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Punjab Chemistry 2023 Paper 2 — Solved Past Paper with Answers

All 17 MCQs from Punjab Chemistry 2023 Paper 2, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT Punjab / UHS past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all Punjab / UHS papers.

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Q1. Mark the correct statement:

  • A. Na+is smaller than Na
  • B. Na+is larger than Na
  • C. Cl- is smaller than Cl
  • D. Cl- ion and Cl are equal

Explanation: When a neutral atom loses one or more electrons, it becomes a positive ion. The size of the atom is decreased in this process because of two reasons. First, the removal of one or more electrons from a neutral atom usually results in the loss of the outermost shell. Second, the removal of electrons causes an imbalance in the proton-electron ratio. Due to the greater attraction of the nuclear charge, the remaining electrons of the ion are drawn closer to the nucleus. Thus, a positive ion is always smaller than the neutral atom from which it is derived. The radius of Na is 157 pm, and the radius of Na+ is 95 pm

Why the other options are wrong
  • B. Wrong
  • C. The anionic size is larger than atomic size
  • D. The anionic size is larger than atomic size

Q2. The mineral (CaSO4 .2H2O) has the gencral name:

  • A. Epsom salt
  • B. Dolomite
  • C. Calcite
  • D. Gypsum

Explanation: The mineral CaSO4.2H2O has the general name gypsum.

Q3. Which element forms an ion with charge +3?

  • A. Beryllium
  • B. Aluminium
  • C. Carbon
  • D. Silicon

Explanation: The formation of an Al3+ ion involves the aluminum atom losing three valence electrons to achieve a stable electron configuration.Aluminum is in Group 3A of the periodic table, meaning it has three valence electrons.To achieve a stable electron configuration, similar to that of a noble gas (like neon), aluminum tends to lose its three valence electrons.Upon losing three electrons, aluminum forms an Al3+ ion.

Q4. Which of the following species has the maximum number of unpsired electrons?

  • A. O2
  • B. O
  • C. O21-
  • D. O22-

Explanation: To determine which species has the maximum number of unpaired electrons, let's analyze the electron configurations of each species:Oxygen (O): Atomic number 8. Electron configuration: 1s² 2s² 2p⁴. It has two unpaired electrons.Oxygen molecule (O2): Molecular oxygen has a double bond, so we need to consider its molecular orbital diagram. Molecular oxygen has two unpaired electrons because it has two antibonding π* orbitals, each containing one unpaired electron.Superoxide ion (O21-) The superoxide ion has one more electron compared to the oxygen molecule. It has an additional electron occupying one of the antibonding π* orbitals, resulting in three unpaired electrons.Peroxide ion (O22-)The peroxide ion has two more electrons compared to the oxygen molecule. It fills both antibonding π* orbitals, resulting in four unpaired electrons.Therefore, the peroxide ion O22- has the maximum number of unpaired electrons, with four unpaired electrons.

Q5. Which of the following hydrogen halide is the weakest acid in solution?

  • A. HI
  • B. HBr
  • C. HF
  • D. HCl

Explanation: In water, hydrogen halides give hydrofluoric, hydrochloric, hydrobromic and hydroiodic acids. Hydrofluoric acid is a weak acid due to limited ionization.The other three acids are very strong acids. The acidic strength increasesin the order.HF<HCl<HBr<HI

Q6. Coordination number of Pr in [Pt Cl (NO2) (NH3)4 | is:

  • A. 2-
  • B. 4
  • C. 1
  • D. 6

Explanation: To determine the coordination number of platinum (Pt) in the complex ion [PtCl(NO2)(NH3)4], we count the number of ligands directly bonded to the central platinum atom.In the given complex ion, we have:One chloride ion (Cl-) directly bonded to platinum.One nitrite ion (NO2-) directly bonded to platinum.Four ammonia molecules (NH3) directly bonded to platinum.So, the total number of ligands bonded to the central platinum atom is 1 (Cl-) + 1 (NO2-) + 4 (NH3) = 6.

Q7. Adouble bond consists of:

  • A. Tw sigma bonds
  • B. One sigma and one Pi bonds
  • C. One sigma and two Pi bonds
  • D. Two Pi bonds

Explanation: A double bond contains one sigma and one pi bond.

Q8. B1 B1 - dichlorocthy! sulphide is commonly known as

  • A. Mustard gas
  • B. Laughing gas
  • C. Phosgene gas
  • D. Bio-gas

Explanation: ‘Mustard gas’ is a chemical used in World War I. The name comes from its mustard like odour. It is not a gas, but a high boiling liquid that is dispersed as a mist of tiny droplets. It is a powerful vesicant i.e., causes blisters. It is 2,2 - dichloro ethyl sulpide.

Q9. Amongst the following the compound that can be most readily sulphonated is:

  • A. Nitrobenzene
  • B. Benzene
  • C. Toluene
  • D. Chlorobenzene

Explanation: In general, the presence of electron-donating groups (such as alkyl groups) on the benzene ring increases the electron density on the ring, making it more reactive towards electrophilic aromatic substitution reactions like sulfonation. On the other hand, electron-withdrawing groups (such as nitro groups) decrease the electron density on the ring, making it less reactive.Benzene: It has no substituents, so it's less reactive than substituted benzenes.Nitrobenzene: The nitro group (-NO2) is a strong electron-withdrawing group. Chlorobenzene: The chlorine atom (-Cl) is a weakly deactivating group.Toluene: Toluene has a methyl group (-CH3) as a substituent. The methyl group is weakly activating, meaning it increases the electron density on the benzene ring, making toluene more reactive towards sulfonation compared to benzene, nitrobenzene, or chlorobenzene.

Q10. When CO2 is made to react with magnesium iodide, followed by acid hydrolysis, the product formed is:

  • A. Propanoic
  • B. Propanoic acid
  • C. Propanal
  • D. Propanol

Explanation: The reaction of carbon dioxide (CO2) with ethyl magnesium bromide (C2H5MgBr) is a classic example of a Grignard reaction. The reaction proceeds as follows:C2H5MgBr+CO2→C2H5CO2MgBrThe reaction is often followed by a protonation step, where the carboxylate salt is treated with a protic acid such as water or dilute acid, resulting in the formation of the corresponding carboxylic acid:C2H5CO2MgBr+H3O+→C2H5COOH+MgBrOH

Q11. The solution which acid is used for manufacture of pickles:

  • A. Acetic acid
  • B. Formic acid
  • C. Benzoic acid
  • D. Butanoic acid

Explanation: Acetic acid is used: i) as a coagulant for latex in rubber industry. ii) in the manufacture of plastics (polyvinyl acetate) rayon (cellulose acetate) and silk. iii) in medicine as a local irritant. iv) as a solvent in the laboratory for carrying out reactions. v) in the manufacture of pickles.

Q12. Which of the following reagent will react with ketones?

  • A. Tollen’s reagent
  • B. Fehling’s reagent
  • C. Benedict's reagent
  • D. Grignard’s reagent

Explanation: Fehling's solution, Benedict's solution and Tollen's reagent are used in identification tests for aldehydes so they do not react with ketones.Grignard reagent can react with ketones and forms tertiary alcohols as a product.

Q13. Which compound will have the maximum repulsion with H2O?

  • A. C2 H5 OH
  • B. CH3CH2CH2 OH
  • C. C6 H6
  • D. CH3 -O- CH3

Explanation: Alcohols are capable of forming hydrogen bonds with water molecules due to the presence of the hydroxyl (-OH) functional group. When alcohol molecules are mixed with water, hydrogen bonding occurs between the hydroxyl groups of alcohol and water molecules.In contrast, benzene does not have functional groups that readily participate in hydrogen bonding, so its interaction with water is primarily governed by weaker dispersion forces. Therefore, benzene is not very soluble in water.Similarly, ethers also lack functional groups that readily participate in hydrogen bonding with water. While some ethers may have slight solubility in water due to dipole-dipole interactions, their interactions with water are generally weaker compared to alcohols.

Q14. Which compound is used as anti-freezing agent in automobile radiator?

  • A. CH3OH
  • B. CH3-O-CH3
  • C. CH2 CH2 CH2 OH
  • D. CH2CH3-O-CH2CH3

Explanation: Methanol is used as a solvent for fats oils, paints, varnishes. It is also used as antifreeze in the radiators of automobiles and for denaturing of alcohol.

Q15. Vegetable fats are:

  • A. Unsaturated fatty acids
  • B. Fessential oils obtained from plants
  • C. Gilycerides of saturated fatty acids
  • D. Gilycerides of unsaturated fatty acids

Explanation: Fats and oils come from a variety of natural sources like animals, plants and marine organisms. Animal fats are located particularly in adipose tissue cells. Butter and ghee are a special type of animal fats which are made form milk. Vegetable oils are chiefly present in seeds and nuts of plants.

Q16. Major nitrogen fertilizers are:

  • A. Urea and ammonium nitrate
  • B. Urea and super phosphate
  • C. Ammonia and DAP
  • D. Diammonium phosphate only

Explanation: Some of the examples of nitrogen fertilizers are: - ammonium sulphate, calcium ammonium nitrate, basic calcium nitrate, calcium cyanamide, ammonia, ammonium nitrate, ammonium phosphate, ammonium chloride and urea.

Q17. The substances which directly kill the unwanted organisms are called:

  • A. Fungicides
  • B. Insecticides
  • C. Pesticides
  • D. Herbicides

Explanation: Pests harm crops and transmit diseases both to human beings and animals. Pesticides are the substances that can directly kill an unwanted organism or otherwise control by interfering with its reproduction process.

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