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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 10 Haloalkanes and Haloarenes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

PSEB 12th Class Chemistry Guide Haloalkanes and Haloarenes InText Questions and Answers

Question 1.
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH₃)₂CHCH(Cl)CH₃
(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₂I
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(Cl)(C₂H₅)CH₂CH₃
(viii) CH₃CH＝C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH＝CHC(Br)(CH₃)₂
(x) p-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-ClCH₂C₆H₄CH₂C(CH₃)₃
(xii) o-Br-C₆H₄CH(CH₃)CH₂CH₃
Answer:

Question 2.
Give the IUPAC names of the following compounds:
(i) CH₃CH(Cl)CH(Br)CH₃
(ii) CHF₂CBrClF
(iii) ClCH₂C ☰ CCH₂Br
(iv) (CCl₃)₃CCl
(v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃
(vi) (CH₃)₃CCH＝CClC₆H₄I-p
Answer:

Question 3.
Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p -Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-l-iodooctane
(v) 2 -Bromobutane
(vi) 4-terf-Butyl-3-iodoheptane
(vii) l-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2 -ene
Answer:

Question 4.
Which one of the following has the highest dipole moment?
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄
Answer:

1. CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C—Cl bonds cancel each other. Hence, its resultant dipole moment is zero.

2. As shown in the above figure, in CHCl₃, the resultant of dipole moments of two C—Cl bonds is opposed by the resultant of dipole moments of one C—H bond and one C—Cl bond. Since the resultant of one C—H bond and one C—Cl bond dipole moments is smaller than two C—Cl bonds, the opposition is to a small extent. As a result, CHC13 has a small dipole moment of 1.08 D.

3. On the other hand, in case of CH₂Cl₂, the resultant of the dipole moments of two C—Cl bonds is strengthened by the resultant of the dipole moments of two C—H bonds. As a result, CH2C12 has a higher dipole moment of 1.60 D than CHCl₃ i.e., CH₂Cl₂ has the highest dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl₄ < CHCl₃ < CH₂Cl₂

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
CO The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or an alkene.

Since, the hydrocarbon does not react with Cl₂ in the dark, it cannot be an alkene but must be a cycloalkane.
As the cycloalkane reacts with Cl₂ in the presence of bright sunlight, to give a single monochloro compound, C₅H₉Cl, therefore all the ten hydrogen atoms of the cycloalkane must be equivalent. Therefore, the cycloalkane is cyclopentane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Answer:
There are four isomers of the compound having the formula C₄H₉Br.
These isomers are given below:

Question 7.
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-l-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.
For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

Question 9.
Which compound in each of the following pairs will react faster in Sn₂ reaction with OH^–?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
Answer:
(i) Since I^– ion is a better leaving group than Br^– ion, hence CH₃I reacts faster than CH₃Br in SN₂ reaction with OH^– ion.

(ii) On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in S_N 2 reactions. Hence, CH₃Cl will react at a faster rate than (CH₃)₃ CCl in a S_N2 reaction with OH^– ion.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1 -Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2, 2, 3-Trimethyl-3-hromopentane.
Answer:
(i) In 1 -bromo-1 -methylcyclohexane, the β-hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.

(ii) All p-hydrogens in 2-chloro-2-methylbutane are not equivalent, hence on treatment with C₂H₅ONa/C₂H₅OH, it gives two alkenes.

(iii) 2, 2, 3-Trimethyl-3-bromopentane has two different sets of p-hydrogen and therefore, in principle, can give two alkenes (I and II). But according to Saytzeff rule, more highly substituted alkene (II), being more stable is the major product.

Question 11.
How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1 -nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propynt
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 12.
Explain why
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride ?
(ii) Alkyl halides, though polar are immiscible with water ?
(iii) Grignard reagents should be prepared under anhydrous conditions ?
Answer:
(i) Because of greater s-character, an sp²-hybrid carbon is more electronegative than an sp³-hybrid carbon. Thus, the sp²-hybrid carbon of C—Cl bond in chlorobenzene has less tendency to release electrons to Cl than an sp³-hybrid carbon of cyclohexyl chloride.

Hence, the C—Cl bond in chlorobenzene is less polar than that in cyclohexyl chloride. In other words, the magnitude of negative charge is less on Cl atom of chlorobenzene than in cyclohexyl chloride. Now, due to delocalisation of lone pairs of electrons of the Cl atom over the benzene ring, C—Cl bond in chlorobenzene acquires some double character while the C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride.

As dipole moment is a product of charge and distance, chlorobenzene has lower dipole moment than cyclohexyl chloride due to lower magnitude of negative charge on the Cl atom and shorter C—Cl distance.

(ii) Alkyl halides, though polar, are immiscible in water because they are unable to form hydrogen bonds with water molecules.

(iii) Grignard reagents are very reactive. They react with moisture present in the apparatus or the starting materials to give hydrocarbons.

Hence, Grignard reagent must be prepared under anhydrous conditions.

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer:
Uses of Freon-12(CCl₂F₂)

1. It is used as a refrigerant in refrigerators and air conditioners.
2. It is also used in aerosol spray propellants such as body sprays, hair sprays.

Uses of DDT (p, p’-dichlorodiphenyltrichloroethane)

1. It is very effective against mosquitoes and lice.
2. It is also used in many countries as insecticide for sugarcane and fodder crops. (But due to its harmful effects, its use has been banned in many contries including U.S.A.

Uses of Carbontetrachloride (CCl₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cAnswer:
2. It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
3. It is used as a solvent in the manufacture of pharmaceutical products. Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

Uses of Iodoform (CHI₃)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Question 14.
Write the structure of the major organic product in each of the following reactions


(v) C₆H₅ONa + C₂H₆Cl →
(vi) CH₃CH₂CH₂OH + SOCl₂ →

(viii) CH₃CH = C(CH₃)₂ + HBr →
Answer:

Question 15.
Write the mechanism of the following reaction

Answer:
The given reaction is

The given reaction is an S_N2 reaction. In this reaction, CN^– acts as the nucleophile and attacks the carbon atom to which Br is attached. CN^– ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 16.
Arrange the compounds of each set in order of reactivity towards S_N2 displacement
(i) 2-Bromo-2-methylbutane, 1 -Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo- 2-methylbutane
(iii) 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, 1-Bromo -2-methylbutane, 1-Bromo-3-methylbutane.
Answer:

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:

In S_N1 reaction, reactivity depends upon the stability of carbocations. carbocation is more stable as compared to . Therefore, C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CHCl.

Question 18.
p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.
Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.

Question 19.
How the following conversions can be carried out?
(i) Propene to propan-l-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl- 1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In an aqueous solution, KOH almost completely ionises to give OH^– ions. OH^– ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO^–) ion, which is a strong base. Thus, it can abstract a hydrogen from the p carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

OH^– ion is a much weaker base than RO^– ion. Also, OH^– ion is highly solvated in an aqueous solution and as a result, the basic character of OH^– ion decreases. Therefore, it cannot abstract a hydrogen from the β carbon.

Question 21.
Primary alkyl halide C₄H₉Br (A) reacted with alcoholic KOH to give compound (B).Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal it gives compound (D), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.
Answer:
There are two primary alkyl halides having the formula, C₄H₉Br. They are n-butyl bromide and isobutyl bromide.

Therefore, compound (A) is either n-butyl bromide or isobutyl bromide. Now, compound (A) reacts with Na metal to give compound (B) of molecular formula, C₈H₁₈ which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence, compound (A) must be isobutyl bromide.

Question 22.
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with aqueous KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether
(vi) methyl chloride is treated with KCN.
Answer:
(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-l-ene takes place. This reaction is a dehydrohalogenation reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.

Chemistry Guide for Class 12 PSEB Haloalkanes and Haloarenes Textbook Questions and Answers

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-butyl-3-iodoheptane
(iv) 1-4-Dibromobut-2-ene
(v) 1-Bromo-4-sec-butyl-2-methylbenzene
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
(i) ClCH₂CH₂CH₂Cl
(ii) ClCH₂CHClCH₃
(iii) Cl₂CHCH₂CH₃
(iv) CH₃CCl₂CH₃

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂ identify the one that on photochemical chlorination yields :
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides
Answer:

All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

(ii)
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Thus, three isomeric products are possible.

(iii)
The equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Question 6.
Arrange each set of compounds in the order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1 -Chlorobutane.
Answer:
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform. Boiling point increases with increase in molecular mass.

(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane. Isopropyl chloride being branched has lower boiling point than 1-Chloropropane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism ? Explain your answer.

Answer:
(i) CH₃CH₂CH₂CH₂Br
Being primary halide, there won’t be any steric hindrance.

(ii)
Being a secondary halide, there will be less crowding around α-carbon than tertiary halide.

(iii)
The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction ?

Answer:
(i)
2-Chloro-2-methylpropane as the tertiary carbocation is more stable than secondary carbocation.

(ii)
2-Chloroheptane as the secondary carbocation is more stable than primary carbocation.

Question 9.
Identify A, B, C, D, E, R and R’ in the following:

Answer:

Punjab State Board PSEB 5th Class Hindi Book Solutions Chapter 1 जब बोलो (कविता) Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Hindi Chapter 1 जब बोलो (कविता) (2nd Language)

Hindi Guide for Class 5 PSEB जब बोलो Textbook Questions and Answers

जब बोलो (कविता) अभ्यास

नीचे गुरुमुखी और देवनागरी लिपि में दिये गये शब्दों को पढ़ो और हिंदी शब्दों को लिखने का अभ्यास करो-

• भिमती = मिसरी
• वप्ले = बोलो
• मच = सच
• भठ = मन

नीचे एक ही अर्थ के लिए पंजाबी और हिंदी भाषा में शब्द दिये गये हैं। इन्हें ध्यान से पढ़ो और हिंदी शब्दों को लिखो-

• व = झुक
• ताहां = गाँठें

सही शब्द चुनकर वाक्य पूरे करो

1. हमें ………………………….. बोलना चाहिए। (रोकर/ हँसकर)
2. हमारी बातचीत में ………………………….. होनी चाहिए। (मिठास/ कड़वाहट)
3. हमें सदा ………………………….. बोलना चाहिए। (सच/ झूठ)
4. हमें ………………………….. अपनी बात कहनी चाहिए। (जल्दी-जल्दी/ सोच-समझकर)
5. हमें अपनी बात ………………………….. कहनी चाहिए। (अकड़कर/ झुककर)

उत्तर :

1. हँसकर
2. मिठास
3. सच
4. सोच – समझकर
5. झुककर।

तुक मिलाओ

1. जब = …………………………..
2. सच = …………………………..
3. झुक = …………………………..
4. बोलो = …………………………..

उत्तर :

1. जब – तब।
2. सच – सच।
3. झुक – रुक।
4. बोलो – तोलो।

पढ़ो और समझो

• हँस + कर = हँसकर
• झुक + कर = झुककर
• रुक + कर = रुककर
• सोच + समझ + कर = सोच-समझकर

वाक्य बनाओ

1. मिसरी-सी
2. सच-सच
3. रच-रच
4. मन की गाँठें

उत्तर :

1. मिसरी – सी – कोयल की मीठी कूक कानों में मिसरी – सी घोलती है।
2. सच – सच – सच – सच बताओ, कल तुम कहाँ थे ?
3. रच – रच – कभी न बातें रच – रच बोलो।
4. मन की गाँठे – हँस – हँस कर अपने मन की गाँठे खोलो।

रचनात्मक अभिव्यक्ति

चित्र देखकर दिए गए शब्दों की सहायता से वाक्य पूरे करो-
आदर बुरी रोकर मिलजुलकर

1. ऊँची आवाज़ में चीखकर बोलना ………………………….. आदत है।
2. हमें आपस में ………………………….. रहना चाहिए।
3. ………………………….. अपनी बात कहने वाले बच्चे किसी को नहीं भाते।
4. हमें बड़ों का ………………………….. करना चाहिए।

उत्तर :

1. बुरी
2. मिल – जुलकर
3. रोकर
4. आदर।

बहुवैकल्पिक प्रश्न

प्रश्न 1.
‘बोलो’ की तुकबन्दी करते हुए शब्द मिलाएं, सही पर गोला लगाओ।
‘तोलो’ सुनो, बुलाओ, भलो
उत्तर :
तोलो

प्रश्न 2.
‘जब’ की तुकबन्दी करते हुए शब्द मिलाएं, सही पर गोला लगाओ।
तब, आब, साहब, ताब
उत्तर :
तब

प्रश्न 3.
‘सच’ की तुकबन्दी करते हुए शब्द मिलाएँ आज, राज, बच, बचना
उत्तर :
बच।

जब बोलो (कविता) Poems

1. जब बोलो, तब हँस कर बोलो,
बातों में मिसरी-सी घोलो।
जब बोलो, तब सच-सच बोलो,
कभी न बातें रच-रच बोलो।

शब्दार्थ –

मिसरी – सी = मिश्री जैसी, मिठास।
सच – सत्य।
रच – रच कर = बना – बना कर।

सरलार्थ – प्रस्तुत पंक्तियों में कवि बच्चों को वाणी में मधुरता और सत्यनिष्ठा को अपनाने की प्रेरणा देते हुए कहता है कि जब भी बोलो तब हँसहँस कर बात करो। मुस्कुरा कर बातें करनी चाहिएं ताकि सुनने वाले को अच्छा लगे। तुम्हारी बातों में मिठास होनी चाहिए। जब भी किसी से बात करो तो सच – सच बोलो कभी भी बना – बना कर बात नहीं करनी चाहिए।

2. जब बोलो, तब झुक कर बोलो,
सोच समझ कर, रुक कर बोलो।
हँस कर मन की गाँठे खोलो,
जब बोलो, तब हँस कर बोलो।

शब्दार्थ – झुककर = विनम्रता से। रुक कर = धीरे – धीरे, शांति से। हँस कर =मुस्करा कर। मन की गाँठें = मन के भेद।

सरलार्थ प्रस्तुत पंक्तियों में कवि बच्चों को विनम्रता और मधुरता को अपनाने की प्रेरणा देते हुए कहता है कि जब भी बोलो विनम्रता से बोलो, बात को सुन कर, सोच – समझकर और धीरे – धीरे, शांत भाव से बोलो। मन पर पडी दविधा की गाँठों को, परतों को, भेदों को, हँसकर खोलना चाहिए। मधुर वाणी बोलनी चाहिए। जब भी बोलो हँसकर बात करनी चाहिए।

जब बोलो (कविता) शब्दार्थ – Meanings

• मिसरी = चीनी से बनी वस्तु
• मिसरी-सी = मिठास से भरी हुई
• रच-रच बोलना = अपने आप बात बनाकर कहना
• मन की गाँठ = मन में छिपी बात

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 1 Physical World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 1 Physical World

PSEB 11th Class Physics Guide Physical World Textbook Questions and Answers

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
The physical world around us is full of different complex natural phenomena so the world is incomprehensible. But with the help of study and observations it has been found that all these phenomena are based on some basic physical laws and so it is comprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science of the validity of this incisive remark.
Answer:
The above statement is true. Validity of this incisive remark can be validated from the example of moment of inertia. It states that the moment of inertia of a body depends on its energy. But according to Einstein’s mass-energy relation (E = mc²), energy depends on the speed of the body.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
It is well known that to win over votes, politicians would make anything and everything possible even when they are least sure of the same. And in science the various natural phenomena can be explained in terms of some basic laws. So as ‘Politics is the art of possible’ similarly ‘Science is the art of the soluble’.

Question 4.
Though India now has a large base in science and technology, which is fast expanding, it is still a long way from realizing its potential of becoming a world leader in science. Name some important factors, which in your view have hindered the advancement of science in India.
Answer:
Some important factors in our view which have hindered the advancement of science in India are:

• Proper funds are not arranged for the development of research work and laboratories. The labs and scientific instruments are very old and outdated.
• Most of the people in India are uneducated and highly traditional. They don’t understand the importance of science.
• There is no proper employment opportunity for the science educated person in India.
• There are no proper facilities for science education in schools and colleges in India.

Question 5.
No physicist has ever “seen” an electron. Yet, all physicists believe in the existence of electrons. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has “seen” one. How will you refute his argument?
Answer:
No physicist has ever seen an electron but there are practical evidences which prove the presence of electron. Their size is so small, even powerful microscopes find it difficult to measure their sizes. But still its effects could be tested. On the other hand, there is no phenomena which can be explained on the basis of existence of ghosts.

Our senses of sight and hearing are very limited to observe the existence of both.
So, there is no comparison between the two given cases.

Question 6.
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?

(a) A tragic sea accident several centimes ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.

(b) After the sea tragedy, fishermen in that area, in a gesture of honor to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. ConseQuestionuently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.

[Note: This interesting illustration taken from Carl Sagan’s ‘The Cosmos’ highlights the fact that often strange and inexplicable facts which on the first sight appear ‘supernatural’ actually turn out to have simple scientific explanations. Try to think out other examples of this kind].
Answer:
Explanation (b) is correct as it is a scientific explanation of the observed fact.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were these advances?
Answer:
More than two centuries ago, England and Western Europe invented steam engine, electricity, theory of gravitation and the explosives. Steam engines helped them in the field of heat and thermodynamics, theory of gravitation in field of motion and making guns and cannons. These progresses brought about industrial revolution in England and Western Europe.
Few of which are given below:

• Steam engine formed on the application of heat and thermodynamics.
• Discovery of electricity helped in designing dynamos and motors.
• Safety lamp which was used safety in mines.
• Invention of powerloom which used steampower was used for spinning and weaving.

Question 8.
It is often said that the world is witnessing now a second industrial revolution, which will transform the society as
radically as did the first. List some key contemporary areas of science and technology, which are responsible for this
revolution.
Answer:
Some of the key contemporary areas of science and technology which may trAnswer:form the society radically are:

• Development of super fast computers.
• Internet and tremendous advancement in information technology.
• Development in Biotechnology.
• Development of super-conducting materials at room temperature.
• Development of robots.

Question 9.
Write in about 1000* words a fiction piece based on your : speculation on the science and technology of the twenty second century.
Answer:
Let us imagine a spaceship moving towards a distant star, 500 light years away. Let it be propelled by current fed into the electric motor consisting of superconducting wires. In space, suppose there is a particular region which has such a high temperature that destroys the superconducting property of electric wires of the motor. At this stage, another spaceship
filled with matter and anti-matter comes to the rescue of the first ship and it (i. e., 1st ship) continues its onward journey.

Another way to put is: Now matter can be changed into energy and energy into matter. A man of 22nd century stands on a plateform of a specially designed machine which energises him and his body disappears in the form of energy. After split of a second, he appears at a place much far away from the previous one just intact.

Question 10.
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous conseQuestionuences for the human society. How, if at all, will you resolve your dilemma?
Answer:
In our view a type of discovery which is of great academic interest but harmful for human society should not be made public because science is for the society, society is not for science.

Question 11.
Science, like any knowledge, can be put to good or had use, depending on the user.Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized:
(a) Mass vaccination against small pox to curb and finally eradicate this disease from the population. (This has already been successfully done in India).
(b) Television for eradication of illiteracy and for mass communication of news and ideas.
(c) Prenatal sex determination
(d) Computers for increase in work efficiency
(e) Putting artificial satellites into orbits around the Earth
(f) Development of nuclear weapons
(g) Development of new and powerful techniQuestionues of chemical and biological warfare.
(h) Purification of water for drinking
(i) Plastic surgery
(j) Cloning
Answer:
(a) Mass vaccination is good as it is used to make the society free from the diseases like Small Pox.

(b) Television for eradication of illiteracy and for mass communication of news and ideas is good as it is a medium which is easily under the reach of common man and also they are very habitual to it.

(c) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it specially with girl child.

(d) Computers for increase in work efficiency is good as using the computer a man can do much more work with greater efficiency and accuracy as it could do without computers.

(e) Putting artificial satellites into orbits around the Earth is a good development as these satellites serve many purposes like Remote Sensing, Weather Forcasting etc. These informations have very high importance for us as we can plan the things in advance.

(f) Development of nuclear weapons is bad as they can be used in mass destruction.

(g) Development of new and powerful techniQuestionues of chemical and biological warfare are bad as they can also be used for mass destruction.

(h) Purification of water for drinking is good as we can save ourseleves from the diseases which we can have due to drinking the contaminated water.

(i) Plastic surgery is good as with the help of it a man or woman can remove the skin defects occuring due to accidents or some other reasons. It has some bad effects too but they are not very considerable.

(j) Cloning is good as far as animals are concerned with the help of it we can develop some special species which can be used to serve some specific purposes. But it is not good for human beings.

Question 12.
India has had a long and unbroken tradition of great scholarship – in mathematics, astronomy, linguistics, logic and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today – among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
Poverty and illiteracy are the two major factors which make people superstitious in India. So to remove the superstitious and obscurantist attitude we have to first overcome these factors. Everybody should be educated, so that one can have scientific attitude. Knowledge of science can be put to use to prove people’s superstitious wrong by showing them the scientific logic behind everything happening in our world.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
Some people in our society have the view that women do not have the innate nature, capacity and intelligence.
To demolish this view there are many examples of women who have proven their abilities in science and other fields.
Madam Curie, Mother Teresa, Indira Gandhi, Marget Thatcher, Rani Laxmi Bai, Florence Nightingale are some examples. So in this era women are definitely not behind man in any field.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P. A. M. Dirac held this view. Criticize this statement.
‘ Look out for some eQuestionuations and results in this book which strike you as beautiful.
Solution:
An equation which agrees with experiment must also be simple and hence beautiful. We have some simple and beautiful equations in physics such as
E = mc² (Energy of light) .
E = hv (Energy of a photon)
KE = 1 / 2 mv ² (Kinetic energy of a moving particle)
PE = mgh (Potential energy of a body at rest)
W = F. d (Work done)
All have the same dimensions. One experiment shows dependency of energy on speed, the other shows dependency on frequency and displacement.
That’s the beauty of equations in physics coming from different experiments.

Question. 15.
Though the statement quoted above may be disputed, most physicists do have a feeling that the great laws of physics are at once simple and beautiful. Some of the notable physicists, besides Dirac, who have articulated this feeling are Einstein, Bohr, Heisenberg, Chandrasekhar and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics (See the Bibliography at the end of this book). Their writings are truly inspiring!
Solution:
There is no doubt that great laws of physics are at once so simple and beautiful and are easy to grasp. For example, let us look at some of these:

• E = mc² is a famous Einstein’s mass energy equivalence relation which has a great impact not only on the various physical phenomena but also on the human lives.
• Plank’s Questionuantum condition i.e., E = hv is also a simple and beautiful eQuestionuation and it is a great law of physics.
• ∆x. ∆p ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) or ∆ E. A t ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) is Heisenberg’s.

Uncertainly Principle which is also very simple, beautiful and interesting. It is a direct conseQuestionuence of the dual nature of matter.

• λ = \(\frac{h}{m v}\) is also a famous eQuestionuation in physics known as de-Broglie euation. It is again simple and beautiful.

Question 16.
Textbooks on science may give you a wrong impression that studying science is dry and all too serious and that scientists are absent minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists like any other group of humans have their share of humorists, and many have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
Answer:
It is not an exercise as such but is a statement of facts. We can add the names of other scientists who were humorists along with being physicists. They are C.V. Raman, Homi Jahangir Bhabha, Einstein and Bohr. India have several politiciAnswer: like M.M. Joshi, V.P. Singh etc. who are physicists. President A.P.J. Kalafn is also a great nuclear scientist.

Punjab State Board PSEB 5th Class Punjabi Book Solutions Chapter 1 ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Punjabi Chapter 1 ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ (1st Language)

ਪਾਠ-ਅਭਿਆਸ ਪ੍ਰਸ਼ਨ-ਉੱਤਰ

I. ਯਾਦ ਰੱਖਣ ਯੋਗ ਗੱਲਾਂ

ਪ੍ਰਸ਼ਨ 1.
“ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ਕਵਿਤਾ ਨੂੰ ਪੜ੍ਹ ਕੇ ਕਿਹੜੀਆਂ ਚਾਰ-ਪੰਜ ਗੱਲਾਂ ਤੁਹਾਨੂੰ ਯਾਦ ਰੱਖਣ ਯੋਗ ਲੱਗੀਆਂ ਹਨ, ਉਨ੍ਹਾਂ ਨੂੰ ਲਿਖੋ ।
ਉੱਤਰ:

1. ਸਾਡੇ ਦੇਸ਼ ਦਾ ਨਾਂ ਹਿੰਦੁਸਤਾਨ ਹੈ, ੲਸ ਨੂੰ ‘ਭਾਰਤ’ ਵੀ ਕਿਹਾ ਜਾਂਦਾ ਹੈ ।
   ਭਾਰਤ ਦੀ ਰਾਜਧਾਨੀ ਦਿੱਲੀ ਹੈ ।
   ਭਾਰਤ ਵਿਚ ਕੁੱਲ 28 ਦੇਸ਼ ਹਨ ।
   ਭਾਰਤ ਦਾ ਰਾਸ਼ਟਰੀ ਝੰਡਾ ਤਿਰੰਗਾ ਹੈ, ਜਿਸ ਦੇ ਤਿੰਨ ਰੰਗ-ਕੇਸਰੀ, ਸਫ਼ੈਦ, ਹਰਾ-ਹਨ ।

II. ਜ਼ਬਾਨੀ.ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਦੇਸਾਂ-ਪਰਦੇਸਾਂ ਵਿਚ ਹਿੰਦੁਸਤਾਨ ਦੀ ਸ਼ਾਨ ਕਿਹੋ-ਜਿਹੀ ਹੈ ?
ਉੱਤਰ:
ਉੱਚੀ ।

ਪ੍ਰਸ਼ਨ 2.
ਹਿੰਦੁਸਤਾਨ ਦੇ ਚਸ਼ਮਿਆਂ ਵਿਚੋਂ ਫੁੱਟਦਾ ਪਾਣੀ ਕਿਹੋ-ਜਿਹਾ ਲਗਦਾ ਹੈ ?
ਉੱਤਰ:
ਚਾਂਦੀ ਰੰਗਾ ।

ਪ੍ਰਸ਼ਨ 3.
ਸਾਨੂੰ ਹਰ ਨਵੀਂ ਸਵੇਰ ਨੂੰ ਕੀ ਵੰਡਣਾ ਚਾਹੀਦਾ ਹੈ ?
ਉੱਤਰ:
ਫੁੱਲਾਂ ਜਿਹੀ ਮੁਸਕਾਨ ।

ਪ੍ਰਸ਼ਨ 4.
ਕਵਿਤਾ ਨੂੰ ਲੈ ਵਿਚ ਗਾਓ ।
ਉੱਤਰ:
(ਨੋਟ-ਵਿਦਿਆਰਥੀ ਆਪ ਹੀ ਗਾਉਣ ਦਾ ਅਭਿਆਸ ਕਰਨ ।).

III. ਸੰਖੇਪ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਅਸੀਂ ਹਿੰਦੁਸਤਾਨ ਦੀ ਮਿੱਟੀ ਨੂੰ ਸੀਸ ਨੇ ਕਿਉਂ ਨਿਵਾਉਂਦੇ ਹਾਂ ?
ਉੱਤਰ:
ਸਿਆਣੇ ਕਹਿੰਦੇ ਹਨ ਕਿ ਆਪਣੇ ਦੇਸ਼ ਦੀ ਮਿੱਟੀ ਮਾਂ ਸਮਾਨ ਹੁੰਦੀ ਹੈ । ੲਸ ਕਰਕੇ ਅਸੀਂ ੲਸ ਨੂੰ ਸੀਸ ਨਿਵਾਉਂਦੇ ਹਾਂ ।

ਪ੍ਰਸ਼ਨ 2.
ਸਾਨੂੰ ਕਿਸ ਪ੍ਰਕਾਰ ਦੀ ਕਮਾੲ ਕਰਨੀ ਚਾਹੀਦੀ ਹੈ ?
ਉੱਤਰ:
ਸਾਨੂੰ ਹੱਕ-ਹਲਾਲ ਦੀ ਕਮਾੲ ਕਰਨੀ ਚਾਹੀਦੀ ਹੈ ।

IV. ਬਹੁਤ ਛੋਟੇ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਤੁਹਾਡੀ ਪਾਠ-ਪੁਸਤਕ ਵਿਚ ਪ੍ਰੋ: ਜੋਗਾ ਸਿੰਘ ਦੀ ਲਿਖੀ ਹੋੲ ਕਿਹੜੀ ਕਵਿਤਾ ਸ਼ਾਮਿਲ ਹੈ ?
ਉੱਤਰ:
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ।

ਪ੍ਰਸ਼ਨ 2.
‘ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ਕਵਿਤਾ ਵਿਚ ਕਵੀ ਕਿਸ ਦੀ ਪ੍ਰਸੰਸਾ ਕਰਦਾ ਹੈ ?
ਉੱਤਰ:
ਆਪਣੇ ਦੇਸ਼ ਹਿੰਦੁਸਤਾਨ ਦੀ ॥

ਪ੍ਰਸ਼ਨ 3.
“ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ਕਵਿਤਾ ਵਿਚ ਕਵੀ ਕਿਸ ਲੲ ਪਿਆਰ ਤੇ ਸਤਿਕਾਰ ਪ੍ਰਗਟ ਕਰਦਾ ਹੈ ?
ਉੱਤਰ:
ਆਪਣੇ ਦੇਸ਼ ਹਿੰਦੁਸਤਾਨ ਲੲ ।

ਪ੍ਰਸ਼ਨ 4.
ਤੁਹਾਡੀ ਪਾਠ-ਪੁਸਤਕ ਵਿਚ ਦੇਸ਼-ਪਿਆਰ ਦੀ ਕਵਿਤਾ ਕਿਹੜੀ ਹੈ ?
ਉੱਤਰ:
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ।

V. ਬਹੁਵਿਕਲਪੀਵਸਤੁਨਿਸ਼ਠ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
“ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ , ਕਵਿਤਾ ਕਿਸ ਕਵੀ ਦੀ ਲਿਖੀ ਹੋੲ ਹੈ ?
(ਉ) ਧਨੀ ਰਾਮ ਚਾਤ੍ਰਿਕ
(ਅ) ਪ੍ਰੋ ਜੋਗਾ ਸਿੰਘ
(ੲ) ਡਾ: ਹਰੀ ਸਿੰਘ ਜਾਚਨ .
(ਸ) ਸਵਰਨ ਹੁਸ਼ਿਆਰਪੁਰੀ ।
ਉੱਤਰ:
(ਅ) ਪ੍ਰੋ: ਜੋਗਾ ਸਿੰਘ

ਪ੍ਰਸ਼ਨ 2.
“ਹਿੰਦੁਸਤਾਨ ਦਾ ਦੂਸਰਾ ਨਾਂ ਕੀ ਹੈ ?
(ਉ) ਪੰਜਾਬ
(ਅ) ਭਾਰਤ
(ੲ) ਸ੍ਰੀ ਲੰਕਾ
(ਸ) ਹਰਿਆਣਾ ।
ਉੱਤਰ:
(ਅ) ਭਾਰਤ ।

ਪ੍ਰਸ਼ਨ 3.
ਭਾਰਤ ਵਿਚ ਕੁੱਲ ਕਿੰਨੇ ਰਾਜ ਹਨ ?
(ਉ) 30
(ਅ) 28
(ੲ) 25
(ਸ) 35.
ਉੱਤਰ:
(ਅ) 28.

ਪ੍ਰਸ਼ਨ 4.
ਭਾਰਤ ਦੀ ਰਾਜਧਾਨੀ ਕਿਹੜੀ ਹੈ ?
(ਉ) ਨਵੀਂ ਦਿੱਲੀ
(ਅ) ਚੰਡੀਗੜ੍ਹ
(ੲ) ਮੁੰਬੲ .
(ਸ) ਕੋਲਕਾਤਾ
ਉੱਤਰ:
(ੳ) ਨਵੀਂ ਦਿੱਲੀ

ਪ੍ਰਸ਼ਨ 5.
ਭਾਰਤ ਦੇ ਝੰਡੇ ਦਾ ਕੀ ਨਾਂ ਹੈ ?
(ਉ) ਤਿਰੰਗਾ
(ਅ) ਯੂਨੀਅਨ ਜੈਕ
(ਬ) ਮੋਰਪੰਖ
(ਸ) ਵਿਕਾਸ ਚਿੰਨ੍ਹ
ਉੱਤਰ:
(ੳ) ਤਿਰੰਗਾ

ਪ੍ਰਸ਼ਨ 6.
ਭਾਰਤ ਦੇ ਝੰਡੇ ਵਿਚ ਕਿੰਨੇ ਰੰਗ ਹਨ ?
(ਉ) ਦੋ
(ਅ) ਤਿੰਨ
(ੲ) ਚਾਰ
(ਸ) ਪੰਜ
ਉੱਤਰ:
(ਅ) ਤਿੰਨ

ਪ੍ਰਸ਼ਨ 7.
ਭਾਰਤ ਦੇ ਪਰਬਤ ਕਿਹੋ ਜਿਹੇ ਹਨ ?
(ੳ) ਉੱਚੇ ਤੇ ਬਰਫ਼ਾਂ ਲੱਦੇ ,
(ਅ) ਉੱਚੇ-ਨੀਵੇਂ
ੲ) ਖੁਸ਼ਕ
(ਸ) ਸੋਨ-ਸੁਨਹਿਰੀ ।
ਉੱਤਰ:
(ੳ) ਉੱਚੇ ਤੇ ਬਰਫ਼ਾਂ ਲੱਦੇ ।

ਪ੍ਰਸ਼ਨ 8.
ਭਾਰਤ ਦੇ ਚਸ਼ਮਿਆਂ ਵਿਚੋਂ ਕਿਹੋ ਜਿਹਾ ਪਾਣੀ ਫੁੱਟਦਾ ਹੈ ?
(ਉ) ਸੋਨ-ਸੁਨਹਿਰੀ
(ਅ) ਚਾਂਦੀ ਰੰਗਾ
(ੲ) ਸ਼ੀਸ਼ੇ ਵਰਗਾ
(ਸ) ਰੋਗ ਨਿਵਾਰਨ ।
ਉੱਤਰ:
ਆ ਚਾਂਦੀ-ਰੰਗਾ ।

ਪ੍ਰਸ਼ਨ 9.
ਭਾਰਤ ਦੀ ਮਿੱਟੀ ਵਿਚੋਂ ਕਿਹੋ ਜਿਹੇ ਦਾਣੇ ਉੱਗਦੇ ਹਨ ?
(ਉ) ਸੁਨਹਿਰੀ
(ਅ) ਰੁਪਹਿਰੀ
(ੲ) ਮੋਤੀਆਂ ਵਰਗੇ
(ਸ) ਰਤਨਾਂ ਵਰਗੇ ।
ਉੱਤਰ:
(ੲ) ਮੋਤੀਆਂ ਵਰਗੇ ।

ਪ੍ਰਸ਼ਨ 10.
ਸਿਆਣੇ ਲੋਕਾਂ ਅਨੁਸਾਰ ਮਿੱਟੀ ਕੀ ਹੈ ?
(ਉ) ਭੈਣ
(ਅ) ਮਾਂ
(ੲ) ਚਾਚੀ
(ਸ) ਤਾੲ ।
ਉੱਤਰ:
(ਅ) ਮਾਂ ।

ਪ੍ਰਸ਼ਨ 11.
ਸਿਆਣੇ ਲੋਕਾਂ ਨੇ ਮਾਂ ਕਿਸ ਨੂੰ ਕਿਹਾ ਹੈ ?
(ਉ) ਦੇਸ਼ ਦੀ ਮਿੱਟੀ ਨੂੰ
(ਅ) ਦੇਸ਼ ਦੀ ਪੂੰਜੀ ਨੂੰ
(ੲ) ਦੇਸ਼ ਦੀ ਹਵਾ ਨੂੰ
(ਸ) ਦੇਸ਼ ਦੀ ਬਨਸਪਤੀ ਨੂੰ ।
ਉੱਤਰ:
(ਉ) ਦੇਸ਼ ਦੀ ਮਿੱਟੀ ਨੂੰ ।

ਪ੍ਰਸ਼ਨ 12.
ਰਿਸ਼ੀਆਂ-ਮੁਨੀਆਂ, ਗੁਰੂਆਂ-ਪੀਰਾਂ ਨੇ ਕਿਸ ਦੀ ਸ਼ਾਨ ਵਧਾੲ ਹੈ ?
(ੳ) ਭਾਰਤ ਦੀ
(ਅ) ਹਰਿਆਣੇ ਦੀ
(ੲ) ਨੇਪਾਲ ਦੀ
(ਸ) ਬੰਗਾਲ ਦੀ ।
ਉੱਤਰ:
(ੳ) ਭਾਰਤ ਦੀ ।

ਪ੍ਰਸ਼ਨ 13.
ਕਿਨ੍ਹਾਂ ਨੇ ਦੁਨੀਆ ਵਿਚ ਭਾਰਤ ਦੀ ਧਾਂਕ ਜਮਾੲ ਹੈ ?
(ਉ) ਪੀਰਾਂ ਨੇ
(ਅ) ਪ੍ਰੇਮੀਆਂ ਨੇ
(ੲ) ਜੋਧਿਆਂ ਨੇ
(ਸ) ਭਗਤਾਂ ਨੇ ।
ਉੱਤਰ:
(ੲ) ਜੋਧਿਆਂ ਨੇ ।

ਪ੍ਰਸ਼ਨ 14.
ਭਾਰਤ ਦੀ ਕਿਹੜੀ ਚੀਜ਼ ਨੂੰ ਦੁਨੀਆ ਮੰਨਦੀ ਹੈ ?
(ਉ) ਵਿੱਦਿਆ ਨੂੰ
(ਅ) ਭਗਤੀ ਨੂੰ
(ੲ) ਸ਼ਕਤੀ ਨੂੰ
(ਸ) ਜੋਤਸ਼ ਨੂੰ ।
ਉੱਤਰ:
(ੳ) ਵਿੱਦਿਆ ਨੂੰ ।

ਪ੍ਰਸ਼ਨ 15.
ਭਾਰਤ ਦੇ ਹਾਲੀ, ਪਾਲੀ ਤੇ ਮਜ਼ਦੂਰ ਕਿਹੋ ਜਿਹੀ ਕਮਾੲ ਕਰਦੇ ਹਨ ?
(ਉ) ਰੋਟੀ ਜੋਗੀ
(ਅ) ਹੱਕ-ਸੱਚ ਦੀ
(ੲ) ਹਰਾਮ ਦੀ
(ਸ) ਥੋੜ੍ਹੀ-ਬਹੁਤ ।
ਉੱਤਰ:
(ਅ) ਹੱਕ-ਸੱਚ ਦੀ ।

ਪ੍ਰਸ਼ਨ 16.
ਭਾਰਤ ਦੇ ਹਾਲੀ, ਪਾਲੀ ਤੇ ਮਜ਼ਦੂਰ ਕਿਸ ਚੀਜ਼ ਤੋਂ ਦੂਰ ਰਹਿੰਦੇ ਹਨ ?
(ਉ) ਮਿਹਨਤ ਤੋਂ
(ਅ) ਲਾਲਚ ਤੋਂ
(ੲ) ਕੂੜ ਤੋਂ
(ਸ) ਸੱਚ ਤੋਂ ।
ਉੱਤਰ:
(ੲ) ਕੁੜ ਤੋਂ।

ਪ੍ਰਸ਼ਨ 17.
ਭਾਰਤ ਦਾ ਹਰ ਗੱਭਰੂ ਕਿਹੋ ਜਿਹਾ ਹੈ ?
(ਉ) ਬਾਂਕਾ
(ਅ) ਸ਼ੁਕੀਨ
(ੲ) ਲਾਲਚੀ
(ਸ) ਕੰਜੂਸ ।
ਉੱਤਰ:
(ੳ) ਬਾਂਕਾ ।

ਪ੍ਰਸ਼ਨ 18.
ਭਾਰਤ ਦੀ ਹਰ ੲਕ ਮੁਟਿਆਰ ਕਿਹੋ ਜਿਹੀ ਹੈ ?
(ੳ) ਰਕਾਨ
(ਅ) ਜਾਨ
(ੲ) ਮਹਾਨ
(ਸ) ਬੇਨਾਮ ।
ਉੱਤਰ:
(ੳ) ਰਕਾਨ ।

ਪ੍ਰਸ਼ਨ 19.
ਹੇਠ ਲਿਖਿਆਂ ਵਿਚੋਂ ਕਵਿਤਾ ਕਿਹੜੀ ਹੈ ?
(ੳ) ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ
(ਅ) ਗੱਤਕਾ
(ੲ) ਫੁਲਕਾਰੀ ਕਲਾ
(ਸ) ਕਹੀ ਹੱਸ ਪੲ ।
ਉੱਤਰ:
(ੳ) ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ।

ਪ੍ਰਸ਼ਨ 20.
“ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ਕਵਿਤਾ ਕਿਹੜੇ ਛੰਦ ਵਿਚ ਲਿਖੀ ਗੲ ਹੈ ?
(ਉ) ਕਬਿੱਤ
(ਆ) ਦੋਹਰਾ
(ੲ) ਸੋਰਠਾ
(ਸ) ਦਵੱੲਆ ।
ਉੱਤਰ:
(ਸ) ਦਵੱੲਆ ।

(ਨੋਟ-ਪੰਜਵੀਂ ਦੇ ਵਿਦਿਆਰਥੀਆਂ ਲੲ . ਕਵਿਤਾ ਦੇ ਛੰਦਾਂ ਬਾਰੇ ਸਮਝਣਾ ਮੁਸ਼ਕਿਲ ਹੋਵੇਗਾ । ੲਸ ਕਰਕੇ ਉਹ ੲਸ ਪੁਸਤਕ ਦੀ ਹਰ ਕਵਿਤਾ ਦੇ ਛੰਦ ਦਾ ਨਾਂ ਯਾਦ ਕਰ ਲੈਣ ਜੋ ਕਿ ਹੇਠ ਲਿਖੇ ਅਨੁਸਾਰ ਹਨ ।)

ਕਵਿਤਾ – ਛੰਦ ,
1. ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ – ਦਵੱੲਆ
2. ਬਾਰਾਂਮਾਹ – ਚੌਪੲ
3. ਆਉ ਰਲ-ਮਿਲ ਰੁੱਖ ਲਗਾੲਏ – ਚੌਪੲ
4. ਚਿੜੀਆ ਘਰ – ਕੋਰੜਾ
5. ਬੋਲੀ ਹੈ ਪੰਜਾਬੀ ਸਾਡੀ – ਕਬਿਤ
6. ਸੱਚੀ ਮਿੱਤਰਤਾ – ਦਵੱੲਆ
7. ਦਾਦੀ ਦੀ ਪੋਤਿਆਂ ਨੂੰ ਨਸੀਹਤ -. ਦਵੱੲਆ
8. ਹਿੰਦ-ਵਾਸੀਆਂ ਨੂੰ ਅੰਤਿਮ ਸੰਦੇਸ਼ – ਬੈਂਤ .
9. ਸਾਰਾਗੜ੍ਹੀ ਦੀ ਲੜਾੲ – ਬੈਂਤ ।

ਪ੍ਰਸ਼ਨ 21.
ਸਤਰ ਪੂਰੀ ਕਰੋ :-‘ ‘
ਉੱਚੇ ਪਰਬਤ ਬਰਫ਼ਾਂ ਲੱਦੇ, ਖੜੇ ਜਿਉਂ ਬੰਨ੍ਹ ਕੇ ਢਾਣੀ ।
ਇਸ ਦੇ ਚਸ਼ਮਿਆਂ ਵਿਚੋਂ ਫੁੱਟਦਾ …………… ।
(ਉ) ਗੰਦਾ-ਮੰਦਾ ਪਾਣੀ
(ਅ) ਸੋਹਣਾ ਸੁਥਰਾ ਪਾਣੀ
(ੲ) ਚਾਂਦੀ ਰੰਗਾ ਪਾਣੀ
(ਸ) ਚਮਕਾਂ ਮਾਰਦਾ ਪਾਣੀ ।
ਉੱਤਰ:
(ੲ) ‘ਚਾਂਦੀ ਰੰਗਾ ਪਾਣੀ ।

ਪ੍ਰਸ਼ਨ 22.
ਸਤਰ ਪੂਰੀ ਕਰੋ :
ੲਸਦੀ ਮਿੱਟੀ ਵਿਚ ਉੱਗਦੇ ਨੇ, ਮੋਤੀਆਂ ਵਰਗੇ ਦਾਣੇ ॥
ਇਹ ਮਿੱਟੀ ਤਾਂ ਮਾਂ ਹੁੰਦੀ ਹੈ ……………
(ੳ) ਆਖਣ ਲੋਕ ਸਿਆਣੇ
(ਅ) ਆਖਣ ਸਭ ਨਿਆਣੇ
(ੲ) ਆਖਣ ਲੁੱਟ ਪੁੱਟ ਜਾਣੇ
(ਸ) ਆਖ਼ਰ ਮਰ ਖਪ ਜਾਣੇ ।
ਉੱਤਰ:
(ੳ) ਆਖਣ ਲੋਕ ਸਿਆਣੇ

ਪ੍ਰਸ਼ਨ 23.
ਸਤਰ ਪੂਰੀ ਕਰੋ :
ੲਕ ਬਾਗ਼ ਵਿਚ ਅਸੀਂ ਹਾਂ ਉੱਗੇ, ਬੂਟੇ ਕੲ ਤਰ੍ਹਾਂ ਦੇ ।
……………. ਪਾਣੀ ਜਿਵੇਂ ਸਰਾਂ ਦੇ ।
(ਉ) ਪਰ ਆਪਸ ਵਿਚ ਲੀਰੋ ਲੀਰ ਹਾਂ
(ਅ) ਪਰ ਆਪਸ ਵਿਚ ਘੁਲੇ-ਮਿਲੇ ਹਾਂ
(ੲ) ਪਰ ਆਪਸ ਵਿਚ ਲੜਦੇ ਰਹਿੰਦੇ
(ਸ) ਪਰ ਆਪਸ ਵਿਚ ਪਿਆਰ ਨਾ ਰੱਖੀਏ ।
ਉੱਤਰ:
(ਅ) ਪਰ ਆਪਸ ਵਿਚ ਘੁਲੇ-ਮਿਲੇ ਹਾਂ ।

ਪ੍ਰਸ਼ਨ: 24.
ਦਿੱਤੇ ਤੁਕਾਂਤਾਂ ਤੋਂ ਕਾਵਿ-ਸਤਰਾਂ ਬਣਾਓ :
…………………… ਢਾਣੀ ।
………………….. ਬਾਣੀ ।
ਉੱਤਰ:
ਉੱਚੇ ਪਰਬਤ ਬਰਫ਼ਾਂ ਲੱਦੇ ਖੜੇ ਜਿਉਂ ਬੰਨ ਕੇ ਢਾਣੀ ।
ਇਸ ਦੇ ਚਸ਼ਮਿਆਂ ਵਿੱਚੋਂ ਫੁੱਟਦਾ ਚਾਂਦੀ ਰੰਗਾ ਪਾਣੀ ।
(ਨੋਟ – ਪ੍ਰਸ਼ਨ 20, 21, 22 ਤੇ 23 ਵਰਗੇ ਪ੍ਰਸ਼ਨਾਂ ਦੇ ਉੱਤਰ ਲੲ ਦੇਖੋ ੲਸ ਪੁਸਤਕ . ਵਿਚ ਅਗਲੇ ਸਫ਼ੇ ॥).

VI. ਵਿਆਕਰਨ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
“ਦੇਸ ਦਾ ਪਰਦੇਸ ਨਾਲ ਜੋ ਸੰਬੰਧ ਹੈ, ਨੂੰ ਉਸੇ ਤਰ੍ਹਾਂ “ਉੱਚੇ ਦਾ ਸੰਬੰਧ ਕਿਸ ਨਾਲ ਹੋਵੇਗਾ ?
(ੳ) ਸੁੱਚੇ
(ਅ) ਭੀੜੇ .
(ੲ) ਨੀਵੇਂ
(ਸ) ਮਾੜੇ ।
ਉੱਤਰ:
(ੲ) ਨੀਵੇਂ ।

ਪ੍ਰਸ਼ਨ 2.
‘ਹਰੇ-ਭਰੇ’ ਨਾਲ ‘ਸੁੱਕੇ-ਸੜੇ ਦਾ ਜੋ ਸੰਬੰਧ ਹੈ, ਉਸੇ ਤਰ੍ਹਾਂ ‘ਮੈਦਾਨ ਦਾ ਸੰਬੰਧ ਕਿਸ ਨਾਲ ਹੋਵੇਗਾ ?
(ਉ) ਟੋਆ-ਟਿੱਬਾ/ਪਰਬਤ
(ਅ ਪੱਧਰ
(ੲ) ਚਰਾਗਾਹ
(ਸ) ਖੇਤ-
ਉੱਤਰ:
(ੳ) ਟੋਆ-ਟਿੱਬਾ/ਪਰਬਤ ।

ਪ੍ਰਸ਼ਨ 3.
ਜੇਕਰ, “ਮਾਂ ਦਾ ਵਿਰੋਧੀ ਸ਼ਬਦ “ਬਾਪ ਹੈ, ਤਾਂ “ਸਿਆਣੇ ਦਾ ਵਿਰੋਧੀ ਕੀ ਹੋਵੇਗਾ ?
(ਉ) ਨਿਆਣੇ/ਮੂਰਖ
(ਅ) ਸਮਝਦਾਰ
(ੲ) ਬੁੱਧੀਹੀਨ
(ਸ) ਬਿਗਾਨੇ ।
ਉੱਤਰ:
(ੳ) ਨਿਆਣੇ/ਮੂਰਖ ।
(ਨੋਟ-ਅਜਿਹੇ ਪ੍ਰਸ਼ਨਾਂ ਦੇ ਉੱਤਰ ਲੲ ਹੇਠ ਲਿਖੇ ਵਿਰੋਧੀ (ਉਲਟੇ ਅਰਥਾਂ ਵਾਲੇ ਸ਼ਬਦ ਯਾਦ ਕਰੋ ॥)

ਵਿਰੋਧੀ ਸ਼ਬਦ

ਪਿਆਰਾ – ਦੁਪਿਆਰਾ
ਸੀਸ – ਚਰਨ
ਵਰਦਾਨ – ਸਰਾਪ
ਘੁਲੇ-ਮਿਲੇ – ਲੀਰੋ-ਲੀਰ/ਵੱਖ-ਵੱਖ
ਵਧਾੲ – ਘਟਾੲ
ਰਿਸ਼ੀ-ਮੁਨੀ – ਚੋਰ-ਉਚੱਕੇ
ਕਾਮੇ – ਵਿਹਲੜ
ਕਰਮੇ – ਮੁਟਿਆਰ
ਗਭਰੂ – ਝੂਠ/ਕੂੜ
ਹੱਕ – ਨਾ ਹੱਕ
(ਨੋਟ-ਅਜਿਹੇ ਪ੍ਰਸ਼ਨਾਂ ਦੇ ਉੱਤਰ ਲੲ ਅਗਲੇ ਸਫ਼ਿਆਂ ਵਿਚ ਦਿੱਤੇ ਵਿਰੋਧੀ ਸ਼ਬਦ ਯਾਦ ਕਰੋ )

ਪ੍ਰਸ਼ਨ 4.
ਕਿਹੜਾ ਸ਼ਬਦ-ਜੋੜ ਸਹੀ ਹੈ ?
(i)
(ੳ) ਪਿਯਾਰਾ
(ਅ) ਪੇਆਰਾ
(ੲ) ਪੇਯਾਰਾ
(ਸ) ਪਿਆਰਾ ।
ਉੱਤਰ:
ਪੇਆਰਾ

(ii)
(ਉ) ਪਰਦੇਸ
(ਅ) ਪ੍ਰਦੇਸ਼
(ੲ) ਪਰਦੇਸ਼
(ਸ) ਪਰਦੇਸ਼ ।
ਉੱਤਰ:
ਪਰਦੇਸ਼

(iii)
(ਉ) ਪਰਬਤ
(ਅ) ਪ੍ਰਬਤ
(ੲ) ਪਰਵਤ
(ਸ) ਪ੍ਰਵਤ ॥
ਉੱਤਰ:
ਪਰਬਤ

(iv)
(ੳ) ਬੰਨ
(ਆ) ਬਨੁ
(ੲ ਬਨ੍ਹ
(ਸ) ਬੰਹ ।
ਉੱਤਰ:
ਬੰਨ੍ਹ

(v)
(ੳ) ਮੈਦਾਨ ‘ਤੇ
(ਆ) ਮਦਾਨ
(ੲ) ਮਦਾਣ
(ਸ) ਮੈਦਾਣ ।
ਉੱਤਰ:
ਮੈਦਾਨ,

(vi)
(ੳ) ਸਿਆਨੇ
(ਅ) ਸਿਆਣੇ
(ੲ) ਸਯਾਨੇ
(ਸ) ਸਿਯਾਨੇ ।
ਉੱਤਰ:
ਸਿਆਣੇ,

(vii)
(ੳ) ਨਿਵਾੲਏ
(ਅ) ਨਿਬਾੲਏ
(ੲ) ਨਵਾੲਏ
(ਸ) ਨਬਾੲਏ ।
ਉੱਤਰ:
ਨਿਵਾੲਏ,

(viii)
(ੳ) ਵਿਦਿਆ
(ਅ) ਵਿੱਦਿਆ
(ੲ) ਬਿਦਿਆ
(ਸ) ਬਿੱਦਿਆ
ਉੱਤਰ:
ਵਿੱਦਿਆ,

(ix)
(ੳ) ਰਹਿਣ
(ਅ) ਰੈਹਣ
(ੲ) ਰੈਣ
(ਸ) ਰੈਹਨ ।
ਉੱਤਰ:
ਰਹਿਣ

(x)
(ੳ) ਗਭਰੂ
(ਅ) ਗੱਭਰੂ
(ੲ) ਗੱਬਰੂ
(ਸ) ਗਭਰੂ ।
ਉੱਤਰ:
ਗੱਭਰੂ ।

ਪ੍ਰਸ਼ਨ 5.
ਹੇਠ ਲਿਖਿਆਂ ਵਿਚੋਂ ਸ਼ਬਦ-ਕੋਸ਼ ਅਨੁਸਾਰ ਕਿਹੜਾ ਸ਼ਬਦ ਪਹਿਲਾਂ ਆਵੇਗਾ ?
(ੳ) ਸਵੇਰ
(ਅ) ਸੀਸ
(ੲ) ਸਭ
(ਸ) ਸਿਆਣੇ ॥
ਉੱਤਰ:
(ੲ) ਸਭ
(ਨੋਟ – ਅਜਿਹੇ ਪ੍ਰਸ਼ਨਾਂ ਦੇ ਉੱਤਰ ਲੲ ਪੜੋ ਅਗਲੇ ਸਫ਼ਿਆਂ ਵਿਚ ‘ਕੋਸ਼ਕਾਰੀ” ਸਿਰਲੇਖ ਹੇਠੇ ਦਿੱਤੀ ਜਾਣਕਾਰੀ ॥)

ਪ੍ਰਸ਼ਨ 6.
‘ਸਭ ਦੇਸਾਂ-ਪਰਦੇਸਾਂ ਦੇ ਵਿਚ, ਉੱਚੀ ੲਸ ਦੀ ਸ਼ਾਨ । ੲਸ ਤੁਕ ਵਿਚ ਕਿਹੜਾ ਸ਼ਬਦ ਪੜਨਾਂਵ ਹੈ ?
(ਉ) ਸਭ
(ਆ) ਦੇਸ਼ਾਂ
(ੲ) ਸ਼ਾਨ
(ਸ) ਇਸ ।
ਉੱਤਰ:
(ਸ) ਇਸ ।
(ਨੋਟ-ਅਜਿਹੇ ਪ੍ਰਸ਼ਨਾਂ ਦੇ ਉੱਤਰ ਲੲ ੲਸ ਪੁਸਤਕ ਦੇ ਅਖ਼ੀਰਲੇ ਸਫ਼ਿਆਂ ਵਿਚ ਦਿੱਤੀ । ਨਾਂਵ, ਪੜਨਾਂਵ, ਵਿਸ਼ੇਸ਼ਣ ਤੇ ਕਿਰਿਆ ਸ਼ਬਦਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ ਪ੍ਰਾਪਤ ਕਰੋ ।

ਪ੍ਰਸ਼ਨ 7.
“ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ਕਵਿਤਾ ਵਿੱਚ ਆਏ ਕੋੲ 10 ਨਾਂਵ ਚੁਣੋ ਅਤੇ ਸੁੰਦਰ ਲਿਖਾੲ ਵਿਚ ਲਿਖੋ :
ਉੱਤਰ:

1. ਹਿੰਦੁਸਤਾਨ .
2. ਦੇਸ
3. ਪਰਬਤ
4.  ਬਰਫ਼ਾਂ
5. ਚਮਿਆਂ
6. ਪਾਣੀ
7. ਮਿੱਟੀ
8. ਮੋਤੀਆਂ
9. ਦਾਣੇ
10. ਲੋਕ ।

ਪ੍ਰਸ਼ਨ 8.
ਹੇਠ ਲਿਖੇ ਸ਼ਬਦਾਂ ਦੀ ਬੋਲ-ਲਿਖਤ ਕਰਵਾੲ ਜਾਵੇ :
ਹਿੰਦੁਸਤਾਨ
ਬਰਫ਼ਾਂ
ਬੰਨ੍ਹ
ਮੋਤੀਆਂ
ਉੱਗਦੇ
ਘੁਲੇ-ਮਿਲੇ
ਮੁਸਕਾਨ
ਵਰਦਾਨ,
ਧਾਂਕ-ਜਮਾੲ ।
ਉੱਤਰ:
(ਨੋਟ – ਵਿਦਿਆਰਥੀ ੲਨ੍ਹਾਂ ਸ਼ਬਦਾਂ ਦਾ ਉਚਾਰਨ ਕਰਦੇ ਹੋਏ ੲਕ-ਦੂਜੇ ਨੂੰ ਸੁੰਦਰ ਲਿਖਾੲ ਕਰ ਕੇ ਲਿਖਣ ਲੲ ਕਹਿਣ )

ਨੋਟ – ਪੰਜਵੀਂ ਦੇ ਵਿਦਿਆਰਥੀਆਂ ਲੲ ਕਵਿਤਾ ਦੇ ਛੰਦਾਂ ਬਾਰੇ ਸਮਝਣਾ ਮੁਸ਼ਕਿਲ ਹੋਵੇਗਾ । ੲਸ – ਕਰਕੇ ਉਹ ੲਸ ਪੁਸਤਕ ਵਿੱਚ ਲਿਖੇ ਹਰ ਛੰਦ ਦਾ ਨਾਂ ਯਾਦ ਕਰ ਲੈਣ ।

ਨੋਟ – ਬਹੁਵਿਕਲਪੀ ਪ੍ਰਸ਼ਨਾਂ ਵਿਚ ਉੱਪਰ ਦਿੱਤੇ ਅਨੁਸਾਰ ਹਰ ਪ੍ਰਸ਼ਨ ਦੇ ਤਿੰਨ-ਚਾਰ ਉੱਤਰ ਦਿੱਤੇ ਹੁੰਦੇ ਹਨ, ਜਿਨ੍ਹਾਂ ਵਿਚ ੲਕ ਠੀਕ ਹੁੰਦਾ ਹੈ ਤੇ ਬਾਕੀ ਗ਼ਲਤ । ਵਿਦਿਆਰਥੀਆਂ ਨੇ ਠੀਕ ਉੱਤਰ ਉੱਤੇ ਜਾਂ ਤਾਂ ਸਹੀ (✓) ਦਾ ਨਿਸ਼ਾਨ ਲਾਉਣਾ ਹੁੰਦਾ ਹੈ ਜਾਂ ਉਸ ਸਹੀ ਉੱਤਰ ਨੂੰ ਲਿਖਣਾ ਹੁੰਦਾ ਹੈ । ੲਸ ਪੁਸਤਕ · ਵਿਚ ਅਗਲੇ ਪਾਠਾਂ ਸੰਬੰਧੀ ਅਜਿਹੇ ਪ੍ਰਸ਼ਨਾਂ ਦਾ ੲੱਕੋੲਕ, ਠੀਕ ਉੱਤਰ ਹੀ ਦਿੱਤਾ ਗਿਆ ਹੈ ਤੇ ਬਾਕੀ ਗ਼ਲਤ ਉੱਤਰ ਨਹੀਂ ਦਿੱਤੇ ਗਏ । ਵਿਦਿਆਰਥੀ ੲਨ੍ਹਾਂ ਨੂੰ ਯਾਦ ਕਰਕੇ ਹੀ ਪ੍ਰੀਖਿਆ ਦੀ ਤਿਆਰੀ ਕਰ ਸਕਦੇ। ਹਨ ।

VII ਕੁੱਝ ਹੋਰ ਜ਼ਰੂਰੀ ਪ੍ਰਸ਼ਨ :

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਲਿਖੀਆਂ ਸਤਰਾਂ ਪੂਰੀਆਂ ਕਰੋ :
(ੳ) ਉੱਚੇ ਪਰਬਤ ਬਰਫ਼ਾਂ ਲੱਦੇ,
ਖੜੇ ਜਿਉਂ ਬੰਨ੍ਹ ਕੇ ਢਾਣੀ ।
………………………….. ।

(ਅ) ੲਸ ਦੀਆਂ ਨਦੀਆਂ ੲਸ ਦੇ ਜੰਗਲ,
ਹਰੇ-ਭਰੇ ਮੈਦਾਨ ।
………………………….. ।

(ੲ) ਇਸ ਮਿੱਟੀ ਵਿੱਚ ਉੱਗਦੇ ਨੇ,
ਮੋਤੀਆਂ ਵਰਗੇ ਦਾਣੇ ।
………………………… ।

(ਸ) ੲਸ ਮਿੱਟੀ ਨੂੰ ਸੀਸ ਨਿਭਾੲਏ,
ੲਹ ਮਿੱਟੀ ਵਰਦਾਨ ।
……………………… ।
ਉੱਤਰ:
(ੳ) ਉੱਚੇ ਪਰਬਤ ਬਰਫ਼ਾਂ ਲੱਦੇ,
ਖੜੇ ਜਿਉਂ ਬੰਨ੍ਹ ਕੇ ਢਾਣੀ ।
ੲਸ ਦੇ ਚਸ਼ਮਿਆਂ ਵਿੱਚੋਂ ਫੁੱਟਦਾ,
ਚਾਂਦੀ ਰੰਗਾ ਪਾਣੀ ।

(ਅ) ੲਸ ਦੀਆਂ ਨਦੀਆਂ, ੲਸ ਦੇ ਜੰਗਲ,
ਹਰੇ-ਭਰੇ ਮੈਦਾਨ ।
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ! ਮੇਰਾ ਪਿਆਰਾ ਹਿੰਦੁਸਤਾਨ !!

(ੲ) ੲਸ ਦੀ ਮਿੱਟੀ ਵਿੱਚ ਉੱਗਦੇ ਨੇ,
ਮੋਤੀਆਂ ਵਰਗੇ ਦਾਣੇ ।
ੲਹ ਮਿੱਟੀ ਤਾਂ ਮਾਂ ਹੁੰਦੀ ਹੈ, ‘
ਆਖਣ ਲੋਕ ਸਿਆਣੇ ।

(ਸ) ੲਸ ਮਿੱਟੀ ਨੂੰ ਸੀਸ ਨਿਵਾੲਏ,
ੲਹ ਮਿੱਟੀ ਵਰਦਾਨ ॥
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ !
ਮੇਰਾ ਪਿਆਰਾ ਹਿੰਦੁਸਤਾਨ !!

ਪ੍ਰਸ਼ਨ 2.
ਸਤਰਾਂ ਪੂਰੀਆਂ ਕਰੋ :
(ਉ) ੲਕ ਬਾਗ਼ ਵਿਚ ਅਸੀਂ ਹਾਂ ਉੱਗੇ,
ਬੂਟੇ ਕੲ ਤਰ੍ਹਾਂ ਦੇ ।
………………………..

(ਅ) ਰਿਸ਼ੀਆਂ, ਮੁਨੀਆਂ, ਗੁਰੂਆਂ, ਪੀਰਾਂ,
ੲਸ ਦੀ ਸ਼ਾਨ ਵਧਾੲ ।
…………………………
…………………………
ਉੱਤਰ:
(ੳ) ੲਕ ਬਾਗ ਵਿਚ ਅਸੀਂ ਹਾਂ ਉੱਗੇ,
ਬੂਟੇ ਕੲ ਤਰ੍ਹਾਂ ਦੇ ।
ਪਰ ਆਪਸ ਵਿਚ ਘੁਲੇ-ਮਿਲੇ ਹਾਂ,
ਪਾਣੀ ਜਿਵੇਂ ਸਰਾਂ ਦੇ ।

(ਅ) ਰਿਸ਼ੀਆਂ, ਮੁਨੀਆਂ, ਗੁਰੂਆਂ, ਪੀਰਾਂ,
ੲਸ ਦੀ ਸ਼ਾਨ ਵਧਾੲ ।
ੲਸ ਦੇ ਜੋਧਿਆਂ ਨੇ ਜੱਗ ਉੱਤੇ,
ਆਪਣੀ ਧਾਂਕ ਜਮਾੲ ।

ਪ੍ਰਸ਼ਨ 3.
ਹੇਠ ਲਿਖੀਆਂ ਕਾਵਿ-ਸਤਰਾਂ ਦੇ ਨਾਲ ਮਿਲਦੀਆਂ ਸਤਰਾਂ ਲਿਖੋ
(ਉ) ਹਰ ੲਕ ਨਵੀਂ ਸਵੇਰ ਵੰਡੀਏ,
ਫੁੱਲਾਂ ਜਿਹੀ ਮੁਸਕਾਨ । ਪ੍ਰੀਖਿਆ 2008)
…………………………… ।

(ਅ) ੲੱਥੋਂ ਦੀ ਵਿੱਦਿਆ ਨੂੰ ਵੀਰੋ !
ਮੰਨਦਾ ਕੁੱਲ ਜਹਾਨ ।
………………………….।

(ੲ) ਇਸ ਦੇ ਹਾਲੀ, ੲਸ ਦੇ ਪਾਲੀ,
ਕਾਮੇ ਤੇ ਮਜ਼ਦੂਰ ।
…………………………..।

(ਸ) ਇਸ ਦਾ ਹਰ ੲਕ ਗੱਭਰੂ ਬਾਂਕਾ,
ਹਰ ਮੁਟਿਆਰ ਰਕਾਨ ।
………………………… ।
ਉੱਤਰ:
(ਉ) ਹਰ ੲੱਕ ਨਵੀਂ ਸਵੇਰ ਵੰਡੀਏ,
ਫੁੱਲਾਂ ਜਿਹੀ ਮੁਸਕਾਨ ।
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ! ਮੇਰਾ ਪਿਆਰਾ ਹਿੰਦੁਸਤਾਨ !!

(ਅ) ੲੱਥੋਂ ਦੀ ਵਿੱਦਿਆ ਨੂੰ ਵੀਰੋ !
ਮੰਨਦਾ ਕੁੱਲ ਜਹਾਨ ।
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ! ਮੇਰਾ ਪਿਆਰਾ ਹਿੰਦੁਸਤਾਨ !!

(ੲ) ਇਸ ਦੇ ਹਾਲੀ, ੲਸ ਦੇ ਪਾਲੀ,
ਕਾਮੇ ਤੇ ਮਜ਼ਦੂਰ ।
ਹੱਕ, ਸੱਚ ਦੀ ਕਰਨ ਕਮਾੲ,
ਰਹਿਣ ਕੂੜ ਤੋਂ ਦੂਰ ।

(ਸ) ੲਸ ਦਾ ਹਰ ੲਕ ਗੱਭਰੂ ਬਾਂਕਾ,
ਹਰ ਮੁਟਿਆਰ ਰਕਾਨ ।
ਮੇਰਾ ਹਿੰਦੁਸਤਾਨ ! ਮੇਰਾ ਪਿਆਰਾ ਹਿੰਦੁਸਤਾਨ !!

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੇ ਸ਼ਬਦਾਂ ਨੂੰ ਵਾਕਾਂ ਵਿੱਚ ਵਰਤੋ :ਪਰਦੇਸ, ਚਾਂਦੀ, ਵਿੱਦਿਆ, ਬਾਗ਼, ਜੋਧੇ ।
ਉੱਤਰ:

1. ਪਰਦੇਸ ਪਿਰਾੲਆ ਦੇਸ਼, ਵਿਦੇਸ| ਬਹੁਤ ਸਾਰੇ ਪੰਜਾਬੀ ਲੋਕੲੰਗਲੈਂਡ, ਅਮਰੀਕਾ, ਕੈਨੇਡਾ ਆਦਿ ਪਰਦੇਸਾਂ ਵਿਚ ਰਹਿੰਦੇ ਹਨ ।
2. ਚਾਂਦੀ (ੲਕ ਬਹੁਮੁੱਲੀ ਚਿੱਟੀ ਚਮਕੀਲੀ ਧਾਤ, ਰੁੱਪਾ)-ਮੇਰੇ ਹੱਥ ਵਿਚ ਚਾਂਦੀ ਦਾ ਕੜਾ ਹੈ ।
3. ਵਿੱਦਿਆ ਪੜ੍ਹਾੲ-ਲਿਖਾੲ)-ਸਕੂਲ-ਕਾਲਜ ਵਿੱਦਿਆ ਦੇ ਮੰਦਰ ਹਨ ।
4. ਬਾਗ਼ ਬਗੀਚਾ, ਪੌਦਿਆਂ, ਫੁੱਲਾਂ-ਫਲਾਂ ਨਾਲ ਸ਼ਿੰਗਾਰੀ ਥਾਂ)-ਅਸੀਂ ਹਰ ਰੋਜ਼ ਸਵੇਰੇ ਸੈਰ ਕਰਨ ਲੲ ਬਾਗ਼ ਵਿਚ ਜਾਂਦੇ ਹਾਂ ।
5. ਜੋਧੇ ਜੰਗ ਲੜਨ ਵਾਲੇ)-ਭਾਰਤੀ ਜੋਧਿਆਂ ਨੇ ਜੰਗ ਵਿਚ ਦੁਸ਼ਮਣਾਂ ਦੇ ਦੰਦ ਖੱਟੇ ਕਰ ਦਿੱਤੇ ।

VIII. ਰਚਨਾਤਮਕ ਕਾਰਜ

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਦਿੱਤੇ ਰਾਸ਼ਟਰੀ ਝੰਡੇ ਦੇ ਚਿਤਰ ਵਿਚ ਰੰਗ ਭਰੋ :

ਉੱਤਰ:
(ਨੋਟ-ਵਿਦਿਆਰਥੀ ਆਪੇ ਹੀ ਕਰਨ ॥

ਪ੍ਰਸ਼ਨ 2.
ਆਪਣੇ ਸਕੂਲ ਦੇ ਮੁੱਖ ਅਧਿਆਪਕ/ਮੁੱਖ ਅਧਿਆਪਕਾ ਨੂੰ ਬਿਮਾਰੀ ਦੀ ਛੁੱਟੀ ਲੈਣ ਲੲ ਅਰਜ਼ੀ ਬੇਨਤੀ-ਪੱਤਰ) ਲਿਖੋ ।
ਉੱਤਰ:
ਨੋਟ-ੲਸ ਪ੍ਰਸ਼ਨ ਦਾ ਉੱਤਰ ਦੇਣ ਲੲ ਦੇਖੋ ਅਗਲੇ ਸਫ਼ਿਆਂ ਵਿਚ ਦਿੱਤਾ ‘ਚਿੱਠੀ-ਪੱਤਰ ਜੋ ਵਾਲਾ ਭਾਗ ॥

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ

ਸ਼ਾਨ – ਵਡਿਆੲ ।
ਢਾਣੀ – ਟੋਲੀ ।
ਚਸ਼ਮਾ – ਧਰਤੀ ਵਿਚੋਂ ਆਪ-ਮੁਹਾਰਾ ਫੁੱਟ ਰਿਹਾ ਪਾਣੀ ।
ਸੀਸ – ਸਿਰ ।
ਵਰਦਾਨ – ਬਖ਼ਸ਼ਿਸ਼ ।
ਸਰਾਂ – ਸਰੋਵਰਾਂ, ਤਲਾਵਾਂ ।
ਮੁਨੀ – ਮੋਨਧਾਰੀ ਸਾਧੂ ।
ਰਿਸ਼ੀ-ਮੁਨੀ – ਧਾਰਮਿਕ ਮਹਾਂਪੁਰਸ਼ ।
ਧਾਂਕ ਜਮਾੲ – ਦਬਦਬਾ ਕਾੲਮ ਕੀਤਾ, ਡੂੰਘਾ ਪ੍ਰਭਾਵ ਪਾੲਆ ।
ਜਹਾਨ – ਦੁਨੀਆ ।
ਹਾਲੀ – ਹਲ ਚਲਾਉਣ ਵਾਲੇ ਕਿਸਾਨ ।
ਪਾਲੀ – ਪਸ਼ੂ ਪਾਲਣ ਵਾਲੇ ।
ਹੱਕ ਸੱਚ ਦੀ – ਧਰਮ ਅਨੁਸਾਰ, ਮਿਹਨਤ ਦੀ ।
ਕੂੜ – ਝੂਠ ।
ਬਾਂਕਾ – ਛੈਲ-ਛਬੀਲਾ, ਸੁੰਦਰ ।
ਰਕਾਨ – ਸੁਘੜ-ਸਿਆਣੀ ॥

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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

PSEB 12th Class Chemistry Guide Coordination Compounds InText Questions and Answers

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
(i) The primary valencies are satisfied by negative ions and equal-to the oxidation state of the metal.

(ii) The secondary valencies can be satisfied by neutral or negative ions. It is equal to the coordination number of the central metal atom and is fixed.

(iii) The ions bound to the central metal ion to secondary linkages have definite spatial arrangements and give geometry to the complex. While primary valency is non-directional.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO⁴ solution mixed with (NH⁴)²SO⁴ solution in 1 : 1 molar ratio forms double salt, FeSO⁴∙(NH⁴)²SO⁴∙6H²O which ionises in the solution to give Fe2+ ions. Hence, it gives the test of Fe²⁺ ions.

CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio forms a complex, with the formula [Cu(NH₃)₄]SO₄. The complex ion, [CU(NH₃)]²⁺ does not ionise to give Cu²⁺ ions. Hence, it does not give the test of Cu²⁺ ion.

Question 3.
Explain with two examples each of the following: Coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
Coordination entity: A coordination entity constitutes usually a central metal atom or ion, to which a fixed number of other atoms or ions or groups are attached by coordinate bonds. A coordination entity may be neutral, positively or negatively charged. For examples : [Ni(CO)₄], [CoCl₃(NH₃)₃], [Co(NH₃)₆]^3+.

Ligand : A ligand is an ion or a small molecule having at least one lone pair of electrons and capable of forming a coordinate bond with central atom or ion in the coordination entity. For example: Cl^–, OH^–, CN^–, CO, NH₃, H₂O etc.

Coordination number : The coordination number of the central atom or ion is determined by the number of a bonds between the ligands and the central atom or ion. n bonds are not consider for the determination of coordination number. The a bonding electrons may be indicated by a pair of dots (:). For example, [Co(:NH₃)₆]³⁺ and [Fe(:CN)₆]^3-.

Coordination polyhedron : The spatial arrangement of the ligands which are directly attached to the central atom or ion called coordination polyhedron.
For example: [Co(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄ ]^2- is square planar.

Homoleptic and heteroleptic : Complexes in which a metal is bound to only one type of donor groups are known as homoleptic.
For example : [Co(NH₃)₆]³⁺, [PtCl₆]^2- .
Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic. ‘
For example : [Co(NH₃)₄Cl₂]⁺, [PdI₂(ONO)₂ (H₂O)₂],

Question 4.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atoms is called unidentate ligand, e.g., Cl^– and NH₃.

A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atoms is called a didentate ligand, e.g., NH₂—CH₂—CH₂—NH₂ and ^–OOC — COO^–.
A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g., CN^–or NC^– and \(\) or : ONO^–.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₂]^2-
(iv) K₃Fe(CN)₆]
(v) [Cr(NH₃)₂Cl₃]
Answer:
(i) x + (-1) + (0) + (0) = + 2 so x = +3 (III)
(ii) x + 2(-1) + 0 = +1 so x = +3 (III)
(iii) x + 4(-1) = -2 so x = +2(11)
(iv) x + 6(-1) = -3 so x = +3 (III)
(v) x + 3(-1) + 0 = 0 so x = +3 (III)

Question 6.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(Il)
(x) Pentaamminenitrito-N-cobalt(lll)
Answer:
(i) [Zn(OH)₄]^2-
(ii) K₂[PdCl₄]
(iii) pt(NH₃)₂Cl₂]
(iv) K₂[Ni(CN)₄]
(v) [Co(ONO) (NH₃)₅]²⁺
(vi) [CO(NH₃)₆]₂ (SO₄)₃
(vii) K₃[Cr(C₂O₄)₃]
(viii) [Pt(NH₃)₆]⁴⁺
(ix) [Cu(Br)₄]^2-
(x) [Co (NO₂) (NH₃)₅]²⁺

Question 7.
Using IUPAC norms write the systematic names of the following:
(i) [CO(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
(iii) [Ti(H₂O)₆]³⁺
(iv) [CO(NH₃)₄Cl(NO₂)]CI
(v) [Mn(H₂O)₆]²⁺
(vi) [NiCl₄]^2-
(vii) [Ni(NH₃)₆]Cl₂
(viii) [Co(en)₃]³⁺
(ix) [Ni(CO)₄]
Answer:
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine) platinum(II) chloride
(iii) Hexaquatitanium(III) ion
(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanganese(II) ion
(vi) Tetrachloridonickelate(II) ion
(vii) Hexamminenickel(II) chloride
(viii) Tris(ethane-1, 2-diamine) cobalt(III) ion
(ix) Tetracarbonylnickel(O)

Question 8.
List various types of isomerism possible for coordination compounds giving an example of each.
Answer:
Two principal types of isomerism are known among coordination compounds :
(A) Sterioisomerism,
(B) Structural isomerism.
Each of which can be further sub-divided as :
(A) Stereoisomerism
(i) Geometrical isomerism : It arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
Example: Pt[(NH₃)₂Cl₂]

(ii) Optical isomerism : It is common in octahedral complexes involving didentate ligands.
Example : [Pt Cl₂(en) ₂]²⁺

Optical isomers (d and l) of cis-[PtCl₂(en)₂]²⁺

(B) Structural isomerism
(i) Linkage isomerism.
Example: [Co(NH₃)₅ (NO₂)]Cl₂
(ii) Coordination isomerism.
Example: [Co(NH₃)₆] [Cr(CN)₆]
(iii) Ionisation isomerism.
Example: [Co(NH₃)₅SO₄]Br and [CO(NH₃)₅ Br]SO₄
(iv) Solvate isomerism.
Example : [Cr(H₂O)₆] Cl₃ (violet) its solvate isomer
[Cr(H₂O)₅Cl]Cl₂. H₂O (grey-green)

Question 9.
How many geometrical isomers are possible in the following coordination entities? ’
(i) [Cr(C₂O₄)₃]^3-
(ii) [Co(NH₃)₃Cl₃]
Answer:
(i) [Cr(C₂O₄)₃]^3-,
No geometric isomer is possible as it is a bidentate ligand.
(ii) [CO(NH₃)₃Cl₃] .
Two geometrical isomers are possible.

Question 10.
Draw the structures of optical isomers of:
(i) [Cr(C₂O₄)₃]^3-
(ii) [PtCl₂(en)₂]²⁺
(iii) [Cr(NH₃)₂ Cl₂ (en)]⁺
Answer:
(i) [Cr(C₂O₄)₃]^3-

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [CoCl₂ (en)₂]⁺
(ii) [Co(NH₃)Cl(en)₂]²⁺
(iii) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(i) [CoCl₂ (en)₂]⁺

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
Three isomers are possible as follows :

Isomers of this type do not show any optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :
(i) a green precipitate with aqueous potassium fluoride, and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄]SO₄. It is a labile complex. The blue colour of the solution is due to [Cu(H₂O)₄]²⁺ ions,

(i) When KF is added, the weak H₂O ligands are replaced by F^– ligands forming [CuF₄]^2- ions, which is a green precipitate.

(ii) When KCl is added, Cl^– ligands replace the weak H₂O ligands forming [CuCl₄]^2- ion, which has bright green colour.

Question 14.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S (g) is passed through this solution?
Answer:
K₂[Cu(CN)₄] is formed when excess of aqueous KCN is added to an aqueous solution of CuSO₄.

As CN^–ions are strong ligands the complex is very stable. It is not replaced by S^2- ions when H₂S gas is passed through the solution and thus no precipitate of CuS is obtained.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)₆]^4-
(ii) [FeFe₆]^3-
(iii) [Co(C₂O₄)₃]^3-
(iv) [CoF₆]^3-
Answer:
(i) [Fe(CN)₆]^4-
In the above coordination complex, iron exists in the +2 oxidation state.
Fe = [Ar] 3d⁶ 4s²
Outer configuration of Fe²⁺ = 3d⁶ 4s⁰
Orbitals of Fe²⁺ ion:

As CN^– is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since, there are six ligands around the central metal ion, the most feasible hybridisation is d²sp³. d²sp³ hybridised orbitals of Fe²⁺ are :

6 electron pairs from CN ions occupy the six hybrid d²sp³ orbitals.
Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]^3-
In this complex, the oxidation state of Fe is + 3.
Fe³⁺ = 3d⁵ 4s⁰
Orbitals of Fe³⁺ ion:

There are 6F^– ions. Thus, it will undergo d²sp³ or sp³d² hybridisation. As F^– is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridisation is sp³d². sp³d² hybridised orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co(C₂O₄)₃]^3-
Cobalt exists in the + 3 oxidation state in the given complex.
Outer configuration of Co = 3d₇ 4s₂
Co₃₊ = 3d₆4s₀
Orbitals of Co₃₊ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d electrons. As there are 6 ligands, hybridisation has to be either sp³d² or d²sp³ hybridisation. sp³d² hybridisation of Co³⁺.

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF₂]^3-
Cobalt exists in the + 3 oxidation state.
Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridisation.
sp³d² hybridised orbitals of Co³⁺ ion are :

Hence, the geometry of complex is octahedral, 6 electron pants.

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Answer:

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (△₀) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy? How does the magnitude of △₀ decide the actual configuration of d-orbitals in a coordi-nation entity?
Answer:
When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) in case of octahedral field.

If △₀ < P, (pairing energy), the 4th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\), thereby forming high spin complexes.

Such ligands for which A 0 < P are called weak field ligands.
If △₀ > P, the 4th electron pairs up in one of the t2g orbitals giving the configuration \(t_{2 g}^{4} e_{g}^{0}\), thus forming low spin complexes. Such ligands for which △₀ > P are called strong field ligands.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
Cr is in the +3 oxidation state i.e., d³ configuration. Also, NH₃ is a weak field ligand that does not cause the pairing of the electrons in the orbital.

Therefore, it undergoes d²sp³ hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)₄]^2-, Ni exists in the + 2 oxidation state i. e., d⁸ configuration.

CN^– is a strong field ligand. It causes the pairing of the 3d electrons. Then, Ni²⁺ undergoes dsp² hybridisation.

As there are no unpaired electrons, it is diamagnetic.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:
In [Ni(H₂O)6]²⁺, \(\mathrm{H}_{2} \ddot{\mathrm{O}}\) is a weak field ligand. Therefore, there are unpaired electrons in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i. e., the possibility of d-d transition is present. Hence, [Ni(H₂O)₆]²⁺ is coloured.

In [Ni(CN)₄]²⁺, the electrons are all paired as CN^– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]^2-. Hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:
In both the complex compounds, Fe is in +2 oxidation state with configuration 3d⁶, i.e., it has four unpaired electrons. In the presence of weak H₂O ligands, the unpaired electrons do not pair up. But in the presence of strong ligand CN^– they get paired up. Then no unpaired electron is left. Due to this, difference in the number of unpaired electrons, both complex ions have different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal carbon in metal carbonyls possesses both CT and π character. The ligand to metal is CT bond and metal to ligand is π bond. The effect of CT bond strengthens the rcbond and vice-versa. This is called synergic effect. This unique synergic provides stability to metal carbonyls.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K₃[CO(C₂O₄)₃]
(ii) cis-[Cr(en)₂Cl₂]Cl
(iii) (NH₄)₂[CoF₄]
(iv) [Mn(H₂O)₆]S0₄
Solution:
(i) K₃[CO(C₂O₄)₃]
The central metal ion is Co.
The oxidation state can be given as :
(+1) × 3 + × + (- 2) × 3 = 0
x – 6 = -3 ⇒ x = + 3
The d orbital occupation for Co³⁺ is \(t_{2 g}^{6} e g^{0}\).
(as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is strong field ligand)
Coordination number of Co = 3 × denticity of C₂O₄
= 3 × 2 (as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is a bidentate ligand) = 6

(ii) cis-[Cr(en)₂Cl₂]Cl
The central metal ion is Cr.
The oxidation state can be given as:
x + 2(0) + 2(-1) + (-1) = 0
x – 2 – 1 = 0
x = + 3
The d orbital occupation for Cr³⁺ is \(t_{2 g}^{3}\).
Coordination number of Cr
= 2 × denticity of en + 2
= 2 × 2 + 2 = 6

(iii) (NH₄)₂[CoF₄]
The central metal ion is Co.
The oxidation state can be given as:
(+1) × 2 + × + (-1) × 4 = 0
x – 4 = -2
x = + 2
The d orbital occupation for Co²⁺ is d⁷ or \(t_{2 g}^{5} e_{g}^{2}\). (as F^– is a weak ligand)
Coordination number of Co = 4

(iv) [Mn(H₂O)₆]S0₄
The central metal ion is Mn.
The oxidation state can be given as:
x + (0) × 6 + (- 2) = 0
x = + 2
The d orbital occupation for Mn is d⁵ or [latext_{2 g}^{3} e_{g}^{2}][/latex].
Coordination number of Mn = 6

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H₂O)₂(C₂O₄)₂] 3H₂O
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) CrCl₃(py)₃
(iv) Cs[FeCl₄]
(v) K₄[Mn(CN)₆]
Answer:
(i) K[Cr(H₂O)₂ (C₂O₄)2 ] ∙ 3H₂O
IUPAC name : Potassium diaquadioxalatochromate (III) hydrate.
Oxidation state of chromium
+1 + x + (0) × 2 + (- 2) × 2 + 3(0) = 0
+ 1 + x – 4 = 0
x = + 3
Electronic configuration of Cr⁺³= 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral


Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(ii) [Co(NH₃)₅Cl]Cl₂
IUPAC name : Pentaammine chlorido cobalt(III) chloride
Oxidation state of Co
x + (0)5 + (-1) + (-1) × 2 = 0
x – 3 =0
x = + 3
Coordination number = 6
Shape: Octahedral.
Electronic configuration of Co³⁺ = 3d⁶ = \(t_{2 g}^{6} e_{g}^{0}\)
The complex does not exhibit geometrical as well as optical isomerism.
Magnetic Moment (μ) = \(\sqrt{n(n+2)}\)BM = \(\sqrt{0(0+2)}\) BM = 0 BM

(iii) CrCl₃(py)₃
IUPAC name : Trichlorido tripyridine chromium (III) Oxidation state of Cr
x + (-1) × 3 + (0)3 = 0
x = + 3
Electronic configuration of Cr = 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral
Stereochemistry

Both isomers are optically active. Therefore, a total of 4 isomers exist.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\) = \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(iv) Cs[FeCl₄]
IUPAC name : Caesium tetrachlorido ferrate (III)
Oxidation state of Fe
+ 1 + x + (-1) × 4= 0
x – 3 = 0
x = + 3
Electronic configuration of Fe = 3d⁵(\(t_{2 g}^{3} e_{g}^{2}\))
Coordination number = 4
Shape : Tetrahedral
The complex does not exhibit geometrical or optical isomerism, (stereo isomerism).
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5+2)}\)
= \(\sqrt{35}\) = 5.92 BM

(v) K₄[Mn(CN)₆]
IUPAC name : Potassium hexacyanomanganate(II)
Oxidation state of Mn
(+1) × 4 + x + (-1) × 6 = 0
x – 2 = 0
x = + 2
Electronic configuration of Mn = 3d⁵ (\(t_{2 g}^{5} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral.
The complex does not exhibit stereo isomerism.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{1(1+2)}\)
= \(\sqrt{3}\)
= 1.732 BM

Question 25.
What is meant by stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:
The stability of a coordination compound in solution refers to the degree of association between the two species involved in the state of equilibrium. The stability of the coordination compound is measured in term of magnitude of stability or formation of equilibrium constant.
M + 4L → ML₄
K = \(\frac{\left[\mathrm{ML}_{4}\right]}{[\mathrm{M}][\mathrm{L}]^{4}}\)
Larger the stability constant, the higher is the proportion of ML₄ that exists in solution.

Factors on which stability of the complex depends are as follows :

1. Charge on the central metal ion : Greater the charge on the central metal ion, greater is the stability of the complex.
2. Nature of the metal ion : Groups 3 to 6 and inner transition element form stable complexes when donor atoms of the ligands are N, O and F. The element after group 6 of the transition metals which have d-orbitals (e.g., Rh, Pd, Ag, Au, Hg, etc.) form stable complexes when the donor atoms of the ligands are heavier members of N, O and F family.
3. Basic nature of the ligand : Greater the basic strength of the ligand, greater is the stability of the complex.
4. Chelate effect: Presence of chelate rings in the complex increases its stability. It is called chelate effect. It is maximum for the 5- and 6- membered rings.
5. Effect of multidentate cyclic ligands : If the ligands happen to be multidentate and cyclic without any steric effect, the stability of the complex is further increased.

Question 26.
What is meant by chelate effect? Give an example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six-membered ring is formed, the effect is called chelate effect. Example, [PtCl₂(en)].

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry
(iv) extraction/metallurgy of metals
Answer:
(i) Role of coordination compounds in biological systems :

• Haemoglobin, the oxygen carrier in blood, is a complex of Fe²⁺ with porphyrin.
• The pigment chlorophyll in plants, responsible for photosynthesis, is a complex of Mg²⁺ with porphyrin.
• Vitamin B₁₂ (cyanocobalamine) the antipemicious anaemia factor, is a complex of cobalt.

(ii) Role of coordination compounds in medicinal chemistry :

• The platinum complex cis-[Pt(NH₃)₂Cl₂] (cis-platin) is used in the treatment of cancer.
• EDTA complex of calcium is used in the treatment of lead poisoning. Ca-EDTA is a weak complex; when it is administered, calcium in the complex is replaced by the lead present in the body and is eliminated in the urine.
• The excess of copper and iron present in animal system are removed by the chelating ligands D-penicillamine and desferroxime B via the formation of complexes.

(iii) Role of coordination compounds in analytical chemistry :
Complex formation is frequently encountered in qualitative and quantitative chemical analysis.
(a) Qualitative analysis
I. Detection of Cu²⁺ is based on the formation of a blue tetraammine copper (II) ion.

II. Ni²⁺ is detected by the formation of a red complex with dimethyl glyoxime (DMG).

III. The separation of Ag⁺ and Hg²⁺ in group I is based on the fact that while AgCl dissolves in NH₃, forming a soluble complex, Hg₂Cl₂ forms an insoluble black substance.

(b) Quantitative analysis : Gravimetric estimation of Ni²⁺ is carried out by precipitating Ni²⁺ as red nickel dimethyl glyoxime complex in the presence of ammonia.

EDTA is used in the complexometric determination of several metal ions such as Ca²⁺, Zn²⁺, Fe²⁺, Co²⁺, Ni²⁺ etc.

(iv) Role of coordination compounds in extraction/metallurgy of metals : Extraction of various metals from their ore involves complex formation. For example, silver and gold are extracted from their ore by forming cyanide complex.

Purification of some metals can be achieved through complex formation. For example in Mond process, impure nickel is converted into [Ni(CO)₄] which is decomposed to yield pure nickel.

Question 28.
How many ions are produced from the complex Co(NH₃)₆ Cl₂ in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Answer:
The correct option is (iii)
Coordination number of cobalt = 6. It ionises in the solution as

Hence, 3 ion are produced.

Question 29.
Amongst the following ions, which one has the highest magnetic moment value?
(i)[Cr(H₂O)₆]³⁺
(ii)[Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Answer:
The oxidation state are: Cr (III), Fe (II) and Zn (II).
Electronic configuration of Cr³⁺ = 3d³, unpaired electrons = 3
Electronic configuration of Fe²⁺ = 3d⁶, unpaired electrons = 4
Electronic configuration of Z²⁺ = 3d¹⁰, unpaired electrons = 0
As μ = \(\sqrt{n(n+2)}\), therefore, (ii) has the highest magnetic moment.

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(i) +1
(ii) +3
(iii) -1
(iv) -3
Solution:
Oxidation number of Co : K[Co(CO)₄]
x+ (4 × 0) = -1; x = -1
Thus, correct answer is (iii).

Question 31.
Amongst the following, the most stable complex is
(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]^3-
(iv) [FeCl₆]^3-
Answer:
In all these complexes, Fe is in +3 oxidation state. However, the complex (iii) is a chelate because three \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions acts as the chelating ligands. Thus, the most stable complex is [Fe(C₂O₄)₃]^3-. Thus, correct answer is (iii).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region of the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
As metal ion is fixed, the increasing CFSE values of the ligands from the spectrochemical series are in the order :
H₂O < NH₃ < \(\mathrm{NO}_{2}^{-}\)
Hence, the energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
As E = \(\frac{h c}{\lambda}\), therefore, the wavelengths absorbed will be in the opposite order,
[Ni(NO₂)₆]^4- < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Chemistry Guide for Class 12 PSEB Coordination Compounds Textbook Questions and Answers

Question 1.
Write the formulas for the following coordination compounds :
(i) Tetraamminediaquacobalt (III) chloride
(ii) Potassiumtetracyanidonickelate(II)
(iii) Tris(ethane-l,2-diammine)chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-l,2-diammine) platinum (IV) nitrate
(vi) Iron(III)hexacyanidoferrate(II).
Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄
(iii) (Cr(en)₃]Cl₃
(iv) [Pt(NH₃)BrCl(NO₂)]^–
(v) [PtCl₂(en)₂] (NO₃)₂
(vi) Fe₄[Fe(CN)₆]₃

Question 2.
Write the IUPAC names of the following coordination compounds:
(i) [CO(NH₃)₆]Cl₃
(ii) [CO(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
Answer:
(i) Hexaamminecobalt(III)chloride
(ii) Pentaamminechloridocobalt(III)chloride
(iii) Potassiumhexacyanoferrate(III)
(iv) Potassiumtrioxalatoferrate (III)
(v) Potassiumtetrachloridopalladate (II)
(vi) Diamminechloridomethylamine platinum(II) chloride.

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers :
(i) K[Cr(H₂O)₂](C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [CO(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Answer:
(i) (a) Both geometrical isomer (cis and traits):

(b) Cis-isomer of this compound can exist as pair of optical is :

(ii) Complex will exist as optical isomers:

The compound will show ionisation as well as linkage isomerism.

(iii) Ionisation isomer :
[Co(NH₃)₅(NO₂)](NO₃)₂,
[Co(NH₃)₅ (NO)₃] (NO₂) (NO₃)
Linkage isomers :
[Co(NH₃)₅ (NO₂)](NO₃)₂;
[CO(NH₃)₅ (ONO)](NO₃)₂

(iv) Geometrical isomerism (cis and trans) :

Question 4.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they give different ions in the solution which can be tested by adding AgNO₃ solution and BaCl₂ solution. If Cl^– dons are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If \(\mathrm{SO}_{4}^{2-}\) ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [Ni(Cl)₂₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
Nickel in [Ni(CN)₄]^2- is in the +2 oxidation state. The formation of [[Ni(CN)₄]^2- may be explained through hybridisation as follows :
Ni atom in the ground state

Since no unpaired electrons is present, the square planar complex is diamagnetic. In [Ni(CN)₄]^2-, Cl^– is a weak field ligand. It is, therefore, unable to pair up the unpaired electrons of the 3d orbital. Hence, the hybridisation involved is sp3 and the shape is tetrahedral. Since all the electrons are unpaired, it is paramagnetic

Question 6.
[Ni(CN)₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In [Ni(CO)₄] Ni is in zero oxidation state whereas in [NiCl₄]^2-, it is in
+ 2 oxidation state. In the presence of strong ligand, CO ligand, the unpaired d electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
In presence of CN^– (a strong ligand), the 3d⁵ electrons pair up leaving only one unpaired electron. The hybridisation is d²sp³ forming an inner orbital complex. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state and has d⁶ electrons. In the presence of NH₃, the 3d electrons pair up leaving two d-orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex. In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d⁸ configuration. The hybridisation involved is sp³d², forming the outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
₇₈Pt lies in group 10 with the configuration 5d⁹6s¹. Thus Pt²⁺ has the configuration :

For square planar shape, the hybridisation is dsp². Hence, the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp² hybridisation.
Thus there is no unpaired electron.

Question 10.
The hexaquomanganese(II) ion contains five impaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Answer:
Mn in the + 2 oxidation state has the configuration 3d⁵. In the presence of H₂O a weak ligand, the distribution of these five electrons is \(t_{2 g}^{3} e_{g}^{2}\)
i.e., all the electrons remain unpaired

However, in the presence of CN^– the distribution of these electrons is \(\), i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) ion, given that β₄ for this complex is 2. 1 × 10¹³.
Solution:
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β₄.
∴ \(\frac{1}{\beta_{4}}\) = \(\frac{1}{2.1 \times 10^{13}}\)
∴ = 4.7 × 10^-14