Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 8.1:

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

(a) NaH₂PO₄ (b) NaHSO₄ (c) H₄P₂O₇ (d) K₂MnO₄

(e) CaO₂ (f) NaBH₄ (g) H₂S₂O₇ (h) KAl(SO₄)₂.12 H₂O

Answer:

(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) 

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.

(g) 

Then, we have

Hence, the oxidation number of S is + 6.

(h) 

Then, we have

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

Hence, the oxidation number of S is + 6.

Question 8.2:

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) KI₃ (b) H₂S₄O₆ (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH

Answer:

(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

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

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) = 0

or, 2x + 4 = 0

or, x = -2

Hence, the O.N. of C is –2.

(e) 

2 (x) + 4 (+1) + 2 (-2) = 0

or, 2x = 0

or, x = 0

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH₃COOH.

Question 8.3:

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)

Answer:

(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.

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.

Answer:

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.

Question 8.5:

Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, –. Suggest structure of these compounds. Count for the fallacy.

Answer:

(i) 

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.

The structure of H₂SO₅ is shown as follows:

Therefore, the O.N. of S is +6.

(ii) 

Here, there is no fallacy about the O.N. of Cr in .

The structure of  is shown as follows:

Here, each of the two Cr atoms exhibits the O.N. of +6.

(iii)

Here, there is no fallacy about the O.N. of N in .

The structure of  is shown as follows:

The N atom exhibits the O.N. of +5.

Question 8.6:

Write the formulae for the following compounds:

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

(c) Tin(IV) oxide                                 (d) Thallium(I) sulphate

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

Answer:

(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

Question 8.7:

Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Answer:

The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the following table.

The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.

While 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?Question 8.8:

Answer:

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.

Question 8.9:

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.

Answer:

(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

.