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Kpk Chemistry 2012 — Solved Past Paper with Answers
All 18 MCQs from Kpk Chemistry 2012, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT KPK / ETEA past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all KPK / ETEA papers.
Attempt this paper interactively →Q1. Half the distance between the centres of the adjacent atoms is called?
- A. Radius
- B. Ionic radius
- C. Molecular radius
- D. Atomic radius✓
Explanation: The distance between the centers of adjacent atoms in a crystal lattice is half the sum of their atomic radii. This distance is important in determining the packing arrangement of atoms in a solid.
Why the other options are wrong
- A. "Radius" alone is a general term for the distance from the center to the outer edge of a circle or sphere. In the context of atoms, the term "atomic radius" is more specific.
- B. Ionic radius refers to the distance from the nucleus of an ion to its outermost electron shell. It is not specifically related to the distance between the centers of adjacent atoms.
- C. Molecular radius is not a commonly used term in chemistry. The size of a molecule is typically described by its bond lengths or van der Waals radii, but it is not directly related to the distance between the centers of adjacent atoms in a crystal lattice.
Q2. Which one is more electrical conductor?
- A. Sl
- B. Al
- C. Mg
- D. Na✓
Explanation: Sodium is a metal and is an excellent conductor of electricity, even better than aluminum (Al) and magnesium (Mg). Sodium is often used in the form of sodium ions in electrolyte solutions, where it allows the flow of electrical current.
Why the other options are wrong
- A. Selenium is a nonmetal and is not a good conductor of electricity.
- B. Aluminum is a metal and is a good conductor of electricity, but it is not as good as sodium (Na).
- C. Magnesium is also a metal and is a good conductor of electricity, but again, it is not as good as sodium (Na).
Q3. Which one is correct according to atomic size?
- A. Cl>l
- B. Mg>Ca
- C. K>Na
- D. Cu>Co✓
Explanation: Copper (Cu) is larger than cobalt (Co) because as you move across a period in the periodic table from left to right, atomic size decreases due to increasing effective nuclear charge pulling the electrons closer to the nucleus. However, there can be exceptions due to factors like electron configuration and electron-electron repulsions, which can lead to anomalies in atomic size trends across a period.
Why the other options are wrong
- A. Iodine (I) is larger than chlorine (Cl) because as you move down a group in the periodic table, atomic size increases due to the addition of more electron shells.
- B. Calcium (Ca) is larger than magnesium (Mg) because as you move down a group in the periodic table, atomic size increases due to the addition of more electron shells.
- C. Sodium (Na) is larger than potassium (K) because as you move down a group in the periodic table, atomic size increases due to the addition of more electron shells.
Q4. Refactory bricks are made by making clay with?
- A. CaCo3
- B. Na2H
- C. MgO✓
- D. Cement
Explanation: Magnesium oxide, also known as magnesia, is a common additive in refractory brick production. It helps improve the high-temperature properties of the bricks, making them more resistant to heat and corrosion.
Why the other options are wrong
- A. Calcium carbonate is not typically used in making refractory bricks. It is a common mineral found in limestone, marble, and chalk, but it is not a typical additive in refractory brick production.
- B. Sodium hydride is a highly reactive compound and is not used in making refractory bricks. It reacts violently with water and is not suitable for this application.
- D. Cement is not typically used in making refractory bricks. Refractory bricks are usually made from ceramic materials like clay, silica, alumina, and magnesia, which are mixed together and then fired at high temperatures to form the bricks.
Q5. In thermite process, Fe2O3 reacts with?
- A. Al✓
- B. Bl
- C. Pb
- D. Fe
Explanation: In the thermite process, aluminum is the reducing agent. It reacts with iron(III) oxide (Fe2O3), which is the oxidizing agent, to produce iron and aluminum oxide. This reaction is highly exothermic, generating a great deal of heat.
Why the other options are wrong
- B. It seems there might be a typo in this option, as "Bl" is not a known element or compound in the context of the thermite reaction. The correct option for the thermite process is aluminum (Al).
- C. Lead is not involved in the thermite process. The thermite reaction specifically involves the reaction between iron(III) oxide and aluminum.
- D. Iron is not a reactant in the thermite process. Instead, it is one of the products produced by the reduction of iron(III) oxide using aluminum.
Q6. The chemical formula for orthoboric acid is?
- A. HBO2
- B. H2B3O5
- C. H3BO3✓
- D. H2B4O7
Explanation: H3BO3 is the correct formula for orthoboric acid. It consists of one boron atom, three hydrogen atoms, and three oxygen atoms.
Why the other options are wrong
- A. This is not the correct formula for orthoboric acid. It does not represent a known compound.
- B. This formula does not correspond to orthoboric acid. It represents tetraboric acid, which is not the same compound.
- D. This formula represents metaboric acid, not orthoboric acid. Metaboric acid has a different chemical structure and composition.
Q7. In halic acid, halogens have an oxidation state of?
- A. +1✓
- B. +3
- C. +5
- D. +7
Explanation: Option A (+1) is correct because it represents the typical oxidation state of halogens in halic acids, while the other options are incorrect as they do not reflect the typical oxidation state of halogens in these hypothetical compounds.
Why the other options are wrong
- B. This oxidation state is not typically found in halic acids. In most cases, halogens in halic acids are considered to have an oxidation state of +1.
- C. While halogens can have an oxidation state of +5 in some compounds (such as in the oxoacids of halogens), in halic acids, the oxidation state is generally considered to be +1.
- D. This oxidation state is typically found in compounds like the oxoacids of halogens (e.g. perchloric acid, HClO4). In halic acids, the oxidation state of halogens is assumed to be +1.
Q8. In bleaching powder an average chlorine available is?
- A. 25-30%
- B. 30-35%✓
- C. 35-40%
- D. 40-45%
Explanation: This range is generally considered the average available chlorine content in bleaching powder and is the most common value cited for this compound.
Why the other options are wrong
- A. This range is slightly below the typical average available chlorine content found in bleaching powder.
- C. This range is slightly above the typical average available chlorine content found in bleaching powder.
- D. This range is significantly higher than the typical average available chlorine content found in bleaching powder and is unlikely to be accurate for this compound.
Q9. Hydrocarbons contain?
- A. Hydrogen and oxygen
- B. Carbon and hydrogen✓
- C. Carbon and oxygen
- D. Carbon and nitrogen
Explanation: Hydrocarbons are composed of carbon and hydrogen atoms, which is why option B (Carbon and hydrogen) is correct.
Why the other options are wrong
- A. This combination would result in a hydroxide, not a hydrocarbon.
- C. This combination would result in an oxide, not a hydrocarbon.
- D. This combination would result in a nitrile, not a hydrocarbon.
Q10. Chromium metal is a member of the row transition series?
- A. 1st
- B. 2nd
- C. 3rd
- D. 4th✓
Explanation: The 4th row transition series includes elements like chromium, molybdenum, and tungsten.
Why the other options are wrong
- A. The 1st row transition series includes elements like scandium, titanium, and vanadium.
- B. The 2nd row transition series includes elements like yttrium, zirconium, and niobium.
- C. The 3rd row transition series includes elements like lanthanum, hafnium, and tantalum.
Q11. Girgnard reagent is prepared by treating?
- A. Alcohol and mg
- B. Alkane and mg
- C. Alkyl halide and mg✓
- D. None of these
Explanation: Option C is the correct combination for preparing Grignard reagents. The alkyl halide reacts with magnesium to form the Grignard reagent.
Why the other options are wrong
- A. This combination does not lead to the formation of Grignard reagents. Grignard reagents are typically not prepared from alcohols.
- B. Alkanes do not react with magnesium to form Grignard reagents. Grignard reagents are not typically prepared from alkanes.
- D. This is incorrect, as Grignard reagents are indeed prepared by treating alkyl halides with magnesium.
Q12. Urea is?
- A. Am ester
- B. Amides✓
- C. Carboxylic acid
- D. Amino acid
Explanation: Urea is classified as an amide. It has a carbonyl group (C=O) bonded to an amine group (-NH2). In urea, the carbonyl group is part of the urea functional group, while the amine group is part of the urea molecule.
Why the other options are wrong
- A. Urea is not an am ester. Am esters have a structure where an amine group is attached to one side of the carbonyl group, while an alkoxyl group is attached to the other side. Urea does not have this structure.
- C. Urea is not a carboxylic acid. Carboxylic acids have a carboxyl group (COOH) as their functional group, which is different from the urea functional group.
- D. Urea is not an amino acid. Amino acids are organic compounds that contain both an amino group (-NH2) and a carboxyl group (COOH) as their functional groups. Urea does not contain a carboxyl group.
Q13. Which one of the following hydrocarbons is alkene?
- A. C4H3
- B. C3H4
- C. C2H8
- D. C2H2✓
Explanation: Explanation will be added soon.
Q14. In the second substituent occupies the?
- A. Para position
- B. Ortho position
- C. Ortho/paraposition✓
- D. Metaposition
Explanation: This option is a combination of ortho and para positions. If one substituent is occupying either the ortho or para position, the second substituent would not be able to occupy the same ortho or para position. Thus, option C is the correct option.
Why the other options are wrong
- A. The para position refers to the position on a benzene ring that is directly opposite to a substituent. In the context of the question, if one substituent is already occupying the para position, the second substituent would not be able to occupy the same para position.
- B. The ortho position refers to the positions on a benzene ring adjacent to a substituent. If one substituent is already occupying an ortho position, the second substituent would not be able to occupy the same ortho position.
- D. The meta position refers to the position on a benzene ring that is two carbons away from a substituent. If one substituent is already occupying a meta position, the second substituent would not be able to occupy the same meta position.
Q15. Hydration of acetylene produces?
- A. Acetone
- B. Ethane
- C. Acetlydehyde✓
- D. None
Explanation: Acetaldehyde, also known as ethanal, is produced from the hydration of acetylene. This reaction is commonly used in industry to produce acetaldehyde.
Why the other options are wrong
- A. Acetone is not produced from the hydration of acetylene. Acetone is commonly produced by the oxidation of isopropanol or cumene.
- B. Ethane is not produced from the hydration of acetylene. Ethane is typically produced from the steam cracking of hydrocarbons.
- D. This option is incorrect as acetylene does undergo hydration to form acetaldehyde.
Q16. Which one of the following is not a fatty acid?
- A. Butanoic acid
- B. Ethanoic acid
- C. Propanoic acid
- D. Phthalic acid✓
Explanation: Phthalic acid is not a fatty acid. It is an aromatic dicarboxylic acid.
Why the other options are wrong
- A. This is a fatty acid with a four-carbon chain (C4H8O2), also known as butyric acid.
- B. It is a two-carbon carboxylic acid, also known as acetic acid.
- C. This is a fatty acid with a three-carbon chain (C3H6O2), also known as propionic acid.
Q17. The size of the smoke particle is less than?
- A. 0.1µm
- B. 0.5µm
- C. 1µm✓
- D. 1.5µm
Explanation: 1 µm is a reasonable estimate for the upper limit of smoke particle size, as smoke particles are generally smaller than 1 µm.
Why the other options are wrong
- A. This is smaller than typical smoke particles, which are generally larger than 0.1 µm.
- B. While some smoke particles may be around this size, many are larger, so this is not a strict cutoff for smoke particle size.
- D. Smoke particles can sometimes be larger than 1.5 µm, so this is not an accurate statement about the size of smoke particles.
Q18. Sources of SO2 in the atmosphere are:
- A. Sea sprays
- B. Households activities
- C. Fossil fuels✓
- D. Municipal wastes
Explanation: Fossil fuels, such as coal, oil, and natural gas, are the primary sources of sulfur dioxide (SO2) in the atmosphere. When these fuels are burned for energy production, sulfur compounds present in them are oxidized to form sulfur dioxide, which is released into the air. The combustion of fossil fuels in power plants, industrial facilities, and vehicles is a major contributor to sulfur dioxide emissions.
Why the other options are wrong
- A. Sea sprays are produced when waves break and release tiny water droplets into the air. While sea sprays can contain various substances, including salts and organic compounds, they are not significant sources of sulfur dioxide (SO2) in the atmosphere.
- B. Household activities can contribute to air pollution through the use of fossil fuels (e.g. heating and cooking with natural gas or oil) and the burning of waste. However, these activities are typically not major sources of sulfur dioxide in the atmosphere compared to industrial and transportation-related emissions.
- D. Municipal wastes, including solid waste from households and commercial establishments, can release sulfur dioxide when burned in incinerators. However, the contribution of municipal wastes to sulfur dioxide emissions is relatively small compared to other sources, such as fossil fuel combustion.
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