NCERT Exemplar: Some Basic Principles & Techniques | Chemistry Class 11 - NEET PDF Download

MULTIPLE CHOICE QUESTIONS 

Q.1. Which of the following is the correct IUPAC name? 
(1) 3-Ethyl-4, 4-dimethylheptane 
(2) 4,4-Dimethyl-3-ethylheptane 
(3) 5-Ethyl-4, 4-dimethylheptane 
(4) 4,4-Bis(methyl)-3-ethylheptane
Ans. (1)
Solution.
In IUPAC naming is different alkyl groups are present, they are written in alphabetical order. Ethyl will be first, followed by methyl. Di, tri, tetra prefixes are not included in alphabetical order.

Q.2. The IUPAC name for  is ________.
(1) 1-hydroxypentane-1,4-dione 
(2) 1,4-dioxopentanol 
(3) 1-carboxybutan-3-one 
(4) 4-oxopentanoic acid
Ans. (4)
Solution.
if more than one functional groups are present in the chain, one will be the main functional group on priority basis and mentioned as a sec. suffix, while the others are written as prefix mentioned as oxo along with locant.

Q.3. The IUPAC name for

(1) 1-Chloro-2-nitro-4-methylbenzene 
(2) 1-Chloro-4-methyl-2-nitrobenzene 
(3) 2-Chloro-1-nitro-5-methylbenzene 
(4) m-Nitro-p-chlorotoluene
Ans. (2)
Solution.
Substituent of the base compound is assigned number 1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order.
As for example:


Q.4. Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?
(1) CH₃ – CH₂ – *CH₂ – CH₃ 
(2) CH₃ – *CH = CH – CH₃ 
(3) CH₃ – CH₂ – C  ≡ *CH
(4) CH₃ – CH₂ – CH  = *CH₂
Ans. (3)
Solution.
Electronegativity of carbon atom depends on its hybridized state and percentage of ‘s’ character. Electronegativity is directly proportional to percentage of ’s’character, sp³ (25% ‘s’ character) < sp² (33% ‘s’ character) < sp (50% ‘s’ character). Carbon linked through two bonds is sp hybridized.

Q.5. In which of the following, functional group isomerism is not possible? 
(1) Alcohols 
(2) Aldehydes 
(3) Alkyl halides 
(4) Cyanides
Ans. (3)
Solution.
Functional isomers have same molecular formula but different functional group. Alcohols are functional isomers of ethers. Aldehydes are functional isomers of ketones. Cyanides are functional isomers of isocyanides. Only alkyl halides do not show functional isomerism.

Q.6. The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is: (1) Distillation 
(2) Crystallisation 
(3) Distillation under reduced pressure 
(4) Steam distillation
Ans. (4)
Solution.
This technique is applied to separate substances which are steam volatile and are immiscible with water. In steam distillation, steam, from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and collected. The compound is later separated from water using a separating funnel.

Q.7. During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results? 
(1) Column chromatography 
(2) Solvent extraction 
(3) Distillation
(4) Thin layer chromatography
Ans. (4)
Solution.
Thin layer chromatography (TLC): It is a method for analyzing mixtures by separating the components of the mixture. TLC can be used to determine the number of components in the mixture, the identity of compounds and the purity of compounds by observing the appearance of a product or the disappearance of a reactant. It can also be used to monitor the progress of a reaction. TLC is a sensitive technique— microgram (0.000001)g quantities can be analyzed by TLC.

Q.8. The principle involved in paper chromatography is 
(1) Adsorption 
(2) Partition 
(3) Solubility 
(4) Volatility
Ans. (2)
Solution.
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper hromatography is a type of partition chromatography.

Q.9. What is the correct order of decreasing stability of the following cations.



(1) II > I > III 
(2) II > III > I 
(3) III > I > II 
(4) I > II > III
Ans. (1)
Solution.
In (I) case, +vely charged C is attached to two alkyl groups and +I effect stabilize carbocation. In (II) case, +R effect of OCH₃ group stabilize carbocation. In (III) case, - I effect of -OCH₃ gp. destabilizes carbocation hence, the order of stability: II > I >III.

Q.10. Correct IUPAC name for  is______________.
(1) 2- ethyl-3-methylpentane 
(2) 3,4- dimethylhexane 
(3) 2-sec-butylbutane 
(4) 2, 3-dimethylbutane
Ans. (2)
Solution.
In longest continuous chain both ethyl groups are included and 3^rd and 4^th carbon have methyl groups present as a substituent.

Q.11. In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?
(1) *CH₃—CH₂—Cl 
(2) *CH₃—CH₂—Mg⁺Cl^– 
(3) *CH₃—CH₂—Br 
(4) *CH₃—CH₂—CH₃
Ans. (1)
Solution.
Order of electronegativity of the attached group is as follows:
Cl > Br > C > Mg.
More electronegative group attached to the C will give more positive charge. Therefore, in (i) case asterisk C will have greatest positive charge.

Q.12. Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?
(1)
(2)
(3)
(4)
Ans. (4)
Solution.
The stabilsation of carboxylate ion depends on dispersal of negative charge. The negative charge is dispersed by two factors, i.e., +R effect of the carboxylate ion and Inductive effect of the halogens. In all the above structures, +R effect is common but halogen atoms are different. Therefore, dispersal of negative charge depends upon halogen atoms. F is most electronegative, in structure (4) two F atoms are present and more dispersal of negative charge is there.

Q.13. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
H₃C—HC = CH₂ + H+ →?
(1) 2° Carbanion 
(2) 1° Carbocation 
(3) 2° Carbocation 
(4) 1° Carbanion
Ans. (3)
Solution.
When the electrophile attacks CH₃ – CH = CH₂, delocalisation of electrons can take place in two possible ways

As 2° carbocation is more stable than 1° carbocation, the first addition is more feasible.

Q.14. Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH3—Br is
(1)
(2)  
(3)  
(4)  
Ans. (2)
Solution.
Br is more electronegative than C, hence heterolytic fission takes place. Electrons displace from carbon to Br. Therefore, CH₃ gets positive charge and Br gets negative charge. Thus, option (2) is correct.

Q.15. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H+ ion to  portion which can be shown as
(1)
(2)
(3)  
(4) All of these are possible
Ans. (2)
Solution.
This effect the π-electrons of the multiple bond are transferred to that atom to

which there agent gets attached.
For example:

MULTIPLE CHOICE QUESTIONS II

In the following questions two or more options may be correct.

Q.16. Which of the following compounds contain all the carbon atoms in the same hybridisation state?
(i) H—C ≡ C—C ≡ C—H
(ii) CH₃—C ≡ C—CH₃
(iii) CH₂ = C = CH₂
(iv) CH₂ = CH—CH = CH₂
Ans. (1,4)
Solution. 
Only in these two compounds all carbon atoms are in same hybridization state i.e., sp and sp² hybridized, respectively

Q.17. In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?

(1)
(2)
(3)
(4)
Ans. (1,3,4)
Solution.
The spatial arrangement of groups or atoms can be checked by doing two interchange and bringing H below the plane of the paper. Find out the sequence of the remaining groups in a particular order whether clockwise or anticlockwise starting from atom with highest atomic number to atom with lower atomic numbers. Hence, option (2) has same spatial arrangement as (A) while in rest three is different.

Q.18. Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?
(1) BF₃, NH₃ , H₂O
(2) AlCl₃, SO₃, NO⁺₂
(3)  
(4)  
Ans. (2,3)
Solution.
Electrophiles are +vely charged or electron deficient species. They are Lewis acids. AlCl₃, SO₃ are Lewis acids,  are +vely charged species

Q.19. Which of the following pairs are position isomers?


(1) I and II
(2) II and III 
(3) II and IV 
(4) III and IV
Ans. (2)
Solution.
In position isomerism, two or more compounds differ in the position of substituent, functional group or multiple bonds but molecular formula is same. In pentanone-2 and pentanone-3, position of ketonic group is different.
CH₃—CH₂—CH₂—CO—CH₃ and CH₃- CH₂-CO-CH₂-CH₃

Q.20. Which of the following pairs are not functional group isomers? 

 
(1) II and III
(2) II and IV
(3) I and IV
(4) I and II
Ans. (1,2)
Solution.
Two or more compounds with same molecular formula but different functional groups are called functional isomers. In the given compounds
I. Aldehydic group.
II. Ketonic group.
III. Ketonic group.
IV. Aldehydic group.
Hence II and III, I and IV are not functional isomers

Q.21. Nucleophile is a species that should have
(1) a pair of electrons to donate
(2) positive charge
(3) negative charge
(4) electron deficient species
Ans. (1,3)
Solution.
Nucleophiles are -vely charged or electron rich (lone pair of electrons) species.
Hence, options (i) and (iii) are correct.

Q.22. Hyperconjugation involves delocalisation of ___________.
(1) electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsaturated system.
(2) electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.
(3) π-electrons of carbon-carbon bond
(4) lone pair of electrons
Ans. (1, 2)
Solution.
Hyperconjugation involves delocalization of a electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. The a electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital. Hyperconjugation is a permanent effect.

SHORT ANSWER TYPE QUESTIONS

Q.23. 





Which of the above compounds form pairs of metamers?
Ans. V and VI, VI and VII and V and VII form pairs of metamers because they have different alkyl chains on either side of ethereal oxygen (-O-).

Q.24. 





Identify the pairs of compounds which are functional group isomers.
Ans. Compounds I to IV, i.e., alcohols, and V to VII, i.e., ethers, are functional group isomers with molecular formula C₄H₁₀O and different functional groups (-OH in I to IV and -O- in V to VII).
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; HI and V, IH and VI; IH and VII; IV and V, IV and VI, IV and VH are functional group isomers.

Q.25. 

Identify the pairs of compounds that represents position isomerism.
Ans. I and II, III and IV, VI and VII are position isomers due to different positions of -OH group and -O- group.

Q.26. 





Identify the pairs of compounds that represents chain isomerism.
Ans. I and III, I and IV, II and III, II and IV.

Q.27. For testing halogens in an organic compound with AgNO₃ solution, sodium extract (Lassaigne’s extract) is acidified with dilute HNO₃. What will happen if a student acidifies the extract with dilute H2SO₄ in place of dilute HNO₃?
Ans. On adding dilute H2SO₄ for testing halogens in an organic compound with AgN03, white precipitate of Ag₂SO₄ is formed. This will interfere with the test of chlorine and this Ag2SO₄ may be mistaken for white precipitate of chlorine as AgCl. Hence, dilute HNO₃ should be used instead of dilute H2SO₄.

Q.28. What is the hybridization of each carbon in H₂C = C = CH₂?
Ans. The given structure is of C₃H₄ in which all three carbon atoms are linked to each other by double bonds. Carbons are marked as Cl, C₂ and C₃. C₁ and C₃ are sp2 hybridized because of having 3σ and 1π bonds. C₂ has 2 σ bonds and 2 π bonds it is sp hybridized.


Q.29. Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Ans. Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
sp³ – 25% s-character, 75% P-character
sp² – 33% s-character, 67% P-character  
sp – 50% s-character, 50% P-character
Hence, the order of electronegativity is sp³ < sp² < sp

Q.30. Show the polarization of carbon-magnesium bond in the following structure.
CH₃—CH₂—CH₂—CH₂—Mg—X
Ans. Carbon is more electronegative than magnesium therefore, Mg has partially positive charge and C has partially negative charge because bonded pair of electrons attracted towards carbon.


Q.31. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

Ans. These are position isomers as the functional groups are attached to different carbon atoms in carbon chain. They are not metamers because the number of C atoms on either side of S—are same:
CH₃-S-CH₂-CH₂-CH₃
Methyl n-propyl thioether.

Methyl isopropyl thioether.

Q.32. Which of the following selected chains is correct to name the given compound according to IUPAC system.


Ans. According to IUPAC nomenclature, the selected longest carbon chain must have maximum functional groups present in the compound. Therefore, only in one selected chain of 4 carbon atoms including both the functional group is corrected one.

In other three, carbon atoms are in selected chain but both the functional groups are not included.

Q.33. In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.
Ans. In DNA and RNA , nitrogen is present in heterocyclic base and also present in ring not as a substituent. Therefore, nitrogen present in ring cannot be converted into (NH₄)₂SO₄. Hence, cannot be estimated by Kjeldahl method.

Q.34. If a liquid compound decomposes at its boiling point, which method (s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.
Ans. Steam distillation can be used for its purification. This method is applied to separate substances which are steam volatile and immiscible with water. Note: Answer Questions 35-38 on the basis of information given below: “Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom, involvement of neighbouring groups in hyperconjugation and resonance.”

Q.35. “Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom involvement of neighbouring groups in hyperconjugation and resonance.”

Draw the possible resonance structures for  and predict which of the structures is more stable. Give reason for your answer.
Ans. Two resonating structures can be of given carbocation:
In structure I,  has + ve charge means octet is not completed, but in structure II,  both the carbon atoms and oxygen atom have octet of electrons hence, more stable.

Q.36. Which of the following ions is more stable? Use resonance to explain your answer.

Ans. Structure A is more stable than B. Carbocation A is more Planar and π electrons from the ring shift to side group CH⁺₂ and stabilized by resonance. Structure (B) is non-planar and does not undergo resonance. Double bond is more stable with in the ring as compared to side chain.

Q.37. The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.

Ans. Triphenylmethyl cation is very stable because +ve charge of methyl carbon is delocalized in three phenyl rings. In each phenyl ring, +ve charge is developed on 2 ortho position and para position, i.e. three resonating structures. Total resonating structures given by triphenylmethyl cation are nine. Hence, it is very stable. These structures can be shown as.


Q.38. Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.
Ans. 2-methylbutane gives 4 different carbocation’s by replacing 1-H atom from different carbons. 2 primary, 1 secondary and 1 tertiary carbocation. Order of stability is as follows:
CH₃ - CH(CH₃) - CH₂ - C⁺H₂  I(1° carbocation)
CH₃ - CH(CH₃) - CH⁺ - CH₃  II(2° carbocation)
CH₃ + C⁺(CH₃) - CH⁺ - CH₃ III(3° carbocation)
⁺CH₂ - CH(CH₃) - CH⁺ - CH₃ IV(1° carbocation)
1°carbocation < 2°carbocation < 3°carbocation
Stability of various carbocations.
I < IV < II < III

Q.39. Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid FeSO₄ and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same Lassaigne’s extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation. Also,  write the chemical equations to explain the formation of compounds of different colours.
Ans. If organic compound contains N and S both, then on fusion with Na metal compound gives NaSCN or NaCN and Na₂S depending on the quantity of Na metal. If Na metal is less, then only NaSCN is formed. In that case L.E on treating with FeSO₄ and H₂SO₄ gives red colour due to the formation of ferric thiocyanide Fe(SCN)₃. In case of NaCN, L.E on treating with FeSO₄ and H₂SO₄ gives Prussian blue colour.
Chemical reactions:

If Na is in excess then :

Prussian blue

Q.40. Name the compounds whose line formulae are given below :
(i)
(ii)
Ans. (i) 3-ethyl, 4-methylhept-5-en-2-one. (Longest chain of carbon atoms selected in such a way that the functional group > C = O gets lowest possible locant.
(ii) 3-nitrocyclohex-l-en (Carbon atoms of the ring are numbered in such a way that double-bonded carbon gets the lowest number followed by the nitro group -NO₂.

Q.41. Write structural formulae for compounds named as
(a) 1 - Bromoheptane
(b) 5 - Bromoheptanoic acid
Ans. (i) CH₃ - CH₂ - CH₂ - CH₂ - CH₂ - CH₂ - CH₂Br
l-Bromoheptane
(ii) CH₃ - CH₂ - CH(Br) - CH₂ - CH₂ - CH₂ - COOH
5-Bromoheptanoic acid

Q.42. Draw the resonance structures of the following compounds;

Ans. (i) CH₂ = CH - Cl ↔ ^-CH₂ - CH = Cl⁺
(ii) CH₂ = CH — CH = CH₂ ↔ ⁺CH₂ - CH = CH - CH^-₂ ↔ CH₂ - CH = CH⁺₂
(iii) CH₂ = CH - CHO ↔ ⁺CH₂- CH=CHO^-

Q.43. Identify the most stable species in the following set of ions giving reasons:

Ans. (i) CH⁺₃ is most stable species because replacement of H atom by Br (-Inductive effect) increases +ve charge on carbon atom and destabilizes the species.
(ii) ^-CCl₃ is most stable because -ve charge on carbon atom is dispersed due to -I effect of Cl. More Cl atoms, more dispersal of -ve charge and more stabilized.

Q.44. Give three points of differences between inductive effect and resonance effect.
Ans. 

Q.45. Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
(i) CH₃OH (ii) R—CONH₂ (iii) CH₃CH = CHCH₂NH₂
Ans. (i) CH₃OH: It does not show resonance because of absence of π- electrons.
(ii) RCONH₂ : It shows resonance due to the presence of lone pair on N atom and π electrons on C = 0 bond. Hence, amide can be represented by three resonating structures.
(iii) CH₃CH = CHCH₂NH₂ : As the lone pair of electrons on N atom, is not conjugated with the π electrons of double bond. Hence, it does not show resonance.

Q.46. Why does SO₃ act as an electrophile?
Ans. SO₃ acts as an electrophile because three highly electronegative oxygen atoms are attached to Sulphur atom in SO₃ which makes Sulphur atom electron deficient. It can be shown by resonance. Sulphur gets +ve charge and acts as an electrophile

Q.47. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.

Ans. Structure I of propenal is more stable because structure having more covalent bonds in a resonating structure, has more stability, hi structure II, the terminal carbon has only six electrons which makes it less stable.
CH₂ = CH — CH = O [Structure I] ↔ ⁺CH₂ — CH = CH — O^- [Structure II]

Q.48. By mistake, an alcohol (boiling point 97°C) was mixed with a hydrocarbon (boiling point 68°C). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
Ans. Mixture of alcohol and hydrocarbon can be separated by simple distillation because both the components have large difference in their boiling points. When temperature is raised to the low boiling liquid i.e., hydrocarbon, the vapours consist only hydrocarbon without any contamination of alcohol.

Q.49. Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.

Ans. (B) is more stable because it does not involve charge separation.

IV. Matching Type Question.

In the following questions more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Q.50. Match the type of mixture of compounds in Column I with the technique of separation/purification given in Column II.

Ans. (a →  3); (b →  6); (c→  2); (d→  1); (e →  4); (f →  5)

Q.51. Match the terms mentioned in Column I with the terms in Column II.

Ans. (a →  3); (b →  6); (c→  2); (d→  1); (e →  4); (f →  5)

Q.52. Match Column I with Column II.

Ans. (a → 3); (b → 5); (c→ 1); (d →2); (e → 4)

Q.53. Match the intermediates given in Column I with their probable structure in

Column II.

Ans. (i) → (a); (ii) → (a); (iii) → (b)

Q.54. Match the ions given in Column I with their nature given in Column II.

Column I	Column II
(i)	(a) Stable due to resonance
(ii)	(b) Destabilised due to inductive effect
(iii)	(c) Stabilised by hyperconjugation
(iv)	(d) A secondary carbocation

Ans. (i) → (a); (ii) → (b); (iii) → (b); (iv) → (c, d)

Column I	Column II
(i)	Stabilize due to resonance. Stability is related to resonating structure.
(ii)	n +CF3, F atoms show -I effect and decreases electron density of C. Hence, instability increases and +CF3 is destabilized.
(iii)	In (CH3) C-, CH3groups show +I effect and increases electron density of C  and destabilized (CH3) C-
(iv)	Secondary carbocation is stabilized due to hyperconjugation and +1 effect of 2CH3 groups.

V. Assertion and Reason Type

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Q.55. Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.
(i) Both A and R are correct and R is the correct explanation of A. 
(ii) Both A and R are correct but R is not the correct explanation of A. 
(iii) Both A and R are not correct. 
(iv) A is not correct but R is correct.
Ans. (i) Both A and R are correct and R is the correct explanation of A.
Explanation: The liquids are having sufficient difference in their boiling points. Liquids having different boiling points vaporize at different temperatures. The vapours are cooled and the liquids so formed are collected separately.

Q.56. Assertion (A) : Energy of resonance hybrid is equal to the average  of energies of all canonical forms.
Reason (R) : Resonance hybrid cannot be presented by a single structure.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (iv) A is not correct but R is correct.
Explanation: Canonical structures always have more energy than resonance hybrid. Resonance hybrids are always more stable than any of the canonical structures. The delocalization of electrons lowers the orbitals energy and gives stability.


Q.57. Assertion (A) : Pent- 1- ene and pent- 2- ene are position isomers.
Reason (R) : Position isomers differ in the position of functional group or a substituent.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (i) Both A and R are correct and R is the correct explanation of A.
Explanation: When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers and this phenomenon is termed as position isomerism. Pent-2-ene and pent-l-ene are position isomers because differ in the position of double bond.

Q.58. Assertion (A): All the carbon atoms in H₂C=C=CH₂ are sp² hybridized
Reason (R): In this molecule, all the carbon atoms are attached to each other by double bonds.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (iv) A is not correct but R is correct.
Explanation: Hybridization of C can be find out by counting σ bonds and π bonds present on C atom.

If C has 3σ bonds, it is sp² hybridized. If C has 2σ bonds, it is sp hybridized.

Q.59. Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carious method.
Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (iii) Both A and R are not correct.
Explanation: Sulphur is estimated by Carius method in the form of white precipitate of BaSO₄ on heating with fuming and BaCl₂. If light yellow solid is obtained means impurities are present. It is filtered washed and then dried to get pure BaSO₄.

Q.60. Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.
Reason (R): The colored components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (i) Both A and R are correct and R is the correct explanation of A.
Explanation: In paper chromatography, a special quality paper known as chromatography paper, is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram.

VI. Long Answer Type

Q.61. What is meant by hybridisation? Compound CH₂ = C = CH₂ contains sp or sp2 hybridised carbon atoms. Will it be a planar molecule?
Ans. The atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridization which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape. In CH₂=C=CH₂ (allene) carbon atom land 3 are sp² hybridized as each one has 3σ bonds while carbon atom 2 has 2σ bonds and it is sp hybridized. Allene molecule as a whole is non-planar.

Q.62. Benzoic acid is a organic compound. Its crude sample can be purified by crystallization from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?
Ans. Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution. Benzoic acid crystallizes from mother liquior.

Q.63. Two liquids (A) and (B) can be separated by the method of fractional distillation. The boiling point of liquid (A) is less than boiling point of liquid (B). Which of the liquids do you expect to come out first in the distillate? Explain.
Ans. If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask. The liquid [A] with low boiling point will distill first.


Q.64. You have a mixture of three liquids A, B and C. There is a large difference in the boiling points of A and rest of the two liquids i.e., B and C. Boiling point of liquids B and C are quite close. Liquid A boils at a higher temperature than B and C and boiling point of B is lower than C. How will you separate the components of the mixture? Draw a diagram showing set up of the apparatus for the process.
Ans. Liquids having different boiling points vaporize at different temperatures. The vapours are cooled and then liquids so formed are collected separately. Liquid A can be separated from B and C because of large difference in boiling point. Liquid B and C have boiling points very close to each other and cannot be separated by simple distillation hence separated by fractional distillation. Liquid B distilled first because the order of boiling points of A, B and C are as follows: B < C < A,


Q.65. Draw a diagram of bubble plate type fractionating column. When do we require such type of a column for separating two liquids. Explain the principle involved in the separation of components of a mixture of liquids by using fractionating column. What industrial applications does this process have?
Ans. If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask. Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile component. By the time the vapours reach to the top of the fractionating column, these are rich in more volatile component.


Q.66. A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification. Explain how is it possible?
Ans. Steam distillation is a special type of separation process for temperature sensitive materials like natural organic compounds. Some organic compounds tend to decompose at higher temperature and normal distillation does not suit this purpose. So, steam/water is added to the apparatus and the temperature of the compounds are depressed, evaporating them at lower temperature. Once the distillation is accomplishing, the vapours are condensed and hence there is the separation of the constituents at ease.