Yearly Archives: 2021

INTEXT Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols :

Solution:
Primary alcohols : (i), (ii) and (iii)
Secondary alcohols : (iv) and (v)
Tertiary alcohols : (vi)

Question 2.
Identify allylic alcohols in the above examples.
Solution:
Allvlir alcohols : (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system.


Solution:

1. 3-Chloromethyl-2-isopropylpentan-l-ol
2. 2, 5-Dimethylhexane-1,3-diol
3. 3-Bromocyclohexanol
4. Hex-l-en-3-ol
5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Solution:

Question 5.
Write structures of the products of the following reactions :

Solution:

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl – ZnCl₂ (b) HBr and (c) SOCl₂.

1.  Butan-1-ol
2. 2-Methylbutan-2-ol

Solution:

Question 7.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol

Solution:

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Solution:
The attachment of the -NO₂ group to the phenol molecule at 0- and p-positions decreases the electron density on oxygen atom. This causes the oxygen atom to pull the bond pair of electrons of the O – H bond towards itself thereby facilitating the release of H as H⁺.

The resonance structures of the phenoxide ions are :

Question 9.
Write the equations involved in the following reactions :

1. Reimer-Tiemann reaction
2. Kolbe’s reaction

Solution:

Question 10.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Solution:

Question 11.
Which of the following is an appropriate set of reactants for the preparation of 1 -methoxy-4- nitrobenzene and why?

Solution:

Question 12.
Predict the products of the following reactions :

Solution:

NCERT Exercises

Question 1.
Write IUPAC names of the following compounds :


Solution:

1. 2,2,4-Trimethylpentan-3-ol
2. 5-Ethylheptan-2,4-diol
3. Butane-2,3-diol
4. Propane-1,2,3-triol
5. 2-Methylphenol
6. 4-Methylphenol
7. 2,5-Dimethylphenol
8. 2,6-Dimethylphenol
9. l-Methoxy-2-methylpropane
10. Ethoxybenzene
11. 1-Phenoxyheptane
12. 2-Ethoxybutane

Question 2.
Write structures of the compounds whose IUPAC names are as follows :

1. 2-Methylbutan-2-ol
2. 1-Phenylpropan-2-ol
3. 3,5-Dimethylhexane-1,3,5-triol
4. 2,3-Diethylphenol
5. 1-Ethoxypropane
6. 2-Ethoxy-3-methylpentane
7. Cyclohexylmethanol
8. 3-Cydohexylpentan-3-ol
9. Cyclopent-3-en-1 -ol
10. 3-Chloromethylpentan-1-ol

Solution:

Question 3.

1. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names.
2. Classify the isomers of alcohols in question 3 (i) as primary, secondary and tertiary alcohols.

Solution:
The isomeric alcohols with molecular formula C₅H₁₂O are :

Question 4.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane.
Solution:
The boiling point of any compound depends on the strength of inter-molecular forces. Stronger is the inter-molecular attraction, higher is the boiling point.

In butane, the molecules interact with each other through weak van der Waals forces. These weak forces can be easily overcome by supplying small amount of heat energy. Thus, they have low boiling point.

In propanol, the molecules are held together by strong hydrogen bonding. These attractive forces operating between molecules are more difficult to break and therefore higher amount of heat needs to be supplied, therefore, the higher boiling point.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Solution:
Organic compounds are soluble in water if they are able to form hydrogen bonds with it. Alcohols are able to establish this interaction by the virtue of their OH group and are therefore soluble in water. On the other hand, other hydrocarbons of comparable mass do not dissolve in water since they cannot form hydrogen bonds.

Question 6.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Solution:
Hydroboration-oxidation is a method of preparation of alcohols from alkenes. The main advantage of this method is the high yield of alcohol obtained. During hydroboration, diborane (BH₃)₂ is made to react with an alkene to form an addition product. This product is then treated with hydrogen peroxide in the presence of sodium hydroxide to give alcohol.
For example,

Question 7.
Give the structure and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Solution:

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Solution:
During steam distillation, it is the lower boiling compound which distills out first. Between ortho- and para-nitrophenol it is the ortho-isomer which will be steam volatile since it has a lower boiling point.

The difference in boiling point between the two isomers can be understood based on the structural difference. In ortho-isomer intra¬molecular hydrogen bonding takes place while in the pttra-isomer, inter-molecular hydrogen bonding takes place.

As a result of the strong forces operating between the molecules of p-isomer, the boiling point is higher and it is not steam volatile.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Solution:

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Solution:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Solution:
The acid catalysed hydration of ethene may be represented as :

Question 12.
You are given benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Solution:
Using the given reagents, phenol may be prepared as :

Question 13.
Show how will you synthesise :

1. 1 -Phenylethanol from a suitable alkene.
2. Cyclohexylmethanol using an alkyl halide by an S_N² reaction.
3. Pentan-1-ol using a suitable alkyl halide.

Solution:

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Solution:


The resonance stabilization provided by the contributing structures (I)-(V) more than compensates for the bond breakage energy of O – H bond and thus causes phenol to be acidic in nature.

No such resonance structures are possible for ethoxide ion and therefore the conversion of ethanol to ethoxide is not favoured under normal conditions. Therefore, ethanol is less acidic than phenol.

Question 15.
Explain why is ortho nitrophenol more acidic than methoxyphenol?
Solution:

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
Solution:
In an electrophilic substitution reaction, an electron deficient species attacks the benzene ring which is e^– rich.

When an -OH group is attached to the benzene ring, by the virtue of its electron releasing nature increases the e^– density of the ring and thus activates it, i.e., makes it a welcome site for electrophiles.

The increase in electron density can be visualised as :

From structures (I) – (V), we find that the attachment of hydroxyl group to benzene has increased the electron density (-ve charge) on the ring carbon atoms (especially C-2, C-4 and C-6). It is therefore said to have activated the ring towards electrophiles which are attracted to the increased electron density.

Question 17.
Give equations of the following reactions :

1. Oxidation of propan-1 -ol with alkaline KMnO₄ solution.
2. Bromine in CS₂ with phenol.
3. Dilute HNO₃ with phenol.
4. Treating phenol with chloroform in presence of aqueous NaOH.

Solution:

Question 18.
Explain the following with an example.
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis
(iv) Unsymmetrical ether.
Solution:
(i) Kolbe’s reaction :
The fact that phenoxide ion is even more reactive than phenol towards incoming electrophiles is made use of in this reaction. Sodium phenoxide is reacted with CO₂ followed by acid treatment to yield o-hydroxybenzoic acid as the major product.

(ii) Reimer-Tiemann reaction :
Treatment of phenol with chloroform in the presence of aqueous alkali introduces a CHO group at the ortho position. Acidification yields salicylaldehyde.

(iii) Williamson synthesis :
In this method, an alkyl halides is reacted with sodium alkoxide.
R-X + R’ – ONa → R-O-R’ + NaX
The reaction involves SN2 attack of an alkoxide ion on 1° RX.

(iv) Unsymmetrical ethers :
Unsymmetrical ethers are organic compounds where the ethereal oxygen atom is attached to two different alkyl or aryl groups, e.g.,
C₂H₅ – O – CH₃, C₆H₅O – C₂H₅, etc.

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Solution:

Question 20.
How are the following conversions carried out?

1. Propene → Propan-2-ol
2. Benzyl chloride → Benzyl alcohol
3. Ethyl magnesium chloride → Propan-1-ol
4. Methyl magnesium bromide → 2-Methylpropan-2-ol.

Solution:

Question 21.
Name the reagents used in the following reactions :

1. Oxidation of primary alcohol to carboxylic acid.
2. Oxidation of primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-tribromophenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.

Solution:

1. Alkaline KMnO₄
2. Pyridinium chlorochromate in chloromethane (CH₂Cl₂)
3. Br₂/H₂O
4. Alkaline KMnO₄
5. Cone. H₂SO₄ or H₃PO₄ at 433-443 K
6. H₂/Ni or NaBH₄ or LiAlH₄

Question 22.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Solution:
The higher boiling point of ethanol may be attributed to the presence of intermolecular hydrogen bonding in it.

Due to such extensive bonding, more energy needs to be supplied to ethanol to break these bonds and move it into the vapour phase.

Question 23.
Give IUPAC names of the following ethers :

Solution:

1. 1 -Ethoxy-2-methylpropane
2. 2-Chloro-l-methoxyethane
3. 4-Nitroanisole
4. 1-Methoxypropane
5. 1-Ethoxy-4,4-dimethylcyclohexane
6. Ethoxybenzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis :

1. 1-Propoxypropane
2. Ethoxybenzene
3. 2-Methyl-2-methoxypropane
4. 1-Methoxyethane

Solution:

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Solution:
The main limitation of Williamson’s ether synthesis lies in its unemployability for preparation of unsymmetrical ethers where the compound contains secondary or tertiary alkyl groups.

e.g., reaction between tert-butyl bromide and sodium methoxide yields an alkene.

This is because the competing elimination reaction predominates over S_N2 and alkene is formed.

Question 26.
How is 1-propoxypropane synthesised from propan-1 -ol? Write mechanism of this reaction.
Solution:

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Solution:
Consider the reaction between propan- 2-ol molecules in the presence of acid.

If an ether is to be formed, another alcohol molecule must carry out a nucleophilic attack on the carbocation as

However, this does not happen because of
(a) the steric hindrance around the carbocation, and
(b) bulky size of the nucleophile which would further cause crowding.
As a result, the carbocation prefers to lose a proton and forms an alkene.

For the same reason 3° alcohols in the presence of acid do not form ethers since 3° alcohols are even more sterically hindered than 2° alcohols.

Question 28.
Write the equation of the reaction of hydrogen iodide with :

1. 1-propoxypropane
2. methoxybenzene; and
3. benzyl ethyl ether.

Solution:

Question 29.
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Solution:
Consider the following resonance structures of aryl alkyl ethers :

(i) From the above structures we find that the presence of the OR group has increased the electron density on the benzene ring and therefore the ring is said to have been activated towards incoming electrophiles.

(ii) From structures (II), (III) and (IV) we find that and the electron density has increased on C-2, C-4 and C-6, i.e., at the ortho and para positions. As a result the electrophile (E ) attaches itself to these e- rich sites and the -OR group is said to have directed the E® to ortho and para positions.

Question 30.
Write the mechanism of the reaction of Hl with methoxymethane.
Solution:
The reaction between methoxymethane and Hl is :

Question 31.
Write equations of the following reactions :

1. Friedel-Crafts reaction – alkylation of anisole.
2. Nitration of anisole.
3. Bromination of anisole in ethanoic acid medium.
4. Friedel-Craft’s acetylation of anisole.

Solution:

Question 32.
Show how would you synthesise the following alcohols from appropriate alkenes.

Solution:

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place :

Give a mechanism for this reaction.
[Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.]
Solution:

 INTEXT Questions

Question 1.
Write the structures of the following compounds :

(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcydohexane
(iii) 4-terf-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) 1 -Bromo-4-sec-butyl-2-methylbenzene

Solution:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with Kl ?
Solution:
H₂SO₄ is a strong oxidising agent. Therefore, when it is used in presence of KI, it tends to convert KI to HI and finally oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Solution:
The structures of all possible dihalogen derivatives of propane are

 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.
Solution:
An alkane with molecular formula C₅H₁₂ can exist in the following isomeric forms :

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


Solution:
The major haloderivatives formed in the given reactions are

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, bromoform, chloromethane, dibromomethane.
(ii) 1 – Chloropropane, isopropyl chloride, 1 – chlorobutane.
Solution:
The boiling points of organic compounds depend on the strength of the intermolecular forces in them. These forces are :
(a) van der Waals forces and
(b) dipole-dipole interactions These forces are dependent on the
(i) molecular mass and
(ii) surface area of the molecules

(i) As the molecular mass of the compound increases, the boiling point also increases. Therefore the correct order is
chloromethane < bromomethane < dibromomethane < bromoform

(ii) Amongst molecules with same mass, it is the size of the molecule that determines the boiling point. Branched compounds are more compact and therefore have less surface area as compared to their straight chain counterparts and therefore lower boiling point. The order of boiling point is
iso-propyl chloride < 1-chloropropane < 1-chlorobutane

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

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
We know that S_N² mechanism involves a transition state wherein both, the incoming nucleophile as well as the leaving group are present around the carbon atom. There are 5 atoms simultaneously bonded to it.

Thus, for such a transition state to be possible, there should be minimum steric hindrance. Hence, 1° alkyl halides are most reactive towards S_N² followed by 2° and finally 3°.
1° RX > 2° RX > 3° RX

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

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
S_N¹ reaction proceeds via the formation of a carbocation intermediate. This intermediate is formed by the cleavage of the C — X bond. More stable is the resultant carbocation faster is the S_N¹ reaction.
Order of stability of carbocation is
3° carbocation > 2° carbocation >1° carbocation

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

Solution:

10. A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Ans. The hydrocarbon with molecular formula C₅H₁₀ can either a cycloalkane or an alkene.Since the compound does not react with Cl₂ in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since 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 cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.

NCERT Exercises

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)₂CH₂l
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(CI)(C₂H₅)CH₂CH₃
(viii) CH₃CH = C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH = CHC(Br)(CH₃)₂
(x) p-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-CICH₂C₆H₄CH₂C(CH₃)₃
(xii) o-BrC₆H₄CH(CH₃)CH₂CH₃

Haloalkanes And Haloarenes Class 12 NCERT Solutions:

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 = C(CI) C₆H₄ l-p
Solution:

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)-1 -iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1 -Bromo-4-sec-butyl-2-methylbenzene
(viii) 1, 4-Dibromobut-2-ene
Solution:
Structures of the given compounds are :

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

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.
Solution:
A number of structural isomers are possible for molecular formula C₅H₁₀. But, the given compound gives a single monochloro derivative when reacted with Cl₂ in sunlight suggests that, all the H-atoms in the compound are equivalent. This is possible only if the compound is a cyclic alkane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Solution:
The possible isomers of C₄H₉Br are

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

Question 8.
What are ambident nucleophiles? Explain with an example.
Solution:
Ambident nucleophiles are nucleophiles that are capable of attacking the substrate (alkyl halide) through two different atoms.

It so ‘happens due to the presence of two nucleophilic centres which arise from the contributing (resonance) structures that are possible for the ion.

e.g., In NO^–₂ ion, there is a lone pair of electrons on N and therefore makes it nucleophilic while oxygen by virtue of the negative charge acts as a nucleophile. Thus, NO^–₂ can attack via O or N atom thereby making it ambidentate.

Question 9.
Which compound in each of the following pairs will react faster in SN2 reaction with OH^– ?
(i) CH₃Br or CH₃l
(ii) (CH₃)₃CCI or CH₃Cl
Solution:
(i) Between CH₃Br and CH₃I, CH₃I will react faster via the S_N² mechanism. In S_N² mechanism, C – X bond breaks and the faster it breaks faster is the reaction.

I- is a better leaving group. Owing to its large size, the C – I bond breaks faster than the C – Br bond and reaction proceeds further at a greater rate.

(ii) The order of reactivity in an S_N² reaction depends on minimal steric hindrance around the carbon involved in the C – X bond. Lesser the steric hindrance felt by the incoming nucleophile, more reactive will be the alkyl halide towards S_N² reaction.
Based on this, CH₃Cl will react faster than (CH₃)₃CCl.

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-bromopentane.
Solution:

Question 11.
How will you bring about the following conversions?

1. Ethanol to but-1 -yne
2. Ethane to bromoethene
3. Propeneto 1-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1 -ene to but-2-ene
9. 1-Chlorobutane ton-octane
10. Benzene to biphenyl

Solution:

Question 12.
Explain why

1. the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions?

Solution:
(i) (a) In order to understand the lower dipole moment of chlorobenzene we need to look into the contributing structures of the molecules.

(b) From the above structures we find that the C – Cl bond in chlorobenzene has a partial double bond character (structure II, III and IV). As a result, the C – Cl bond length here is shorter than the C – Cl single bond but longer than the C – Cl double bond.

(c) Also evident is the positive charge on Cl atom which reduces the partial negative (δ-) charge which it is expected to carry by the virtue of its electronegativity.

(d) Consequently, the dipole moment, which is a product of bond length and partial negative charge on Cl atom, reduces. However, in cyclohexyl chloride this does not happen. It is an alkyl halide and carbon is purely sp3 hybridised and C – Cl bond has the bond length of a single bond and 8“ appearing on Cl is also higher, thus, the greater dipole moment.

(ii) Only those compounds which can form hydrogen bonds with water are miscible with it. Alkyl halides, though polar due to the presence of electronegative halogen atom, are immiscible since they cannot form hydrogen bonds.

(iii) Grignard reagents R – Mg – X is a class of highly reactive compounds which can extract a proton even from water molecule. They thus, turn into the corresponding alkanes and render any other desired reaction ineffective.

This is why Grignard reagents are prepared in the absolute absence of water (anhydrous conditions), (e.g.,)

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Solution:
(i) Freon 12 (CCl₂F₂) is

• used in aerosol propellants
• refrigeration
• air-conditioning.

(ii) DDT (p, p’- dichlorodiphenyltrichloroethane) is

• used as an insecticide,
• mainly used against mosquitoes.

(iii) Carbontetrachloride (CCl₄) is used

• in manufacture of refrigerants and propellants for aerosol cans
• in synthesis of chlorofluorocarbons
• as degreasing agent
• as cleansing agent
• as a solvent in laboratories

(iv) Iodoform (CHI₃) is used as an antiseptic.

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

Solution:

Question 15.
Write the mechanism of the following reaction

Solution:

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement :

1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methylbutane, 2-Bromo-2- methylbutane, 3-Bromo-2-methylbutane
3. 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methyl butane, 1-Bromo-3- methylbutane.

Solution:
S_N² reaction proceeds via the formation of transition state where the carbon atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to form, the steric interactions have to be minimum. Therefore, the most favourable substrates for S_N² reactions are 1° alkyl halides followed by 2° and 3° alkyl halide. Order of reactivity towards S_N² : 1° > 2° > 3° > aryl halide.

In example (iii), although all the given alkyl halides are 1° but the steric hindrance around the carbon bearing the -Br atom decides the order of reactivity. More the number of bulky groups around this carbon lower will be its reactivity towards S_N².

Question 17.
Out of C₆ H₅ CH₂ Cl and C₆H₅ CHClC₆ H₅, which is more easily hydrolysed by aqueous KOH?
Solution:
C₆ H₅ CHClC₆ H₅ is hydrolysed faster.
(a) Hydrolysis of an alkyl halide is an example of nucleophilic substitution reaction. In case of aryl halides this follows the S_N¹ pathway i.e., via the formation of carbocation.
(b) C₆ H₅ CH₂ Cl or benzyl chloride gives

(c) Out of I & II, carbocation II is more stable. The reason is the presence of two phenyl rings attached to the carbon carrying the positive charge.
(d) As a result, the delocalisation of the +ve charge is greater and the carbocation is more stable. Due to this, (II) is formed faster and the corresponding halide is hydrolysed with greater ease as compared to benzyl chloride.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
Solution:
The para-isomers have high melting points and solubility as compared to their ortho and meta isomers due to symmetry of para-isomers that fits into crystal lattice better than ortho and para isomers.

Question 19.
How can the following conversions be carried out?

1. Propene to propan-1 -ol
2. Ethanol to but-1 -yne
3. 1 -Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bromonitrobenzene
6. Benzyl alcohol to 2-phenylethanoic acid
7. Ethanol to propanenitrile
8. Aniline to chlorobenzene
9. 2-Chlorobutane to 3,4-dimethylhexane
10. 2-Methyl-1 -propene to 2-chloro-2- methylpropane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-Chloropropaneto 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphenyl
19. ferf-Butyl bromide to isobutyl bromide
20. Aniline to phenylisocyanide

Solution:

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.
Solution:
Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.

But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.

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.
Solution:

Question 22.
What happens when

1. n-butyl chloride is treated with alcoholic KOH,
2. bromobenzene is treated with Mg in the presence of dry ether,
3. chlorobenzene is subjected to hydrolysis,
4. ethyl chloride is treated with aqueous KOH,
5. methyl bromide is treated with sodium in the presence of dry ether,
6. methyl chloride is treated with KCN?

Solution:


Chlorobenzene is highly unreactive towards nucleophilic substitution. However, it can be hydrolysed to phenol by heating in aqueous sodium hydroxide solution at a temperature of 623 K and 300 atm pressure. The presence of an electron withdrawing group increases the reactivity of haloarenes.

INTEXT Questions

Question 1.
Write the formulas for the following coordination compounds :

1. Tetraamminediaquacobalt (lll) chloride
2. Potassium tetracyanonickelate (ll)
3. Tris(ethane-1,2-diamine) chromium (lll) chloride
4. Amminebromidochloridonitrito-N- platinate (ll)
5. Dichloridobis(ethane-1,2-diamine) platinum (IV) nitrate
6. Iron (lll) hexacyanoferrate (ll)

Solution:

Question 2.
Write the IUPAC names of the following coordination compounds :

Solution:

1. Hexaamminecobalt (III) chloride
2. Pentaamminechloridocobalt (III) chloride (HO₃)₂
3. Potassium hexacyanoferrate (III)
4. Potassium trioxalatoferrate (III)
5. Potassium tetrachloridopalladate (II)
6. Diamminechlorido (methylamine) platinum (II) chloride

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :


Solution:

Question 4.
Give evidence that [CO(NH₃)5CI] SO₄ and [CO(NH₃)₅ SO₄]Cl are ionisation isomers.
Solution:
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents :

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCI₄]^2- ion with tetrahedral geometry is paramagnetic.
Solution:

Question 6.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Solution:
In Ni(CO)₄, Ni is in zero oxidation state whereas in NiCl₄^2-, it is in +2 oxidation state. In the presence of 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.
Solution:
In presence of CN^–, (a strong ligand), the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d² sp³ forming inner orbital complex. In the presence of H₂O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³ d² 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.
Solution:
in the presence of NH₃, the 3d electrons pair up leaving two d orbitals empty to be involved in d² sp³ hybridisation forming inner orbital complex in case of [CO(NH₃)₆]³⁺.
In Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d³ configuration, the hybridisation involved is sp³ d² forming outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planer [Pt(CN)₄]^2- ion.
Solution:
For square planer shape, the hybridisation is dsp² . Hence, the unpaired electrons in 5d orbital pair up to make one f-orbital empty for dsp² hybridisation. Thus there is no unpaired electron.

Question 10.
The hexaaquomanganese (ll) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Solution:
Mn(II) has 3d⁵ electronic configuration. Water is a weak field ligand and therefore ∆₀ is small. Therefore, the hexaaqua complex will be high spin complex containing 5 unpaired electrons. On the other hand, CN^– is a strong field ligand and therefore, ∆₀ is large. Therefore, in its cyano complex, the electrons pair up and have only one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)₄²⁺ ion, given that β₄ for this complex is 2.1 × 10¹³.
Solution:

NCERT Exercises

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Solution:
The main postulates are :

1. In coordination compounds metals show two types of linkages (valencies)-primary and secondary.
2. The primary valencies are normally ionisable and are satisfied by negative ions.
3. The secondary valencies are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valency is equal to the coordination number and is fixed for a metal.
4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial .arrangements corresponding to different coordination numbers.

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?
Solution:
FeSO₄ solution when mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio forms a double salt FeSO₄(NH₄)₂SO₄-6H₂O which when dissolved in water dissociates into simple ions to give tests for its constituent ions. When CuSO₄ is mixed with aqueous ammonia a complex ion [Cu(NH₃)₄]²⁺ is formed which does not give Cu²⁺ in the solution.

Question 3.
Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Solution:
(i) Coordination entity : A coordination entity constitutes a central metal atom or ion bonded loa fixed number of ions or molecules. For example, [COCl₃(NH₃)₃] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are [Ni(CO)₄], [PtCl₂(NH₃)₂], [Fe(CN)₆]^4-, [CO(NH₃)₆]³⁺.

(ii) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^–, small molecules such as H₂O or NH₃, larger molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even macromolecules, such as proteins.

(iii) Coordination number : The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl₆]^2- and [Ni(NH₃)₄]²⁺, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C₂O₄)₃]^3- and [CO(en)₃]³⁺,the coordination number of both, Fe and Co, is 6 because C₂O₄^2-and en (ethane-1,2-diamine) are bidentate ligands.

(iv) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄]^2- is square planar.

(v) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH₃)₆]³⁺, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor groups, e.g., [CO(NH₃)₄Cl₂]⁺, are known as heteroleptic.

Question 4.
What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Solution:
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl^– , H₂O.

Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H₂NCH₂CH₂NH₂), oxalate (C₂O₄^2-) ion.

Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO₂^– and SCN^– ion.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities :

Solution:

Question 6.
Using IUPAC norms write the formulas for the following :

1. Tetrahydroxozincate (ll)
2. Potassium tetrachloridopalladate (ll)
3. Diamminedichloridoplatinum (ll)
4. Potassium tetracyanonickelate (ll)
5. Pentaamminenitritio-O-cobalt (lll)
6. Hexaamminecobalt (lll) sulphate
7. Potassium tri(oxalato)chromate (lll)
8. Hexaammineplatinum (IV)
9. Tetrabromidocuprate (ll)
10. Pentaamminenitrito-N-cobalt (lll)

Solution:

Question 7.
Using IUPAC norms write the systematic names of the following :

Solution:

1. Hexaamminecobalt (III) chloride
2. Diamminechloridomethylamine platinum (II) chloride
3. Hexaaquatitanium (III) ion
4. Tetraamminechloridonitrito-N-cobalt (III) chloride
5. Hexaaquamanganese (II) ion
6. Tetrachloridonickelate (II) ion
7. Hexaammine nickel (II) chloride
8. Tris (ethane-1,2-diamine) cobalt(III) ion
9. Tetracarbonylnickel (0)

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
Solution:
(i) Geometrical isomerism : This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX₂L₂] (X and L are unidentate), the two identical ligands may be arranged adjacent to each other in a cis- isomer, or opposite to each other in a trans isomer.

Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma₃b₃] like [CO(NH₃)₃(NO₂)₃]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.

(ii) Optical isomerism : Optical isomerism is common in octahedral complexes involving bidentate ligands. In a coordination entity of the [PtCl₂(en)₂]²⁺, only the ds-isomer shows optical activity.

(iii) Linkage isomerism : Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS^–, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH₃)₅(NO₂)]Cl₂, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO₂).

(iv) Coordination isomerism : This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH₃)₆] [Cr(CN)₆], in which the NH₃ ligands are bound to CO³⁺ and the CN^– ligands to Cr³⁺. In its coordination isomer [Cr(NH₃)₆][CO(CN)₆], the NH₃ ligands are bound to Cr³⁺ and the CN^– ligands to CO³⁺.

(v) Ionisation isomerism : This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [CO(NH₃)₅SO₄]Br and [CO(NH₃)₅Br]SO₄.

(vi) Solvate isomerism : This form of isomerism is known as ‘hydrate isomerism in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H₂O)₆]Cl₃ (violet) and its solvate isomer [Cr(H₂O)₅Cl]Cl₂ H₂O (grey green).

Question 9.
How many geometrical isomers are possible in the following coordination entities?

1. [Cr(C₂O₄)₃]^3-
2. [CO(NH₃)₃CI₃]

Solution:

1. Zero
2. Two – Facial and meridional.

Question 10.
Solution:

Question 11.
Draw all the isomers (geometrical and optical) of :

Solution:

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(CI)(py)] and how many of these will exhibit optical isomerism?
Solution:

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Solution:

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?
Solution:

CN^– is a strong field ligand, therefore, [Cu(CN)₄]^2- is highly stable and has large value of stability constant. On passing H₂S, CuS is not formed because this coordination entity does not give Cu²⁺ ion.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory :

Solution:

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Solution:
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with metal atom at the origin.

As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.

The orbitals lying along the axes (d_z2 and d_x2 – y₂) get repelled more strongly than d_xy, d_yz and d_zx orbitals which have lobes directed between the axes.

The d_z2 and d_x2-y2 orbitals get raised in energy and d_xy d_yz, d_xz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets : the lower energy orbitals set, t_2g and the higher energy orbitals set, e_g. The energy is separated by ∆₀

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Solution:
The crystal field splitting, ∆₀, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below :

Such a series is termed as spectrochemical series.

If ∆₀ < P, the fourth electron enters one of the eg orbitals giving the configuration t₂³_g g e¹_g. Ligands for which ∆₀ < P are known as weak field ligands and form high spin complexes. If ∆₀ > P, it becomes more energetically favourable for the fourth electron to occupy a t_2g orbital with configuration t_2g ⁴e⁰_g. Ligands which produce this effect are known as strong field ligands and form low spin complexes.

Question 18.
What is crystal field splitting energy? How does the magnitude of ∆₀ decide the actual configuration of d-orbitals in a coordination entity?
Solution:

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Solution:

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Solution:
[Ni(H₂O)₆]2+has unpaired electrons due to weak H₂O ligands which absorb light from visible region and radiate complementary colour i.e., green whereas [Ni(CN)₄]^2- does not have any unpaired electron due to strong CN^– ligand, therefore, does not absorb light from visible region hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Solution:


Both ligands show different magnitude of crystal field splitting energy due to different nature hence, absorb different wavelengths ans show different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Solution:

The metal-carbon bond in metal carbonyls possess both s and p character. The M-C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :

Solution:

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state,electronicconfigurationandcoordination number. Also give stereochemistry and magnetic moment of the complex :

Solution:

Question 25.
What is meant by stability of the coordination compound in solution? State the factors which govern stability of complexes.
Solution:
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :
 then, larger the stability constant, the higher is the proportion of ML₄ that exists in solution. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules, L and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows :

Factors affecting stability of complexes :

1. Smaller the size of cation, greater will be the stability of complex e.g., Fe³⁺ forms more stable complex than Fe²⁺.
2. Greater the charge on central metal ion, more stable will be the complex e.g., Pt⁴⁺ forms more stable complex than Pt²⁺.
3. Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH₃.

Question 26.
What is meant by the chelate effect1 Give an example.
Solution:
When a di-or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands. Example : EDTA, DMG, etc.

Question 27.
Discuss briefly giving an example in each case the role of coordination compound in :

1. biological systems
2. medicinal chemistry
3. analytical chemistry
4. extraction/metallurgy of metals.

Solution:
(i) Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin, the red pigment 1 of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B₁₂, cyanocobalamine, the anti- pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).

(ii) There is growing Interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iron are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds.

EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. Examples are: ds-platin and related compounds.

(iii) Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), a-nitroso- β-naphthol, cupron, etc.

(iv) Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]- in aqueous solution. Gold can be separated in metallic form from this solution by the addition of zinc.

Question 28.
How many ions are produced from the complex CO(NH₃)₆CI₂ in solution?

(a) 6
(b) 4
(c) 3
(d) 2

Solution:

Question 29.
Amongst the following ions which one has the highest magnetic moment value?

(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [Zn(H₂O)₆]²⁺

Solution:

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(a) +1
(b) +3
(c) -1
(d) -3
Solution:
(c) : + 1 + x + 4(0) = 0
∴ x = – 1.

Question 31.
Amongst the following, the most stable complex is
(a) [Fe(H₂O)₆]³⁺
(b) [Fe(NH₃)₆]³⁺
(c) [Fe(C₂O₄)₃]^3-
(d) [FeCI₆]^3-
Solution:
(c) : Since C₂O₄^2- is a bidentate ligand, it forms the most stable complex.

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following :

Solution:
In spectrochemical series the order of the given ligands is H₂O < NH₃ < NO₂– . Hence, the wave length of light will be absorbed in the opposite order since E = 

INTEXT Questions

Question 1.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Solution:
Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence it is a transition element.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1. Why?
Solution:
In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d-series, electrons from the d-orbitals are always involved in the formation of metallic bonds.

Question 3.
Which of the 3d-series of the transition metals exhibits the largest number of oxidation states and why?
Solution:
Manganese (Z = 25), as its atom has the maximum number of unpaired electrons.

Question 4.
The E°(M²⁺/M) value for copper is positive (+0.34 V). What is possibly the reason for this? (Hint: consider its high ∆_aH° and low ∆_hydH°)
Solution:
The high energy to transform Cu_(s) to Cu²⁺_(aq) is not found balanced by its hydration energy. Hydration energy and lattice energy of Cu²⁺ is more than Cu.

Question 5.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Solution:
Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d – configurations (e.g., d°, d⁵, d¹⁰ are exceptionally stable).

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Solution:
Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.

Question 7.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Solution:
Cr²⁺ is stronger reducing agent than Fe²⁺ Reason: d⁴ → d³ occurs in case of Cr²⁺ to Cr²⁺. But d⁶ → d⁵ occurs in case of Fe²⁺ to Fe²⁺.
In a medium (like water) d’1 is more stable as compared to d⁵.

Question 8.
Calculate the ‘spin only’ magnetic moment of M²⁺_(aq) ion (Z = 27).
Solution:

Question 9.
Explain why Cu⁺ ion is not stable in aqueous solutions?
Solution:
Cu⁺ in aqueous solution undergoes disproportionation, i.e.,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
The stability of Cu²⁺_(aq) rather than Cu⁺_(aq) is due to the much more negative ∆_hyd H°of Cu²⁺_(aq) than Cu⁺, which more than compensates for the second ionisation enthalpy of Cu.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Solution:
The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the seriest.

NCERT Exercises

Question 1.
Write down the electronic configuration of

1. Cr³⁺
2. Pm³⁺
3. Cu⁺
4. Ce⁴⁺
5. CO²⁺
6. Lu²⁺
7. Mn²⁺
8. Th⁴⁺

Solution:

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation of their +3 state?
Solution:
Mn²⁺ has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰. To go to Mn³⁺ state the electron has to be taken out from stable d⁵ orbital which is half filled and requires very high ionisation energy. In case of Fe²⁺ ion, the third electron is taken out from 3d⁶ configuration which results in more stable 3d⁵ configuration.

Question 3.
Explain briefly how +2 state becomes more and more stable in the first half of the first row With increasing atomic number the transition elements with increasing atomic numbers?
Solution:
With increasing atomic number the effective nuclear charge increases after losing two electrons from s-orbital. The ionic size decreases which results in more stability. The slability is less in the beginning due to too few electrons to lose or share.

Question 4.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Solution:
Sc³⁺ has stable electronic configuration (vacant d-orbital), therefore Sc³⁺ is more stable than Sc⁺.

Fe³⁺ is more stable than Fe²⁺ due to half filled d-orbitals.

Mn²⁺ is more stable than Mn³⁺ due to half filled d-orbitals.

V⁵⁺ is more stable (due to vacant d-orbital) than V³⁺.

Question 5.
What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d³, 3d⁵, 3d⁸ and 3d⁴ ?
Solution:

1. 3d³ : Stable oxidation state will be +5 due to outer electronic configuration 3d³4s² (+2, +3, +4, +5).
2. 3d⁵: Stable oxidation state will be +2 and +7 due to outer electronic configuration 3d⁵ 4s² (+2, +3, +4, +6, +7).
3. 3d⁸: Stable oxidation state will be +2 due to outer electronic configuration 3d⁸ 4s² (+2, +3, +4).
4. 3d⁴ : Stable oxidation state will be +3 and +6 due to outer electronic configuration 3d⁴ 4s¹. There is no d⁴ configuration in ground state, as it becomes 3d⁵ 4s¹

Question 6.
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Solution:
Manganese shows oxidation state of +7 in its oxometal anion MnO₄^– which is equivalent to its group number 7. Cr in Cr₂O₇^2- and CrO₄^2- show oxidation state +6 which is equivalent to its group number 6.

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Solution:
The overall decrease in atomic and ionic radii from lanthanum to lutetium is a unique feature in the chemistry of the lanthanoids. It has far reaching consequences in the chemistry of the third transition series of the elements.

This contraction is attributed to the imperfect shielding of one electron by another in the same sub-shell. However, the shielding of one 4f electron by another is less than one d electron by another with the increase in nuclear charge along the series. There is fairly regular decrease in the sizes with increasing atomic number.

The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their occurrence together in nature and for the difficulty faced in their separation.

Question 8.
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Solution:
General characteristics of transition elements.

1. Electronic configuration – (n – 1) d^1-10 ns^1-2
2. Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.
3. Atomic and ionic size – Ions of same charge in a given series show progressive decrease in radius with increasing atomic number.
4. Oxidation state – Variable ; ranging from + 2 to + 7.
5. Paramagnetism – The ions with unpaired electrons are paramagnetic.
6. Ionisation enthalpy – Increases due to increase in molecular charge.
7. Formation of coloured ions – Due to unpaired electrons.
8. Formation of complex compounds – Due to small size and high charge density of metal ions.
9. They possess catalytic properties – Due to their ability to adopt multiple oxidation states.
10. Formation of interstitial compounds.
11. Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or any stable oxidation state and they are placed between s and p-block elements. Zn, Cd and Hg have fully filled d° configuration in their ground state hence may not be regarded as the transition elements.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?.
Solution:
The electronic configuration of the transition elements is (n – 1 )d^1-10 ns^1-2. (n-1) stands for penultimate shell and d-orbitals may have one to ten d electrons and n denotes valence s or the outermost shell which can have one or two electrons. Hence the basic difference in electronic configuration of transition metals is that their penultimate shell is incomplete and progressively filled and not the valence shell.

Question 10.
What are the different oxidation states exhibited by the lanthanoids?
Solution:
In lanthanoids +3 oxidation state is predominant. However occasionally +2 and +4 ions in the solution or in solid compounds are also obtained, e.g. Ce⁴⁺, Tb⁴⁺, Eu²⁺, Yb²⁺, etc. +2 and +4 oxidation states are exhibited due to extra stability of empty, half – filled or fully filled f – subshells.

Question 11.
Explain giving reasons :

1. Transition metals and many of their compounds show paramagnetic behaviour.
2. The enthalpies of atomisation of the transition metals are high.
3. The transition metals generally form coloured compounds.
4. Transition metals and their many compounds act as good catalysts.

Solution:
(i) Paramagnetism arises from the presence of unpaired electrons, each such electron has magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment,is determined by the number of unpaired electrons and is calculated by using the ‘spin’ only’ formula, i.e., µ =  B.M

Where n is the number of unpaired electrons and µ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magneton (BM).

(ii) Because of large number of unpaired electrons in their atoms they have stronger interatomic interactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.

(iii) Due to presence of unpaired electrons and d-d transitions, the transition metals are generally coloured. When an electron from a lower energy d orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand.

(iv) The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in r Catalytic Hydrogenation) are some of the examples.Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst.

Question 12.
What are interstitial compounds? Why are such compounds well known for transitions metals ?
Solution:
Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent, for example, TiC, Mn₄N, Fe₃H, VH_0.56 and TiH1.7, etc.

These do not correspond to any normal oxidation state of the metal. Because of the nature of their composition, these compounds are referred to as interstitial compounds.The principal physical and chemical characteristics of these compounds are as follows :

1. They have high melting points, higher than those of pure metals.
2. They are very hard, some borides approach diamond in hardness.
3. They retain metallic conductivity
4. They are chemically inert.

Question 13.
How is the variability in oxidation states of transition metals different from that of the non-transition metals ? Illustrate with examples.
Solution:
The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., V^II, V^III, V^IV, V^V. This is in contrast with the variability of oxidation sates of non transition elements where oxidation states normally differ by a unit of two.

In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO₃ and WO₃ are not.

Question 14.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Solution:
Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows :
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7-2H2O can be crystallised.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. If pH of potassium dichromate is increased it is converted to yellow potassium chromate.
2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with

1. iodide
2. iron(ll) solution and
3. H₂S.

Solution:
K₂Cr₂O₇ is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from +6 to +3. The oxidising action can be represented as follows :

Question 16.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (ii) ions (ii) SO₂ and (iii) oxalic acid ? Write the ionic equations for the reactions.
Solution:
Potassium permanganate (KMnO₄) is prepared by the fusion of a mixture of pyrolusite (MnO2), potassium hydroxide and oxygen, first green coloured potassium manganate is formed.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.
Electrolytically :

Question 17.
For M²⁺/M and M³⁺/M²⁺ systems the P values for some metals are as follows
Use this data to comment upon

1. the stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and
2. the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution:

1. From the E° values for M³⁺/M²⁺ for Cr, Mn and Fe, it is very clear that Fe³⁺ is more stable than Mn³⁺ in an acidic medium, whereas less stable than Cr³⁺ because of the higher reduction potential in comparison to Cr³⁺/Cr²⁺ and lower reduction potential than Mn³⁺/Mn²⁺.
2. Reduction potential for Mn²⁺/Mn is most negative and therefore, it will be most easily oxidised and ease of getting oxidised will be Mn > Cr > Fe.

Question 18.
Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and CO³²⁺. Give reasons for each.
Solution:
The configuration of the given metal ions can be given as

Question 19.
Compare the stability of +2 oxidation state for the elements of the first transition series.
Solution:
The common oxidation state of 3d series elements is + 2 which arises due to participation of only 4s electrons. The tendency to show highest oxidation state increases from Sc to Mn, then decreases due to pairing of electrons in 3d subshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). At the other end of the series, oxidation state of Zn is +2 only.

Question 20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to
(i) electronic configuration
(ii) atomic and ionic sizes
(iii) oxidation state and
(iv) chemical reactivity.
Solution:
(i)Electronicconfiguration : Lanthanoids have general electronic configuration of [Xe] 4f^1-14 5d^0-1 6s² and actinoids have general electronic configuration of [Rn]5f^1-14 6d^0-1 7s2. Thus, lanthanoids belong to 4 f-series whereas actinoids belong to 5 f-series.

(ii) Atomic and ionic sizes : The atomic size of lanthanoids decreases from lanthanum to lutetium. Though the decrease is not regular, in case of atomic radii, the decrease in the ionic size (M³⁺) is regular. Decrease in size between two successive elements is higher in actinoids due to poor screening by 5f electrons.

(iii) Oxidation state : The most common oxidation state of lanthanoids is +3 while actinoids show more variable oxidation states than lanthanoids ranging from +3 to +7. The tendency of showing greater range of oxidation states can be attributed to the fact that the 5f 6d and 7s levels are of comparable energies and larger distance of 5fas compared to 4f from the nucleus.

(iv) Chemical reactivity : Actinoids are far more reactive than lanthanoids. They react with non-metals at moderate temperatures whereas lanthanoids react at high temperatures. Most actinoids are attacked by HCl but are slightly affected by HNO₃ due to formation of a protective layer of oxide and alkalies give no reaction.

Lanthanoids liberate hydrogen from dilute acids and burn in halogens to form halides.

Question 21.
How would you account for the following?
(i) Of the d⁴ species, Cr²⁺ is strongly reducing while manganese(lll) is strongly oxidising.
(ii) Cobalt(ll) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.
Solution:
(i) Both Cr2+ and Mn3+ have d4 configuration, Cr2+ is reducing since its configuration is converted to d3 from d4. d3 has half filled t28, configuration with higher stability. Mn3+ is oxidising since in changing from d4 to d5 the configuration becomes half filled which has extra stability,

(ii) Co(II) gets oxidised to Co(III) in presence of complexing agent because Co(III) is more stable than Co(II). Most of the strong field ligands cause pairing of electrons forming diamagnetic octahedral complexes which are very stable due to very large crystal field stabilization energy.

(iii) d¹ configuration is very unstable in ions because after losing one more electron it will become more stable due to vacant d-orbital. All elements with d¹ configuration are either reduced or undergo disproportionation, e.g.,

Question 22.
What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Solution:
Disproportionation reaction involves the oxidation and reduction of the same substance. The two examples of disproportionation reaction are

1. 2Cu⁺ → Cu²⁺ + Cu
2. 3MnO₄^2- + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O

Question 23.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Solution:
Copper exhibits +1 oxidation state in the first series of transition metals because when one electron is lost, the configuration becomes stable due to fully filled d¹⁰ configuration.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions : Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which ono of these is the most stable in aqueous solution ?
Solution:

Question 25.
Give examples and suggest reasons for the following features of the transition metal chemistry :
(i) The lowest, oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Solution:
(i) Lowest oxidation compounds of transition metals are basic due to their ability to get oxidised to higher oxidation states. Whereas the higher oxidation state of metal and compounds gets reduced to lower ones and hence acts as acidic in nature.
e.g. MnO is basic whereas Mn₂O₇ is acidic.

(ii) Due to high electronegativities of oxygen and fluorine, the oxides and fluorides of transition metals exhibit highest oxidation state.
e.g. OSF₆,V₂O₅

(iii) In oxoanions of metals, the metals from bonds with oxygen and hence are present in their highest oxidation states. For example : Cr forms CrO₄^2- and Cr2O₇^2-, both contain chromium in +6 oxidation state.

Permanganate ion, MnO₄^– contains Mn in its highest oxidation state of +7.

Question 26.
Indicate the steps in the preparation of

1. K₂Cr₂O₇ from chromite ore
2. KMnO₄ from pyrolusite ore.

Solution:

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Solution:
An allov is a blend of metals prepared by mixing the components. Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of other. Misch metal is an alloy which contains some of the lanthanoid metals. It contains 95% lanthanoid metals, 5% iron and traces of S, C, Ca and Al. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 28.
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers oftheinnertransition element: 29, 59, 74, 95,102,104.
Solution:
Lanthanoids and actinoids are called inner transition elements because inner f-orbitals are progressively filled and the last electron goes to anti penultimate f orbital.

Elements with atomic number 59, 95, 102 are inner transition metals because they belong to lanthanoids and actinoids.

Question 29.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Solution:
The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z = 103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult.

There is a greater range of oxidation states, which is attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements, The actinoids resemble the lanthanoids in having more compounds in +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse. Because the distribution of oxidation states among the actinoids is so uneven and so different for the earlier and latter elements. It is unsatisfactory to review their chemistry in terms of oxidation states.

Question 30.
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Solution:
Lawrencium (Lr) is the last element of actinoids. Its outer electronic configuration is 5f¹⁴ 6d¹ 7s² and its possible oxidation state is +3.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Solution:
The electronic configuration of Ce³⁺ is 4f¹

Question 32.
Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behaviour with the electronic configurations of these elements.
Solution:
All lanthanoids show Ln³⁺ oxidation state. But some of them show +2 and +4 states also in solution and solid form like Ce⁴⁺, Eu²⁺, Yb²⁺, Tb⁴⁺, etc. The variable oxidation state is related to electronic configuration due to extra stability of half filled, fully filled or empty orbitals, e.g., Ce⁴⁺ has 4f⁰, Eu²⁺ has 4f⁷ Tb⁴⁺ has 4f⁷ and Yb²⁺ has 4f¹⁴ configuration.

Question 33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.
Solution:
Refer answer number 20.

Question 34.
Write the electronic configurations of the elements with the atomic numbers 61,91,101, and 109.
Solution:

Question 35.
Compare the general characteristics of the first series of transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points
(i) electronic configurations
(ii) oxidation states
(iii) ionisation enthalpies and
(iv) atomic sizes.
Solution:
(i) Electronic configurations : In 1^st transition series, 3d-orbitals are progressively filled whereas in 2^nd transition series, 4d-orbitals are progressively filled and in 3^rd transition series, 5d-orbitals are progressively filled.

(ii) Oxidations states : Elements show variable oxidation states in both the series. The highest oxidation state is equal to total number of electrons in ‘s’ as well as ‘d’ orbitals. The number of oxidation states shown are less in 5d transition series than 4d series. In 3d series +2 and +3 oxidation states are common and they form stable complexes in these oxidation states. In other series OsO₄ and PtF₆ are formed which are quite stable in higher oxidation state.

(iii) Ionisation enthalpies : The ionisation enthalpies in each series generally increases gradually from left to right. The ionisation enthalpy of 5d series do not differ appreciably due to lanthanoid contraction.

(iv) Atomic sizes : The atomic sizes of 4d and 5d-series do not differ appreciably due to lanthanoid contraction. The atomic radii of second and third series are larger than 3d series.

Question 36.
Write down the number of 3d electrons in each of the following ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, CO²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Solution:

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Solution:

1. In first transition series lower oxidation state is more stable whereas in heavier transition elements higher oxidation states are more stable.
2. The ionisation enthalpy of 5d transition series is higher than 3d and 4d transition series.
3. M-M bonding is most common in heavier transition metals but less in first series.
4. The elements of first transition series do not form complexes with higher coordination number of 7 and 8.
5. The elements of first series can form high spin or low spin complexes depending upon strength of ligands but elements of other series form low spin complexes irrespective of strength of ligands.

Question 38.
What can be inferred from the magnetic moment values of the following complex species?

Solution: