Showing posts with label July 2023. Show all posts
Showing posts with label July 2023. Show all posts

HSC Maharashtra Board Class 12 Biology Question Paper Solutions July 2023

Board Question Paper: July 2023

BIOLOGY

Time: 3 Hrs. Max. Marks: 70

General Instructions:
  1. The question paper is divided into four sections.
  2. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying one mark each.
    • For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet, e.g., (A) ……. / (B) ……. / (C) ……. / (D) ……. etc. No mark/s shall be given if ONLY the correct answer or alphabet of the correct answer is written.
    • In case of MCQ, evaluation will be done for the first attempt only.
    • Q. No. 2 Contains Eight very short answer type of questions carrying one mark each.
  3. Section B: Q. No. 3 to 14 are short answer type of questions carrying two marks each. (Attempt any Eight)
  4. Section C: Q. No. 15 to 26 are short answer type of questions carrying three marks each. (Attempt any Eight)
  5. Section D: Q. No. 27 to 31 are long answer type of questions carrying four marks each. (Attempt any Three)
  6. Begin the answer of each section on a new page.
SECTION – A

Q.1. Select the correct alternatives and write the answers:

i. If members of two populations have difference in the structure of reproductive organs, then this type of isolation is called _______.

  • (A) ethological
  • (B) seasonal
  • (C) mechanical
  • (D) habitat
Answer: (C) mechanical

ii. The primary precursor of Indole-3-Acetic acid is _______.

  • (A) Tryptophan
  • (B) Phenyl alanine
  • (C) Mevalonic acid
  • (D) Methionine
Answer: (A) Tryptophan

iii. If only one DNA molecule is subjected to PCR and the time required for each cycle is three minutes, then after five cycles, how many DNA molecules are obtained?

  • (A) 10
  • (B) 15
  • (C) 32
  • (D) 64
Answer: (C) 32
Explanation: Number of DNA molecules = \(2^n\), where \(n\) is the number of cycles. \(2^5 = 32\).

iv. The specific gravity of CSF is _______.

  • (A) 1.005
  • (B) 1.02
  • (C) 1.502
  • (D) 1.81
Answer: (A) 1.005

v. Cardiac output of a person is 5400 ml and heart rate 72 per min. What will be his stroke volume?

  • (A) 65 ml
  • (B) 74 ml
  • (C) 75 ml
  • (D) 78 ml
Answer: (C) 75 ml
Explanation: Stroke Volume (SV) = Cardiac Output (CO) / Heart Rate (HR) = 5400 / 72 = 75 ml.

vi. Detritus food chain starts from _______.

  • (A) producers
  • (B) dead organic matter
  • (C) parasite
  • (D) photosynthesis
Answer: (B) dead organic matter

vii. Plants absorb _______ water.

  • (A) gravitational
  • (B) capillary
  • (C) combined
  • (D) hygroscopic
Answer: (B) capillary

viii. The organisms having tolerance for wide range of salinity are called _______.

  • (A) stenothermal
  • (B) euryhaline
  • (C) stenohaline
  • (D) eurythermal
Answer: (B) euryhaline

ix. Hisardale is a new breed of sheep developed by crossing _______.

  • (A) Bikaneri ram and Marino ewe
  • (B) Bikaneri ram and Bikaneri ewe
  • (C) Marino ram and Bikaneri ewe
  • (D) Marino ewe and Marino ram
Answer: (C) Marino ram and Bikaneri ewe

x. Perforins are secreted by _______.

  • (A) Helper T-cells
  • (B) Cytotoxic T-cells
  • (C) Suppressor T-cells
  • (D) Memory T-cells
Answer: (B) Cytotoxic T-cells

HSC Biology

Q.2. Answer the following questions:

i. Mention the chromosome number having the mutated gene for β-Thalassemia.

Answer: The mutated gene for β-Thalassemia is located on chromosome 11.

ii. Which organ produces calcitriol?

Answer: The Kidney produces calcitriol.

iii. If the megaspore mother cell has 26 chromosomes, what will be the total number of chromosomes in endosperm of the same plant?

Answer: Megaspore Mother Cell (2n) = 26, so haploid (n) = 13.
Endosperm is triploid (3n).
Total chromosomes = 3 × 13 = 39 chromosomes.

iv. Define the term − Facilitated diffusion.

Answer: Facilitated diffusion is the passive transport of substances across the cell membrane from a region of higher concentration to lower concentration with the help of specialized carrier proteins or channels, without the expenditure of metabolic energy (ATP).

v. Give reason − Energy pyramid is always upright.

Answer: The energy pyramid is always upright because energy flows from one trophic level to the next with a significant loss of energy as heat at each step (according to the 10% law). Therefore, the amount of energy decreases at successive trophic levels.

vi. What will be the base sequence on the template strand of DNA, which codes for methionine?

Answer: Methionine is coded by the mRNA codon AUG. The template strand of DNA is complementary to the mRNA codon. Therefore, the base sequence on the template strand is TAC.

vii. Deficiency of which element causes Brown heart disease in plants?

Answer: Deficiency of Boron causes Brown heart disease in plants.

viii. Where are the cells of Rauber situated in the blastocyst of human embryo?

Answer: The cells of the trophoblast which are in contact with the inner cell mass (embryonal knob) in the blastocyst are called the cells of Rauber.
SECTION – B

Attempt any EIGHT of the following questions:

Q.3. i. Give one example each of:

a. Autosomal dominant traits

b. Autosomal recessive traits

Answer: a. Autosomal dominant traits: Widow's peak / Huntington's disease.
b. Autosomal recessive traits: Phenylketonuria (PKU) / Cystic fibrosis / Sickle cell anaemia.

ii. If a carrier woman marries a colorblind man, what will be the phenotype of their progeny? Show in the form of a chart.

Answer: Parents: Carrier Woman (\(X^C X^c\)) × Colorblind Man (\(X^c Y\))

Gametes: (\(X^C\), \(X^c\)) and (\(X^c\), \(Y\))

Progeny Chart:
Gametes \(X^c\) \(Y\)
\(X^C\) \(X^C X^c\) (Carrier Daughter - Normal Vision) \(X^C Y\) (Normal Son)
\(X^c\) \(X^c X^c\) (Colorblind Daughter) \(X^c Y\) (Colorblind Son)
Phenotypic Ratio: 1 Normal daughter (carrier) : 1 Colorblind daughter : 1 Normal son : 1 Colorblind son.

Q.4. Sketch the appropriate diagrams showing following chromosomal aberrations:

i. deletion

ii. inversion

Answer:
Diagram showing Deletion and Inversion chromosomal aberrations [Please draw a diagram showing a chromosome segment being lost (Deletion) and a segment rotating 180 degrees (Inversion).]
i. Deletion: A segment of the chromosome (e.g., segment D) is lost.
ii. Inversion: A segment of the chromosome breaks, rotates 180°, and rejoins (e.g., B-C-D becomes D-C-B).

Q.5. Observe the following diagram of double circulation and identify A, B, C and D:

Pulmonary veins
Superior and Inferior vena cava
Heart
Right atrium ↓ Right ventricle
Left atrium ↓ Left ventricle
Right Ventricle → [ A ][ D ] → Pulmonary veins
Left Ventricle → [ B ][ C ] → Vena Cava
Answer:
  • A: Pulmonary artery
  • B: Dorsal Aorta (Systemic Aorta)
  • C: Body organs / Systemic capillaries
  • D: Lungs (Pulmonary capillaries)

Q.6. i. Select the names of fresh-water fishes from the given list:

Sardinella, Rastrelliger, Cirrhina, Harpadon, Labeo

Answer: The fresh-water fishes are: Cirrhina and Labeo.

ii. Write the economic importance of Lac (Any Two).

Answer: 1. Lac is used in the manufacture of bangles, toys, and woodwork.
2. It is used in the preparation of inks, polishes, and sealing wax.

Q.7. i. Define the term − Ecological succession.

Answer: The gradual and predictable change in the species composition of a given area is called ecological succession.

ii. What is the reason of eutrophication?

Answer: Eutrophication is caused by the nutrient enrichment of water bodies, particularly with nitrogen and phosphorus (from agricultural runoff, sewage, etc.), which leads to excessive growth of algae (algal bloom) and subsequent depletion of oxygen.

Q.8. i. Mention any two ill-effects of UV-rays on human beings.

Answer: 1. It causes skin cancer (melanoma).
2. It causes inflammation of the cornea (snow blindness) or cataract.

ii. Give significance of Ecosan.

Answer: Ecological sanitation (Ecosan) is a sustainable approach for handling human excreta using dry composting toilets. It transforms human waste into a natural fertilizer, thereby recycling nutrients and preventing water pollution.

Q.9. What are oral vaccines? Enlist the benefits of oral vaccines.

Answer: Oral vaccines are immunizing agents administered through the mouth.
Benefits:
1. They are easy to administer and painless (no needles required).
2. They are cost-effective and suitable for mass immunization campaigns (e.g., Pulse Polio).
3. They stimulate mucosal immunity (IgA production) in the gut.

Q.10. What is vernalization? Give the advantages of vernalization.

Answer: Vernalization: It is the dependence of certain plants on exposure to low temperatures, either quantitatively or qualitatively, to induce flowering.
Advantages:
1. It prevents precocious reproductive development late in the growing season.
2. It enables plants to have sufficient time to reach maturity.
3. It can induce early flowering in some plants.

Q.11. Enlist the causes of biodiversity losses.

Answer: The causes of biodiversity losses ("The Evil Quartet") are:
1. Habitat loss and fragmentation.
2. Over-exploitation.
3. Alien species invasions.
4. Co-extinctions.

Q.12. Give role of hormones Relaxin and Inhibin.

Answer: Relaxin: It is secreted by the ovary (corpus luteum) and placenta. It relaxes the pubic symphysis and dilates the cervix to facilitate the birth of the child (parturition).
Inhibin: It is secreted by Sertoli cells in males and Granulosa cells in females. It inhibits the secretion of FSH (Follicle Stimulating Hormone) from the pituitary gland to regulate spermatogenesis or follicular development.

Q.13. Sketch and label the diagram of a stoma showing kidney − shaped guard cells.

Answer:
[Students are expected to draw a diagram of a stomatal apparatus]
Labels to include:
- Kidney-shaped Guard cells
- Stomatal pore (aperture)
- Chloroplasts (inside guard cells)
- Subsidiary cells (surrounding guard cells)
- Epidermal cells

Q.14. Explain in brief the process of southern blotting and hybridization in DNA fingerprinting.

Answer: Southern Blotting: The DNA fragments separated by gel electrophoresis are transferred from the gel to a synthetic membrane, such as nitrocellulose or nylon membrane. This process preserves the arrangement of DNA fragments.
Hybridization: The membrane containing the DNA fragments is exposed to radioactive DNA probes (single-stranded DNA sequences complementary to the VNTRs). These probes bind (hybridize) to specific complementary DNA sequences on the membrane. The hybridized fragments can then be detected using autoradiography.
SECTION – C

Attempt any EIGHT of the following questions:

Q.15. i. Define − Palaeontology.

Answer: Palaeontology is the branch of biology that deals with the study of fossils.

ii. Give any four points of significance of palaeontology.

Answer: 1. It provides direct evidence of evolution.
2. It helps in constructing the phylogeny (evolutionary history) of organisms.
3. It helps in the study of extinct organisms and their habits.
4. It connects links between different groups of organisms (e.g., Archaeopteryx connects reptiles and birds).
5. It helps in the study of the geological time scale.

Q.16. i. What is pollination?

Answer: Pollination is the transfer of pollen grains from the anther to the stigma of a flower.

ii. Differentiate between Anaemophily and Entomophily with reference to:

  • a. pollinating agent
  • b. stigma
  • c. nectar
  • d. fragrance
Answer:
Feature Anaemophily (Wind Pollination) Entomophily (Insect Pollination)
a. Pollinating agent Wind Insects
b. Stigma Large, feathery, and exposed to trap pollen. Sticky, often situated deep within the flower.
c. Nectar Absent Present (produced to attract insects)
d. Fragrance Absent Present (often sweet-smelling)

Q.17. i. Differentiate between hypotonic and hypertonic solutions.

Answer: Hypotonic solution: A solution having a lower concentration of solutes (and higher water potential) compared to the cell sap.
Hypertonic solution: A solution having a higher concentration of solutes (and lower water potential) compared to the cell sap.

ii. Mention the effect of exo-osmosis and endo-osmosis on shape of the cell.

Answer: Exo-osmosis: Causes the cell to shrink (plasmolysis in plant cells, crenation in animal cells).
Endo-osmosis: Causes the cell to swell and become turgid.

iii. Give one difference between symplast and apoplast pathway.

Answer: Symplast pathway: Water moves through the living parts of the cell (protoplasm) connected by plasmodesmata. It is slower.
Apoplast pathway: Water moves through the non-living parts (cell walls and intercellular spaces). It is faster.

Q.18. i. Mention the control measures to prevent ascariasis.

Answer: 1. Maintaining personal hygiene (washing hands before eating).
2. Washing vegetables and fruits thoroughly before consumption.
3. Proper disposal of human excreta and preventing soil pollution.

ii. With appropriate terms, complete the following chart and rewrite it.

Answer:
Sr. No. Name of disease Name of pathogen
a. Amoebiasis (Amoebic Dysentery) Entamoeba histolytica
b. Typhoid Salmonella typhi
c. Filariasis (Elephantiasis) Wuchereria bancrofti
d. Malaria Plasmodium species

Q.19. i. Define − Adaptation.

Answer: Adaptation is any attribute of the organism (morphological, physiological, or behavioural) that enables the organism to survive and reproduce in its habitat.

ii. Explain any two adaptations in Opuntia and Seal.

Answer: Opuntia (Desert Plant): 1. Leaves are modified into spines to reduce water loss through transpiration.
2. Stems are modified into flattened, green structures (phylloclades) to perform photosynthesis and store water.
Seal (Aquatic Mammal in cold climate): 1. They possess a thick layer of fat (blubber) below their skin that acts as an insulator to reduce the loss of body heat.

Q.20. Match the disease resistant varieties given in Column-I with the crops in Column-II and rewrite:

Answer:
Column-I (Variety) Column-II (Crop)
i. Himgiri b. Wheat
ii. Pusa shubhra c. Cauliflower
iii. Pusa sadabahar a. Chilli

Q.21. i. Explain the qualitative and quantitative aspects of growth phenomenon.

Answer: Quantitative aspect: Growth involves an increase in parameters like mass, volume, surface area, number of cells, or dry weight, which can be measured.
Qualitative aspect: Growth involves development, differentiation, and change in the form or function of cells and tissues leading to maturity.

ii. Explain the phase of cell maturation.

Answer: In the phase of cell maturation (differentiation), the cells that have elongated attain their maximum size. They undergo structural and physiological differentiation to perform specific functions. The cell walls become thickened, and the protoplasm undergoes modifications.

Q.22. i. What is co-dominance?

Answer: Co-dominance is a condition where both alleles of a gene pair in a heterozygote are fully expressed, with neither one being dominant or recessive to the other.

ii. If a red colored female cattle is crossed with a white male cattle, what will be the appearance of progeny in F2 generation? Show the genotypes with the help of a chart.

Answer: Parents: Red (RR) × White (WW)
F1 Generation: All Roan (RW)
F2 Generation (Selfing F1): RW × RW

Chart:
Gametes R W
R RR (Red) RW (Roan)
W RW (Roan) WW (White)
Appearance of Progeny in F2:
Red : Roan : White = 1 : 2 : 1.

Q.23. i. Give any two involuntary vital functions of medulla oblongata.

Answer: 1. Regulation of heart beat (Cardiac center).
2. Regulation of respiration (Respiratory center).
(Other functions: Vasomotor activities, peristalsis).

ii. Mention two functions of spinal cord.

Answer: 1. Conduction of sensory and motor impulses to and from the brain.
2. Acting as a center for spinal reflexes.

Q.24. i. Define − Transcription.

Answer: Transcription is the process of copying genetic information from one strand of the DNA into RNA.

ii. Write anticodons for the following triplet codons:

AUG, GAG, CUA, CCU

Answer:
  • AUG → UAC
  • GAG → CUC
  • CUA → GAU
  • CCU → GGA

Q.25. i. Mention the position of the following in human heart:

a. Eustachian valve

b. Bicuspid valve

Answer: a. Eustachian valve: Guards the opening of the Inferior Vena Cava into the right atrium.
b. Bicuspid valve (Mitral valve): Located between the left atrium and the left ventricle.

ii. Differentiate between open and closed circulation with reference to:

a. blood pressure

b. exchange of material

Answer: a. Blood pressure: In open circulation, blood flows at low pressure. In closed circulation, blood flows at high pressure.
b. Exchange of material: In open circulation, exchange occurs directly between blood and cells/tissues. In closed circulation, exchange occurs through the walls of capillaries.

Q.26. Draw a neat and proportionate diagram of Graafian follicle and label oocyte and antrum. Explain its structure in brief.

Answer:
[Students are expected to draw the Graafian Follicle]
Labels: Secondary Oocyte, Antrum, Theca externa, Theca interna, Membrana Granulosa, Cumulus oophorus, Zona pellucida.
Structure Explanation:
The Graafian follicle is a mature ovarian follicle. It consists of an outer protective layer called Theca externa and an inner vascular layer called Theca interna. Inside this, there are follicular cells forming the Membrana Granulosa. A large fluid-filled cavity called the Antrum (filled with liquor folliculi) is present. The secondary oocyte is situated eccentrically and is surrounded by a group of cells called Cumulus oophorus (or Discus proligerus). The oocyte is covered by a non-cellular layer called Zona pellucida.
SECTION – D

Attempt any THREE of the following questions:

Q.27. i. What is placenta?

Answer: The placenta is a temporary structural and functional unit formed by the intimate connection between the foetal and maternal tissues, which facilitates the supply of oxygen and nutrients to the embryo and the removal of carbon dioxide and excretory wastes from it.

ii. Give reason − Placenta is considered as a temporary endocrine gland.

Answer: The placenta produces several hormones necessary for the maintenance of pregnancy, such as Human Chorionic Gonadotropin (hCG), Human Placental Lactogen (hPL), estrogens, progesterone, and relaxin. Since it secretes hormones directly into the blood, it acts as a temporary endocrine gland.

iii. Give significance of hCG.

Answer: Human Chorionic Gonadotropin (hCG) maintains the corpus luteum and stimulates it to secrete progesterone, which is essential for maintaining the endometrium and thus the pregnancy. Its presence in urine is the basis for pregnancy tests.

Q.28. Describe in brief the structural and hormonal changes during ovarian cycle.

Answer: The ovarian cycle consists of the following phases:

1. Menstrual Phase (Days 1-5):
If fertilization does not occur, the corpus luteum degenerates. Progesterone and estrogen levels fall. This triggers the breakdown of the endometrium.
2. Follicular Phase (Days 5-13):
Under the influence of FSH (Follicle Stimulating Hormone) from the pituitary, primary follicles in the ovary develop into mature Graafian follicles. The developing follicles secrete Estrogen, which helps in the regeneration of the endometrium.
3. Ovulatory Phase (Day 14):
Estrogen levels peak, stimulating a surge in LH (Luteinizing Hormone). This LH surge causes the rupture of the Graafian follicle and the release of the secondary oocyte (ovulation).
4. Luteal Phase (Days 15-28):
Under the influence of LH, the remaining cells of the ruptured follicle transform into the Corpus Luteum. The Corpus Luteum secretes large amounts of Progesterone (and some estrogen), which maintains the endometrium for implantation. If fertilization does not occur, the Corpus Luteum degenerates into a white scar called Corpus Albicans.

Q.29. i. Name the formed elements which are useful in blood coagulation. Give its normal number per cubic milimeter (mm³) in human blood.

Answer: The formed elements useful in blood coagulation are Platelets (Thrombocytes).
Normal number: 1.5 to 3.5 lakhs / mm³ (150,000 to 350,000 / mm³).

ii. Comment on the shape and secretion of the above mentioned formed elements.

Answer: Shape: They are cell fragments, irregular or rounded in shape, and non-nucleated.
Secretion: They release clotting factors like Thromboplastin (Platelet factor 3).

iii. Explain in brief the mechanism of blood coagulation.

Answer: 1. At the site of injury, platelets rupture and release Thromboplastin.
2. Thromboplastin initiates the formation of the enzyme complex Prothrombinase (or Thrombokinase).
3. Prothrombinase converts inactive Prothrombin into active Thrombin in the presence of Calcium ions (\(Ca^{++}\)).
4. Thrombin converts soluble Fibrinogen protein into insoluble Fibrin threads.
5. These fibrin threads form a mesh that traps blood cells, forming a clot (coagulum).

Q.30. i. Give full form of the cloning vectors BAC and YAC.

Answer: BAC: Bacterial Artificial Chromosome.
YAC: Yeast Artificial Chromosome.

ii. Write the appropriate palindrome for Eco RI and indicate by an arrow its recognition sequence.

Answer: Sequence:
5' — G ↓ A A T T C — 3'
3' — C T T A A ↑ G — 5'
(The arrows indicate the cut site between G and A).

iii. Give any four uses of gene therapy.

Answer: 1. Treatment of genetic disorders like SCID (Severe Combined Immunodeficiency) caused by ADA deficiency.
2. Treatment of Cystic Fibrosis.
3. Treatment of Haemophilia.
4. Treatment of certain cancers (using suicide genes).
5. Treatment of Parkinson’s disease.

Q.31. Explain any four contrivances to prevent self pollination in plants with an appropriate example of each type.

Answer: 1. Unisexuality (Dicliny): Plants produce unisexual flowers (either male or female). This prevents self-pollination. Example: Papaya (Dioecious), Maize (Monoecious).

2. Dichogamy: In bisexual flowers, anthers and stigma mature at different times.
- Protandry: Anthers mature first. Example: Sunflower, Disc florets.
- Protogyny: Stigma matures first. Example: Michelia, Gloriosa.

3. Prepotency: Pollen grains of another flower germinate more rapidly over the stigma than the pollen grains from the same flower. Example: Apple.

4. Heterostyly (Heteromorphy): Plants produce flowers with different lengths of styles and stamens, preventing pollen from reaching the stigma of the same flower. Example: Primula (Primrose).

5. Herkogamy: Physical barriers prevent self-pollination. Example: Calotropis (Pentangular stigma positioned above anthers).

6. Self-sterility (Self-incompatibility): Pollen grains fail to germinate on the stigma of the same flower due to genetic mechanisms. Example: Tobacco, Thea.
Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Question Paper Page No. 5 Question Paper Page No. 6 Question Paper Page No. 7 Question Paper Page No. 8 Maharashtra Board Resources

Maharashtra Board HSC Chemistry Question Paper Solution July 2023 - Complete Answer Key

Board Question Paper: July 2023

Subject: Chemistry | Max. Marks: 70 | Time: 3 Hrs.

SECTION ‒ A [10 Marks]
Q.1. Select and write the correct answer for the following multiple choice type of questions:
(i) Anisole on heating with concentrated HI gives _______.
  • (a) Iodobenzene
  • (b) Phenol + Methanol
  • (c) Iodobenzene + Methanol
  • (d) Phenol + Iodomethane
Answer: (d) Phenol + Iodomethane
Explanation: The bond between oxygen and the methyl group breaks because the phenyl-oxygen bond has partial double bond character due to resonance. Thus, phenol and methyl iodide are formed.
(ii) Which solution shows positive deviation from Raoult’s law?
  • (a) Phenol and Aniline
  • (b) Chloroform and Acetone
  • (c) Ethanol and Acetone
  • (d) Chloroform and Ethanol
Answer: (c) Ethanol and Acetone
Explanation: In pure ethanol, molecules are hydrogen-bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds, increasing vapor pressure (Positive Deviation).
(iii) The coordination number of cobalt in [CoCl2(en)2]+ is
  • (a) 6
  • (b) 4
  • (c) 2
  • (d) 0
Answer: (a) 6
Explanation: 'en' (ethylenediamine) is a bidentate ligand (coordinates through 2 sites) and Cl is monodentate. Total coordination = (2 × 2) + (2 × 1) = 6.
(iv)
Benzene to Benzaldehyde Reaction Scheme The name of above reaction is _______. (Reaction shown: Benzene + CO, HCl, Anhydrous AlCl3, High Pressure → Benzaldehyde)
  • (a) Etard reaction
  • (b) Friedel Craft acylation reaction
  • (c) Stephen reaction
  • (d) Gatterman-Koch reaction
Answer: (d) Gatterman-Koch reaction
(v) Which is an example of thermoplastic polymer?
  • (a) Bakelite
  • (b) Polystyrene
  • (c) Nylon 6, 6
  • (d) Urea formaldehyde resin
Answer: (b) Polystyrene
(vi) Nichrome is an alloy of _______.
  • (a) Cu, Sn
  • (b) Cu, Ni
  • (c) Ni, Cr
  • (d) Fe, Cr
Answer: (c) Ni, Cr
(vii) Identify ‘A’ in the following reaction:
A + 2Na $\xrightarrow{\text{dry ether}}$ Biphenyl + 2NaCl
  • (a) Bromobenzene
  • (b) 1, 4-dichlorobenzene
  • (c) Naphthalene
  • (d) Chlorobenzene
Answer: (d) Chlorobenzene
Explanation: This is the Fittig reaction. The byproduct is 2NaCl, indicating the halogen in reactant 'A' must be Chlorine. Hence, A is Chlorobenzene.
(viii) Which amine does NOT react with Hinsberg reagent?
  • (a) Ethanamine
  • (b) N-ethylethanamine
  • (c) N, N-diethylethanamine
  • (d) 2-methyl-propan-2-amine
Answer: (c) N, N-diethylethanamine
Explanation: Tertiary amines (like N,N-diethylethanamine) do not have replaceable hydrogen atoms on the nitrogen, so they do not react with benzenesulfonyl chloride (Hinsberg reagent).
(ix) The dissociation constant of NH4OH is \(1.8 \times 10^{-5}\). The degree of dissociation in its 0.01 M solution is _______.
  • (a) 0.04242
  • (b) 0.4242
  • (c) 0.004242
  • (d) 4.242
Answer: (a) 0.04242
\(\alpha = \sqrt{\frac{K_b}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{10^{-2}}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{18 \times 10^{-4}} = 4.242 \times 10^{-2} = 0.04242\)
(x) Half-life of a first order reaction is 30 minutes at 300 K. The value of its rate constant, K is _______.
  • (a) \(2.31 \text{ min}^{-1}\)
  • (b) \(0.0231 \text{ min}^{-1}\)
  • (c) \(0.231 \text{ min}^{-1}\)
  • (d) \(2.310 \times 10^{-3} \text{ min}^{-1}\)
Answer: (b) \(0.0231 \text{ min}^{-1}\)
\(K = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \text{ min}^{-1}\)

HSC Chemistry

Q.2. Answer the following questions: [8 Marks]
(i) Write the name of radioactive element in group 16.
Answer: Polonium (Po).
(ii) Write the structure of glycine.
Answer: \( \text{H}_2\text{N} - \text{CH}_2 - \text{COOH} \)
(iii) Write the unit of cell constant.
Answer: \( \text{m}^{-1} \) or \( \text{cm}^{-1} \).
(iv) Write the number of particles present in FCC per unit cell.
Answer: 4 particles.
(v) Name the \(\gamma\)-isomer of BHC.
Answer: Lindane (or Gammexane).
(vi) Write the IUPAC name of isobutyraldehyde.
Answer: 2-Methylpropanal.
(vii) Which alloy is used in Fischer Tropsch process in the synthesis of gasoline?
Answer: Cobalt-Thorium alloy (Co-Th) or Fe-based catalyst.
(viii) Three moles of an ideal gas are expanded isothermally from \(15 \text{ dm}^3\) to \(20 \text{ dm}^3\) at constant external pressure of 1.2 bar. Estimate the amount of work in Joules.
Answer:
\(W = -P_{ext} \Delta V\)
\(P_{ext} = 1.2 \text{ bar}\)
\(\Delta V = V_2 - V_1 = 20 - 15 = 5 \text{ dm}^3\)
\(W = -1.2 \times 5 = -6 \text{ bar dm}^3\)
Convert to Joules: \(1 \text{ bar dm}^3 = 100 \text{ J}\)
\(W = -6 \times 100 = -600 \text{ J}\)
Work done by the system is 600 Joules.
SECTION ‒ B [16 Marks]
(Attempt any EIGHT of the following questions)
Q.3. Write four postulates of Werner theory of coordination complexes.
Answer:
  1. Coordination compounds possess two types of valencies: primary (ionizable) and secondary (non-ionizable).
  2. Primary valency corresponds to the oxidation state and is satisfied by negative ions.
  3. Secondary valency corresponds to the coordination number and is satisfied by negative ions or neutral molecules.
  4. Secondary valencies have a fixed spatial arrangement (directionality) around the central metal ion, giving the complex a definite geometry.
Q.4. Why fluorine shows anomalous behaviour?
Answer: Fluorine shows anomalous behavior due to:
  1. Extremely small size of the atom and fluoride ion.
  2. High electronegativity (most electronegative element).
  3. Absence of d-orbitals in its valence shell, restricting its covalency to 1.
  4. Low F-F bond dissociation enthalpy.
Q.5. What is the mass of Cu metal produced at the cathode during the passage of 5 ampere current through CuSO4 solution for 6000 seconds. Molar mass of Cu is 63.5 g mol-1.
Answer:
Reaction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
Mole ratio (n) = 2 moles of electrons per mole of Cu.
Using Faraday's Law: \( W = \frac{I \cdot t \cdot M}{n \cdot F} \)
Given: \( I = 5 \text{ A}, t = 6000 \text{ s}, M = 63.5 \text{ g/mol}, F = 96500 \text{ C/mol} \)
\( W = \frac{5 \times 6000 \times 63.5}{2 \times 96500} \)
\( W = \frac{30000 \times 63.5}{193000} \)
\( W = \frac{1905000}{193000} \approx 9.87 \text{ g} \)
Mass of Cu produced = 9.87 g.
Q.6. How is glucose prepared from sucrose?
Answer: Glucose is prepared from sucrose by hydrolysis. Sucrose is boiled with dilute HCl or H2SO4 in alcoholic solution.
\( \underset{\text{Sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{Glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{Fructose}}{\text{C}_6\text{H}_{12}\text{O}_6} \)
Glucose is separated from fructose by crystallization (as glucose is less soluble in alcohol).
Q.7. Derive integrated rate law for zero order reaction.
Answer: Consider a zero order reaction \( A \rightarrow \text{Products} \).
The rate is given by: \( \text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k \)
Rearranging: \( d[A] = -k dt \)
Integrating both sides within limits \([A]_0\) at \(t=0\) and \([A]_t\) at \(t=t\):
\( \int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \)
\( [A]_t - [A]_0 = -k(t - 0) \)
\( [A]_t = -kt + [A]_0 \) or \( k = \frac{[A]_0 - [A]_t}{t} \)
Q.8. The normal boiling point of ethyl acetate is 77.06 °C. A solution of 50 g of a non-volatile solute in 150 g of ethyl acetate boils at 84.27 °C. Evaluate the molar mass of solute if Kb for ethyl acetate is 2.77 °C kg mol-1.
Answer:
Elevation in boiling point, \( \Delta T_b = T_b - T_b^0 \)
\( \Delta T_b = 84.27 - 77.06 = 7.21^\circ \text{C} \)
Formula: \( M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} \)
Given:
\( W_2 \) (mass of solute) = 50 g
\( W_1 \) (mass of solvent) = 150 g
\( K_b = 2.77 \)
\( \Delta T_b = 7.21 \)
\( M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150} \)
\( M_2 = \frac{138500}{1081.5} \approx 128.06 \text{ g mol}^{-1} \)
Q.9. How is phenol prepared from cumene?
Answer: This is the commercial method. It involves two steps:
  1. Oxidation: Cumene (isopropylbenzene) is oxidized by air in the presence of Co-naphthenate catalyst to form Cumene hydroperoxide.
  2. Decomposition: Cumene hydroperoxide is treated with dilute sulfuric acid to decompose into Phenol and Acetone.
\( \text{Cumene} + \text{O}_2 \rightarrow \text{Cumene hydroperoxide} \xrightarrow{\text{dil H}_2\text{SO}_4} \text{Phenol} + \text{Acetone} \)
Q.10. Why do d-block elements form coloured compounds?
Answer: d-block elements form colored compounds due to:
  1. Presence of unpaired d-electrons: Ions with partly filled d-orbitals (d1 to d9 configuration).
  2. d-d transition: When ligands approach the metal ion, the d-orbitals split into two sets of different energies (crystal field splitting). Electrons absorb energy from the visible region to jump from lower energy d-orbitals to higher energy d-orbitals. The complementary color of the absorbed light is observed.
Q.11. Write a note on: Wolf-Kishner reduction reaction.
Answer: Wolf-Kishner reduction is used to convert the carbonyl group (\(>\text{C=O}\)) of aldehydes and ketones into a methylene group (\(>\text{CH}_2\)).
The aldehyde or ketone is heated with hydrazine (\(\text{NH}_2\text{NH}_2\)) and potassium hydroxide (KOH) in a high boiling solvent like ethylene glycol.
\( >\text{C=O} + \text{NH}_2\text{NH}_2 \xrightarrow{-\text{H}_2\text{O}} >\text{C=N-NH}_2 \text{ (Hydrazone)} \xrightarrow{\text{KOH, glycol, heat}} >\text{CH}_2 + \text{N}_2 \)
Q.12. How is Nylon 6, 6 prepared?
Answer: Nylon 6, 6 is prepared by the condensation polymerization of Hexamethylenediamine and Adipic acid.
Equimolar amounts of both monomers are mixed to form 'Nylon salt', which upon heating under high pressure and temperature loses water molecules to form Nylon 6, 6.
\( n(\text{HOOC}-(\text{CH}_2)_4-\text{COOH}) + n(\text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2) \rightarrow [-\text{CO}-(\text{CH}_2)_4-\text{CONH}-(\text{CH}_2)_6-\text{NH}-]_n + 2n\text{H}_2\text{O} \)
Q.13. Derive Ostwald’s dilution law equation for weak acid.
Answer: Consider a weak acid HA. Let C be the total concentration and \(\alpha\) be the degree of dissociation.
Equilibrium: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \)
Initial conc: \( C \quad 0 \quad 0 \)
Equilibrium conc: \( C(1-\alpha) \quad C\alpha \quad C\alpha \)
Applying Law of Mass Action: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)
\( K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \)
For weak acids, \(\alpha\) is very small (\(\alpha \ll 1\)), so \(1-\alpha \approx 1\).
\( K_a = C\alpha^2 \) or \( \alpha = \sqrt{\frac{K_a}{C}} \)
Q.14. What is Grignard reagent? How it is prepared?
Answer: Definition: Grignard reagent is an organometallic compound with the general formula \( \text{R-Mg-X} \) (Alkyl magnesium halide).
Preparation: It is prepared by the reaction of an alkyl halide (RX) with pure magnesium metal (Mg) in the presence of dry ether.
\( \text{R-X} + \text{Mg} \xrightarrow{\text{dry ether}} \text{R-Mg-X} \)
SECTION ‒ C [24 Marks]
(Attempt any EIGHT of the following questions)
Q.15. Calculate the standard enthalpy of the reaction:
\(\text{SiO}_{2(s)} + 3\text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)} + 2\text{CO}_{(g)}\)
From the following reactions:
(i) \(\text{Si}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SiO}_{2(s)}, \Delta_r H^\circ = -911 \text{ kJ}\)
(ii) \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}, \Delta_r H^\circ = -221 \text{ kJ}\)
(iii) \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}, \Delta_r H^\circ = -65.3 \text{ kJ}\)
Answer:
To get the target equation:
1. Reverse eq (i): \(\text{SiO}_{2(s)} \rightarrow \text{Si}_{(s)} + \text{O}_{2(g)}\) ; \(\Delta H = +911 \text{ kJ}\)
2. Keep eq (ii) as is: \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}\) ; \(\Delta H = -221 \text{ kJ}\)
3. Keep eq (iii) as is: \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}\) ; \(\Delta H = -65.3 \text{ kJ}\)

Add these three modified equations:
\((\text{SiO}_2) + (2\text{C} + \text{O}_2) + (\text{Si} + \text{C}) \rightarrow (\text{Si} + \text{O}_2) + (2\text{CO}) + (\text{SiC})\)
Cancelling common terms (\(\text{Si}, \text{O}_2\)):
\(\text{SiO}_{2(s)} + 3\text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)} + 2\text{CO}_{(g)}\)

\(\Delta_r H^\circ = (+911) + (-221) + (-65.3)\)
\(\Delta_r H^\circ = 911 - 286.3 = +624.7 \text{ kJ}\)
Q.16. Write a note on Hofmann bromamide degradation. Convert benzene diazonium chloride into benzene.
Answer: Hofmann Bromamide Degradation:
This reaction is used to prepare primary amines from amides. The amide is treated with bromine and aqueous or ethanolic solution of NaOH or KOH. The amine formed has one carbon atom less than the original amide.
\( \text{R-CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{R-NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O} \)
Conversion of Benzene Diazonium Chloride to Benzene:
Benzene diazonium chloride is reduced to benzene by treating it with hypophosphorous acid (H3PO2) or ethanol.
\( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{H}_3\text{PO}_3 + \text{HCl} \)
Q.17. Write any three advantages and disadvantages of nanoparticles and nanotechnology.
Answer: Advantages:
  1. Revolutionized electronics (faster, smaller devices).
  2. Advances in medicine (targeted drug delivery, cancer treatment).
  3. Development of self-cleaning surfaces and stronger, lighter materials.
Disadvantages:
  1. Potential health hazards (lung damage similar to asbestos) if inhaled.
  2. Environmental pollution (nanopollutants are hard to remove).
  3. High cost of production and lack of knowledge regarding long-term effects.
Q.18. Write molecular formula and structure of: (i) Sulphuric acid (ii) Peroxy monosulphuric acid (iii) Thiosulphuric acid
Answer: (i) Sulphuric Acid: \( \text{H}_2\text{SO}_4 \)
Structure: S is central atom bonded to two -OH groups and double bonded to two Oxygen atoms. (Tetrahedral geometry).

(ii) Peroxy monosulphuric acid (Caro's Acid): \( \text{H}_2\text{SO}_5 \)
Structure: Contains a peroxy linkage (-O-O-). S=O (two), S-OH (one), S-O-OH (one).

(iii) Thiosulphuric acid: \( \text{H}_2\text{S}_2\text{O}_3 \)
Structure: Similar to sulphuric acid but one double bonded Oxygen is replaced by Sulphur (S=S).
Q.19. Explain optical activity of 2-chlorobutane.
Answer: 2-chlorobutane (\(\text{CH}_3-\text{CH(Cl)}-\text{CH}_2\text{CH}_3\)) contains an asymmetric (chiral) carbon atom, which is attached to four different groups: -H, -Cl, -\(\text{CH}_3\), and -\(\text{C}_2\text{H}_5\).
Due to the presence of the chiral center and the lack of a plane of symmetry, the molecule is non-superimposable on its mirror image.
Therefore, it exists in two enantiomeric forms (dextro and levo) which rotate plane-polarized light in opposite directions, making it optically active.
Q.20. Write different oxidation states of manganese. Why +2 oxidation state of manganese is more stable?
Answer: Oxidation states: Manganese (Mn, Z=25) exhibits oxidation states from +2 to +7 (+2, +3, +4, +5, +6, +7).
Stability of +2 state: The electronic configuration of Mn is \([\text{Ar}] 3d^5 4s^2\). By losing 2 electrons from the 4s orbital, it forms \( \text{Mn}^{2+} \) which has the configuration \([\text{Ar}] 3d^5\). The 3d orbital is exactly half-filled, which provides extra stability due to symmetry and high exchange energy. Hence, the +2 state is very stable.
Q.21. Prepare the following by using methyl magnesium iodide: (i) Ethanol (ii) Propan-2-ol (iii) 2-methylpropan-2-ol
Answer: Methyl magnesium iodide is \( \text{CH}_3\text{MgI} \).
(i) Ethanol: React with Methanal (Formaldehyde) followed by hydrolysis.
\(\text{HCHO} + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{OH}\)
(ii) Propan-2-ol: React with Ethanal (Acetaldehyde) followed by hydrolysis.
\(\text{CH}_3\text{CHO} + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{CH(OH)}-\text{CH}_3\)
(iii) 2-methylpropan-2-ol: React with Propanone (Acetone) followed by hydrolysis.
\(\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_3\text{C-OH}\)
Q.22. Define: Ebullioscopic constant. Derive the relation between freezing point depression and molar mass of solute.
Answer: Definition: Ebullioscopic constant (Kb) is defined as the elevation in boiling point produced when 1 mole of a non-volatile solute is dissolved in 1 kg (1000 g) of solvent.

Derivation (Freezing Point Depression):
The depression in freezing point (\(\Delta T_f\)) is directly proportional to the molality (m) of the solution.
\( \Delta T_f \propto m \)
\( \Delta T_f = K_f \cdot m \) ...(1) (where \(K_f\) is Cryoscopic constant)
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W_2/M_2}{W_1/1000} = \frac{W_2 \times 1000}{M_2 \times W_1} \)
Substituting m in eq (1):
\( \Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1} \)
Rearranging for molar mass \( M_2 \):
\( M_2 = \frac{K_f \times W_2 \times 1000}{\Delta T_f \times W_1} \)
Q.23. Define: Buffer solution. Write any four applications of buffer solution.
Answer: Definition: A buffer solution is a solution which resists drastic change in its pH when a small amount of strong acid or strong base is added to it or upon dilution.
Applications:
  1. Biological Systems: Blood pH is maintained at ~7.36-7.42 by bicarbonate buffer system.
  2. Agriculture: Soil pH is maintained for specific crop growth using buffers (carbonates, phosphates).
  3. Industry: Used in paper, dye, ink, and paint industries to maintain specific pH during manufacturing.
  4. Medicine: Used in the preparation of injections (e.g., Penicillin preparations are stabilized by buffers).
Q.24. An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell?
Answer:
Formula: \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \)
Given:
\( M = 27 \text{ g/mol} \)
\( a = 405 \text{ pm} = 4.05 \times 10^{-8} \text{ cm} \)
\( \rho = 2.7 \text{ g/cm}^3 \)
\( N_A = 6.022 \times 10^{23} \)

Step 1: Calculate \( a^3 \)
\( a^3 = (4.05 \times 10^{-8})^3 \approx 66.4 \times 10^{-24} \text{ cm}^3 \)

Step 2: Solve for Z
\( Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \)
\( Z = \frac{2.7 \times (66.4 \times 10^{-24}) \times (6.022 \times 10^{23})}{27} \)
\( Z = \frac{2.7 \times 66.4 \times 6.022 \times 0.1}{27} \)
\( Z = \frac{107.96}{27} \approx 3.99 \approx 4 \)

Since Z = 4, the cubic unit cell is Face Centered Cubic (FCC) or ccp.
Q.25. On the basis of valence bond theory explain the nature of bonding in [Ni(Cl)4]2- complex ion.
Answer: 1. Oxidation state: Ni is in +2 oxidation state. Configuration: \([\text{Ar}] 3d^8 4s^0\).
2. Ligand nature: Cl- is a weak field ligand, so it does not cause pairing of electrons in 3d orbitals.
3. Hybridization: To accommodate 4 ligands, Ni2+ uses one 4s and three 4p orbitals to undergo sp3 hybridization.
4. Geometry: sp3 hybridization corresponds to Tetrahedral geometry.
5. Magnetic property: There are 2 unpaired electrons in 3d orbitals, so the complex is Paramagnetic.
Q.26. Convert: (i) Acetic acid to acetamide (ii) Acetyl chloride to acetic anhydride (iii) Sodium acetate to methane
Answer: (i) Acetic acid to Acetamide: Reaction with ammonia followed by heating.
\( \text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{COONH}_4 \xrightarrow{\Delta, -\text{H}_2\text{O}} \text{CH}_3\text{CONH}_2 \)
(ii) Acetyl chloride to Acetic anhydride: Reaction with sodium acetate.
\( \text{CH}_3\text{COCl} + \text{CH}_3\text{COONa} \rightarrow (\text{CH}_3\text{CO})_2\text{O} + \text{NaCl} \)
(iii) Sodium acetate to Methane: Decarboxylation using soda lime (NaOH + CaO) and heat.
\( \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO, }\Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3 \)
SECTION ‒ D [12 Marks]
(Attempt any THREE of the following questions)
Q.27. Define isomorphism. Write Arrhenius equation. Derive an expression to determine activation energy for two different temperatures T1 and T2.
Answer: Isomorphism: The phenomenon where two or more different substances exist in the same crystalline structure is called isomorphism (e.g., NaF and MgO).
Arrhenius Equation: \( k = A \cdot e^{-E_a/RT} \)
(Where k = rate constant, A = frequency factor, Ea = activation energy, R = gas constant, T = temperature).

Derivation: Taking natural log on both sides: \( \ln k = \ln A - \frac{E_a}{RT} \)
At temperature T1: \( \ln k_1 = \ln A - \frac{E_a}{RT_1} \) ...(1)
At temperature T2: \( \ln k_2 = \ln A - \frac{E_a}{RT_2} \) ...(2)
Subtract eq (1) from eq (2): \( \ln k_2 - \ln k_1 = (-\frac{E_a}{RT_2}) - (-\frac{E_a}{RT_1}) \)
\( \ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2}) \)
Or in log base 10: \( \log_{10}(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} [\frac{T_2 - T_1}{T_1 T_2}] \)
Q.28. What are interhalogen compounds? Write any two general characteristics of interhalogen compounds. Draw the Fischer projection formula for \(\alpha-D-(+)\) glucose. Write reaction involved in the formation of Teflon.
Answer: Interhalogen Compounds: Compounds formed by the combination of two or more different halogen atoms (e.g., ICl, BrF3).
Characteristics: 1. They are essentially covalent compounds.
2. They are generally more reactive than pure halogens (except Fluorine) because the X-X' bond is weaker than the X-X bond.
Teflon Formation: Polymerization of Tetrafluoroethylene.
\( n(\text{CF}_2=\text{CF}_2) \xrightarrow{\text{Catalyst, High P}} -(\text{CF}_2-\text{CF}_2)_n- \)
Fischer Projection of \(\alpha-D-(+)\) Glucose:
(Structure description: C1 has OH on right, C2 OH on right, C3 OH on left, C4 OH on right, C5 OH on right, C6 is CH2OH).
Q.29. Describe the construction and working of Standard Hydrogen Electrode. Write any two difficulties in setting SHE.
Answer: Construction: It consists of a platinum plate coated with platinum black (to increase surface area). This plate is suspended in a solution of 1 M H+ ions (1 M HCl). Pure dry hydrogen gas is bubbled through the solution at 1 bar pressure at 298 K. The glass tube has a side arm for H2 inlet and exit.
Working: It acts as a reference electrode with potential defined as 0.0 V.
  • If it acts as Anode (Oxidation): \( \text{H}_{2(g)} \rightarrow 2\text{H}^+_{(aq)} + 2e^- \)
  • If it acts as Cathode (Reduction): \( 2\text{H}^+_{(aq)} + 2e^- \rightarrow \text{H}_{2(g)} \)
Difficulties: 1. It is difficult to maintain exactly 1 M concentration of H+ ions. 2. It is difficult to maintain pressure of hydrogen gas at exactly 1 bar. 3. The Pt electrode gets easily poisoned by impurities.
Q.30. Write any two statements of first law of thermodynamics. For a certain reaction \(\Delta H^\circ\) is -224 kJ and \(\Delta S^\circ\) is -153 Jk-1. At what temperature the change over from spontaneous to nonspontaneous will occur?
Answer: Statements of First Law: 1. Energy can neither be created nor destroyed, but can be converted from one form to another. 2. The total energy of the universe remains constant.
Calculation: Given: \( \Delta H = -224 \text{ kJ} = -224000 \text{ J} \), \( \Delta S = -153 \text{ J K}^{-1} \). The changeover occurs at equilibrium where \( \Delta G = 0 \). \( \Delta G = \Delta H - T\Delta S \)
\( 0 = \Delta H - T\Delta S \)
\( T = \frac{\Delta H}{\Delta S} \)
\( T = \frac{-224000}{-153} \)
\( T \approx 1464 \text{ K} \)
Q.31. Define: (i) Gangue (ii) Ionization isomer (iii) Aromatic ketones. Write the use and environmental effect of methylene chloride.
Answer: Definitions:
(i) Gangue: The earthy impurities (sand, clay, rock, silica, etc.) associated with the mineral/ore in the earth's crust.
(ii) Ionization Isomer: Isomers that have the same molecular formula but give different ions in solution (exchange of ligands between coordination sphere and ionization sphere).
(iii) Aromatic Ketones: Ketones in which the carbonyl carbon is attached to at least one aryl (aromatic) group (e.g., Acetophenone).

Methylene Chloride (Dichloromethane, CH2Cl2):
Use: Commonly used as a solvent in paint removers, propellant in aerosols, and process solvent in drug manufacturing.
Environmental Effect: It harms the human central nervous system. Direct contact causes skin burns. It is a potential occupational carcinogen. High levels in air cause dizziness and nausea.
End of Question Paper Solution
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