: Homogeneous vs Heterogeneous, or Interstitial vs Substitutional).
Draw labelled diagram of H₂ – O₂ fuel cell. Write two applications of fuel cell.
- Anode: Porous carbon containing catalyst (H₂ gas inlet).
- Cathode: Porous carbon containing catalyst (O₂ gas inlet).
- Electrolyte: Hot aqueous KOH solution in the center.
- Reactions: \( 2H_2 + 4OH^- \to 4H_2O + 4e^- \) (Anode), \( O_2 + 2H_2O + 4e^- \to 4OH^- \) (Cathode).
Applications:
1. Used as a primary source of electrical energy in space programs (e.g., Apollo space program).
2. Used in electric vehicles and power generators due to high efficiency and zero pollution.
Explain formation of [CoF₆]³⁻ complex with respect to:
i. Hybridisation
ii. Magnetic properties
iii. Inner / outer complex
iv. Geometry
Fluoride (F⁻) is a weak field ligand, so no pairing of electrons occurs.
i. Hybridisation: \( sp^3d^2 \) (Uses 4s, 4p, and outer 4d orbitals).
ii. Magnetic properties: Paramagnetic (Due to 4 unpaired electrons in 3d).
iii. Type: Outer orbital complex (High spin complex).
iv. Geometry: Octahedral.
What is Pseudo first order reaction? Derive integrated rate law equation for zero order reaction.
Derivation for Zero Order:
Consider reaction \( A \rightarrow P \).
Rate = \( -\frac{d[A]}{dt} = k[A]^0 = k \)
\( d[A] = -k dt \)
Integrate both sides: \( \int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \)
\( [A]_t - [A]_0 = -kt \)
\( [A]_t = [A]_0 - kt \)
\( k = \frac{[A]_0 - [A]_t}{t} \)
Explain Aldol condensation of ethanal.
Step 1: Formation of Aldol (3-hydroxybutanal).
\( 2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \)
Step 2: Dehydration on heating.
\( CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO \)
Product: But-2-enal (Crotonaldehyde).
Explain anomalous behaviour of oxygen in group 16 with respect to:
i. Atomicity
ii. Magnetic property
iii. Oxidation state
ii. Magnetic property: Oxygen is paramagnetic (due to unpaired electrons in antibonding orbitals), while others are diamagnetic.
iii. Oxidation state: Oxygen shows -2 oxidation state mostly (and -1 in peroxides, +2 in \( OF_2 \)). It does not show higher positive oxidation states (+4, +6) due to absence of d-orbitals, unlike Sulfur etc.
Write chemical reactions for the following conversions:
i. Acetic acid into acetic anhydride
ii. Acetic acid into ethyl alcohol
Write IUPAC name and structure of methylphenylamine.
Dehydration using \( P_2O_5 \) and heat.
\( 2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O \)
ii. Acetic acid to Ethyl alcohol:
Reduction using \( LiAlH_4 \).
\( CH_3COOH \xrightarrow{LiAlH_4/ether, H_3O^+} CH_3CH_2OH \)
Methylphenylamine:
Structure: \( C_6H_5-NH-CH_3 \) (Aniline ring with N attached to Methyl).
IUPAC Name: N-Methylbenzenamine (or N-Methylaniline).
SECTION − D
Attempt any THREE of the following questions: [12]
Show that, time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Give electronic configuration of Gd (Z = 64).
Write the name of nano structured material used in car tyres to increase the life of tyres.
For 1st order: \( t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t} \)
For 90% completion: \( [A]_t = 10\% \) of \( [A]_0 = 0.1 [A]_0 \).
\( t_{90} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} (1) \)
For 99.9% completion: \( [A]_t = 0.1\% \) of \( [A]_0 = 0.001 [A]_0 \).
\( t_{99.9} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} (3) \)
Comparing: \( t_{99.9} = 3 \times t_{90} \). Hence Proved.
2. Gadolinium (Gd, Z=64):
Config: [Xe] \( 4f^7 5d^1 6s^2 \).
3. Nanomaterial in Tyres:
Carbon Black (nanoparticles).
Derive relationship between ∆H and ∆U for gaseous reaction.
Define: Vulcanization
What is peptide bond?
\( H = U + PV \). Change: \( \Delta H = \Delta U + \Delta(PV) \).
For ideal gas at constant T: \( \Delta(PV) = \Delta(nRT) = \Delta n_g RT \).
\( \therefore \Delta H = \Delta U + \Delta n_g RT \).
2. Vulcanization: The process of heating raw rubber with sulfur and an accelerator at 373K-415K to form cross-links between polymer chains, improving elasticity, strength, and stability.
3. Peptide Bond: The amide linkage (\( -CO-NH- \)) formed between the carboxyl group of one amino acid and the amino group of another amino acid with the elimination of a water molecule.
Silver crystallizes in fcc structure. If edge length of unit cell is 400 pm, calculate density of silver (Atomic mass of Ag = 108).
Write a note on Haloform reaction.
Given: FCC lattice (\( z=4 \)), \( a = 400 \, \text{pm} = 4 \times 10^{-8} \, \text{cm} \), \( M = 108 \, \text{g/mol} \).
Formula: \( \rho = \frac{z \times M}{a^3 \times N_A} \)
\( \rho = \frac{4 \times 108}{(4 \times 10^{-8})^3 \times 6.022 \times 10^{23}} \)
\( \rho = \frac{432}{64 \times 10^{-24} \times 6.022 \times 10^{23}} \)
\( \rho = \frac{432}{64 \times 0.6022} = \frac{432}{38.54} \approx 11.21 \, \text{g/cm}^3 \).
2. Haloform Reaction:
Reaction given by methyl ketones (\( CH_3-CO-R \)) or alcohols with \( CH_3-CH(OH)- \) group. When treated with Halogen (\( Cl_2/Br_2/I_2 \)) and alkali (NaOH), they form haloform (\( CHX_3 \)) and sodium salt of carboxylic acid. Example: Iodoform test (formation of yellow ppt of \( CHI_3 \)).
Define: Diastereoisomers.
Give cis and trans isomers of [Co(NH₃)₄Cl₂]⁺.
What is reference electrode?
Give reason: Bleaching action of ozone is also called dry bleach.
2. Isomers of [Co(NH₃)₄Cl₂]⁺:
- Cis: Two Cl⁻ ligands are adjacent (90°).
- Trans: Two Cl⁻ ligands are opposite (180°).
3. Reference Electrode: An electrode whose potential is arbitrarily fixed or exactly known and is used to measure the potential of other electrodes (e.g., Standard Hydrogen Electrode).
4. Reason (Dry Bleach): Ozone bleaches by oxidation due to the liberation of nascent oxygen (\( O_3 \to O_2 + [O] \)). Unlike Chlorine which requires moisture (water) to form HOCl to bleach, ozone can bleach dry oils and starch effectively through direct oxidation.
Write Dow process for preparation of Phenol. What is the action of bromine water on phenol?
Give reason: Group 16th elements have lower ionisation enthalpy compared to group 15th elements.
Write two uses of dioxygen.
\( Ph-Cl + NaOH \to Ph-ONa \xrightarrow{H^+} Ph-OH \).
2. Action of Bromine water: Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol.
3. Reason (IE Trend): Group 15 elements have extra stable half-filled p-orbital configuration (\( ns^2 np^3 \)). Group 16 has \( ns^2 np^4 \). Removing an electron from Group 16 results in a stable half-filled configuration, requiring less energy than breaking the stable configuration of Group 15.
4. Uses of Dioxygen:
- Used in respiration by living organisms.
- Used in oxy-acetylene welding.
- Used in steel manufacturing.
Target Publications - Board Question Paper March 2023 Solution
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