inorganic

IPE

SECOND YEAR

`

KUKATPALLY CENTRE

INORGANIC CHEMISTRY

VA GROUP ELEMENTS

1. Discuss the structures of the oxides of

N

₂

A: Nitrogen forms several oxides

N O, NO, N O , NO or N O and N O

_{2 2 3 2 2 4 2 5}. The structures of these oxides are as follows:

*

Nitrous oxides

(

N O

₂

)

is a linear molecule with the following resonance structures.

N

N – O

N = N = O

*

Nitric oxide

(

NO

)

is an odd electron molecule containing unpaired electron. It is a resonance hybrid of the following structures.

N = O

N = O

N = O

*

Dinitrogen trioxide

(

N O

_{2 3}

)

may have one of the following two structures.

O

O = N

N = O

(Symmetric)

N – N

O

O

O

(Unsymmetric)

*

Nitrogen dioxide

(

NO

₂

)

is an odd electron molecule and paramagnetic in nature in gaseous state due to the presence of unpaired electron. It is a resonance hybrid of the following structures.

N

O

O

N

O

O

N

O

O

N

O

O

In the solid state it exist as dimmer

N O

_{2 4}. It is diamagnetic in nature.

N

N

O

O

O

O

N

N

O

O

O

O

*

Nitrogen Pentoxide

(

N O

_{2 5}

)

: Its structure is

N

N

O

O

O

O

O

2. Write an essay on the preparation and structures of any three oxides of Nitrogen. A:

*

Nitrous oxide

(

N O

₂

)

: It can be prepared by heating ammonium nitrate.

4 3 2 2

NH NO

¾¾®

D

N O

+

2H O

Properties:

a) It is a colourless, neutral gas

b) When inhaled it produces laughing hysteria so it is called laughing gas. Structure:

N

N – O

N = N = O

*

Nitric oxide or Nitrogen oxide

(

NO

)

: It is prepared by the catalytic oxidation of ammonia.

Catalyst

3 2 2

Properties:

a) It is a colourless, neutral gas.

b) It is an odd electron molecule with paramagnetic character in gaseous state. c) It is diamagnetic in nature in solid state.

Structure:

N = O

N = O

N = O

*

Nitrogen dioxide

(

NO

₂

)

: It is prepared by the thermal decomposition of lead nitrate.

(

2

)

2 2 2

2Pb NO

¾¾®

D

2PbO 4NO

+

+

O

Properties:

a) It is an odd electron molecule. b) It is brown coloured gas.

On cooling it converts into solid state. In the solid state it is colourless and diamagnetic. In solid state it do not contain unpaired electron

Structure:

N

O

O

N

O

O

N

O

O

N

O

_O

In the solid state it exist as dimmer

N O

_{2 4}. It is diamagnetic in nature.

N

N

O

O

O

O

N

N

O

O

O

O

3. Write an essay on the structural aspects of the phosphorous series of acids. A: Phosphorous forms three oxyacids in phosphorous series of acids.

*

Orthophosphorous acid

H PO

_{3 3}

*

Metaphosphorous acid

HPO

₂

*

Hypophosphorous acid

H PO

_{3 2}

a) In all oxyacids, phosphorous is involved in

sp

3 hybridization and is surrounded by the atoms or groups tetrahedrally.

b) In all these oxyacids atleast one –OH group is in bond with phosphorous.

c) The hydrogen atoms in –OH groups are ionizable and are responsible for the acidic nature. d) The basicity of an oxyacid is equal to the number of OH groups in that acid.

e) The phosphorous series of acids contain

P H

-

bonds and

P OH

-

bonds. The reducing property of these acids is due to this

P H

-

bonds.

Structures:

*

Orthophosphorous acid

H PO

_{3 3}: Its basicity is two. It contains two –OH groups and one P – H bond.

P

O

H

OH

OH

*

Metaphosphorous acid

HPO

₂: Its basicity is one. Generally it exists as cyclic compound. The structure of monomer is

HO P

- =

O

*

Hypophosphorous acid

H PO

_{3 2}: Its basicity is one. It contains one –OH group and two P – H bonds.

P

H

H

OH

OH

4. Write an essay on the preparation and properties of (a) Hypophosphorous acid (b) Hypophosphoric acid (c) Metaphosphoric acid

A: (a) Hypophosphorous acid

(

H PO

_{3 2}

)

:

Preparation: It is prepared in the laboratory by heating white phosphorous with dilute solution of

( )

₂

Ba OH

( )

(

)

4 ₂ 2 2 2 ₂ 3

2P

+

3Ba OH

+

6H O

®

3Ba H PO

+

2PH

From this barium salt

H PO

_{3 2} is obtained.

Properties: It is a mono basic acid. It is very strong reducing agent in basic medium and is oxidized to

3 3

H PO

.

(b) Hypophosphoric acid

(

H P O

_{4 2 6}

)

:

Preparation: When phosphorous oxide is exposed to moist air hypophosphoric acid is formed. Properties: It is a tetra basic acid. Oxidation number of phosphorous in hypophosphoric acid is +4 (c) Metaphosphoric acid

(

HPO

₃

)

:

Preparation: It is prepared by heating orthophosphoric acid at 870 K.

870 K

3 3 3 2

H PO

¾¾¾®

HPO

+

H O

Properties: It gives a glassy transparent solid. So it is called glacial phosphoric acid. It is a monobasic acid. Its salts are called metaphosphates. Its salts exist as cyclic metaphosphates.

5. Describe Haber’s process for the preparation of

NH

₃. Give a Neat diagram of the plant. Lable it.

A: Ammonia can be manufactured by Haber’s process. In this method ammonia is manufactured by the direct reaction between nitrogen and hydrogen.

2 2 3

N

+

3H

®

2NH ; H

D = -

93.63 kJ

*

The above reaction is reversible reaction

*

The forward reaction is exothermic reaction.

*

When the reactants converts into products the volume decreases, as the number of product molecules are less than the total number of reactant molecules.

Synthesis of NH3 by Haber’s Process:

or

At ordinary temperature the reaction occurs slowly. Accoridng Le Chatelier’s principle the following conditions are more favourable for more yield of ammonia.

(i) Low temperature (ii) High pressure (iii) Catalyst

Ammonia is manufactured at about 725 – 775 K; using a pressure of 200 to 300 atmospheres. Iron powder mixed with molybdenum is used as catalyst. Molybdenum acts as promoter to iron catalyst. A mixture of potassium and aluminium oxides

(

K O and Al O

_{2 2 3}

)

also act as promoter to the iron catalyst.

Pure and dry

N

₂ and

H

₂ mixture in 1 : 3 ratio by volume are made to react at a pressure of 200 to 300 atmospheres. Then it is passed into a catalytic chamber. These gases react at a pressure of 200 to 300 atmosphere. Then it is passed into a catalytic chamber. These gases react in the presence of catalyst at 725-775 K. Ammonia is formed to the extent of about 10%. It is separated from the unreacted

N

₂ and

H

₂

by condensation.

6. Detail the Ostwald’s process for the manufacture of

HNO

₃. Give balanced equations wherever possible. A: Ostwald’s process for the manufacture of

NH

₃: Ammonia mixed with air in 1 : 7 or 1:8 ratio is passed over

hot platinum gauze catalyst. Then ammonia is oxidized to NO.

Pt. gauze

3 2 _1155K 2

or Manufacture of HNO3 – Ostwald’s process:

The NO formed in the above reaction is cooled and mixed with oxygen to convert into

NO

₂. The

NO

₂ is then passed into warm water under pressure in the presence of excess of air to give

HNO

₃

2 2

2NO O

+

®

2NO

2 2 2 3

4NO

+

2H O O

+

®

4HNO

The acid formed is 61% concentrated. This is further concentrated in three states. Stage 1: 61%

HNO

₃ is distilled until 68%

HNO

₃ is obtained.

Stage 2: 68%

HNO

₃ is mixed with conc.

H SO

_{2 4} and distilled to get 98% acid.

Stage 3: 98%

HNO

₃ is cooled in a freezing mixture. Crystals of pure

HNO

₃ separate out.

7. Describe an industrial method for the preparation of superphosphate of lime. Why is it converted into “triple phosphate of lime”?

A: Superphosphate of lime is a mixture of calcium dihydrogen phosphate

é

_ë

Ca H PO

(

_{2 4 2}

)

ù

_û

and Gypsum

(

CaSO .2H O

4 2

)

It is manufactured by treating the powdered phosphate rock with calculated amount of sulphuric acid.

(

)

(

)

(

)

3 4 ₂ 2 4 2 2 4 ₂ 4 2

Ca

PO

+

2H SO

+

4H O

®

Ca H PO

+

2 CaSO .2H O

Manufacture of Superphosphate of lime:

The phosphate rock is powdered. It is taken in a cast iron mixer. Calculated amount of sulphuric acid is added and mixed well. This reaction mixture is dumped in one of the dens

D

₁ and

D

₂ through the values

1 2

V and V

. The reaction is allowed to take place for about 24 – 36 hours. The impurities like carbonate and fluoride are removed in the form of

CO

₂ and HF gases. The hard mass formed is powdered and sold as superphosphate of lime.

The

CaSO

₄ in the superphosphate of lime is an insoluble waste product and is of no value to the plants. To avoid this waste product formation, it is converted into triple superphosphate by using phosphoric acids.

(

)

(

)

3 4 ₂ 2 4 2 4 ₂

Ca

PO

+

4H PO

®

3Ca H PO

The triple phosphate completely dissolves in water i.e., no waste product formed.

8. What are the similar features of ammonia and phosphine? Write a comparative note on them.

A: Both ammonia and phosphine can be prepared by similar methods. They can be prepared by the hydrolysis of their binary compounds either with water or dilute acid.

Ex: _{3 2 2}

( )

₃ 2

Mg N

+

6H O

®

3Mg OH

¯ +

2NH

-( )

3 2 2 ₂ 3

Ca P

+

6H O

®

3Ca OH

+

2PH

-Comparative notes on their general properties:

*

Stability of

NH

₃ is greater than

PH

₃ because N – H bond is stronger than P – H bond.

*

Both contain one lone pair on central atom, so they can act as Lewis bases. But

NH

₃ is stronger Lewis base than

PH

₃

*

Both can act as reducing agents but

NH

₃ is weaker reducing agent then

PH

₃ because of more stability.

*

Both

NH

₃ and

PH

₃ are soluble in water but the solubility of

NH

₃ in water is more than

PH

₃.

*

NH

₃ and

PH

₃ are volatile and colourless gases but

NH

₃ is less volatile because it can form inter

molecular hydrogen bonds.

*

The hydrogen atoms in

NH

₃ and

PH

₃ can be substituted by alkyl group like

CH

₃ and halogens like Cl. The ease of substitution decrease from

NH

₃ to

PH

₃

*

NH

₃ can form salts with any acid forming salts like

NH Cl.PH

_{4 3} can react with very strong acids like HI.

*

Both

NH

₃ and

PH

₃ are pyramidal in shape. In

NH

₃ nitrogen is involved in

sp

3 hybridization but in

3

PH

the pure p-orbitals participate in bonding. The bond angle in

PH

₃ (

93 36 '

°

) is less than in

(

)

3

NH 107°

9. In what way the structures of trioxides and pentoxides of

N

₂ and

P

₄ differ. Draw their structures and show. A: Trioxides and Pentoxides of nitrogen and phosphorous are chemically similar but differ in their structures.

The oxides of nitrogen exists as monomers

N O

_{2 3} and

N O

_{2 5} while the oxides of phosphorous

P O

_{4 6} and

4 10

P O

exists as dimers. Their structures are given below.

Sl.No. Nitrogentrioxide structure Nitrogen pentoxide structure

1.

O

N

N

O

O

asymmetrical form

N

N

O

O

O

O

O

O

O = N

N = O

(Symmetric)

N

N

O

O

O

O

O

P

P

P

P

O

O

O

_O

O

O

(dotted line does

not represent bond)

P

P

P

P

O

O

O

_O

O

O

Phosphorous

Oxygen

O

O

O

O

O

10. Write balanced equations for the reactions of

H O

₂ with

a)

P O

_{4 6} b)

P O

_{4 10} c)

PCl

₃ d)

PCl

₅ A:

P O

_{4 6}

+

6H O

₂

®

4H PO

_{3 3}

(

Phosphorous acid

)

(

)

4 10 2 3 4

P O

+

6H O

®

4H PO

Orthophosphoric acid

(

)

4 10 2 3 little

P O

+

2H O

®

4HPO

Metaphosphoric acid

(

)

3 2 3 3

PCl

+

3H O

®

3HCl H PO

+

Phosphorous acid

(

)

5 2 3 4

PCl

+

4H O

®

5HCl H PO

+

Orthophosphoric acid

11. Write a note on the Cyanamide process. Mention any three important uses of

NH

₃.

A: Cyanamide process: Ammonia can be manufactured synthetically by cyanamide process. In this method the calcium is made to react with nitrogen gas in an electric furnace at 1273 – 1373 K. In this reaction a mixture of

2

CaCl

and

CaF

₂ acts as catalyst. Then a mixture of calcium cyanamide and graphite known as nitrolium is produced. When super heated steam is passed over calcium cyanamide at 453 K ammonia is produced.

1273 1373 K 2 2 2 Graphite Calcium Cyanamide

CaCl

+

N

¾¾¾¾¾®

-

CaCN

+

C

453 K 2 2 3 3

CaCN

+

3H O

¾¾¾®

CaCO

+

2NH

Uses of Ammonia:

*

Ammonia is used in the manufacture of fertilizers like ammonium sulphate, urea, calcium ammonium nitrate etc.

*

Ammonia is used as refrigerant.

*

Ammonia is used in the manufacture of sodium carbonate by Solvay process.

*

Ammonia is used in the manufacture of nitric acid and explosives like ammonium nitrate. 12. Discuss the principle underlying the manufacture of

HNO

₃

A: Nitric acid can be manufactured by two methods:

*

Birkland and Eyde Method – Principle:

a) An electric arc converts

N

₂ and

O

₂ of air into NO.

2 2

N

+

O

®

2NO

b) NO is oxidized to

NO

₂ by air.

2 2

2NO O

+

®

2NO

c)

NO

₂ is dissolved in water in the presence of oxygen to get

HNO

₃

2 2 2 3

4NO

+

2H O O

+

®

4HNO

*

Ostwald’s process-Principle

a) Ammonia is oxidized by air when

NH

₃ and air in 1 : 7 or 1 : 8 ratio passed over platinum gauze catalyst.

Pt. guaze

3 2 _{1155 K} 2

b) NO is then oxidized to

NO

₂ by air.

2 2

2NO O

+

®

2NO

c)

NO

₂ is dissolved in water in the presence of oxygen to get

HNO

₃

2 2 2 3

4NO

+

2H O O

+

®

4HNO

13. What makes the difference between the two following reactions? Phosphate rock treated with a) Chambers acid b) excess of phosphoric acid write the equation.

A: a) When phosphate rock reacts with chambers acid superphosphate of lime is formed.

(

)

(

)

3 4 ₂ 2 4 2 2 4 ₂ 4 2

Ca

PO

+

2H SO

+

4H O

®

Ca H PO

+

2CaSO .2H O

+

Heat

Superphosphate of lime contains insoluble waste product

CaSO

₄. So the percentage of phosphorous in superphosphate of lime become less.

b) To avoid the waste product gypsum in superphosphate of lime the phosphate rock is treated with phosphoric acid

(

H PO

_{3 4}

)

. Then triple phosphate of lime or triple superphosphate is formed.

(

)

(

)

3 4 ₂ 3 4 2 4 ₂

Ca

PO

+

4H PO

®

3Ca H PO

The triple phosphate completely dissolves in water i.e., no waste product formed. SAQ

1. What is allotropy? Explain in N and its congeners?

A: If the same element exists in two or more physical states having nearly similar chemical properties but different physical properties. It is known as allotropy.

Except bismuth all the VA group elements exhibit allotropy. Solid

N

₂ exists in

a

and

b

forms.

Phosphorous exists in many allotropic forms such as white P, red P, scarlet P,

a

- black,

b

- black and violet. Arsenic exists as metallic or grey arsenic, non-metallic or yellow arsenic, black arsenic.

Antimony exist in three allotropic forms

*

crystalline, metallic variety or common variety.

*

non-metallic or yellow or

a

-antimony

*

explosive antimony.

2. What is catenation? How does it vary in group 15?

A: Combining capacity of the atoms of same element to form long chains is called catenation. Catenation capacity depends on bond energy. In the group from top to bottom atomic size increases. So bond length increases and bond energy decreases, hence catenation capacity decreases. Since nitrogen is smaller atom it has more catenation power. So it can form

H N

₂

-

NH

₂ and

N H

₃ having two and three atoms in a chain respectively. Phosphorous can form

P H

_{2 4} having two P atoms in the chain. Other elements of VA group don’t exhibit catenation power.

3. Draw the structures of

P O

_{4 6} and

P O

_{4 10}. In what respect do thy resemble each other, what happen when they react with water?

A:

P

P

P

P

O

O

O

_O

O

O

(dotted line does

not represent bond)

P

P

P

P

O

O

O

_O

O

O

Phosphorous

Oxygen

O

O

O

O

O

In both

P O

_{4 6} and

P O

_{4 10} the four phosphorous atoms are arranged in tetrahedral shape. In both

P O

_{4 6} and

4 10

P O

six oxygen atoms are acting as bridges between phosphorous atoms forming

P O P

- -

bonds. In

P O

_{4 10} one oxygen atom is in dative bond with phosphorous atoms.

When

P O

_{4 6} react with water orthophosphorous acid is formed.

4 10 2 3 4

P O

+

6H O

®

4H PO

4. Write balanced equations for the formation of

NCl

₃ and

PCl

₃. Give their hydrolysis reactions. A:

NCl

₃ can be prepared by the action of excess chlorine on ammonia.

3 2 3

NH

+

3Cl

®

NCl

+

3HCl

3

PCl

can be prepared by the direct reaction between phosphorous and chlorine.

4 2 3

P

+

6Cl

¾¾®

D

4PCl

Hydrolysis:

3

NCl

hydrolyses in water forming ammonia and hypochlorous acid

3 2 3

NCl

+

3H O

®

NH

+

3HOCl

3

PCl

hydrolyses in water forming phosphorous acid hydrochloric acid.

3 2 3 3

PCl

+

3H O

®

H PO

+

3HCl

5. In a compound

MX

₅, M is any V group element except Nitrogen why (X) is halogen? A: Except nitrogen all the VA group elements can form penta halides of the type

MX

₅

Ex:

PCl

₅,

AsCl

₅,

SbCl

₅,

BiF

₅ etc. Nitrogen do not contain d-orbitals, in its valency shell but other VA group elements contain vacant d-orbitals in their valency shell. So when excited one of the ns electron can go into the d-orbital.

Ground state electronic configuration

ns

np

nd

Excited state electronic configuration

ns

np

nd

Since there are five unpaired electrons in the excited state except nitrogen other VA group elements can form five covalent bonds with halogens. So they can form pentahalides of the type

MX

₅. Nitrogen cannot form pentahalides because of the absence of d-orbitals in its valency shell.

6. Which of the acids of phosphorous does not show monomeric state but cyclic structure? Name any polymer of the acid.

A: Metaphosphoric acid do not exist as monomer. It exist as cyclic polymer. Ex: Cylcic metaphosphate.

O

P

P

O

P

O

O

O

O

O

O

O

7. A sample of most ammonia is to be dried. What method do you suggest?

A: Moisture from ammonia can be removed by passing over dry lime or CaO. But it cannot be dried over Conc.

2 4

H SO

, fused

CaCl

₂ or

P O

_{2 5} as they react with ammonia.

8. Write the formula of superphosphate of lime. Why is it converted into triple phosphate?

A: The common formula of superphospate of lime is

Ca H PO

(

_{2 4}

)

₂

+

2 CaSO .2H O

(

_{4 2}

)

. It acts as good fertilizer. The

CaSO

₄ in its insoluble waste product and its presence has no significance to the plants. To avoid this waste product

CaSO

₄ superphosphate is changes into triple phosphate which completely dissolves in water.

9. How is superphosphate of lime formed? Give an equation. Explain why the product is a hard mass? A: Superphosphate of lime can be prepared by treating powdered phosphate rock with calculated quantity of

2 4

H SO

(

)

(

)

(

)

3 4 ₂ 2 4 2 2 4 ₂ 4 2

Super phosphate of lime

Ca

PO

+

2H SO

+

4H O

®

Ca H PO

+

2 CaSO .2H O

VSAQ

1. Write the composition of “phosphate rock”.

A:

Ca

₃

(

PO

_{4 2}

)

2. Give any two examples to show negative oxidation state of nitrogen. A: Ammonia

NH

₃

-

3

; Hydrazine

N H

_{2 4}

-

2

; Hydroxyl amine

NH OH

₂

-

1

3. Why is NO paramagnetic in nature? When does it become diamagnetic?

A: NO, is an odd electron molecule containing odd number of electrons. The total number of electrons in No molecule is 15. When all electrons are paired one electron remains unpaired. So NO is paramagnetic. But when temperature decreases it dimerises and become diamagnetic.

4. How is dinitrogen tetroxide formed? Give equation.

A: When the temperature of

NO

₂ is decreased it dimerises to convert into dinitrogen tetroxide.

2 2 4

2NO

¾¾

®

N O

5. How many oxygens surround a phosphorous in phosphorous pentoxide?

A: In phosphorous pentoxide each phosphorous atom is surrounded by four oxygen atoms. 6. Write the equations for the hydrolysis of

NCl

₃. How does it differ from hydrolysis of

PCl

₃?

A:

NCl

₃ hydrolysis in water giving

NH

₃ and HOCl while

PCl

₃ hydrolysis in water giving orthophosphorous acid and HCl.

3 2 3

NCl

+

3H O

®

NH

+

3HOCl

3 2 3 3

PCl

+

3H O

®

H PO

+

3HCl

7. What are the orbitals of P that are involved in the formation of

PCl

₅? A: In the formation of

PCl

₅,

sp d

3 hybrid orbitals are involved.

8. What is nitrolim? How is it formed?

A: Nitrolium is a mixture of calcium cyanamide

(

CaCN

₂

)

and graphite. When

N

₂ is passed over calcium carbide containing

CaCl

₂ or

CaF

₂ as catalyst at 1273 – 1372 K nitrolim is formed.

1273 1378

2 2 2

Calcium cynamide Grpahite

CaC

+

N

¾¾¾¾®

-

CaCN

+

C

9. What drying agent is suitable to dry

NH

₃?

A: Ammonia can be dried by passing over dry lime (CaO) because it does not react with ammonia. 10. What is the function of

CaCl

₂ in cyanamide process?

A:

CaCl

₂ acts as catalyst in cynamide process during

CaC

₂ react with

N

₂ to form nitrolim. 11. Which oxides of

N

₂ are neutral oxides?

A: Among the oxides of nitrogen, nitric oxide (NO) and nitrous oxide

(

N O

₂

)

are neutral.

12. Which of the two oxides

N O

_{2 5} and

P O

_{2 5} is better dehydrating agent? Give an example for the same reaction.

A:

P O

_{4 10} is a strong dehydrating agent than

N O

_{2 5}

Ex: Dehydrating of

HNO

₃ with

P O

_{4 10} gives

N O

_{2 5}

3 4 10 2 5 3

4HNO

+

P O

®

2N O

+

4HPO

13. Give reasons for the chemical inactivity of nitrogen at ordinary conditions.

A:

N

₂ molecule contains triple bond

(

N

º

N

)

. To break the triple bond large amount of energy

(

1

)

945.4 kJ mol

- is required. Due to this high bond dissociation energy nitrogen is apparently inactive under normal conditions.

14. What is the stability order of VA group hydrides? Explain the gradation in the reducing property of these hydrides?

A: The stability order of the hydrides of VA group hydrides is

NH

₃

>

PH

₃

>

AsH

₃

>

SbH

₃

>

BiH

₃. As the stability decrease they dissociate easily and can act as strong reducing agents. So the order of reducing power of VA group hydrides is

NH

₃

<

PH

₃

<

AsH

₃

<

SbH

₃

<

BiH

₃.

15. Why does nitrogen does not form pentahalides?

A: To form pentahalides the ns electron should be excited to nd orbital. In the case of nitrogen there is no d orbital in its valency shell i.e., second orbit. So nitrogen cannot form pentahalides.

16. Can

NCl

₅ be prepared by direct union of the elements? Why or why not?

A: No.

NCl

₅ cannot be prepared by the direct union of elements because nitrogen do not react with chlorine directly. Further due to the absence of d orbitals in its valency shell nitrogen cannot form

NCl

₅

17. Write the structure of

HNO

₃

A: The structure of nitric acid is the resonance hybrid of the following structures.

N

O

_O

OH

N

O

O

OH

Nitric Acid

18.

P O

_{2 5} is strong dehydrating agent. Why is it not used to dry

NH

₃?

A:

P O

_{2 5} is acidic and

NH

₃ is basic. They bond react to form salt. So

NH

₃ cannot be dried using

P O

_{2 5}

19. How do you convert

NH NO

_{4 3} into

NH

₃ give reactions?

A: When ammonium nitrate is heated with a base ammonia gas will be liberated.

4 3 3 3 2

NH NO

+

NaOH

®

NaNO

+

NH

+

H O

When

NH NO

_{4 3} is heated with alkaline solution of Zinc, it is converted into ammonia.

4 3 2 2 3 2

NH NO

+

4Zn 8NaOH

+

®

4Na ZnO

+

2NH

+

3H O

20. How do you prepare hypophosphorus acid in the laboratory?

A: When white phosphorous is boiled with dilute solution of barium hydroxide barium hypophosphite will be formed.

( )

(

)

4 2 2 2 2 2 3

2P

+

3Ba OH

+

6H O

®

3Ba H PO

+

2PH

To the barium hypophosphite solution if dil. Sulphuric acid is added hypophosphorous acid will be formed.

(

2 2

)

₂ 2 4 4 3 2

Ba H PO

+

H SO

®

BaSO

+

2H PO

4

BaSO

being insoluble can be removed by filtration.

21. How many Valence shell electrons are utilized by each phosphorous atom in

P

₄ molecule?

A: From the structural representation of

P

₄, we can infer that each phosphorous shares three of its valence electrons with other phosphorous atoms. A lone pair of electrons is seen on each phosphorous.

22. Why is nitrogen is a diatomic gaseous molecule while phosphorous is a tetra atomic solid?

A: Nitrogen atoms are small in size and can approach very close to one another. This facilities the lateral overlap of the p-orbitals to form

p

-bonds. In phosphorous only single bonds are formed due to the larger sizes of phosphorus atoms. Multiple bond formation and through it acquiring octet of electrons is not possible. Hence

P

₄ molecules are formed.

23.

PH

₃ is quite stable in Air. But it catches fire when heated to

150 C

°

. Why?

A:

PH

₃ frequently contains

P H

_{2 6}(disphosphine) in trace amounts as impurity. This catches fire on heating in air.

24. Write the names of the compounds formed by the union of

PH

₃ and

AsH

₃ separately with HI.

A: The reaction must be similar to the union of

NH

₃ with HI giving ammonium iodide. Therefore,

PH

₃ gives

4

25. If pure P-orbitals in As or Sb overlap with the S-orbitals of hydrogens, the bond angle is expected to be

90°

, why is it

91 .48 '

°

in their hydrides?

A: The HMH bond angle in

AsH

₃ and

SbH

₃ would be expected to be

90°

But due to repulsions between M-H bonds, the angle increases to

91 .49 '

°

26. Many penta halides of VA group elements are known. NO hydrides of

MH

₅ exist. Why?

A: To attain the pentavalent state, d-orbitals must be used. Hydrogen is not sufficiently electronegative to make the d-orbitals effective by contraction.

27. Identify the oxidant and reductant in the given reaction

3HNO

₂

®

HNO

₃

+

2NO

+

H O

₂

A:

HNO

₂ acts as both the oxidant and reductant.

HNO

₂ as a reductant changes to

HNO

₃.

HNO

₂ as an oxidant changes to NO.

28. What is the change in oxidation state of nitrogen in auto oxidation, auto reduction of

HNO

₂? A:

3HNO

₂

®

HNO

₃

+

2NO

+

H O

₂

In

HNO

₂ to

HNO

₃, the change of oxidation state of nitrogen is from +III to +V. From

HNO

₂ to NO the change in oxidation state is from +III to +II

29. What is the difference between tautomeric and resonance structures?

A: In tautomers the skeleton of atoms in the structures differs

(

Ex : HNO

₂

)

. In resonance, the skelton of atoms does not change (Ex: Benzene)

30. What is nitration mixture?

A: A mixture (1:1) of

HNO

₃ (Conc.) and Conc.

H SO

_{2 4} is known as nitration mixture. This is used in nitration reaction.

(

)

H SO2 4

6 6 3 _{60 C} 6 5 2 2

Nitrobenzene

VIA GROUP ELEMENTS

1.

How is Ozone prepared in the laboratory? Give any three oxidation reactions of

O

₃?

A: Principle: Ozone is prepared by passing silent electric discharge through pure cold and dry oxygen.

( ) Silent Electric ( )

2 g disch arg e 3 g

3O

¾¾¾¾¾®

2O

, H

D = +

284.5 kJ

Siemen’s and Brodie’s ozonizers are used to prepare ozone in the Laboratory.

Siemen’s Ozonizer:

*

This ozonizer consists of two coaxial glass tubes sealed at one end.

*

The inner sides of the inner tube and the outer sides of the outer tube are coated with tin foils.

*

These tin foils are connected to the terminals of a powerful induction coil.

*

Cold and dry oxygen is passed through the annular space from one end.

*

Oxygen undergoes silent electric discharge partially and 10% Ozone is formed.

*

Ozone and oxygen mixture known as ozonized oxygen is collected from the other end. Oxidising properties of Ozone:

*

Ozone oxidizes black lead sulphide (PbS) to white lead sulphate

(

PbSO

₄

)

3 4 2

PbS 4O

+

®

PbSO

+

4O

*

Hydrogen chloride is oxidized to chlorine by ozone.

3 2 2 2

2HCl O

+

®

H O Cl

+

+

O

*

Ozone oxidizes moist potassium iodide

( )

KI

to iodine

( )

I

₂

2 3 2 2

2KI H O O

+

+

®

2KOH

+

O

+

I

2.

How is ozone is prepared in Brodie’s method? Write any three reduction reactions of

O

₃ with equations. A: Preparation of Ozone, Brodie’s method:

Principle: Ozone is prepared by subjecting cold dry oxygen gas to the silent electric discharge using dilute sulphuric acid as conducting medium and Cu wires as electrodes.

( ) ( )

2 g 3 g

3O

+

68K.Cal.

®

2O

*

Brodie’s ozonizer consists of double wall U tube in which dilute

H SO

_{2 4} is placed.

*

Copper wires are immersed in dilute

H SO

_{2 4}

*

Copper wires are connected to the terminals of powerful induction coil.

*

Cold and dry oxygen is passed through annular space between the two walls of U tube.

*

Oxygen undergoes silent electric discharge and 15% of oxygen is converted into ozone.

Reducing properties of ozone:

*

Ozone reduces hydrogen peroxide to water

2 2 3 2 3

H O

+

O

®

H O

+

2O

2 3 3

BaO

+

O

®

BaO

+

2O

*

Ozone reduces silver oxide to metallic silver

2 3 2

Ag O O

+

®

2Ag

+

2O

3.

How do you prepare hypo in the laboratory? Giving proper equations detail the reactions of hypo. A: Crystalline hydrated sodium thiosulphate

(

Na S O .5H O

_{2 2 3 2}

)

is known as “hypo”

Laboratory preparation:

In the laboratory hypo is prepared by the following methods.

*

By boiling alkaline or neutral sodium sulphite solution with flowers of sulphur.

2 3 excess 2 2 3

Na SO

+

S

®

Na S O

*

By the oxidation of sodium sulphide or sodium polysulphide with air.

Heat in

2 5 2 _air 2 2 3

2Na S

+

3O

¾¾¾®

2Na S O

+

6S

*

By treating sodium sulphide solution with sulphur dioxide.

2 2 2 2 3

2Na S 3SO

+

®

2Na S O

+

S

Reactions of Hypo:

*

Action of heat: On heating hypo undergoes thermal decomposition to give

H S,SO

_{2 2} and S. Hypo looses all the molecules of water (water by crystallization) when heated to about 488 K.

*

Reaction with dilute acids: When hypo reacts with dilute acids like HCl or

H SO

_{2 4} to give

SO

₂ and S.

2 2 3 2 2

Na S O

+

2HCl

®

2NaCl H O SO

+

+

+

S dilute

*

Reaction with

AgNO

₃ solution: When hypo reacts with

AgNO

₃ solution, two kinds of reactions may take place.

a) When dilute hypo is added to

AgNO

₃ solution, a white precipitate of

Ag S O

_{2 2 3} is formed which readily changes to a black solid

(

Ag S

₂

)

. The reactions are:

(

)

2 2 3 3 3 2 2 3

Na S O

+

2AgNO

®

2NaNO

+

Ag S O

¯

white ppt

(

)

2 2 2 3 2 4 2

H O

+

Ag S O

®

H SO

+

Ag S

¯

Black ppt

b) When concentrated hypo is added to

AgNO

₃ solution a white precipitate

(

Ag S O

_{2 2 3}

)

is obtained first. The precipitate readily dissolves in excess of sodium thiosulphate due to the formation of complex compound.

(

)

2 2 3 3 2 2 3 3

Na S O

+

2AgNO

®

Ag S O

¯ +

2NaNO

White ppt.

(

) (

) (

)

2 2 3 2 2 3 3 2 _{3 2}

Ag S O

+

3Na S O

®

2Na

é

_ë

Ag S O

ù

_û

Sodium argentothiosulphate

complexcompound

*

Reaction with iodine:

Sodium thiosulphate reacts with iodine to give sodium tetrathionate

(

Na S O

_{2 4 6}

)

2 2 3 2 2 4 6

2Na S O

+ ®

I

2NaI

+

Na S O

This reaction is used in volumetric analysis to estimate iodine.

*

Reaction with exposed photographic film or AgBr:

a) In photography the fixing is done by washing the film with hypo solution.

b) The silver bromide (or the silver halide) on the film reacts with sodium thiosulphate to give a complex compound.

(

)

2 2 3 3 2 _{3 2}

AgBr

+

2Na S O

®

Na

é

_ë

Ag S O

ù

_û

+

NaBr

*

Reaction with moist Cl2: Hypo reacts with moist

Cl

₂ to give

Na SO

_{2 4} and HCl.

2 2 3 2 2 2 4

Sodium sulphate

Na S O

+

Cl

+

H O

®

Na SO

+ +

S 2HCl

*

Reaction with salts: Sodium thiosulphate reacts with ferric chloride, cupric chloride or auric chloride etc. And converts them into complex thiosulphates.

4.

Write the reactions of the following with

O

₃.

i)

C H

_{2 2} ii)

C H

_{2 4} iii)

C H

_{6 6}

Give the structures of the products in each case. What happens if these products are treated with

Zn / H O

₂

A: Reactions of O3:

*

With Acetylene

(

C H

_{2 2}

)

: Ozone when treated with acetylene, an addition compound called acetylene ozonide is obtained. This on hydrolysis in presence of zinc, glyoxal is produced.

HC CH + O

3

HC

CH

O

O

O

Zn

H

2

O

CHO

CHO

+ H

2

O

2

(Acetylene ozonide)

(Glyoxal)

*

With Ethylene

(

C H

_{2 4}

)

: Ozone when treated with ethylene, an addition compound called ethylene ozonide is obtained. This on hydrolysis in presence of zinc, formaldehyde is produced.

H

2

C

CH

2

+ O

3

CH

CH

2

O

O

O

Zn

H

2

O

2HCHO + H

2

O

2

(Ethylene ozonide)

(Formaldehyde)

*

With Benzene

(

C H

_{6 6}

)

: Ozone on treating with benzene an addition compound called benzene tri ozonide is obtained, which on hydrolysis gives 3 moles of glyoxal.

+ 3O

3

O

O

O

O

O

O

O

O O

Zn

3H

2

O

CHO

3CHO + 3H

2

O

2

(Glyoxal)

SAQ

1.

What are the bond angles in

H O

₂ and

H S

₂ ? Why are they different?

A:

*

1) In

H O

₂ oxygen undergoes

sp

3 hybridization. Due to the presence of two lone pair of electrons shape of

H O

₂ molecule is angular or V-shape with H – O – H bond angle

104 28 '

°

2) Due to the greater repulsion between lone pair and lone pair of electrons than lone pair and bond pair, the tetrahedral angle is decreased to

104 28 '

°

*

1) In

H S

₂ only pure “p” orbitals of sulphur are involved in bond formation. Hence the bond angle is less than of

104 28 '

°

and it is only

92 30 '

°

2)

H S

₂ is angular or V-shaped molecule.

2.

What are the structures of

SO and SO

_{2 3}? Explain them in terms of VBT.

A:

*

Structure of

SO

₂:

SO

₂ is angular molecule. The O S O bond angle is

119 30 '

°

. In

SO

₂ sulphur atom undergoes

sp

2 hybridization in first excited state 3s,

3p

_x and

3p

_y orbitals undergo

sp

2

hybridization.

3p

_z and one d orbital are unhybridised and are used in the formation of

p

p - p

p

and

S

O

O

S

O

O

(

px-py

)

(

)

z z p -d

(

)

(

px-py

)

z z p -d o 1.43 A o 1.43 A

*

Structure of

SO

₃: In gaseous state

SO

₃ has a planar triangular structure. The O S O bond angle is

120°

. In

SO

₃ sulphur atom undergoes

sp

2 hybridization in second excited state. 3s,

3p

_x and

3p

_y

orbitals underto

sp

2 hybridization. One

3p

_z and two 3d orbitals are unhybridized state and are used in the formation of one

p

p - p

p

and two

d

p - p

p

bonds.

S

O

O

O

3.

Give the structures of

( )

i SF

₄

( )

ii SF

₆. Explain them. A: Structure of

SF

₄:

*

The structure of

SF

₄ is trigonal bipyramidal with one equatorial position occupied by a lone pair.

*

Sulphur in

SF

₄ undergoes

sp d

3 hybridization.

*

S utilizes four hybrid orbitals for bonding while the fifth orbital accomidates a lone pair of electrons.

F

F

F

F

S Structure of

SF

₆:

*

The shape of

SF

₆ is octahedral or square bipyramidal.

*

Sulphur in

SF

₆ undergoes

sp d

3 2 hybridization.

*

All hybrid orbitals are used in bonding to form 6 sigma bonds between S and F atoms.

F

F

F

F

F

F

S

VSAQ

1.

Write the structure of gaseous sulphur molecule at low temperatures. A:

O = S

2.

What is allotropy? Give the allotropes of oxygen?

A: Existence of an element in two or more physical forms is called as allotropy. Allotropes of oxygen and

O

₂

and

O

₃.

3.

Write the names of the allotropic forms of S.

A: Sulphur exists in many allotropic forms. The important forms are

a

or rhombic sulphur,

b

or monoclinic sulphur,

g

or monoclinic sulphur, plastic or

c

sulphur.

4.

At room temperature

H O

₂ is a liquid while

H S

₂ is a gas. Explain.

A: At room temperature water exists as liquid because of inter molecular hydrogen bonding. While

H S

₂ exist as a gas because of the absence of intermolecular hydrogen bonding.

5.

What are the bond angles in

H O

₂ and

H S

₂ ? Why they differ in their bond angles?

A: Bond angles in

H O

₂ and

H S

₂ are

104 36 '

°

and

92 30 '

°

respectively. In

H O

₂ molecule oxygen undergoes

3

sp

hybridization where as in

H S

₂ pure p-orbitals are involved in bond formation.

6.

What is tailing of mercury? What chemical changes takes place in this process?

A: On passing

O

₃, mercury looses it metallic luster and meniscus and it sticks on the glass walls due to formation of mercurous oxide

(

Hg O

₂

)

. This is known as tailing of mercury. The chemical change takes place in this process is

2Hg

+

O

₃

®

Hg O O

₂

+

₂

7.

Give an example of a reaction which consumes all of the atoms of oxygen in

O

₃. A:

3SO

₂

+

O

₃

®

3SO

₃

In this reaction

O

₃ is completely consumed.

8.

What happens when hypo reacts with

AgNO

₃?

A:

*

When diluted hypo reacts with

AgNO

₃ solution, it gives a white ppt. of silver thiosulphate which on hydrolysis gives a black ppt. of silver sulphide.

_{2 2 3 3 2 2 3 3}

dil. hypo Silver thiosulphate

Na S O

+

2AgNO

®

Ag S O

+

2NaNO

( ) 2 2 3 2 2 2 4 Silver sulphide Black.ppt

Ag S O

+

H O

®

Ag S

¯ +

H SO

*

When concentrated hypo reacts with

AgNO

₃ solution, it gives a complex compound.

2 2 3 3 2 2 3 3

Na S O

+

2AgNO

®

Ag S O

+

2NaNO

(

)

2 2 3 2 2 3 3 2 _{3 2}

Sodium argento thiosulphate

3Na S O

+

Ag S O

®

2Na

é

_ë

Ag S O

ù

_û

9.

How is hypo useful in photography?

A: Hypo is used as fixing agent in photography.

(

)

2 2 3 2 2 _{3 2} Sodium argento thiosulphate (Complex compound)

AgBr

+

2Na S O

®

Na

é

_ë

Ag S O

ù

_û

+

NaBr

10.

Mention two advantages of contact process over other processes. A:

*

H SO

_{2 4} obtained is extremely pure and concentrated.

*

The impurities can be tested and the reactants can be recycled.

11.

What is the reaction of

O

₃ with PbS? Give the equation. A:

O

₃ oxidizes black PbS to white

PbSO

₄

3 4 2

PbS 4O

+

®

PbSO

+

4O

12.

What is antichlor? Give example.

A: The reagent used to remove excess of chlorine in textile industry is called antichlor.

2 2 3 2 2 2 4

Na S O

+

Cl

+

H O

®

Na SO

+

2HCl S

+

In this hypo is used as antichlor agent.

13.

Write any two uses of ozone?

VII GROUP ELEMENTS

1.

Describe Whytlaw-Gray method for the preparation of fluorine?

A:

*

Fluorine is prepared by the electrolysis of fused potassium hydrogen fluoride

(

KHF

₂

)

. The electrode reactions are:

KHF

2

Fusion

K

+

+ H

+

+ 2F

-

Electrolysis

at cathode

at anode

2H

+

+ 2e

-

H

2

2F

-

F

2

+ 2e

-1

2

3

4

5

8

7

6

9

1. Fused

KHF

₂ 2. Heating coil 3. Grpahite (anode) 4. Copper diaphragm 5. Fluorspar stopper 6. Fluorine 7. Copper cell (cathode) 8.

H

₂

9. Inlet for HF Manufacture:

*

In this method, electrolysis is carried out in an electrically heated copper cell.

*

The copper vessel serves as cathode also.

*

Anode is made of graphite. The anode is surrounded by a copper diaphragm perforated at the bottom.

*

This diaphragm prevents the mixing of

H

₂ and

F

₂ which reacts explosively if they come into contact.

*

F

₂, liberated at the anode, is passed through the U-tube containing sodium fluoride.

*

Hydrogen fluoride vapours accompanying fluorine as impurity, are removed by NaF.

(

NaF HF

+

®

NaHF

2

)

*

H

₂ is liberated at the cathode.

*

The gaskets used in the cell are coated with Teflon to prevent corrosion of the parts.

2.

Write the chemical properties of

F

₂ with relevant equations? A:

*

Reaction with water:

2

F

reacts with water and gives ozonized oxygen.

2 2 2

2F

+

2H O

®

4HF O ;

+

3F

₂

+

3H O

₂

®

6HF O

+

₃

*

Reaction with alkalies:

a) When

F

₂ reacts with cold, dilute NaOH gives sodium fluoride and oxygen difluoride

(

OF

₂

)

.

2 2 2

2NaOH

+

2F

®

2NaF OF

+

+

H O

b) When

F

₂ reacts with hot, concentrated NaOH gives sodium fluoride and oxygen.

2 2 2

4NaOH

+

2F

®

4NaF O

+

+

2H O

*

Reaction with other halides:

Fluorine oxidizes all other halide ions to the corresponding halogens.

2 2

F

+

2KCl

®

2KF Cl

+

*

Reaction with inert gases:

Heavier inert gases like Kr and Xe form compounds with fluorine.

2 2

Xe 3F

+

®

XeF

Fluorine oxidises potassium hydrogen sulphate to potassiumper sulphate

(

K S O

_{2 2 8}

)

. 2 4 2 2 8

F

+

2KHSO

®

K S O

+

2HF

*

Reaction with H2S: Fluorine oxidizes

H S to SF

_{2 6} 2 2 6

H S 4F

+

®

2HF SF

+

*

Reaction with non metals:

Except oxygen and nitrogen, other non-metals directly combine with fluorine and give binary compounds.

2 6

S 3F

+

®

SF

*

Reaction with metals:

All the metals (including noble metals like Au, Pt, etc.) form metal fluorides.

2 2

Cu

+

F

®

CuF

3.

What is the principle of preparing

Cl

₂ in the laboratory? Describe Nelson’s method for its. A: Principle:

Cl

₂ is manufactured by electrolysis of brine solution.

Manufacture:

*

Nelson cell consists of U-shaped porous steel vessel lined inside with asbestos.

*

It serves as cathode.

*

This vessel is suspended in a rectangular iron tank.

*

Brine solution (10% NaCl) is taken in the vessel.

*

A carbon rod is dipped in the Brine solution which acts as Anode.

*

Steam is passed into the cell. On electrolysis, the following reactions take place.

( )aq ( )aq ( )aq

2NaCl

®

2Na

+

+

2Cl

-At cathode:

2H O 2e

₂

+

®

2OH

_{( )aq}

+

H

_{2 g( )}

At anode:

2Cl

_{( )aq}

®

Cl

_{2 g( )}

+

2e

*

Cl

₂ gas liberated at anode is collected and is compressed in steel cylinders.

*

Sodium ions penetrate through the asbestos paper lining and reach the cathode.

*

Here sodium ions combine with

OH

- to form NaOH. In this process,

H

₂ and NaOH are important byproducts.

2Na

+

+

2OH

-

®

2NaOH

4.

Write the structures of all the oxyacids of

Cl

₂.

A: NAME FORMULA OXIDATION STATE OF

CHLORINE BASICITY

Hypochlorous acid HClO +1 1

Chlorous acid

HClO

₂ +3 1

Chlorine acid

HClO

₃ +5 1

Structures of oxyacids of chlorine:

*

Hypochlorous acid: HClO

Cl

OH

The chlorine atom in this acid undergoes

sp

3 hybridisation. This conjugate base of Hypochlorous acid is Hypochlorite ion

(

ClO

-

)

*

Chlorous acid:

HClO

₂

Cl

OH

O

The chlorine atom in this acid undergoes

sp

3 hybridization. The chlorine atom has

2s

bonds one

p

bond

(

p

p - p

d

)

and two lone pairs.

The conjugate base of chlorous acid is chlorite ion

(

ClO

₂-

)

which is angular in shape with a bond angle of

111°

.

*

Chloric acid:

HClO

₃

Cl

OH

O

O

The chlorine atom in this acid undergoes

sp

3 hybridization. The chlorine atom has

3s

bonds,

2p

bonds (both are

p

p - p

d

) and one lone pair.

The conjugate base of chloric acid is chlorate ion

(

ClO

₃-

)

which is pyramidal in shape with a bond angle of

106°

*

Perchloric acid:

HClO

₄

Cl

OH

O

O

O

The chlorine atom in this acid undergoes

sp

3 hybridization. The chlorine atom has

4s

bonds, three

p

bonds (all

d

p - p

p

) and no lone pairs.

The conjugate base of perchlorine acid is perchlorate ion

(

ClO

₄-

)

which is tetrahedral in shape with a bond angle of

109.5°

*

The acid strength of different oxyacids of chlorine increases with an increase in the oxidation state of the chlorine.

2 3 4

HOCl

HClO

HClO

HClO

acid strength increases

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5.

How is bleaching power prepared industrially? Give any of its four chemical properties with equations. A: Industrial preparation of bleaching powder (Baechmann’s plant):

*

Bachmann’s plant consists of a vertical iron tower provided with a hopper at the top and inlets for chlorine and hot air slightly above the base.

*

The tower is fitted with a number of horizontal shelves at regular heights. Each shelf is fitted with a rotating rake.

*

Dry slaked lime is introduced into the tower through the hopper at the top.

*

The slaked line moves downwards with the help of the rotating rakes and comes in contact with the current of chlorine rising upwards.

*

Slaked lime reacts with chlorine and gets converted into bleaching power which is collected in the container placed at the bottom.

*

The hot air drives away unreacted chlorine

( )

_{2 2 2}

2

Ca OH

+

Cl

®

CaOCl

+

H O

Chemical properties of bleaching powder:

*

Reaction with water:

With cold water, bleaching powder gives chloride and hypochlorite ion.

2 2

CaOCl

®

Ca

+

+

Cl

-

+

ClO

-In hot water, it undergoes auto oxidation and gives chloride and chlorate ions.

*

(a) Reaction with insufficient amount of dilute acids:

When small amounts of dilute acid is added, bleaching powder liberates oxygen.

2 2 4 2 4 2

2CaOCl

+

H SO

®

CaCl

+

CaSO

+

2HCl O

+

(b) Reaction with excess of dilute acids:

On treating bleaching powder with excess of dilute acid, chlorine is liberated. This liberated chlorine is known as “available chlorine”.

2 2 4 4 2 2

CaOCl

+

H SO

®

CaSO

+

H O Cl

+

-*

Effect of a catalyst:

Bleaching powder decomposes to give

O

₂ in the presence of a catalyst

CoCl

₂. 2

CoCl

2 2 2

2CaOCl

¾¾¾®

2CaCl

+

O

-*

Oxidising property:

Bleaching powder oxidizes lead salts to lead dioxide and ethanol to acetaldehyde.

2

2 2 2

Pb

+

+

2CaOCl

®

PbO

+

2CaCl

( )

CaOCl2

3 2 3 2

Ethyl alcohol Acetaldehyde

CH CH OH

+

O

¾¾¾¾

®

CH CHO H O

+

6.

Write all the chemical properties of bleaching powder. Give equations. How is it useful to man. A:

*

On long standing, bleaching powder undergoes auto-oxidation and changes into chloride and

chlorate.

(

)

2 2 _{3 2}

*

Reaction with water:

With cold water, bleaching powder gives chloride and hypochlorite ion.

2 2

CaOCl

®

Ca

+

+

Cl

-

+

ClO

-In hot water, it undergoes auto oxidation and gives chloride and chlorate ions.

*

Reaction with insufficient amount of dilute acids:

When small amounts of dilute acid is added, bleaching powder liberates oxygen.

2 2 4 2 4 2

2CaOCl

+

H SO

®

CaCl

+

CaSO

+

2HCl O

+

*

Reaction with excess of dilute acids:

On treating bleaching powder with excess of dilute acid, chlorine is liberated. This liberated chlorine is known as “available chlorine”.

2 2 4 4 2 2

CaOCl

+

H SO

®

CaSO

+

H O Cl

+

-*

Effect of a catalyst:

Bleaching powder decomposes to give

O

₂ in the presence of a catalyst

CoCl

₂. 2

CoCl

2 2 2

2CaOCl

¾¾¾®

2CaCl

+

O

-*

Oxidising property:

Bleaching powder oxidizes lead salts to lead dioxide and ethanol to acetaldehyde.

2

2 2 2

Pb

+

+

2CaOCl

®

PbO

+

2CaCl

( )

CaOCl2

3 2 3 2

Ethyl alcohol Acetaldehyde

CH CH OH

+

O

¾¾¾¾

®

CH CHO H O

+

Uses

*

It is used in the sterilization of water, as bleaching agent, as an oxidizing agent and in the preparation of chloroform.

SAQ

1.

How does

F

₂ react with (i)

H O

₂ , (ii) NaOH. Give equations for them. A:

*

Reaction with

H O

₂ :

2

F

reacts with water and gives ozonized oxygen.

2 2 2

2F

+

2H O

®

4HF O

+

;

3F

₂

+

3H O

₂

®

6HF O

+

₃

*

Reaction with NaOH:

a) When

F

₂ reacts with cold, dilute NaOH gives sodium fluoride and oxygen difluoride

(

OF

₂

)

.

2 2 2

2NaOH

+

2F

®

2NaF OF

+

+

H O

b) When

F

₂ reacts with hot, concentrated NaOH gives sodium fluoride and oxygen.

2 2 2

4NaOH

+

2F

®

4NaF O

+

+

2H O

2.

Give the reaction of

Cl

₂with the following:

(i)

SO

₂ (ii) NaOH (iii) Iron metal

A:

*

Reaction with

SO

₂:

2

Cl

when reacts with

SO

₂ under the influence of sunlight and gives

SO Cl

_{2 2}(sulphuryl chloride).

sunlight

2 2 2 2

SO

+

Cl

¾¾¾¾

®

SO Cl

*

Reaction with NaOH:

a) When chlorine reacts with cold, dilute NaOH gives sodium chloride and sodium hypochlorite (NaOCl).

Cl

₂

+

2NaOH

®

NaCl

+

NaOCl H O

+

₂

b) When chlorine reacts with hot, concentrated NaOH gives sodium chloride and sodium chlorate.

(

NaClO

₃

)

.

3Cl

₂

+

6NaOH

®

5NaCl

+

NaClO

₃

+

3H O

₂

*

Reaction with Iron metal:

Chlorine reacts with iron metal and gives ferric chloride

(

FeCl

₃

)

2 3

3.

What is “Available Chlorine”? Give chemical equation(s) which determine the same.

A: Available chlorine is the amount of

Cl

₂ set free when bleaching powder is treated with excess of dilute

2 4

H SO

or

CO

₂

A good sample of bleaching powder contain about 35-38% available chlorine.

2 2 4 4 2 2

CaOCl

+

H SO

®

CaSO

+

H O Cl

+

-2 2 3 2

CaOCl

+

CO

®

CaCO

+

Cl

-VSAQ

1.

Why is the E.A. of

Cl

₂ greater than that of

F

₂?

A: Fluorine has unexpectedly low electron affinity than chlorine. This is due to very small size of the fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p subshell of fluorine and thus, the incoming electron does not feel much attraction. Therefore, its electron affinity is less.

2.

Write an equation for

F

₂ reaction with

KHSO

₄ and tell the nature of the chemical change. A:

F

₂ oxidizes potassium hydrogen sulphate to potassium persulphate.

2 4 2 2 8

F

+

2KHSO

®

K S O

+

2HF

3.

Write the balanced equation(s) for the reaction of

Cl

₂ with

NH

₃

A: a) With excess of

Cl

₂, nitrogen trichloride

(

NCl

₃

)

and HCl are formed.

3 2 3

NH

+

3Cl

®

NCl

+

3HCl

b) With excess of

NH

₃, ammonium chloride and nitrogen are formed.

3 2 4 2

8NH

+

3Cl

®

6NH Cl

+

N

-4.

1 mole of

NH

₃ is mixed with 8 moles of

Cl

₂ in a reaction vessel. Write the equation for the reaction. A: Since chlorine is taken in excess amount, 1 mole of Ammonia reacts with 3 moles of chlorine.

3 2 3

NH

+

3Cl

®

NCl

+

3HCl

5.

What is the reaction between bleaching powder and excess of dil.

H SO

_{2 4}?

A: On treating bleaching powder with excess of dilute

H SO

_{2 4}, chlorine is liberated. This liberated chlorine is known as “available chlorine”

2 2 4 4 2 2

CaOCl

+

H SO

®

CaSO

+

H O Cl

+

-6.

Give any two uses of bleaching powder.

A: a) It is used in the sterilization of water b) It is used as a bleaching agent for cotton and paper pulp.

7.

Give the uses of florine. A:

*

It is used in rocket fuels.

*

HF is used in etching of glass.

*

NaF,

Na AlF

_{3 6} are useful insecticides.