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Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4Question 12.9:

Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?

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Electron Movement, Electron Displacement Effects in Organic Compounds Read now to understand this topic better »

NO2 group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO2 group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O2NCH2CH2O– is expected to be more stable than CH3CH2O–.

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2OWrite the formulae for the following compounds:

(a) Mercury(II) chloride (b) Nickel(II) sulphate

(c) Tin(IV) oxide (d) Thallium(I) sulphateQuestion 12.4: Question 12.7:

Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :

(a) 2,2,4-Trimethylpentane

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid

(c) Hexanedial

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Organic Compounds : Shapes and Structures Read now to understand this topic better »

(a) 2, 2, 4–trimethylpentaneExplain why alkyl groups act as electron donors when attached to a π system.

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Resonance Structure, Resonance Effect, Electromeric Effect, and Hyperconjugation Read now to understand this topic better »

When an alkyl group is attached to a π system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the C–H bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a sp3 –s sigma bond orbital with an empty p orbital of the π bond of an adjacent carbon atom.

The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the π electrons, making the molecule more stable.

Condensed formula:

(CH3)2CHCH2C (CH3)3

Bond line formula:

(b) 2–hydroxy–1, 2, 3–propanetricarboxylic acid

Condensed Formula:

(COOH)CH2C(OH) (COOH)CH2(COOH)

Bond line formula:

The functional groups present in the given compound are carboxylic acid (–COOH) and alcoholic (–OH) groups.

(c) Hexanedial

Condensed Formula:

(CHO) (CH2)4 (CHO)

Bond line Formula:

The functional group present in the given compound is aldehyde

(–CHO).

Give the IUPAC names of the following compounds:

(a)

(b) Identify the functional groups in the following compounds

(a)

(b)

(c)

(d)

(e)

(f) Cl2CHCH2OH

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IUPAC (International Union of Pure and Applied Chemistry) System of Nomenclature of Organic Compounds Read now to understand this topic better »

(a)

3–phenyl propane

(b)

2–methyl–1–cyanobutane

(c)

2, 5–dimethyl heptane

(d)

3–bromo–3–chloroheptane

(e)

3–chloropropanal

(f) Cl2CHCH2OH

1, 1–dichloro–2–ethanol

(e) Iron(III) sulphate (f) Chromium(III) oxide

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Oxidation Number Read now to understand this topic better »

(a) Mercury (II) chloride:

HgCl2

(b) Nickel (II) sulphate:

NiSO4

(c) Tin (IV) oxide:

SnO2

(d) Thallium (I) sulphate:

Tl2SO4

(e) Iron (III) sulphate:

Fe2(SO4)3

(f) Chromium (III) oxide:

Cr2O3

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Oxidation Number Read now to understand this topic better »

(a)

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

Then, we have

Hence, the oxidation number of P is +5.

(b)

Then, we have

Hence, the oxidation number of S is + 6.

(c)

Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

(d) 2K(s) + F2(g) → 2K+F– (s)

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

AnswerDraw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(a) C6H5OH (b) C6H5NO2 (c) CH3CH = CH – CHO

(d) C6H5CHO (e) (f)

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Resonance Structure, Resonance Effect, Electromeric Effect, and Hyperconjugation Read now to understand this topic better »

(a) The structure of C6H5OH is:

The resonating structures of phenol are represented as:

(b) The structure of C6H5NO2 is:

The resonating structures of nitro benzene are represented as:

(c) CH3CH = CH – CHO

The resonating structures of the given compound are represented as:

(d) The structure of C6H5CHO is:

The resonating structures of benzaldehyde are represented as:

(e) C6H5CH2⊕

The resonating structures of the given compound are:

(f) CH3 CH = CH CH2⊕

The resonating structures of the given compound are:

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Oxidation Number Read now to understand this topic better »

(a)

Let us write the oxidation number of each element involved in the given reaction as:

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.

(b)

Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.

(c)

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

(d)

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to –1 in KF i.e., F2 is reduced to KF.

Hence, the above reaction is a redox reaction.

(e)

The oxidation number of each element in the given reaction can be represented as:

Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

Then, we have

Hence, the oxidation number of P is + 5.

(d)

Then, we have

Hence, the oxidation number of Mn is + 6.

(e)

Then, we have

Hence, the oxidation number of O is – 1.

(f)

Then, we have

Hence, the oxidation number of B is + 3. Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.

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Organic Compounds : Shapes and Structures Read now to understand this topic better »

The bond line formulae of the given compounds are:

(a) Isopropyl alcohol

(b) 2, 3–dimethyl butanal

(c) Heptan–4–one

(g)

Then, we have

Hence, the oxidation number of S is + 6.

(h)

Then, we have

(a) KI3

In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to find the oxidation states.

In a KI3molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI3molecule, the O.N. of the two I atoms forming the I2molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(b) H2S4O6

Question 8.4:

Fluorine reacts with ice and results in the change:

H2O(s) + F2(g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction.

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Types of Redox Reactions Read now to understand this topic better »

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to –1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

Question 12.6:

Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2

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Classification of Organic Compounds Read now to understand this topic better »

The first five members of each homologous series beginning with the given compounds are shown as follows:

(a)

H–COOH : Methanoic acid

CH3–COOH : Ethanoic acid

CH3–CH2–COOH : Propanoic acid

CH3–CH2–CH2–COOH : Butanoic acid

CH3–CH2–CH2–CH2–COOH : Pentanoic acid

(b)

CH3COCH3 : Propanone

CH3COCH2CH3 : Butanone

CH3COCH2CH2CH3 : Pentan-2-one

CH3COCH2CH2CH2CH3 : Hexan-2-one

CH3COCH2CH2CH2CH2CH3 : Heptan-2-one

(c)

H–CH=CH2 : Ethene

CH3–CH=CH2 : Propene

CH3–CH2–CH=CH2 : 1-Butene

CH3–CH2–CH2–CH=CH2 : 1-Pentene

CH3–CH2–CH2–CH2–CH=CH2 : 1-Hexene

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c)

On taking the O.N. of O as –2, the O.N. of Fe is found to be . However, O.N. cannot be fractional.

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

(d)

2 (x) + 6 (+1) + 1 (-2) = 0While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

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Redox Reactions: Oxidation and Reduction Reactions Read now to understand this topic better »

In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to –2.

Therefore, SO2 can act as an oxidising as well as a reducing agent.

In hydrogen peroxide (H2O2), the O.N. of O is –1 and the range of the O.N. that O can have is from 0 to –2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H2O2 can act as an oxidising as well as a reducing agent.

In ozone (O3), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to –2. Therefore, the O.N. of O can only decrease in this case. Hence, O3 acts only as an oxidant.

In nitric acid (HNO3), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to –3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO3 acts only as an oxidant.

Consider the reactions:

(a) 6 CO2(g) + 6H2O(l) → C6 H12 O6(aq) + 6O2(g)

(b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)

Why it is more appropriate to write these reactions as:

(a) 6CO2(g) + 12H2O(l) → C6 H12 O6(aq) + 6H2O(l) + 6O2(g)

(b) O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox

reactions.

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Balancing of Redox Reactions Read now to understand this topic better »

(a)The process of photosynthesis involves two steps.

Step 1:

H2O decomposes to give H2 and O2.

Step 2:

The H2 produced in step 1 reduces CO2, thereby producing glucose (C6H12O6) and H2O.

Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.

The path of this reaction can be investigated by using radioactive H2O18 in place of H2O.

(b) O2 is produced from each of the two reactants O3 and H2O2. For this reason, O2 is written twice.

The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O2 reacts with the O produced in the first step, thereby producing H2O and O2.

The path of this reaction can be investigated by using or .

or, 2x + 4 = 0

or, x = -2

Hence, the O.N. of C is –2. Question 12.5:

Which of the following represents the correct IUPAC name for the compound