Chemistry MCQs for NEET — Practice Questions with Answers

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The extent to which reactions proceed to reach chemical equilibrium can be classified into three groups. Which group describes reactions where concentrations of reactants and products are comparable at equilibrium?

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Explanation

The NCERT text lists three groups: '(i) The reactions that proceed nearly to completion...', '(ii) The reactions in which only small amounts of products are formed...', and '(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.'

What happens to the rates of forward and reverse reactions during the approach to chemical equilibrium?

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Explanation

The NCERT text (Fig. 6.2 and accompanying description) explains: 'With passage of time, there is accumulation of the products C and D and depletion of the reactants A and B (Fig. 6.2). This leads to a decrease in the rate of forward reaction and an increase in the rate of the reverse reaction. Eventually, the two reactions occur at the same rate and the system reaches a state of equilibrium.'

Alkanes exhibit which type of isomerism due to free rotation along C-C sigma bonds?

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Explanation

The NCERT summary states: 'Alkanes show conformational isomerism due to free rotation along the C–C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart.'

Which of the following compounds can exhibit geometrical isomerism?

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Explanation

The NCERT text explains: 'Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond.' For geometrical isomerism, each carbon of the double bond must be attached to two different groups. In but-2-ene ($CH_3-CH=CH-CH_3$), both carbons of the double bond are attached to a hydrogen and a methyl group, allowing for cis and trans isomers. Pent-1-ene ($CH_2=CH-CH_2-CH_2-CH_3$) has one carbon of the double bond attached to two hydrogen atoms, preventing geometrical isomerism.

Consider the compounds with molecular formula $C_3H_6O$. Which type of isomerism can be observed between propanal and propanone?

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Explanation

The NCERT text states: 'Functional group isomerism: Two or more compounds having the same molecular formula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism. For example, the molecular formula C3H6O represents an aldehyde and a ketone.' Propanal is an aldehyde and propanone is a ketone, both having the formula $C_3H_6O$ but different functional groups.

Which of the following statements about isomers is correct?

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Explanation

The NCERT text defines isomerism: 'The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Such compounds are called as isomers.' This difference in properties arises due to differences in their structures.

Identify the type of carbon atoms (1°, 2°, 3°, 4°) in 2,2-dimethylpropane (neopentane).

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Explanation

The NCERT defines carbon types: 'Tertiary carbon is attached to three carbon atoms and neo or quaternary carbon is attached to four carbon atoms.' In 2,2-dimethylpropane ($C(CH_3)_4$), the central carbon atom is bonded to four other carbon atoms (making it a quaternary or 4° carbon), and each of the four methyl groups has a carbon atom bonded to only one other carbon atom (making them primary or 1° carbons).

How many chain isomers does $C_6H_{14}$ have?

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Explanation

The problem in the NCERT text (Problem 9.1) asks to 'Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14. Also write their IUPAC names.' The solution lists five isomers for $C_6H_{14}$: n-Hexane, 2-Methylpentane, 3-Methylpentane, 2,3-Dimethylbutane, and 2,2-Dimethylbutane. These are all chain isomers.

Why is rotation around a C=C bond restricted, leading to geometrical isomerism?

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Explanation

The NCERT text explains: 'Rotation around C=C bond is not free. It is restricted... This illustrates that the restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries of such compounds. The stereoisomers of this type are called geometrical isomers.' This restricted rotation is a characteristic feature of the π-bond.

Which type of isomerism is depicted by structures I and II for $C_4H_{10}$ in the given context?

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Explanation

The context says, 'Structures I and II possess same molecular formula but differ in their boiling points and other properties... Structures I and II are isomers of butane... Since difference in properties is due to difference in their structures, they are known as structural isomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers.' Structure I ($CH_3-CH_2-CH_2-CH_3$) is n-butane (continuous chain) and Structure II ($CH_3-CH(CH_3)-CH_3$) is 2-methylpropane (branched chain), both are chain isomers of $C_4H_{10}$.

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