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Punjab Chemistry 2021 Paper 1 — Solved Past Paper with Answers
All 16 MCQs from Punjab Chemistry 2021 Paper 1, 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.
Attempt this paper interactively →Q1. Which of the following halogen is weak oxidizing agent?
- A. Cl2
- B. F2
- C. I2✓
- D. Br2
Explanation: Iodine is relatively less electronegative compared to chlorine and fluorine. While it can still act as an oxidizing agent, it is weaker than chlorine and fluorine in this regard. Iodine tends to gain electrons less readily than chlorine and fluorine, making it a weaker oxidizing agent.
Why the other options are wrong
- A. Chlorine is a strong oxidizing agent because it has a high electronegativity and readily accepts electrons. In chemical reactions, chlorine tends to gain electrons to form chloride ions (Cl⁻), thereby oxidizing other substances by causing them to lose electrons.
- B. Fluorine is the most electronegative element, meaning it has a strong tendency to attract electrons. As a result, fluorine is an extremely potent oxidizing agent. It readily gains electrons to form fluoride ions (F⁻), making it highly reactive and capable of oxidizing other substances by causing them to lose electrons.
- D. Bromine is also a strong oxidizing agent, although not as strong as fluorine. It has a high electronegativity and readily accepts electrons to form bromide ions (Br⁻). Bromine's oxidizing properties are weaker than fluorine but stronger than iodine.
Q2. Which of the following is a typical transition element?
- A. Sc
- B. Y
- C. Co✓
- D. Ra
Explanation: Y (yttrium) is a typical transition element because it belongs to the transition metals group in the periodic table. Transition elements are characterized by their partially filled d-orbitals in one or more of their oxidation states.
Why the other options are wrong
- A. Sc (scandium) belongs to group IIIB and elements of this group are called non- typical transition elements as they do not exhibit properties of transition elements.
- B. Y (yttrium) belongs to group IIIB and elements of this group are called non- typical transition elements as they do not exhibit properties of transition elements.
- D. Randium is not a typical transition element. Radium is an alkaline earth metal, belonging to group 2 of the periodic table, and it does not exhibit the characteristic properties of transition elements such as the presence of partially filled d-orbitals.
Q3. The state of hybridization of carbon atom in methane is:
- A. s*p3✓
- B. s*p2
- C. s*p2
- D. dsp
Explanation: This option refers to sp³ hybridization. In methane, each carbon atom undergoes sp³ hybridization, where one s orbital and three p orbitals combine to form four equivalent sp³ hybrid orbitals. These orbitals are then used to form sigma bonds with the four hydrogen atoms, resulting in a tetrahedral geometry around the carbon atom.
Why the other options are wrong
- B. This option describes sp² hybridization. However, this hybridization is not applicable to methane. In sp² hybridization, one s orbital and two out of the three available p orbitals combine to form three equivalent sp² hybrid orbitals. These orbitals lie in a trigonal planar arrangement and are utilized to form sigma bonds with the three hydrogen atoms. For example, sp² hybridization is found in molecules like ethene (C₂H₄) where each carbon atom forms a double bond with one another.
- C. This option describes sp² hybridization. However, this hybridization is not applicable to methane. In sp² hybridization, one s orbital and two out of the three available p orbitals combine to form three equivalent sp² hybrid orbitals. These orbitals lie in a trigonal planar arrangement and are utilized to form sigma bonds with the three hydrogen atoms. For example, sp² hybridization is found in molecules like ethene (C₂H₄) where each carbon atom forms a double bond with one another.
- D. This option refers to a hybridization involving d orbitals, which is not relevant for the carbon atom in methane. dsp hybridization typically occurs in transition metal compounds where the central atom utilizes one s orbital, one d orbital, and two p orbitals to form four equivalent hybrid orbitals. This hybridization is not observed in simple hydrocarbons like methane.
Q4. Formula of chloroform is:
- A. CH3Cl
- B. CCl4
- C. CH2Cl2
- D. CHCl3✓
Explanation: CH3Cl is chloromethane.CH2Cl2 is dichloromethane.CHCl3 is trichloromethane (chloroform).CCL4 is tetrachloromethane (Carbon Tetrachloride).
Q5. The electrophile in aromatic sulphonation is:
- A. H2SO4
- B. HSO4
- C. SO+3✓
- D. SO3
Explanation: Sulfur trioxide or its protonated derivative is the actual electrophile in this electrophilic aromatic substitution.
Why the other options are wrong
- A. Sulfur trioxide or its protonated derivative is the actual electrophile in this electrophilic aromatic substitution.
- B. Sulfur trioxide or its protonated derivative is the actual electrophile in this electrophilic aromatic substitution.
- D. Sulfur trioxide or its protonated derivative is the actual electrophile in this electrophilic aromatic substitution.
Q6. Elimination bimolecular reaction involves:
- A. First order kinetics
- B. Second order kinetics✓
- C. Zero order kinetics
- D. Third order kinetics
Explanation: In a second order reaction, the rate of the reaction is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. Elimination bimolecular reactions often follow second order kinetics because they involve the collision of two molecules to form the products.
Why the other options are wrong
- A. In a first order reaction, the rate of the reaction is directly proportional to the concentration of only one reactant. Elimination reactions typically involve the simultaneous loss of two substituents, making them unlikely to follow first order kinetics.
- C. In a zero order reaction, the rate of the reaction is independent of the concentration of the reactant(s). This is not typically observed in elimination reactions.
- D. Third order kinetics involve the rate being proportional to the cube of the concentration of a reactant or the product of the concentrations of three reactants. This is not commonly observed in elimination reactions.
Q7. Which compound shows hydrogen bonding?
- A. C2*H6
- B. C2*H5*Cl
- C. CHOSCH
- D. C2*H5*OH✓
Explanation: Ethanol contains carbon, hydrogen, and oxygen atoms. The hydrogen atoms in ethanol are directly bonded to an oxygen atom. Oxygen is highly electronegative, creating a large dipole moment between the hydrogen and oxygen atoms. This polarity allows ethanol molecules to form hydrogen bonds with each other, making ethanol the correct option.
Why the other options are wrong
- A. Ethane consists only of carbon and hydrogen atoms bonded together by single covalent bonds. Hydrogen bonding occurs when hydrogen is directly bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. Since ethane lacks such atoms, it cannot exhibit hydrogen bonding.
- B. Ethyl chloride contains a chlorine atom along with carbon and hydrogen atoms. While chlorine is electronegative, it is not as electronegative as oxygen or nitrogen, which are typically involved in hydrogen bonding. Therefore, ethyl chloride does not participate in hydrogen bonding.
- C. This formula represents dimethyl sulfoxide (DMSO). DMSO contains sulfur and oxygen atoms, both of which are electronegative. However, in DMSO, the hydrogen atoms are not directly bonded to either sulfur or oxygen in a manner conducive to hydrogen bonding. Therefore, dimethyl sulfoxide does not exhibit hydrogen bonding.
Q8. Percentage of water in Formalin is:
- A. 52%
- B. 8%
- C. 40%
- D. 60%✓
Explanation: Formalin is a solution of formaldehyde gas dissolved in water, typically containing around 60% water and 37-40% formaldehyde.
Why the other options are wrong
- A. Formalin typically contains more water than this.
- B. Formalin usually contains much more water than this.
- C. Formalin consists of approximately 60% water, not 40%.
Q9. Which of the following will have the highest boiling point?
- A. Methanal
- B. Ethanal
- C. Propanal
- D. 2-Hexanone✓
Explanation: 2-Hexanone has the molecular formula C₆H₁₂O. It's larger than the previous three compounds and has a more complex structure. Due to its larger size and more extensive molecular interactions, it has a significantly higher boiling point compared to the other options, approximately 129°C.
Why the other options are wrong
- A. Methanal (also known as formaldehyde) has the molecular formula CH₂O. It's a small molecule with a boiling point of approximately -21°C.
- B. Ethanal (also known as acetaldehyde) has the molecular formula C₂H₄O. It has a slightly higher molecular weight compared to methanal and a boiling point of approximately 20°C.
- C. Propanal (also known as propionaldehyde) has the molecular formula C₃H₆O. It's larger than ethanal, and its boiling point is approximately 49°C.
Q10. Which of the following ester gives apricot flavour?
- A. Amyl acetate
- B. Benzyl acetate✓
- C. Amyl butyrate
- D. Otyl acetate
Explanation: Benzyl acetate is indeed responsible for the characteristic apricot flavor. It is naturally occurring in many fruits, including apricots.
Why the other options are wrong
- A. This ester is often associated with banana flavor rather than apricot. It's commonly found in artificial banana flavorings.
- C. This ester is typically associated with a fruity aroma, but it's more commonly found in apple or pear flavors rather than apricot.
- D. This ester is often used in artificial orange flavorings, not apricot.
Q11. The solution of which acid is used for seasoning of food?
- A. Formic acid
- B. Acetic acid✓
- C. Benzoic acid
- D. Butanoic acid
Explanation: Acetic acid is correct because it is commonly used for seasoning food. Acetic acid, also known as vinegar when diluted, is a weak acid that adds a sour flavor to foods and is widely used in cooking and food preparation.
Why the other options are wrong
- A. Acetic acid is correct because it is commonly used for seasoning food. Acetic acid, also known as vinegar when diluted, is a weak acid that adds a sour flavor to foods and is widely used in cooking and food preparation.
- C. Benzoic acid is incorrect because it is primarily used as a preservative in food and is not commonly used for seasoning.
- D. Butanoic acid is incorrect because it is not commonly used for seasoning food. It has a strong odor and is used more in the production of flavors and fragrances.
Q12. Through how many zones does the charge pass in a rotary kiln?
- A. 6
- B. 3
- C. 4✓
- D. 2
Explanation: Option C (4) correctly indicates that the charge passes through four zones in a rotary kiln. These zones include drying, preheating, calcining, and cooling, covering the essential stages of material processing in the kiln.
Why the other options are wrong
- A. This option suggests that the charge passes through six zones in a rotary kiln. However, this is not accurate. Rotary kilns typically have four main zones: drying, preheating, calcining, and cooling.
- B. This option implies that the charge passes through three zones in a rotary kiln. However, this is not sufficient to cover the key stages of the process. Four zones are generally recognized as necessary for the complete processing of materials in a rotary kiln.
- D. This option suggests that the charge passes through only two zones in a rotary kiln, which is insufficient. Rotary kilns typically require multiple zones to facilitate the drying, heating, chemical reactions, and cooling of materials.
Q13. Keeping in view the size of atoms, which order is the correct one?
- A. Mg > Sr✓
- B. Ba > Mg
- C. Lu > Ce
- D. C|>I
Explanation: Magnesium (Mg) has a smaller atomic size than Strontium (Sr) because as you move across a period (from left to right) in the periodic table, atomic size generally decreases due to increasing effective nuclear charge pulling the electrons closer to the nucleus.
Why the other options are wrong
- B. This is incorrect because Barium (Ba) typically has a larger atomic size than Magnesium (Mg) as you move down a group (from top to bottom) in the periodic table, atomic size generally increases due to additional electron shells.
- C. This is incorrect because Lutetium (Lu) typically has a smaller atomic size than Cerium (Ce) because as you move across a period (from left to right) in the periodic table, atomic size generally decreases due to increasing effective nuclear charge pulling the electrons closer to the nucleus.
- D. This option is not clear as it's written in a non-standard format. However, it seems to imply that Carbon (C) has a larger atomic size than Iodine (I), which is generally not the case. Carbon is smaller than Iodine.
Q14. Which element belongs to group IVA of the periodic table?
- A. Ba
- B. I
- C. Pb✓
- D. O
Explanation: Lead (Pb) is the correct element belonging to group IVA of the periodic table.
Why the other options are wrong
- A. Barium (Ba) belongs to group IIA of the periodic table, not group IVA.
- B. Iodine (I) is a halogen and belongs to group VIIA of the periodic table.
- D. Oxygen (O) is a non-metal and belongs to group VIA of the periodic table.
Q15. Which of the following catalyst is used in contact process:
- A. FeO3
- B. V2O5✓
- C. SO3
- D. Ag2O
Explanation: The contact process is used for the production of sulfuric acid (H2SO4) from sulfur dioxide (SO2). In this process, vanadium pentoxide (V2O5) is commonly used as a catalyst.
Why the other options are wrong
- A. Iron(III) oxide is not typically used as a catalyst in the contact process.
- C. Sulfur trioxide is one of the products of the contact process, not the catalyst.
- D. Silver(I) oxide is not used as a catalyst in the contact process.
Q16. The anhydride of HCIO4 is:
- A. CIO3
- B. CIO2
- C. C*l2*O5
- D. C*l2*O7✓
Explanation: This is the correct anhydride of perchloric acid (HClO₄). It is formed when perchloric acid loses water (H₂O), leaving behind the anhydride, dichlorine heptoxide.
Why the other options are wrong
- A. This compound is not the anhydride of perchloric acid (HClO₄). Chlorine trioxide is a distinct compound with a different chemical formula and properties.
- B. Similarly, chlorine dioxide is not the anhydride of perchloric acid. It is a separate compound with its own chemical characteristics.
- C. This compound is also not the anhydride of perchloric acid. Dichlorine pentoxide is a different substance with its own formula and properties.
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