SURE SHOT 4 MARKS Junior inter CHEMISTRY IPE QUESTIONS

 

SURE SHOT 4 MARKS 

Junior inter CHEMISTRY IPE QUESTIONS

1 Explain the difference in properties of diamond and graphite on the basis of their structure.

ANSWER:

GRAPHITE STRUCTURE:

Graphite has layered lattice structure. Every carbon in graphite is involved in sp2 hybridization and in bond with three carbon atoms forming σ bonds.

The p-orbitals which do not participate in hybridization 

are perpendicular to the plane of each layer and overlap 

laterally forming π bonds delocalized over the plane.

Each layer is a hexagonal net of carbon atoms and may 

be regarded as a fused system of benzene rings

DIAMOND STRUCTURE:

In diamond every carbon is in sp3

hybridization and 

linked to four carbon atoms by single bonds.Hence in diamond every carbon is surrounded by four other carbon atoms tetrahedrally and covalently bonded to them resulting in the formation of giant molecule.

The C – C bond length in diamond is 154 pm and CCC bond angle is 109º28’.

2 What do you understand by 

(1) Allotropy (2) Inertpair effect (3) Catenation

ANSWER:

(1) Allotropy: The occurrence of the same element in two or more different physical forms having more or less similar chemical properties but different physical properties is called allotropy.The different forms of the element are called allotropes and are formed due to the difference in the arrangement of atoms. Allotropes may be crystalline or amorphous 

(2)INERT PAIR EFFECT: The two s-electrons in the outer shell tend to remain paired and are not participating in compound formation. This pair of electrons is called inert pair and the effect is called inert pair effect

(3) Catenation: Combining capacity of the atoms of the same element to form long chains, branched chains and cyclic compounds is called catenation power.

3 a) Diamond has high melting pint - Explain.

b) What do you mean by Auto protolysis. Give the equation to represent the auto protolysis of water.

ANSWER:

(a)Diamond is the hardest material so far known. In diamond each carbon is linked to four other carbon atoms by strong covalent bonds and more energy is required to break the bonds.So the melting point of diamond is very high(3600°C).

(b)

Loss of H+ by a molecule on its own is called Autoprotolysis.

4 Explain with suitable examples (a) Electron deficient

b) Electron rich (c) Electron precise hydride

ANSWER: 

(a) Molecular hydrides having less number of electrons for writing the conventional Lewis structure are called 

electron-deficient hydrides, e.g., B2H6.

(b) Molecular hydrides formed by the elements of IV or 14 group elements in which all the valence electrons are involved for bond formation are known as electron-precise hydrides.

• Molecular hydrides which contain more valence electrons, 

than the required for bond formation are called electron-

rich hydrides. They contain lone pairs, 

e.g., NH3, H2O.

5 Explain the terms

a)Hard water b) Soft water c) Calgon method for the removal of hardness of water

ANSWER:

(a) Hard Water:

• The water which does not give ready and permanent lather with soap is called hard water.

• Soap is sodium or potassium salt of oleic, pamitic or stearic acids.

• Hardness of water is due to the presence of 

bicarbonates, sulphates and chlorides of magnesium 

and calcium.

• The lather formed by normal sodium or potassium soap is coagulated by magnesium or calcium because they form insoluble salts.

• Temporary hardness is due to the presence of 

bicarbonates of magnesium and calcium.

• Permanent hardness is due to the presence of 

chlorides and sulphates of magnesium and calcium.

(b) Water which gives ready and permanent lather with soap is soft water.

(c) Calgon Process: 

In calgon process permanent hardness is removed by using sodium hexametaphosphate which is known as 

calgon.

Calgon reacts with Mg2+ and Ca2+ ions to form inactive complex anions which do not react with soap. This is known as sequestration (rendering active species into inactive species).