Chemistry MCQ Asked in Previous Year’s WBCS Preliminary Exams [130 MCQs]

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Chemistry MCQ: Asked in WBCS Prelims 2017

Q29. Conversion of CH3C=CH to CH3CH=CH2
A. Lindlar catalyst
B. H2/Pd
C. NaBH4
D. LiAlH4

Answer –A. Lindlar catalyst
Explanation- The conversion of CH3C=CH (methylacetylene) to CH3CH=CH2 (propene) is achieved by using a Lindlar catalyst, which is a palladium catalyst poisoned with lead or quinoline. This catalyst allows for selective hydrogenation of the alkyne bond, stopping at the alkene stage.


Q30. Conversion of RBr to RMgBr requires
A. Mg /dry ether / N2 – atmosphere
B. Mg/ moist ether /N2 – atmosphere
C. Mg/ ethanol /N2 -atmosphere
D. Mg / dry ether /O2 – atmosphere.

Answer –A. Mg/dry ether/N2 atmosphere
Explanation- The conversion of RBr (alkyl bromide) to RMgBr (Grignard reagent) requires the use of magnesium (Mg) in dry ether (anhydrous ethyl ether) under a nitrogen (N2) atmosphere. This reaction allows for the formation of an organomagnesium compound, the Grignard reagent.


Q31. The fastest SN1 reaction is of the following:
A. MeO — CH2 –Cl
B. Me — CH2 -Cl
C. Me –C –CH2 –Cl
D. Ph — CH2 –CH2 – Cl

Answer – A. MeO—CH2—Cl (methyl chloride)
Explanation-The fastest SN1 reaction occurs with the substrate MeO—CH2—Cl (methyl chloride). SN1 reactions are nucleophilic substitution reactions in which the rate-determining step involves the formation of a carbocation intermediate. In this case, the methyl chloride substrate provides a more stable carbocation, resulting in a faster reaction.


Q32. The strongest electropositive element is
A. Cs
B. Li
C. Mg
D. K

Answer – A. Cs (Cesium)
Explanation- Cesium (Cs) is the strongest electropositive element among the options listed. Electropositivity refers to the tendency of an element to lose electrons and form positive ions. Cesium has the lowest electronegativity and the largest atomic size among the options, making it the most electropositive.


Q33. The geometry of SF4 (Sulfur tetrafluoride) is
A. square planar
B. tetrahedral
C. octahedral
D. see-saw

Answer –D. see-saw
Explanation-SF4, or sulfur tetrafluoride, has a see-saw molecular geometry. In this arrangement, the central sulfur atom is bonded to four fluorine atoms, resulting in a trigonal bipyramidal electron pair geometry. However, due to the presence of one lone pair of electrons on the sulfur atom, the fluorine atoms are not evenly distributed around the central atom. The see-saw molecular geometry occurs when one of the equatorial positions in the trigonal bipyramidal arrangement is occupied by a lone pair. As a result, the molecule takes on a distorted shape, resembling a see-saw. The lone pair of electrons causes the fluorine atoms to be pushed closer together, resulting in an uneven distribution.


Q34. A reaction is spontaneous when
A. Delta G = -ve
B. Delta H = -ve
C. Delta S = +ve
D. Delta S = -ve

Answer –A. Delta G = -ve
Explanation- A reaction is spontaneous when the change in Gibbs free energy (ΔG) is negative. A negative ΔG indicates that the reaction can proceed spontaneously, releasing energy. The other options, ΔH = -ve (enthalpy change) and ΔS = +ve (entropy change), can provide information about the thermodynamics of a reaction but do not solely determine spontaneity.


Q35. The material used for safety from nuclear radiation is
A. Copper
B. Platinum
C. Iron
D. Lead

Answer – D. Lead
Explanation- Lead (Pb) is commonly used as a material for safety from nuclear radiation. Lead has high density and a high atomic number, which allows it to effectively absorb and attenuate various forms of radiation, including gamma rays and X-rays. The thickness of lead serves as a barrier, reducing exposure to harmful radiation.


Chemistry MCQ: Asked in WBCS Prelims 2016

Q36. Which among the following is called ‘Laughing Gas’ popularly?
A. Nitric oxide
B. Nitrous oxide
C. Nitrogen pentaoxide
D. Nitrogen

Answer –B. Nitrous oxide
Explanation- Nitrous oxide is popularly known as “Laughing Gas.” It is a chemical compound with the formula N2O. The name “Laughing Gas” originates from the euphoric and anesthetic effects it can produce when inhaled in moderate doses.


Q37. Which among the following gas was leaked during the Bhopal gas tragedy in 1984?
A. Methyl isocyanides
B. Methyl isocyanate
C. Methyl isochloride
D. Methyl isochlorate

Answer –B. Methyl isocyanate
Explanation- During the Bhopal gas tragedy in 1984, the gas leaked was Methyl isocyanate (MIC). The tragedy occurred at the Union Carbide India Limited (UCIL) pesticide plant in Bhopal, India. Due to a series of safety failures and maintenance issues, a large amount of MIC gas was released into the atmosphere, resulting in one of the worst industrial disasters in history.


Q38. Which metal is present in chlorophyll?
A. Iron
B. Zinc
C. Aluminium
D. Magnesium

Answer –D. Magnesium
Explanation- Magnesium (Mg) is the metal present in chlorophyll. Chlorophyll is a pigment found in plants and is responsible for absorbing light energy during photosynthesis. The central atom in chlorophyll’s structure is magnesium, which plays a crucial role in capturing sunlight and converting it into chemical energy.


Q39. In the manufacture of Vanaspati ghee from vegetable oils, which among the following gas is used?
A. Nitrogen
B. Carbon dioxide
C. Hydrogen
D. Neon

Answer – C. Hydrogen
Explanation-In the manufacture of Vanaspati ghee from vegetable oils, hydrogen gas (H2) is used. This process involves the hydrogenation of vegetable oils to produce a semi-solid fat known as Vanaspati ghee. Hydrogen gas acts as a reducing agent in the hydrogenation process.


Q40. Which among the following is responsible for the depletion of the Ozone layer?
A. Carbon monoxide
B. Carbon dioxide
C. Chlorofluorocarbon
D. Mercuric oxide-

Answer – C. Chlorofluorocarbon
Explanation- Chlorofluorocarbons (CFCs) are primarily responsible for the depletion of the ozone layer. CFCs were widely used in aerosol propellants, refrigerants, and solvents. When released into the atmosphere, CFCs can rise to the stratosphere, where they undergo photodissociation, releasing chlorine atoms that catalytically destroy ozone molecules.


Q41. The term PVC used in the plastic industry stands for :
A. Phospho Vinyl Chloride
B. Poly Vinyl Carbonate
C. Poly Vinyl Chloride
D. Phospho Vanadium Chloride

Answer – C. Poly Vinyl Chloride
Explanation- The term PVC used in the plastic industry stands for Poly Vinyl Chloride. PVC is a versatile synthetic polymer widely used in various applications, including pipes, cables, flooring, and construction materials, due to its durability, low cost, and chemical resistance.


Chemistry MCQ: Asked in WBCS Prelims 2015

Q42. Mg2+ is isoelectronic with
A. Ca2+
B. Na+
C. Zn2+
D. Cu+

Answer – C. Zn2+
Explanation- Isoelectronic species have the same number of electrons. In the case of Mg2+, it has a 2+ charge, indicating that it has lost two electrons. Magnesium (Mg) has an atomic number of 12, so it normally has 12 electrons. By losing two electrons to become Mg2+, it achieves the electron configuration of the noble gas neon (Ne), which has 10 electrons. Among the given options, Zn2+ (zinc ion) is isoelectronic with Mg2+. Zinc (Zn) has an atomic number of 30, so it normally has 30 electrons. By losing two electrons to become Zn2+, it also attains the electron configuration of neon (Ne) with 10 electrons.


Q43. Hybridization of C2 and C3 of CH3-2CH-4CH-5CH3
A. sp sp3
B. sp2 sp
C. sp2 sp2
D. sp sp

Answer –C. sp2 sp2
Explanation- o determine the hybridization of each carbon atom in the molecule CH3-2CH-4CH-5CH3, let’s examine each carbon atom individually. Starting from the left side of the molecule: C2 is bonded to two hydrogen atoms and two carbon atoms (C1 and C3). It forms a single bond with C1 and C3, and a sigma (σ) bond with each hydrogen atom. Since C2 is bonded to three other atoms, it undergoes sp2 hybridization to form three sp2 hybrid orbitals. These orbitals are involved in bonding with the other atoms. C3 is bonded to three carbon atoms (C2, C4, and C5) and one hydrogen atom. It forms a single bond with each of these atoms, resulting in a total of four sigma (σ) bonds. Since C3 is bonded to four other atoms, it undergoes sp2 hybridization to form three sp2 hybrid orbitals. Therefore, the hybridization of C2 is sp2 and the hybridization of C3 is also sp2. Thus, the correct answer is C. sp2 sp2.


Q44. The compound that cannot be kept in a glass vessel because it reacts with glass is
A. HNO3
B. HCl
C. HF
D. HBr

Answer – C. HF – Hydrofluoric acid (HF)
Explanation- Hydrofluoric acid is known to react with glass. This is due to its ability to form stable compounds with silicon dioxide (SiO2) present in glass. When hydrofluoric acid comes into contact with glass, it can slowly dissolve the silica network, leading to the formation of silicon tetrafluoride (SiF4) and water. This reaction can cause damage to the glass vessel and compromise its structural integrity. On the other hand, HNO3 (nitric acid), HCl (hydrochloric acid), and HBr (hydrobromic acid) can generally be stored in glass vessels without significant reactions. Glass is resistant to these acids and does not undergo substantial chemical reactions or dissolution when in contact with them.


Q45. How many grams of NaOH are needed to make 100 ml. of a 0.5 M solution of NaOH?
(Atomic weight : Na = 23, 0 = 16, H = l)
A. 2
B. 20
C. 4
D. 1
C. 4 –

Answer – A. 2
Explanation-To calculate the grams of NaOH needed to make a 0.5 M solution in 100 ml, we can use the formula:
Grams = Molarity (M) × Volume (L) × Molecular Weight

Given:
Molarity (M) = 0.5 M
Volume (L) = 0.1 L (since 100 ml is equal to 0.1 L)
Molecular Weight of NaOH = 23 + 16 + 1 = 40 g/mol
Grams = 0.5 M × 0.1 L × 40 g/mol = 2 g

Therefore, 2 grams of NaOH are needed to make a 0.5 M solution of NaOH in 100 ml.


Q46. Chemically Aspirin is :
A. Acetyl salicylic acid
B. Sodium salicylate
C. Methyl salicylate
D. Ethyl salicylate

Answer –A. Acetyl salicylic acid
Explanation- Chemically, aspirin is known as acetylsalicylic acid. It is a common medication used for pain relief, fever reduction, and anti-inflammatory purposes. Acetylsalicylic acid is derived from salicylic acid by acetylation, where an acetyl group (-COCH3) is added to the hydroxyl group (-OH) of salicylic acid.


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