Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 11.26:

(a) Classify following oxides as neutral, acidic, basic or amphoteric:

CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, Tl₂O₃

(b) Write suitable chemical equations to show their nature.

Answer:

(1) CO = Neutral

(2) B₂O₃ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.

(3) SiO₂ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium silicate.

(4) CO₂ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.

(5) Al₂O₃ = Amphoteric

Amphoteric substances react with both acids and bases. Al₂O₃ reacts with both NaOH and H₂SO₄.

(6) PbO₂ = Amphoteric

Amphoteric substances react with both acids and bases. PbO₂ reacts with both NaOH and H₂SO₄.

(7) Tl₂O₃ = Basic

Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium chloride.

Question 11.27:

In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.

Answer:

Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3. However, heavier members of this group also display the +1 oxidation state. This happens because of the inert pair effect. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals.

Thallium, like aluminium, forms compounds such as TlCl₃ and Tl₂O₃. It resembles alkali metals in compounds Tl₂O and TlCl.

Question 11.28:

When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

Answer:

The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

Question 11.29:

What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?

Answer:

(a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is ns² np¹ and their group valency is +3. However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the ns² electrons by the d– and f– electrons. As a result of the poor shielding, the ns² electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation

The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S.

Question 11.30:

A certain salt X, gives the following results.

(i) Its aqueous solution is alkaline to litmus.

(ii) It swells up to a glassy material Y on strong heating.

(iii) When conc. H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates out.

Write equations for all the above reactions and identify X, Y and Z.

Answer:

The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a weak acid. Also, when X is strongly heated, it swells to form substance Y. Therefore, X must be borax.

When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material Y. Hence, Y must be a mixture of sodium metaborate and boric anhydride.

When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.

Question 11.31:

Write balanced equations for:

(i) BF₃ + LiH →

(ii) B₂H₆ + H₂O →

(iii) NaH + B₂H₆ →

(iv) H₃BO₃

(v) Al + NaOH →

(vi) B₂H₆ + NH₃ →

Answer:

(iv)

Question 11.32:

Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.

Answer:

Caron dioxide

In the laboratory, CO₂ can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

CO₂ is commercially prepared by heating limestone. The reaction involved is as follows:

Caron monoxide

In the laboratory, CO is prepared by the dehydration of formic acid with conc. H₂SO₄, at 373 K. The reaction involved is as follows:

CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

Question 11.33:

An aqueous solution of borax is

(a) neutral (b) amphoteric

(c) basic (d) acidic

Answer:

(c) Borax is a salt of a strong base (NaOH) and a weak acid (H₃BO₃). It is, therefore, basic in nature.

Question 11.34:

Boric acid is polymeric due to

(a) its acidic nature (b) the presence of hydrogen bonds

(c) its monobasic nature (d) its geometry

Answer:

(b) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.

Question 11.35:

The type of hybridisation of boron in diborane is

(a) sp (b) sp² (c) sp³ (d) dsp²

Answer:

(c) Boron in diborane is sp³ hybridised.

Question 11.36:

Thermodynamically the most stable form of carbon is

(a) diamond (b) graphite

(c) fullerenes (d) coal

Answer:

(b) Graphite is thermodynamically the most stable form of carbon.

Question 11.37:

Elements of group 14

(a) exhibit oxidation state of +4 only

(b) exhibit oxidation state of +2 and +4

(c) form M^2– and M⁴⁺ ion

(d) form M²⁺ and M⁴⁺ ions

Answer:

(b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.

If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.Question 11.38:

Answer:

(i)

(ii)