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p-BLOCK ELEMENTS
Syllabus :
Groupwise study of the the p-block elements Group - 13 : Preparation,
properties and uses of boron and aluminium; Structure, properties and uses of
borax, boric acid, diborane, boron trifluoride, aluminium chloride and alums.
Group - 14 : Tendency for catenation; Structure, properties and uses of
allotropes and oxides of carbon, silicon tetrachloride, silicates, zeolites and
silicones.
Group - 15 : Properties and uses of nitrogen and phosphorus; Allotrophic
forms of phosphorus; Preparation, properties, structures and uses of
ammonia, nitric acid, phosphine and phosphorus halides, (PCl3, PCl5);
Structures of oxides and oxoacids of nitrogen and phosphorus.
Group - 16 : Preparation, properties, structures and uses of hydrogen and ozone;
Allotropic forms of sulphur; Preparation, properties, structures and uses of
sulphur dioxide, sulphuric acid (including its industrial preparation);
structures of oxoacids of sulphur.
Group - 17 : Preparation, properties and uses of chlorine and hydrochloric
acid; Trends in the acidic nature of hydrogen halides; Structures of Interhalogen
compounds and oxides and oxoacids of halogens.
Group - 18 : Occurrence and uses of noble gases; Structures of fluorides and
oxides of xenon.
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C O N C E P T S (p-Block)
C1A The Boron Family :
(i)
Boron is mainly occurs in orthoboric acid (H3BO3), borax Na2B4O7·10H2O and kernite
(Na2B4O7 · 4H2O). Bauxite (Al2O3·2H2O) and cryolite (Na3AlF6) are the important minerals of
aluminium.
(ii)
Atomic radius of Ga is less than that of Al.
(iii)
The observed discontinuity in the ionisation enthalpy values between Al and Ga and between In
and Tl are due to inability of d- and f-electrons, which have low screening effect, to compensate
the increase in nuclear charge.
(iv)
Chemical property of Boron family :
(a)
The relative stability of +1 oxidation state in the order of Al < Ga < In < Tl
(b)
Electron deficient molecule act as a Lewis acids. This behaviour decreases down the
group as size increases.
(c)
Reactivity in air is

2E(s )  3O 2 (g )  2E 2 O 3 (s )

2E(s )  N 2 (g )  2EN(s)
(E - elements)
(d)
Boron does not react with acids and bases. Aluminium reacts with acids and bases as
follows :
2 Al(s )  6HCl(aq ) 
 2 Al 3  (aq )  6Cl  (aq )  3H 2 (g )
2Al(s)  2NaOH(aq)  6H 2 O(l) 
 2Na  [Al(OH)4 ] (aq)  3H 2 (g )
sodium tetrahydroxy aluminate( II)
(e)
Reaction with halogen as follows :
2E(s )  3X 2 (s ) 
 2EX 3 (s)
(X = F, Cl, Br, I)
C1B
Compound of Boron :
(a)
Borax dissolves in water to give an alkaline solution
Na 2 B 4 O 7  7H 2 O 
 2NaOH  4H 3 BO 3
Borax
(b)
orthoboric acid
When borax is heated it convert into B2O3 as follows :


Na 2 B 4 O 7 ·10H 2 O  Na 2 B 4 O 7  2NaBO 2  B 2O 3
sodium
metaborate
(c)
Boric
anhydride
Orthoboric acid can be prepare by :
Na 2 B 4 O 7  2HCl  5H 2 O 
 2NaCl  4B(OH ) 3
ortho boric acid
(d)
H3BO3 heated to give B2O3


H 3 BO 3  HBO 2  B 2 O 3
(e)
Diborane can be prepared by :
(i)
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4BF3  3LiAlH 4 
 2B 2 H 6  3LiF  3AlF3
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(f)
(ii)
2NaBH 4  I 2 
 B 2 H 6  2NaI  H 2
(iii)
2BF3  6NaH  B 2 H 6  6NaF
450 k
Some of reaction of Boron are :
(i)
B 2 H 6  3O 2 
 B 2 O 3 (aq )  6H 2
(ii)
B 2 H 6 (g )  6H 2 O( l ) 
 2B(OH ) 3 (aq )  6H 2
(iii)
B 2 H 6  2NMe 3 
 2B 3 ·NMe 3
(iv)
B 2 H 6  2CO 
 2BH 3 · CO
(v)
3B 2 H 6  6NH 3 
 3[BH 3 ( NH 3 ) 2 ] [BH 4 ]  2B 3 N 3 H 6  12H 2

borazine
(g)
Borazine is inorganic benzene.
(h)
The structure of diborane is :
(i)
Borohydrides can be prepare by B2H6 as :
2MH  B 2 H 6 
 2M  [BH 4 ]
[M = Li or Na]
Detailed explanation of compound of Boron :
Boron Hydrides
The boron hydrides are sometimes called boranes by analogy with the alkanes. They fall into two series :
1.
BnH(n + 4) (called nido-boranes)
2.
A less stable series BnH(n + 6) (called arachno-boranes)
Use of Diborane (B2H6)
It is used to prepare the higher boranes, and is an important reagent in synthetic organic chemistry. Diborane
is used as a powerful electrophilic reducing agent for certain functional groups.
R — C  N  RCH2NH2
R — NO2  RNH2
R — CHO  RCH2OH
Preparation
1.
Mg 3B2  H 3PO 4  mixture of boranes mainly B4 H10 heat
 B2 H 6
magnesium
boride
Ortho
Phosphoric
Acid
diborane
0
,150 C
750
atmosphere


 B2H6 + Al2O3
2.
B2O3 + 3H2 + 2Al
3.
C
2BF3  6 NaH 180

 B2 H 6  6 NaF
0
gas
gas
There are several convenient laboratory preparations :
a.
Reducing the etherate complexes of the boron halides with Li[AlH4].
ether
4[Et2O . BF3] + 3Li[AlH4] 
 2B2H6 + 3Li[AlF4] + 4Et2O
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b.
Reacting Na[BH4] and iodine in the solvent diglyme. Diglyme is a polyether CH3OCH2CH2OCH2CH2OCH3.
in diglyme solution
 B2H6 + H2 + 2NaI
2Na[BH4] + I2    
c.
in diglyme
Reducing BF3 with Na[BH4]  
 2B2H6 + 3Na[BF4] + 4Et2O
Method (c) is particularly useful when diborane is required as a reaction intermediary. It is produced in situ,
and used without the need to isolate or purify it.
Properties
Diborane is a colourless gas, and must be handled with care as it is highly reactive. It catches fire
spontaneously in air and explodes with dioxygen. The heat of combustion is very high. In the laboratory it
is handled in a vacuum frame. Since it reacts with the grease used to lubricate taps, special taps must be
used. it is instantly hydrolysed by water, or aqueous alkali. At red heat the boranes decompose to boron and
hydrogen.
B2H6 + 3O2  2B2O3 + 3H2O
H = –2165 kJ mol–1
B2H6 + 6H2O  2H3BO3 + 3H2
Reactions of the Boranes :
Hydroboration
A very important reaction occurs between B2H6 (or BF3 + NaBH4) and alkenes and alkynes.
½B2H6 + 3RCH = CHR  B(CH2 – CH2R)3
½B2H6 + 3RC  CR  B(RC = CHR)3
The reactions are carried out in dry ether under an atmosphere of dinitrogen because B2H6 and the products
are very reactive. The alkylborane products BR3 are not usually isolated. They may be converted as
follows :
1.
to hydrocarbons by treatment with carboxylic acids,
2.
to alcohols by reaction with alkaline H2O2, or
3.
to either ketones or carboxylic acids by oxidation with chromic acid (H2CrO4).
The complete process is called hydroboration, and results in cis-hydrogenation, or cis-hydration. Where the
organic molecule is not symmetrical, the reaction follows the anti-Markovnikov rule, that is B attaches to
the least substituted C atom.
BR3 + 3CH3COOH  3RH + B(CH3COO)3
hydrocarbon
B(CH2 · CH2R)3 + H2O2  3RCH2CH2OH + H3BO3
2 CrO 4
H

2 CrO 4
(CH3 · CH2)3 – B H
 CH3COOH
carboxylic acid
2O 2
(CH3 · CH2)3 – B + CO diglyme
 [(CH3 · CH2)3 – CBO]2 H
 [CH3 · CH2]3 – COH
Hydroboration is a simple and useful process for two main reasons :
1.
The mild condition required for the initial hybride addition.
2.
The variety of products which can be produced using different reagents to break
the B – C bond.
H.C. Brown won the Nobel Prize for Chemistry in 1979 for work on these organoboron compounds.
Reaction with ammonia
All the boranes act as Lewis acids and can accept electron pairs. Thus they react with amines, forming
simple adducts. They also react with ammonia, but the products depend on the conditions :
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B2H6 + 2(Me)3N  2[Me3N · BH3]
excess NH 3
B 2 H 6  NH 3 low
temperatur
 e  B 2 H 6 ·2 NH 3
excess NH 3
high

( BN ) x
temperature
boron nitride
ratio 2 NH 3 :1B 2 H 6
higher
 temperatur

e  B3 N 3 H 6
borazine
The compound B2H6 · 2NH3 is ionic, and comprises [H3N  BH2  NH3]+ and [BH4]– ions. On heating, it
forms borazine which is also known as Inorganic Benzene.
Boric Acid
Preparation
1.
 Na2SO4 + 4H3BO3
Na2B4O7 + H2SO4 + 5H2O 
2.
 2NaCl + 4H3BO3
Na2B4O7 + 2HCl + 5H2O 
3.
Mainly from native calcium borate, colemanite. Sulphur dioxide is passed through powdred colemanite
suspension in boiling water; calcium sulphite and basic acid are formed - the former dissolves in excess of
sulphur dioxide forming calcium bisulphite and the boric acid crystallises out on cooling.
 2CaSO3 + 6H3BO3
2CaO.3B2O3 + 2SO2 + 9H2O 
 2Ca (HSO3)2
2CaSO3 + 2H2O + 2SO2 
Properties :
1.
B2O3 is made conveniently by dehydrating boric acid :
0
C
H 3 BO3 100

 HBO 2 red
heat
 B 2O 3
orthoboric
acid
metaboric
acid
boron
sesquioxid e
B2O3 is a typical non-metallic oxide and is acidic in its properties. It is the anhydride of orthoboric acid, and
it reacts with basic (metallic) oxides, forming salts called borates or metaborates, for example :
CoO + B2O3  Co(BO2)2
cobalt metaborate (blue colour).
However, it is possible to formce B2O3 to behave as a basic oxide by reacting with very strongly scidic
compounds. Thus with P2O5 boron phosphate is formed.
B2O3 + P2O5  2BPO4
2.
Orthoboric acid H3BO3 is soluble in water, and behaves as a weak monobasic acid. It does not donate
protons like most acids, but rather it accepts OH—. It is therefore a Lewis acid, and is better written as
B(OH)3.
B(OH)3 + 2H2O
H3O+ + [B(OH)4]—
pK = 9.25
[H3BO3]
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Polymeric metaborate species are formed at higher concentrations, for example :
H3O+ + [B3O3(OH)4]— + H2O
3B(OH)3
pK = 6.84
[3H3BO3]
3.
Acidic properties of H3BO3 or B(OH)3
Since B(OH)3 only partially reacts with water to form H3O+ and [B(OH)4]—, it behaves as a weak acid. Thus
H3BO3 or (B(OH)3) cannot be titrated satisfactory with NaOH, as a sharp end point is not obtained. If
certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added to the titration
mixture, then B(OH)3 behaves as a strong monobasic acid. It can now be titrated with NaOH, and the end
point is detected using phenolphtalien as indicator (indicator changes at pH 8.3-10.0)
B(OH)3 + NaOH
Na[B(OH)4]
NaBO2 + 2H2O
sodium metaborate
The added compound must be a cis-diol, to enhance the acidic properties in this way. (This mean that it has
OH group on adjacent carbon atoms in the cis configuration). The cis-diol forms very stable complexes
with the [B(OH)4]— formed by the forward reaction above, thus effectively removing it from solution. The
reaction is reversible. Thus removal of one of the product at the right hand side of the equation upsets the
balance, and the reaction proceeds completely to the right. Thus all the B(OH)3 reacts with NaOH: in effect
it acts as a strong acid in the presence of the cis-diol.
Structures of boric acid :
Thus orthoboric acid contains triangular BO33— units. In the solid the B(OH)3 units are hydrogen bonded
together into two-dimensional sheets with almost hexagonal symmetry. The layers are quite a large distance
apart (3.18 Å) and thus the crystal breaks quite easily into very fine particles.
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Figure : Hydrogen bonded structure of orthoboric acid
Borax
The most common metaborate is borax Na2[B4O5(OH)4] · 8H2O. Borax is usually written as Na2B4O7 ·
10H2O. It is actually made from two tetrahedra and two triangular units joined as shown in figure and
should be written Na2[B4O5(OH)4] · 8H2O.
Preparation :
1.
Orthoboric acid, H3BO3 on neutralization with Na2CO3 gives borax.
Na2CO3 + 4H3BO3  Na2B4O7 + 6H2O + CO2
2.
Colemanite, Ca2B6O11 is converted into borax by boiling it with concentrated solution of Na2CO3.
Ca 2 B6 O11  2 Na 2 CO 3 
 Na 2 B4 O 7  2CaCO 3  2 NaBO 2
borax
sodium
metaborate
Properties :
The solution of borax is alkaline in nature and hence it is a useful primary standard for titrating against
acids.
(Na2[B4O5(OH)4] · 8H2O) + 2HCl  2NaCl + 4H3BO3 + 5H2O
One of the products H3BO3 is itself a weak acid. Thus the indicator used to detect the end point of this
reaction must be one that is unaffected by H3BO3. Methyl orange is normally used, which changes in the pH
range 3.1–4.4.
One mole of borax reacts with two moles of acid. This is because when borax is dissolved in water both
B(OH)3 and [B(OH)4]— are formed, but only the [B(OH)4]— reacts with HCl.
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[B4O5(OH)4]2— + 5H2O
2B(OH)3 + 2[B(OH)4]—
2[B(OH)4]— + 2H3O+  2B(OH)3 + 4H2O
Borax is also used as a buffer since its aqueous solution contains equal amount of weak acid and its salt.
Qualitative analysis of boron compounds
When borates are treated with HF (or with concentrated H2SO4 and CaF2) the volatile compound BF3 is
formed. If the BF3 gas produced is introduced into a flame (for example a Bunsen flame) the flame gives a
characteristic green coloration.
conc.
H2SO4 + CaF2  2HF + CaSO4
H3BO3 + 3HF  2BF3 + 3H2O
An alternative test is to make the ester methyl borate B(OCH3)3. The suspected borate sample is mixed with
concentrated H2SO4 to form H3BO3, and warmed with methyl alcohol in a small evaporating basin.
B(OH)3 + 3CH3OH  B(OCH3)3 + 3H2O
The concentrated H2SO4 removes the water formed. The mixture is then set on fire. Methyl borate is
volatile, and colours the flame green.
Fluoboric acid
H3BO3 dissolves in aqueous HF, forming fluoboric acid HBF4.
H3BO3 + 4HF  H+ + [BF4]— + 3H2O
Fluoboric acid is a strong acid. The [BF4]— ion is tetrahedral, and fluoborates resemble perchlorates ClO4—
and sulphates in crystal structure and solubility (KClO4 and KBF4 are both not very soluble in water).
Trihalides of Boron
The boron halides are covalent. BF3 is gaseous, BCl3 liquid and BI3 is solid. BF3 is covalant but AlF3, GaF3,
InF3, TlF3 are ionic. The other halides are largely covalent of Al, Ga, In, Tl when anhydrous.
Preparation :
heat
1.
B2O3 + 3CaF2 + conc. 3H2SO4  2BF3 + 3CaSO4 + 3H2O
2.
B2O3 + 6NH4BF4  8BF3 + 6NH3 + 3H2O
3.
B2O3 + 6HF + 3H2SO4  2BF3 + 3H2SO4 · H2O
4.
heat
H 2O
2 H 2SO 4
Na 2 [ B 4 O 5 (OH ) 4 ]  12 HF 
[ Na 2 O( BF3 ) 4 ] 

 4BF3  2 NaHO 4  H 2 O
Properties :
1.
BF3 is very useful for promoting certain organic reactions, e.g. :
Friedel – Crafts reactions such as alkylations and acylations. In these the BF3 is used up in the reaction, and
so is not strictly catalytic.
C6H6 + C2H5F + BF3  C6H5 · C2H5 + H+ + [BF4]—
2.
The boron halides are all hydrolysed by water. BF3 hydrolyses incompletely and forms fluoborates. This is
because the HF first formed reacts with the H3BO3.
4BF3 + 12H2O  4H3BO3 + 12HF
12HF + 3H3BO3  3H+ + 3[BF4]— + 9H2O
4BF3 + 3H2O  H3BO3 + 3H+ + 3[BF4]—
The other halides hydrolyse completely, giving boric acid.
BCl3 + 3H2O  H3BO3 + 3HCl
3.
BX3 is electron-deficient (octet of B incomplete) and thus behaves as a Lewis acid.
Structure of BF3
The shape of the BF3 molecule is a planar triangle with bond angles of 1200. The bond lengths in BF3 are
1.30 Å. The bond energy is very high, which is higher than for any single bond. The shortness and strength
of the bonds is interpreted in terms of a p – p interaction, that is the bonds possess some double bond
character.
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C1C Compounds of Aluminium :
Alumina (Aluminium oxide)
Alumina Al2O3 exists principally in two crystalline forms called -Al2O3 or corundum, and -Al2O3, and in
addition there is a fibrous form.
Preparation :
1.
-Al2O3 is made by dehydrating Al(OH)3 below 4500C,
0
0
C
,1000 C
Al(OH ) 3 450

   Al 2 O3 

   Al 2 O 3

2.
In the lab, it is prepared by igniting aluminium hydroxide, aluminium sulphate or ammonium alum.
2Al(OH)3  Al2O3 + 3H2O
Al2(SO4)3  Al2O3 + 3SO3
(NH4)2SO4.Al2(SO4)3.24H2O  2NH3 + Al2O3 + 4SO3 + 25H2O
3.
Aluminium has a very strong affinity for oxygen. Al may be used in the thermite reduction of less stable
metal oxides. 3Mn3O4 + 8Al  4Al2O3 + 9Mn
Properties :
1.
Corundum is unaffected by acids. The crystal structure of corundum is hexagonally close-packed oxygen
atoms, with two thirds of the octahedral holes filled by Al3+ ions.
2.
In contrast to -Al2O3, -Al2O3 dissolves in acids, absorbs water, and is used for chromatography. It is
amphoteric in nature.
Al2O3 + 6HCl  2AlCl3 + 3H2O
Al2O3 + 2NaOH  2NaAlO2 + H2O
3.
Alumina is white, but it can be coloured by the addition of Cr2O3 or Fe2O3.
Aluminium Chloride
Preparation :
2Al + 6HCl  2AlCl3 + 3H2
Properties :
1.
Anhydrous AlCl3 (and to a lesser extent AlBr3) is used as the ‘catalyst’ in a variety of Friedel-Crafts type of
reactions for alkylations and acylations.
C6H5 · H + CH3CH2Cl + AlCl3  C6H5 · CH2CH3 + H+ + [AlCl4]—
This is not true ‘catalytic’ action, as the AlCl3 is used up, and the formation of [AlCl4]— or [AlBr4]— is an
essential part of the reaction. Acylations are similar :
C6H5 · H + RCOCl + AlCl3  RCOC6H5 + H+ + [AlCl4]—
AlCl3 is also used to catalyse the reaction to make ethyl bromide (which is used to make the petrol additive
PbEt4).
CH2 = CH2 + HBr  C2H5Br
2.
AlCl3 exists as dimer (Al2Cl6) in inert (non-polar) solvent as well as in vapour state.
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However, when the halides dissolve in water, the high enthalpy of hydration is sufficient to break the
covalent dimer into [M · 6H2O]3+ and 3X— ions. At low temperatures AlCl3 exists as a close packed lattice
of Cl— with Al3+ occupying octahedral holes. On heating Al2Cl6 species are formed.
3.
Aqueous solution of AlCl3 is acidic due to hydrolysis :
AlCl3 + 3H2O  Al(OH)3 + 3HCl
4.
Al forms monohalides in the gas phase at elevated temperatures, e.g.
high
AlCl3  2AI temperatur
 e  3AlCl
This compound is not very stable, and is covalent.
Alums
Alums are the double sulphates having general formula : X2SO4.M2(SO4)3.24H2O
X = monovalent cation such as Na+, K+, NH4+ etc.
M = trivalent cation such as Al3+, Cr3+, Fe3+ etc.
when alum contains aluminium as trivalent cation then it is named after monovalent cation.
e.g.
K2SO4.Al2(SO4)3.24H2O
potash alum
Na2SO4.Al2(SO4)3.24H2O
Soda alum.
When trivalent cation is not aluminium then alum is named after both, monovalent as well as trivalent
cation.
(NH4)2SO4.Fe2(SO4)3.24H2O
-
ferric ammonium alum.
Preparation of Potash alum
It is prepared by boiling powdered alum stone, K2SO4.Al2(SO4)3.4Al(OH)3 with dil sulphuric acid and
filtered. The filtrate is mixed with a requisite quantity of potassium sulphate and crystallized.
K2SO4.Al2(SO4)3.4Al(OH)3 + 6H2SO4  K2SO4 + 3Al2(SO4)3 + 12H2O
K2SO4 + Al2(SO4)3 + 24H2O  2[KAl(SO4)2 . 12H2O]
Qualitative analysis of aluminium
In qualitative analysis Al(OH)3 is precipitated as a white gelatinous substance when NH4OH is added to the
solution (after previously removing acid-insoluble sulphides with H2S). Fe(OH)3, Cr(OH)3 grey-green or
grey-blue. Zn(OH)2 is white, like Al(OH)3, but it is not gelatinous. Zn(OH)2 dissolves in excess NH4OH,
whereas Al(OH)3 does not. A confirmatory test for aluminium is the formation of a red precipitate from
Al(OH)3 and the dye aluminon.
Amphoteric behaviour – aluminates
Al(OH)3 is amphoteric. It reacts principally as a base. It reacts with acids to form salts that contain the
[Al(H2O)6]3+ ion. However, Al(OH)3 shows some acidic properties when it dissolves in NaOH, forming
sodium aluminate.
excess NaOH
 NaAl(OH)4
Al(OH)3   
sodium
NaAlO2 · 2H2O aluminate
the structure of the aluminate ion changes with both pH and concentration :
1.
Between pH 8 and 12 the ions polymerize using OH bridges and each aluminium is octahedrally
coordinated.
2.
In dilute solutions above pH values of 13, a tetrahedral [Al(OH)4]— ion exists.
3.
In concentrated solutions above 1.5 M and at pH values greater than 13 the ion exists as a dimer :
[(HO)3Al — O — Al(OH)3]2–
Difference between Boron and the other Elements
Boron differs significantly from the other elements in Group 13, mainly because the atoms are very small.
It is always covalent, and it is non-metallic. In addition, boron shows a diagonal relationship with slilicon
in Group 14.
1.
B2O3 is an acidic oxide, like SiO2. This is in contrast to Al2O3, which is amphoteric.
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2.
H3BO3, which may be written B(OH)3, is acidic, whilst Al(OH)3 is amphoteric.
3.
Simple borates and silicate ions can polymerize, forming isopolyacids. Aluminium forms no analogous
compounds.
4.
The hydrides of B are gaseous, readily hydrolysed and spontaneously inflammable. In contrast aluminium
hydride is a polymeric solid (AlH3)n. SiH4 is gaseous, readily hydrolysed and inflammable.
5.
Apart from BF3, the halides of B and Si hydrolyse readily and vigoruosly. The aluminium halides are only
partly hydrolysed in water.
C2A Carbon Family :
(i)
Ionization energy slightly increases from Sn to Pb because the poor shieldering effect of
intervening d- and -f orbitals and increase in size of the atom.
(ii)
Chemical property of boron family
(a)
Tendency to show +2 oxidation state increases in the sequence Ge < Sn < Pb.
(b)
The monooxides CO – Neutral, GeO – acidic, SnO, PbO – amphoteric
(c)
The dioxides, CO2, SiO2, GeO2 – a acidic, SnO2, PbO2 – amphoteric
(d)
Carbon silicon and germinium and lead are not affected by water, whereas tin forms

dioxide Sn  2H 2 O 
SnO 2  2H 2
(iii)
(iv)
(v)
(e)
PbI4 does not exist because the reaction does not release enought energy to unpair 6s2
electrons.
(f)
Because of thermal and chemical stability, GeX4 is more stable than GeX2 whereas
PbX2 is more stable than PbX4.
(g)
CCl4 not hydrolysed because of unavailibity of d-orbitals.
Anomalous Behaviour of carbon :
(a)
Anomalous behaviour of carbon due to its smaller size, higher electronegativity, higher
ionisation enthalpy and unavailability of d-orbitals.
(b)
Carbon can form p-p multiple bonds with itself and with other atom of small size and
high electronegativity for e.g. C = C, C = S, C = N.
(c)
The tendency of carbon to link with one another through covalent bonds known as
catenation.
(d)
The order of catenation is C > > Si > Ge  Sn.
Allotropes of carbon :
(a)
Carbon forms mainly 3 allotropes i.e. Diamond, Graphite and fullerenes.
(b)
Diamond has a crystalline lattice and undergoes sp3 hybridisation. It is a hardest
substance on earth and use to sharping hard tools, in making dyes etc.
(c)
Graphite has layered structure and it undergoes sp2 hybridisation because electrons are
mobile, therefore it conduct electricity. It is used as a dry lubricant in machine running
at high temperature.
(d)
Fullerene are made by the heating of graphite in an electric are in the presence of inert
gases such as helium or argon. Fullerences are the only pure form of carbon because
they have smooth structure without having ‘dangling’ bonds. All the carbon atoms are
equal and they undergo sp2 hybridisation. It has ball type structured with 60 vertices.
(e)
Thermodynamically graphite is most stable, therefore fH of graphite is taken as zero.
Uses of carbon :
The composites are used in products such as tennis rackets, fishing rods etc. Being good
conductor, graphite is used for electrodes in batteries and industrial electrolysis. Carbon black is
used as block pigment in black ink and as filler in automobile tyres. Diamond is a precious stone
and used in jewellery.
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C2B
Compounds of Carbon :
(a)
 1273 K
Water gas or Synthesis gas : C(s )  H 2 O(g ) 473

 CO(g )  H 2 (g )
water gas
(b)
K
Producer gas : 2C(s )  O 2 (g )  4N 2 (g ) 1273
 2 CO(g )  4N 2 (g )
producer gas
(c)
CO is powerful reducing agent and reduces almost all metal oxides other than those of the alkali
and alkaline earth metals, aluminium and a few transition metals. For e.g.

Fe 2O 3 (s )  3CO(g )  2Fe(s )  3CO 2 (g )

ZnO(s )  CO(g )  Zn(s)  CO 2 (g )
(d)
The highly poisonous nature of CO arises because of its ability to form a complex with
haemoglobin.
(e)
CO2 prepared by :

CH 4 (g )  2O 2 (g )  CO 2 (g )  2H 2 O(g )
CaCO 3 (s )  2HCl(aq ) 
 CaCl 2 (aq )  CO 2 (g )  H 2 O( l )
(f)
CO2 is removed by photo-synthesis as :
hv
6CO 2  12H 2O  C 6 H 12O 6  6O 2  6H 2 O
chlophyll
(g)
CO2 can be obtained as a solid in the form of dry ice, used as a refrigerant for ice-cream and
frozen food.
(h)

The resonance structures for CO2 are : : O
 C  O :  : O  C  O :  :O  C  O
.. :
..
..
..
..
..
Detailed explanation of compounds of carbon :
Carbides
Compounds of carbon and a less electronegative element are called carbides. This excludes compounds
with N, P, O, S and the halogens from this section.
Carbides are of three main types :
1.
ionic or salt-like
2.
interstitial or metallic
3.
covalent
Salt-like carbides
It is covenient to group these depending on whether the structure contains C, C2 or C3 anions. Aluminium
carbide Al4C3 is a pale yellow solid formed by heating the elements in an electric furnace, the structure of
Al4C3 is complex. It is misleading to formulate the structure as 4Al3+ and 3C4– as such a high charge
separation is unlikely. Both Be2C and Al4C3 are called methanides because they react with H2O, yielding
methane.
4Al + 3C  Al4C3, Al4C3  12H2O  4Al(OH)3 + 3CH4
Carbides with a C2 unit are well known. By far the most important compound is CaC2. This is made
commercially by strong heating lime and coke :
CaO + 3C  CaC2 + CO
H = +466 kJ mol–1
These carbides react exothermically with water, liberating ethyne (formerly called acetylene)., so they are
called acetylides.
CaC2 + 2H2O  Ca(OH)2 + HC  CH
CaC2 is an important chemical intermediate and is used as an industrial scale to produce calcium
cyanamide. Cyanamide is used as a nitrogenous fertilizer, and to make urea and melamine.
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CaC2 + N2
C
1100

 Ca (NCN) + C
The acetylides have a NaCl type of lattice, with Ca2+ replacing Na+ and C22– replacing Cl—In CaC2, SrC2 and
BaC2 the elongated shape of the (C  C)2– ions causes tetragonal distortion of the unit cell, that is it
elongates the unit cell in one direction.
Two carbides of magnesium Mg2C3 contains a C3 unit, and on hydrolysis with water it yields propyne
CH3 – C  CH.
Mg2C3+4H2O  2Mg(OH)2 + CH3 – C  CH
Oxygen Compounds of Cabron
Carbon forms more oxides than the other elements, and these oxides differ from those of the other elements
because they contain p – p multiple bonds between C and O. Two of these oxides, CO and CO2, are
extremely stable and important. Three are less stable : C3O2, C5O2 and C12O9. Others which are even less
stable include graphite oxide, C2O and C2O3.
Carbon monoxide CO
CO is a colourless, odourless, poisonous gas. It is formed when C is burned in a limited amount of air. In the
laboratory it is prepared by dehydrating forming acid with concentrated H2SO4.
H . COOH + H2SO4  CO + H2O
CO can be detected because it burns with a blue flame. It also reduces an aqueous PbCl2 solution to metallic
Pd, and when passed through a solution of I2O5 it liberates I2, i.e. it reduces I2O5 to I2. The latter reaction is
used to estimate CO quantitatively. The I2 is titrated with Na2S2O3.
PdCl2 + CO + H2O  Pd + CO2 + 2HCl
5CO + I2O5  5CO2 + I2
CO is toxic because it forms a complex with haemoglobin in the blood, and this complex is more stable than
oxy-haemoglobin. This prevents the haemoglobin in the red blood corpuscles from carrying oxygen round
the body. This causes an oxygen deficiency, leading to unconsciousness and then death. CO is sparingly
soluble in water and is a neutral oxide. CO is an important fuel, because it evolves a considerable amount
of heat when it burns in air.
2CO + O2  2CO2
H0 = –565kJ mol–1
The following are all important industrial fuels :
1.
Water gas : an equimolecular mixture of CO and H2.
2.
Producer gas : a mixture of CO and N2.
3.
Coal gas : a mixture of CO, H2, CH4 and CO2 produced while distilling. Coal gas is used for cooking and
heating. It was known as town gas.
Water gas is made by blowing steam through red or white hot coke.
red heat
C + H2O  CO + H2
(water gas)
H0 = +131 kJ mol–1
S0 = +134 kJ mol–1
The water gas reaction is strongly endothermic (G = H – TS). Thus the coke cools down, and at
intervals the steam must by turned off and air blown through to reheat the coke. It is particularly good fuel,
i.e., it has a high calorific value, because both CO and H2 burn and evolve heat.
CO is a good reducing agent, it can reduce many metal oxides to the metal.
blast furnace
Fe2O3 + 3CO   2Fe + 3CO2
CuO + CO  Cu + CO2
CO is an important ligand. It can donate an electron pair to many transition metals, forming carbonyl
compounds.
Ni + 4CO
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Fe + 5CO
0
C under pressure
200

  Fe (CO)5
2Fe(CO)5 photolysis

 Fe2(CO)9 + CO
CrCl6 + 3Fe(CO)5 heat
 Cr(CO)6 + 3FeCl2 + 9CO
The bonding in CO may be represented as three electron pairs shared between the two atoms :C  O:
It is better represented using the molecular orbital theory
2p2 ,
* 2p0y
1s2 , *1s2 , 2s2 , * 2s2  2y 2p2x , *2p0x , 
0
2pz ,
*2pz
     
 increasing energy
The carbon-metal bond in carbonyls may be represented as the donation of an electron pair from carbon to
the metal M  C  O. This original  bond is weak. A stronger second bond is formed by back bonding,
sometimes called dative  bonding. This arises from sideways overlap of a full dxy orbital on the metal with
the empty antibonding *2py orbital of the carbon, thus forming a  M  C bond. The total bonding is thus
M = C = O. The filling, or partial filling, of the antibonding orbital on C reduces the bond order of the
C — O bond from the triple bond in CO towards a double bond. This is shown by the increase in C — O
bond length from 1.128 Å in CO to about 1.15 Å in many carbonyls.
Carbon dioxide CO2
CO2 is a colourless, odourless gas. The main industrial source is as a by-product from the manufacture of
hydrogen for making ammonia.
CO + H2O
CO2 + H2
CH4 + 2H2O  CO2 + 4H2
It is also recovered from fermentation processes in breweries, from the gases evolved from calcining
limestone in lime kilns and from the flue gases from coal-burning electric power stations. The CO2 is
recovered by absorbing it in either aqueous Na2CO3 or ethanolamine.
yeast under
C 6 H12 O 6 anacrobic

 2C 2 H 5OH  2CO 2
conditions
strong heat
 CaO + CO2
CaCO3  
It is obtained in small amounts by the action of dilute acids on carbonates. It can also be made by burning
carbon in excess of air.
CaCO3 + 2HCl  CaCl2 + CO2 + H2O
C + O2  CO2
Recovery of CO2
cool
Na2CO3 + CO2 + H2O
 2NaHCO3
hot
Girbotol process
2HOCH 2 CH 2 NH 2 + CO2 + H2O
ethanola min e
30 600 C
100  1500 C
(HOCH2CH2NH3)2CO3
CO2 gas can be qualified under pressure between –570C and +310C. About 80% is sold in liquid form, and
20% as solid. The solid is produced as while snow by expanding the gas from cylinders. (Expansion causes
cooling). This is compacted into blocks and sold. Solid CO2 sublimes directly to the vapour state (without
going through the liquid state) at –780C under atmospheric pressure. Over half the CO2 produced is used as
a refrigerant. Solid CO2 is called ‘dry ice’ or ‘cardice’, and is used to freeze meat, frozen foods and ice
cream, and in the laboratory as a coolant. Over a quarter is used to carbonate drinks (Coca-Cola, lemonade,
beer etc.). Other uses include the manufacture of urea, as an inert atmosphere, and for neutralizing alkalis.
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(Urea is the most widely used nitrogeneous fertilizer and is also for making formaldehyde urea resins.)
1800 C
CO 2  2 NH 3 pressure
 NH 4 CO 2 NH 2  CO( NH 2 ) 2  H 2O
ammonium
urea
carbamate
Small scale uses of CO2 include use in fire extinguishers, blasting in coal mines, as an aerosol propellant,
and for inflating life-rafts.
CO2 gas is detected by its action on lime water Ca(OH)2 or baryta water Ba(OH)2, as a white insoluble
precipitate of CaCO3 or BaCO3 is formed. If more CO2 is passed through the mixture, the cloudiness
disappears as the soluble bicarbonate is formed.
Ca (OH ) 2  CO 2  CaCO 3  H 2 O
white
precipitat e
CaCO3  CO 2  H 2 O  Ca (HCO3 ) 2
so lub le
CO2 is an acidic oxide, and reacts with bases, forming salts. It dissolves in water but it is only slightly
hydrated to carbonic acid H2CO3, and the solution contains few carbonate or bicarbonate ions. A hydrate
CO2 . 8H2O can be crystallized at 00C under a pressure of 50 atmospheres CO2.
CO2 + H2O
H2CO3
Carbonic acid has never been isolated, but it gives rise to two series of salts, hydrogencarbonates
(otherwise called bicarbonates), and carbonates.
CO2 can also act as a ligand, and it form a few complexes such as [Rh(CO2)Cl(PR3)3] and [Co(CO2)(PPh3)3].
In the first complex the C atom in CO2 is bonded to the metal. In the second complex the CO2 acts as a
bidentate ligand with one C atom and one O atom bonded to the metal and the CO2 molecule is bent.
The structure of CO2 is linear O — C — O, Both C — O bonds are the same length. In addition to  bonds
between C and O, there is a three centre four-electron  bond
covering all three atoms. This adds two
bonds to the structure in addition to the two  bonds. Thus the C — O bond order is two.
Biologically, carbon dioxide is important in the process of photosynthesis, where the green parts of plants
manufacture glucose sugar. Ultimately all animal and plant life depends on this process.
6CO 2  6H 2 O sunlight
 C 6 H12 O 6  6O 2
glu cos e
The reverse reaction occurs during the process of respiration, where animals and plants release energy.
C6H12O6 + 6O2  6CO2 + 6H2O + energy
C2C Compounds of Silicon :
(a)
Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is
covalency bounded in a tetrahedral manner to four oxygen atoms.
(b)
Some reactions of SiO2 are :
SiO 2  2NaOH 
 Na 2 SiO 3  H 2 O
SiO 2  4HF 
 SiF4  2H 2 O
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(c)
Silicons are a group of organosilicon polymers which have (–R2SiO-) as a repeating uint.
(d)
The chain length of polymer can be controlled by adding (CH3)3SiCl which blocks the ends.
(e)
Silicons are used as sealant, greases, electrical insulators and for water proofing of fabrices.
(f)
The basic structure unit of silicates is SiO44–. Two man-made silicates are glass and cement.
(g)
If aluminium atoms replace few silicon atoms in three dimensional network of silicon dioxide,
overall structure known as aluminosilicates aquired a negative charge. Cations such as Na+, K+
or Ca2+ balance the negative charge. This type of structured known as zealites, which used in
petrochemical industries for cracking of hydrocarbons.
Detailed explanation of compounds of silicon :
Silicates
About 95% of the earth’s crust is composed of silicate minerals, aluminosilicate clays, or silica.
Soluble silicates
Silicates can be prepared by fusing an alkali metal carbonate with sand in an electric furnace at about
14000C.
0
SiO 2
C
Na 2CO 3 1400

 CO 2  Na 2 O 
 Na 4SiO 4 , ( Na 2SiO 3 ) n and others
The product is a soluble glass of sodium or potassium silicate. It is dissolved in hot water under pressure,
and is filtered from any insoluble material, this is known as water glass. They are used in liquid detergent
preparation to keep the pH high. Soluble silicates must not be used if the water is hard, or they will react
with Ca2+ to form insoluble calcium silicate.
Principles of silicate structures
The majority of silicate minerals are very insoluble, because they have an infinite ionic structure and
because of the great strength of the Si—O bond. This made it difficult to study their structures, and physical
properties such as cleavage and the hardness of rocks were originally studied.
1.
The electronegativity difference between O and Si, 3.5 – 1.8 = 1.7, suggests that the bonds are almost 50%
ionic and 50% covalent.
2.
The structure may therefore be considered theoretically by both ionic and covalent methods. The radius
ratio Si4+ : O2— is 0.29, which suggests that Si is four-coordinate, and is surrounded by four O atoms at the
corners of a tetrahedron. This can also be predicted from the use of the 3s and three 3p orbitals by Si for
bonding. Thus silicates are based on (SiO4)4— tetrahedral units.
3.
The SiO4 tetrahedra may exist as discrete units, or may polymerize into larger units by sharing corners, that
is by sharing O atoms.
4.
The O atoms are often close-packed, or nearly close-packed. Close packed structure have tetrahedral and
octahedral holes, and metal ions may occupy either octahedral or tetrahedral sites depending on their size.
Most metal ions are the right size to fit one type of hole, through Al3+ can fit into either. Thus Al can replace
either a metal in one of the holes, or a silicon atom in the lattice. This is particularly important in the
aluminosilicates.
Classification of Silicates
1.
Orthosilicates
A wide variety of minerals contain discrete (SiO4)4—, tetrahedra, that is they share no corners, for example
ZrSiO4.
Figure : Structure of orthosilicates.
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Zircon ZrSiO4 is used as a gemstone as it can be cut to look like a diamond, but is much cheaper. Zircon is
much softer than diamond. Ziron has a coordination number of 8. The structure is not close-packed.
2.
Pyrosilicates
Two tetrahedral units are joined by sharing the O at one corner, thus giving the unit (Si2O7)6—.Example :
Thortveitite Sc2[Si2O7], Hemimorphite Zn4(OH)2[Si2O7] . H2O
Figure : Structure of pyrosilicates Si2O76–.
3.
Cyclic silicates
If two oxygen atoms per tetrahedron are shared, ring structure may be formed of general formula (SiO3)n2n–
. Rings containing three, four, six and eight tetrahedral units are known,but those with three and six are the
most common. The cyclic ion Si3O96– occurs in wollastonite Ca3[Si3O9] and in benitoite BaTi[Si3O9]. The
Si6O1812— unit occurs in beryl Be3Al2[Si6O18].
Figure : Structure of cyclic silicates Si3O96— and Si6O1812—.
4.
Chain silicates
Simple chain silicates or pyroxenes are formed by the sharing of the O atoms on two corners of each
tetrahedron with other tetrahedra. This gives the formula (SiO3)n2n–, e.g. in spodumene LiAl[(SiO3)2]. Diopsite
CaMg [(SiO3)2].
Figure : Structure of pyroxenes (SiO3)n2n–.
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Double chains can be formed when two simple chains are joined together by shared oxygens. These minerals are called amphiboles, and they are will known. There are several ways of forming double chains,
giving formulae (Si 2O5)n2n–, (Si4O11)n6n–, (Si6O17)n10– and others. The most numerous and best known
amphiboles are the asbestos minerals. Amphiboles always contain hydroxyl groups, which are attached to
the metal ions.
Figure : Structure of amphiboles (Si4O11)n6n–.
The Si—O bonds in the chains are strong and directional. Adjacent chain are held together by the metal to
the chains. Thus pyroxenes and amphiboles cleave readily parallel to the chains, forming fibres. For this
reason they are called fibrous minerals. The cleavage angle for pyroxenes is 890, and for amphiboles 560.
This is used as a means of identifying the minerals. These angles are related to the size of the
cross-sectional trapezium of the chains and the way in which they are packed together.
Asbestos minerals come from two different groups of silicates :
1.
The amphiboles.
2.
The sheet silicates.
The amphiboles include crocidolite Na2Fe3IIFe2III [Si8O22] (OH)2, which is called blue asbestos and others
derived from it by isomorphous replacement, for example amosite or brown asbestos (Mg, FeII)7[Si8O22](OH)2.
The mineral chrysotile Mg3(OH)4[Si2O5] is called white asbestos, and this is derived from serpentine, and
is a sheet silicate.
Sheet silicates (phyllo-silicates)
When SiO4 units share three corners the structure formed is an infinite two-dimensional sheet of empirical
formula (Si2O5)n2n–.
Structures with simple planar sheets are rare. A large number of sheet silicates are made up of either two or
three layers joined together. These include :
1.
Clay minerals (kaolinite, pyrophyllite, talc)
2.
White asbestos (chrysotile, biotite)
3.
Micas (muscovite and margarite)
4.
Montmorillonites (Fullers earth, bentonite and vermiculite)
Figure : Structure of sheet silicates (Si2O5)n2n–.
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Three-dimensional Silicates
Sharing of all of the four corners of a SiO4 tetrahedron results in a three dimensional lattice of the formula
SiO2 (quartz, tridymite etc). These contain no metal ions. Replacement of one quarter or one half of the Si
atoms are quite common. Such replacements resuls in three groups of minerals.
1.
Feldspars
2.
Zeolites
3.
Ultramarines.
Feldspars are the most important rock forming materials e.g. granite is made up of feldspar with some mica
and quartz.
Zeolites are used as ion-exchange materials. This is the reason that sodium zeolites Na2[Al2Si3O10]2H2O
are used to soften the hard water. By passing the hard water through a column of zeolite or hydrated sodium
aluminium silicate called permutit, NaAlSiO4 . 3H2O. Due to ion-exchange, sodium in the silicate / zerolite
gets exchanged for Ca2+/Mg2+ ions and the water gets softened.
2NaAlSiO4 + Ca2+
Ca(AlSiO4)2 + 2Na+, 2NaAlSiO4 + Mg2+
Mg (AlSiO)4 + 2Na+
The zeolite/permutit can be regenerated by treatment with sodium chloride solution.
Organosilicon Compounds and the Silicones
Si — C bonds are almost as strong as C — C bonds. Thus silicon carbide SiC is extremely hard and stable.
Many thousands of organosilicon compounds containing Si — C bonds have been made. However, the vast
range of organic compounds is not replicated by silicon for three main reasons.
1.
Silicon has little tendency to bond to itself (catenate) whilst carbon has a strong tendency to do so. The
largest chains formed by Si are contained in Si16F34 and Si8H18, but these compounds are exceptional. This
is related to the weakness of Si — Si bonds in contrast to the strength of C — C bonds.
2.
Silicon does not form p — p double bonds, whilst carbon does not readily.
3.
Silicon forms a number of compounds containing p – d double bonds in which the silicon atom uses d
orbitals.
Preparation of organosilicon compounds
There are several ways of forming Si — C bonds :
1.
By a Grignard reaction
SiCl4 + CH3MgCl  CH3SiCl3 + MgCl2
CH3SiCl3 + CH3MgCl  (CH3)2SiCl2 + MgCl2
(CH3)2SiCl2 + CH3MgCl  (CH3)3SiCl + MgCl2
(CH3)3SiCl + CH3MgCl  (CH3)4Si + MgCl2
2.
Using an organolithium compound
4LiR + SiCl4  SiR4 + 4LiCl
R may be alkyl or aryl.
3.
Rochow ‘Direct Process’. Alkyl or aryl halides react directly with a fluidized bed of silicon in the presence
0
of large amounts (10%) of a copper catalyst. Si + 2CH3Cl
280  300 C
Cu
catalyst


 (CH3)2SiCl2
Silicones
The silicones are a group of organosilicon polymers.
The complete hydrolysis of SiCl4 yields SiO2, which has a very stable three-dimensional structure. The
fundamental research of F. S. Kipping on the hydrolysis of alkyl substituted chlorosilanes led, not to the
expected silicon compound analogous to a ketone but to long-chain polymers called silicones.
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The starting materials for the manufacture of silicons are alkyl or aryl substituted chlorosilanes. Methyl
compounds are mainly used, though some phenyl derivatives are used as well. Hydrolysis of
dimethyldichlorosilane (CH3)2SiCl2 gives rise to straight chain polymers and, as an active OH group is left
at each end of the chain, polymerization continues and the chain increases in length. (CH3)2SiCl2 is
therefore a chain building unit. Normally, high polymers are obtained.
Hydrolysis under carefully controlled conditions can produce cyclic structures, with rings containing three,
four, five or six Si atoms :
Hydrolysis of trimethylmonochlorosilane (CH3)3SiCl yields (CH3)3SiOH trimethylsilanol as a volatile liquid, which can condense, giving hexamethyldisiloxane. Since this compound has no OH groups, it cannot
polymerize any further.
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hexamethyldisiloxane
If Some (CH3)3SiCl is mixed with (CH3)2SiCl2 and hydrolysed, the (CH3)3SiCl will block the end of the
straight chain produced by (CH3)2SiCl2. Since there is no longer a functional OH group at this end of the
chain, it cannot grow any more at this end. Eventually the other end will be blocked in a similar way. Thus
(CH3)3SiCl is a chain stopping unit, and the ratio of (CH3)3SiCl and (CH3)2SiCl2 in the starting mixture will
determine the average chain size.
The hydrolysis of methyl trichlorosilane RSiCl3 gives a very complex cross-linked polymer.
In a similar way addition of a small amount of CH3SiCl3 to the hydrolysis mixture produces a few
cross-links, or provides a site for attaching other molecule. By controlled mixing of the reactants, any given
type of polymer can be produced.
Properties and Uses
Silicones are strongly water repellent, good electrical insulators have non-stick properties and anti-foaming
properties. Their strength and inertness are related to two factors :
1.
Their stable silica-like skeleton of Si—O—Si—O—Si. The Si—O bond energy is very high
(502 kJ mol–1).
2.
The high strength of the Si—C bond.
Their water repellency arises because a silicone chain is surrounded by organic side grous, and looks like
an alkane from the outside.
Silanes
Silicon forms a limited number of saturated hydrides, SinH2n + 2, called the silanes. Monosilane SiH4 is the
only silicon hybride of importance.
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Preparation
1.
heat in absence air
2Mg + Si     Mg2Si
Mg2Si + H2SO4  SiH4
2.
(40%)
More recently monosilane has been prepared by reducing SiCl4 with Li[AlH4], LiH or NaH in ether
solution at low temperatures.
SiCl4 + Li[AlH4]  SiH4 + AlCl3 + LiCl
Si2Cl6 + 6LiH  Si2H6 + 6LiCl
Si3Cl8 + 8NaH  Si3H8 + 8NaCl
3.
Silanes may also be prepared by direct reaction by heating Si or ferrosilicon with anhydrous HX or RX in
the presence of a copper catalyst.
Si + 2HCl  SiH2Cl2
Si + 3HCl  SiHCl3 + H2
Si + 2CH3Cl  CH3SiHCl2 + C + H2
Halides of Silicon
Preparation :
The silicon halides can be prepared by heating either Si or SiC with the appropriate halogen.
Properties :
1.
SiF4 is readily hyrolysed by alkali.
SiF4 + 8OH—  SiO44– + 4F– + 4H2O
[In marked contrast to the inertness of CF4, CCl4 and the Freons, SiF4 is readily hydrolysed]
2.
In the case of the tetrafluoride, a secondary reaction occurs between the resultant HF and the unchanged
SiF4, forming the hexafluorosilicate ion [SiF6]2–.
SiF4 + 2HF  2H+ + [SiF6]2–
The [SiF6]2– ion is usually formed from SiO2 and aqueous HF
SiO2 + 6HF  2H+ + [SiF6]2– + 2H2O
The [SiF6]2– complex is stable in water and alkali, but the others in the group are less stable. [SnF6]2– are
hydrolysed by alkali, and [PbF6]2– is hydrolysed by both alkali and water.
[SiF6]2– gives an octahedral structure (sp3d2 hybridization)
3.
The silocon halides are repidly hydrolysed by water to give silicic acid.
SiCl4 + 4H2O  Si(OH)4 + 4HCl
Large quantity of SiCl4 are hydrolysed at a high temperature (in an oxy-hydrogen flame) giving very finely
powdered SiO2 rather than Si(OH)4.
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C O N C E P T S (p-Block) XIIth
C1A Group 15 (General properties of Group 15)
I.
Atomic radius : increases down the group because of increase of number of shells. From As to Bi only a
small increase in radius is observed because of the presence of inner d and f-orbital which having poor
shielding effect and hence effective nucleus change increases.
II.
Ionisation enthalpy : (a) IE decreases down the group because of increase of atomic size
First ionisation energy of group 15 elements higher than group-14 and group-16 because of stable config.
half filled orbitals i.e., stable config.
III.
Electronegativity : decreases down the group because of increase of atomic size.
IV.
Metallic Character : increases down the group because of increase of atomic size and decrease if IE.
V.
Melting Point : increases from N to As but decreases from As to Bi. Lower m.p. of Sb and Bi are due to
their tendency to form 3-covalent bonds instead of 5 due to inert pair effect.
VI.
Allotropy : All elements show except nitrogen. Allotropic forms (white, red and black) As and Sb forms
two (white and grey).
VII.
Catenation : Tendency for catenation increases from N to P. N has some extent to shows catenation
because of triple bond.
VIII. Oxidation State : –3, +3, +5
(i)
The tendency to show – 3 o.s. decreases down the group due to decrease in EN, which is due to increase in
atomic size, therefore increase in metallic character.
(ii)
The stability of +5 o.s. decreases whereas +3 o.s. increases down the group due to inert pair effect. Thus the
electrons present in the ‘s’ subshell do not participate in bond formation and only the p-electrons
participate.
(iii)
Bi3+ is more stable than Bi5+ due to inert pair effect i.e., why Bi5+ is a good oxidising agent. The only well
characterisied Bi5+ is BiF5.
(iv)
Nitrogen shows o.s. from –3 to +5.
–3 o.s. acts as reducing agent while +5 o.s. acts as oxidising agent while +1 to +4 shows disproportionation
in acid solution.
( 3 )
e.g.
( 5 )
( 2 )
3HNO 2 
 HNO 3  H 2O  2NO
( 3 )
( 5 )

( 3 )
H 3 PO 3  H 3 PO 4  PH 3
( 1)
( 2 )
( 3 )
( 4 )
( 4 )
( 5 )
( 3 )
N 2O, NO, N 2O 3 , NO 2 , N 2 O 4 , N 2O 5 , NH 3
(v)
The nitrogen can show maximum covalency upto 4 not more than 4 because of absence of d-orbitals. While
rest elements can show higher covalency due to presence of vacant d-orbitals e.g. PF6–.
(vi)
Nitrogen does not shows pentavalent because of absence of d-orbitals.
C1B
Anomalous behaviour of Nitrogen :
Reasons : (a) small size (b) high EN, IE (c) absence of d-orbitals and hence cannot expand its covalency
more than 4. (d) has tendency to form p-p multiple bonds with itself and other elements having small size
and high EN like carbon and oxygen.
C1C Allotropy of Phosphorous :
Phosphorous exists in 3 main allotropic forms : (1) white phosphorous (2) red phosphorous (3) black
phosphorous.
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(1)
(2)
White Phosphorous :
(i)
It is formed by condensation from the gaseous or liquid states.
(ii)
It is a waxy solid which is insoluble in water but highly soluble in CS2 and benzene.
(iii)
It is highly reactive and spontaneously indigestive and hence it is stored under water.
(iv)
It glows in dark and this property gives the element its name PHOSPHOROUS (green for “light
bringing”)
(v)
It consists of discrete P4 molecules in which the four phosphorous atoms are at the corners of a
tetrahedron and each phosphorous atoms is covalently linked to the other 3 phosphorous atoms.
(vi)
It is highly toxic.
Red Phosphorous :
(i)
When white phosporous is heated to 570 K in an inert atmosphere for several days it gets
converted into red phosphorous.
(ii)
It has higher m.p. (870 K) and greater density (~ 2.16 gcm–3) than white phosphorous.
(iii)
It is very less reactive than white phosphorous
(iv)
It is safer and easier to handle
(v)
It is essentially non-toxic
(vi)
It is amorphous and has a polymeric structure
(3)
Black Phosphorous :
(i)
It is obtained by heating white phosphorous at 470 K and under high pressure. A series of phase of block
phosphorous are formed. One of these phases consists of an extended layer structure in which each P is
bound to three neighbours by single bonds.
(ii)
Thermodynamically it is most stable form of phosphorous. It is least reactive and crystalline in nature.
C1D Comparison of nitrogen with phosphorus
(i)
N-N bond is weaker than P-P bond due to high interelectronic repulsion of the non-bonding electrons.
(ii)
Nitrogen cannot form d-d and d-p bonds only form p - p due to absence of d-orbitals. Others can
form d-p bond e.g. R3P = 0 and d-d bond with transition element when their compounds act as ligands
e.g. P(C6H5)3, As(C6H5)3.
C1E
Compounds : (1) Hydrides : (MH3) (sp3 hybridisiation)
(i)
Thermal Stability : Moving down in the group it decreases because increase of size of central atom (M)
and hence increase of M-H bond length and hence decrease of bond strength and bond dissociation energy.
NH3 > PH3 > AsH3 > SbH3 > BiH3
(ii)
Reducing Character : Except NH3 all hydrides are strong (good) reducing agent and hence react with
metal ions (Ag+, Cu2+) to give phosphides, arsenides or antimanides.
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Order of Reducing Character : NH3 < PH3 < AsH3 < SbH3 < BiH3 because of increase of atomic size and
hence increase of M - H bond length, decrease of bond dissociation. energy thus bond weakens from NH3
to BiH3. Bi-H bond can be easily broken.
(iii)
Bond Angle :
NH 3  PH 3  AsH 3  SbH 3  BiH 3
1070
940
920
910
900
EN of N is highest  lone pairs will be towards nitrogen, so there is high e– density around the N-atom and
hence more repulsion between bond pairs  bond angle increases.
As EN , e– density also decreases down the group  repulsive interaction between bond pair electrons
also decreases and bond angle decreases.
(iv)
Boiling Point : BiH3 > SbH3 > NH3 > AsH3 > PH3
NH3 : exception high boiling point because of intermolecular H-bond.
PH3 to BiH3 B.P. increases because of increase of atomic size and hence increase of vanderwall’s
interaction.
(v)
Basicity : group-15 hydrides are lewis base because of presence of lone pairs of electrons on the central
atom and hence has tendency to donate a pair of e–. Order of basicity is NH3 > PH3 > AsH3 > SbH3 > BiH3
i.e., basicity decreases down the group because nitrogen atom has the smallest size among the hydrides 
the lone pair e–s is concentrated on a small region so e– density per unit volume is more. As we move down
the group the size of central atom keeps on increases  then the e– is distributed over large volume as a
result e– density decreases. Thus the e– donar capacity i.e. basic strength decreases.
2.
Halides of Group 15 : (a) MX3 (b) MX5
Trihalides (MX3) : sp3 hybridised, covalent except BiF3 which is ionic in nature.
Pentahalides (MX5) : sp3d (Trigonal bipyramidal in shape)
(i)
Nitrogen does not form pentahalides because of absence of d-orbitals.
(ii)
They are lewis acids because of presence of vacant d-orbitals accept a pair of e–. i.e., PCl5 + Cl–  [PCl6]–
(iii)
PCl5 is covalent in gaseous and liquid phase as it has a trigonal bipyramidal geometry while in solid state it
is ionic as it exist as [PCl4]+ [PCl6]– i.e., tetrahedral cation and octahedral anion.
(iv)
In PCl5, 3 P-Cl bonds are equitorial & 2 P-Cl bonds are axial. Axial bonds are longer due to greater
repulsion from equitorial bonds and hence bond length are not equivalent and bonds become weaker and
hence easily dissociated into PCl3 and Cl2 so PCl5 is quite reactive.
(v)
PCl3 fumes in moist air because of production of HCl with water PCl3 + 3H2O  H3PO3 + 3HCl
3.
Oxides of Group 15 :
(i)
All group-15 elements form tri-oxides and pentoxides. Pentoxides are more acidic than trioxides.
(ii)
Acidic character decreases down the group and basic character increases.
N 2 O 3 , P2 O 3 , As 2 O 3 , Sb 2 O 3 , Bi 2 O 3
Acidic
Amphoteric
Basic
Reason : because the size of N is very small. It has a strong positive field in a very small area  it attracts
the e–s of water’s O-H bond to itself and releases H+ ions easily.
Oxides of Nitrogen : N 2O, NO, N 2 O 3 , N 2O 4 , N 2 O 5
Neutral
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Oxides of Nitrogen :
Name
Formula
o.s.
Common Methods of
Physical Appearance
Preparation
and Chemical Nature

1.
Nitrogen (I) oxide N2O
+1
NH 4 NO 3  N 2 O  2H 2O
colourless, neutral gas
2.
Nitrogen (II)
+2
NaNO2 + 2FeSO4 + 3H2SO4 
colourless, neutral gas
3.
NO
oxide (nitrogen
Fe2(SO4)3 + 2NaHSO4 + H2O
monoxide)
+ 2NO
Nitrogen (III)
250 k
N2O 3
+3
2NO  N 2O 4  2N 2 O 3
NO2
+4
2Pb( NO 3 ) 2  4NO 2  2PbO Brown gas, acidic
N2O 4
+4
2NO2
N2O4
colourless, solid, acidic
N2O 5
+5
4HNO3 + P4O10 
colourless, solid, acidic
blue solid, acidic
oxide
4.
Nitrogen (IV)

oxide
5.
Nitrogen (IV)
oxide
6.
Nitrogen (V)
oxide
HPO3 + 2N2O5
Structure of oxides of nitrogen :
N2O :
NO :
N2O3 :
NO2
N2O4
N2O5
Oxides of Phosphorous : (P2O3, P2O5) but exist as a dimer form. P to Bi oxides exist as the dimer form
because of reluctance of p - p multiple bonds leads to cage like structure i.e, M4O6 and M4O10
M4 + 3O2  M4O6
M4 + 5O2  M4O10
(M = P, As, Sb, Bi)
(excess air)
Bi4O10 is unstable
P4O6 + H2O  H3PO3
(acidic oxides)
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P4O10 + H2O  H3PO4
(acidic oxides)
4.
Preparation of N2 (Dinitrogen) :
Preparation
(i)
NH4Cl (aq) + NaNO2 (aq)  N2(g) + 2H2O(l) + NaCl (aq)
small amount of impurities like NO and HNO3 are removed by passing the gas through aqueous sulphuric
acid containing potassium dichromate.
(ii)
Thermal decomposition of ammonium dichromate :

( NH 4 ) 2 Cr2O 7  N 2  4H 2 O  Cr2 O 3
(iii)
Thermal decomposition of sodium or barium azide

Ba( N 3 ) 2  Ba  3N 2

Properties :
5.
6Li  N 2  2Li 3 N

3Mg  N 2  Mg 3 N 2
NH3 (Ammonia) :
Preparation
(i)

NH 2CONH 2  H 2 O  NH 3  CO 2  H 2 O
urea
(ii)
NH 4 Cl  Ca(OH ) 2 
 NH 3  CaCl 2  H 2 O
( NH 4 ) 2 SO 4  NaOH 
 NH 3  H 2 O  Na 2 SO 4
(iii)
Haber-Bosch process
The most important commercial process is the Haber - Bosch process. Fritz Haber discovered how to make
N2 and H2 combine directly in the laboratory. He was awarded the Nobel Prize for Chemistry in 1918. Carl
Bosch was a chemical engineer who developed the plant to make ammonia using this reaction on an
industrial scale. He too was awarded the Nobel Prize for Chemistry in 1931 for his work on high pressure
reactions.
The reaction is reversible, and Le Chatelier’s principle suggests that a high pressure and low temperature
are required to drive the reaction to the right, and thus form NH3. A low temperature gives a higher percentage conversion to NH3, but the reaction is slow in reaching equilibrium, and a catalyst is required. In
practice the condition used are 200 atmospheres pressure, a temperature of 380 - 4500C and a catalyst of
promoted iron. It is more economic to use a higher temperature, so that equilibrium will be reached much
faster, even though this gives a lower percentage conversion. At a temperature of about 4000C a 15%
conversion is obtained with a single pass over the catalyst. The gas mixture is cooled to condense liquid
NH3, and the unchanged mixture of N2 and H2 gases is recycled. The plant is made of steel alloyed with Ni
and Cr.
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The catalyst is made by fusing Fe3O4 with KOH and a refractory material such as MgO, SiO2 or Al2O3. This
is broken into small lumps and put into the ammonia convertor, where the Fe3O4 is reduced to give small
crystals of iron in a refractory matrix. This is the active catalyst.
The actual plant is more complicated than this one-stage reaction implies, since the N2 and H2 must be made
before they can be converted to NH3. The cost of H2 is of great importance for the economy of the process.
Originally the H2 required was produced by electrolysis of water. This was expensive, and a cheaper method
using coke and water was then used (water gas, producer gas). Nowadays the H2 is produced from
hydrocarbons, either naptha or CH4, by reacting with steam at 7500C with a Ni catalyst. All traces of S must
be removed since these poison the catalyst.
CH4 + 2H2O
CH4 + H2O
CO2 + 4H2
CO + 3H2
Some air is added. The O2 burns with some of the H2, thus leaving N2 to give the required reaction ratio N2
: H2 of 1 : 3.
( 4 N 2  O 2 ) 2H 2
4N2 + 2H2O
air
CO must also be removed as it too poisons the catalyst.
CO + H2O
CO2 + H2
Finally the CO2 is removed in a scrubber by means of a concentrated solution of K2CO3, or ethanolamine.
Ammonia is used as a fertilizer. Other uses include the following :
1.
Making HNO3, which can be used to make NH4NO3 (fertilizer), or explosive such as nitroglycerine,
nitrocellulose and TNT . HNO3 can be used for many other purposes.
2.
Making caprolactam, which on polymerization forms nylon-6 (see hydroxylamine).
3.
Making hexamethylenediamine which is used in making nylon-6-6, polyurethanes and polyamides.
4.
Making hydrazine and hydroxylamine.
5.
Liquid NH3 is often used as a cheaper and more convenient way of transporting H2 than cylinders of
compressed H2 gas. The H2 is obtained from NH3 by heating over a catalyst of finely divided Ni or Fe.
6.
Ammonia has been used as the cooling liquid in refrigerators. It has a very high heat of vaporization, and
convenient boiling and freezing points. With the environmental concern over using Freons in refrigerators,
this use of NH3 could increase.
Properties of ammonia :
(i)
2FeCl 3 (aq)  3NH 4OH(aq) 
 Fe 2 O 3 . xH 2O(&) 3NH 4 Cl
brown ppt .
ZnSO 4 (aq )  2NH 4OH (aq ) 
 Zn(OH ) 2  ( NH 4 ) 2 SO 4
white
(ii)
NH3 is a weak base. It has a lone pair of e–.
Cu 2 (aq)  4NH 3 (aq)
[ Cu ( NH 3 ) 4 ]2  ( aq )
deep blue
AgCl  (s)  2NH 3
[ Ag( NH 3 ) 2 ]Cl(aq )
colourless
Ammonia is a good complexing agent because of presence of lone pair of electron it act as a legand and
forms complexes
PH3 (Phosphine) : because of presence of lone pair, and it acts as ligand and forms complexes.
6.
PH3 (Phosphine) : Preparation
(i)
Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
Ca3P2 + 6HCl  3CaCl2 + 2PH3
(ii)
P4 (white) + 3NaOH + 3H2O  PH3 + 3NaH2PO2
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(iii)
Lab preparation :
PH 4 I  KOH
Phosphoniu m
iodide

 KI  H 2 O  PH 3
Properties :
(i)
Weak base : PH3 + HBr or HI  PH4Br
(ii)
3CuSO4 + 3PH3  Cu3P2 + 3H2SO4
(iii)
3HgCl2 + 2PH3  Hg3P2 + 6HCl
Uses : used in Holeme’s signals and smoke screens.
7.
Halides of Group 15 PCl3 (Phosphorous trichloride)
Trihalides
All the possible trihalides of N, P, As, Sb and Bi are known. The nitrogen compounds are the least stable.
Though NF3 is stable, NCl3 is explosive.
The trihalides are predominantly covalent and, like NH3, have a tetrahedral structure with one position
occupied by a lone pair. The exceptions are BiF3 which is ionic and the other halides of Bi and SbF3 which
are intermediate in character.
The trihalides typically hydrolyse readily with water, but the products vary depending on the element :
NCl3 + 4H2O  NH4OH + 3HOCl
PCl3 + 3H2O  H3PO3 + 3HCl
AsCl3 + 3H2O  H3AsO3 + 3HCl
SbCl3 + H2O  SbO+ + 3Cl— + 2H+
BiCl3 + H2O  BiO+ + 3Cl— + 2H+
They also react with NH3.
PCl3 + 6NH3  P(NH2)3 + 3NH4Cl
NF3 behaves differently from the others. It is unreactive, rather like CF4, and does not hydrolyse with water,
dilute acids or alkali. It does react if sparked with water vapour.
NF3 has little tendency to act as a donor molecule. The molecule is tetrahedral with one position occupied
by a lone pair, and the bond angle F — N — F is 102030’. However, the dipole moment is very low (0.23
Debye units) compared with 1.47D for NH3. The highly electronegative F atoms attract electrons, and these
moments partly cancel the moment from the lone pair, and this reduces both the dipole moment and its
donor power.
PCl3 is the most important trihalide, and 250000 tonnes/year are produced commercially from the elements.
Some PCl3 is used to make PCl5.
PCl3 + Cl2 (or S2Cl2)  PCl5
PCl3 is widely used in organic chemistry to convert carboxylic acids to acid chlorides, and alcohols to alkyl
halides.
PCl3 + 3RCOOH  3RCOCl + H3PO3
PCl3 + 3ROH  3RCl + H3PO3
PCl3 can be oxidized by O2 or P4O10 to give phosphorus oxochloride POCl3.
2PCl3 + O2  2POCl3
6PCl3 + P4O10 + 6Cl2  10POCl3
POCl3 is used in large amounts in the manufacture of trialkyl and triaryl phosphates (RO)3PO.
Triethyl phosphate
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Pentahalides
Nitrogen is unable to form pentahalides because the second shell contains a maximum of eight electrons,
i.e. four bonds. The subsequent elements have suitable d orbitals, and form the following pentahalides :
PF5
PCl5
AsF5
(AsCl5)
SbF5
SbCl5
PBr5
PI5
BiF5
3PCl5 + 5AsF3  3PF5 + 5AsCl3
PCl3 + Cl2 (in CCl4)  PCl5
2As2O3 + 10F2  4AsF5 + 3O2
2Sb2O3 + 10F2  4SbF5 + 3O2
2Bi + 5F2  2BiF5
These molecules have a trigonal bipyramid shape in the gas phase, as expected from the VSEPR theory for
five pairs of electrons.
Figure : Structure of gaseous phosphorus pentachloride.
The trigonal bipyramid is not a regular structure. Electron diffraction on PF5 gas shows that some bond
angles are 900 and others are 1200, and the axial P — F bond lengths are 1.58 Å whilst the equatorial
P — F lengths are 1.53 Å. The axial and equatorial F atoms are thought to interchange their positions in less
time than that needed to take the NMR. The interchange of axial and equatorial positions is called
‘pseudorotation’.
PF5 remains covalent and keeps this structure in the solid state. However, PCl5 is close to the ionic-covalent
borderline, and it is covalent in the gas and liquid states, but is ionic in the solid state. PCl5 solid exists as
[PCl4]+ and [PCl6]— : the ions have tetrahedral and octahedral structures respectively. In the solid, PBr5
exists as [PBr4]+Br—, and PI5 appears to be [PI4]+ and I— in solution.
PCl5 is the most important pentahalide, and it is made by passing Cl2 into a solution of PCl3 in CCl4.
Complete hydrolysis of the pentahalides yields the appropriate -ic acid. Thus PCl5 reacts violently with
water :
PCl5 + 4H2O  H3PO4 + 5HCl
phosphoric acid
If equimolar amounts are used, the reaction is more gentle and yields phosphorus oxochloride POCl3.
PCl5 + H2O  POCl3 + 2HCl
PCl5 is used in organic chemistry to convert carboxylic acids to acid chlorides, and alcohols to alkyl
halides.
PCl5 + 4RCOOH  4RCOCl + H3PO4 + HCl
PCl5 + 4ROH  4RCl + H3PO4 + HCl
It reacts with P4O10, forming POCl3, and with SO2, forming thionyl chloride SOCl2.
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6PCl5 + P4O10  10POCl3
PCl5 + SO2  POCl3 + SOCl2
PCl5 also reacts with NH4Cl, forming a variety of phosphonitrilic chloride polymers.
nPCl5 + nNH4Cl  (NPCl2)n + 4nHCl
(ring compounds n = 3 – 8)
and
(chain compounds)
Cl4P · (NPCl2)n · NPCl3
8.
Oxoacids of Nitrogen and Phosphorous
1.
Nitrogen oxacids are HNO2, HNO3
2.
Preparation of HNO3
(i)
Lab preparation : NaNO3 + H2SO4  NaHSO4 + HNO3
(ii)
Ostwald’s process
(a)
catalytic oxidation of NH3 by atmospheric oxygen :
Pt / Rb
4NH3(g) + 5O2(g)      4NO(g) + 6H2O(g)
catalyst , 500K , 9 bar
(b)
2NO(g) + O2(g)
2NO2(g)
(c)
NO 2  H 2O 
 HNO 3 
( 68%)
(d)
NO
( can recycle to form HNO 3 )
Further concentration upto 98% is done by dehydration with concentrated H2SO4.
Properties :
(i)
Planar Structure :
(ii)
Strong acid : HNO3 + H2O  H3O+ + NO3–
(iii)
Concentration HNO3 is a strong oxidising agent :
Cu + HNO3(dil)  Cu(NO3)2 + 2NO + H2O
4Zn + 10HNO3(dil)  4Zn(NO3)2 + N2O + 5H2O
Cu + HNO3(conc.)  Cu(NO3)2 + 2NO2 + H2O
Zn + 4 conc. HNO3  Zn(NO3)2 + 2NO2 + 2H2O
with non-metals : (conc. HNO3)
I2 + 10HNO3  2HIO3 + 10NO2 + 4H2O
C + HNO3  CO2 + 2H2O + 4NO2
S8 + 48HNO3  8H2SO4 + 48NO2 + 16H2O
P4 + 20HNO3  4H3PO4 + 20NO2 + 4H2O
(iv)
Brown ring test :
NO3– + 3Fe2+ + 4H+  NO + 3Fe3+ + 2H2O
[Fe( H 2O) 6 ]2  NO  [Fe( H 2O) 5 ( NO)]2  H 2O
Brown ring
C2A Group-16 (Chalcogens) (ns2np4)
General Properties
(i)
IE, EN, Electron Gain Enthalpy (Heg) decreases down the group because of increase of atomic size.
Exception : O has less negative Heg than S because of small size of O.
(ii)
Metallic character increases from O to Po because of increase of atomic size and decrease of IE.
(iii)
Non-metallic character decreases due to decrease in EN.
(iv)
Melting point & Boiling point : Increased down the group because of increase of atomic number. Oxygen
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has much lower m.p. and b.p. than S because oxygen exist as diatomic (O2) form while S is octaotomic (S8)
(v)
Oxidation state :
( 2 )
( 1)
( 1 / 2 )
(a)
oxygen shows - 2; except in OF2 , O 2 F2 , KO 2 , H2O2 (peroxide-1)
(b)
S shows –2, +4, +6; Se, Te show +4, +6 with F and oxygen.
(c)
stability of –2 o.s. decreases down the group because increase in atomic size and decrease in EN.
(d)
The stability of +6 o.s. decreases and +4 o.s. increases due to inert pair effect.
Oxygen does not shows +4 and +6 o.s. while sulphur shows because of absence of d-orbital to oxygen and
presence of vacant d-orbitals to S.
(v)
Multiple bonds : formation tendency decreases down the group because of increase of atomic size they
become unstable.
C2B
Compounds of Group - 16
1.
Hydrides :
H2O liquid (sp3 bent shape) (because of H-bond)
H2S (gas), H2Se, H2Te, H2PO
(i)
Bond angle : H 2O  H 2S  H 2Se  H 2 Te
104.50
92 0
910
900
H2O bond angle is higher because O has the highest EN  lone pair will be towards O and hence bond
pairs will move away while going down in the group EN decrease and atomic size increases due to which
bond pair-bond pair. repulsion decreases.
(ii)
Boiling point :
(iii)
Acidic character : H 2 O  H 2 S  H 2 Se  H 2 Te (weak diprotic acid)
H 2O
 HeTe  H 2 Se  H 2 S
higher intermolec ular increase in vanderwaal 's forces
H  bond
neutral
increase in acidic character
Reasons : As the size of the central atom increases in the order O < S < Se < Te, the distance between the
central atom and hydrogen also increases. As a result the bond length increases the bond dissociation
energy decreases and bond cleavage become more and more easy. Therefore, the acidic strength of the
hydrides increases down the group.
(iv)
Thermal Stability : H2O > H2S > H2Se > H2Te > H2PO
Reason : On going down the group, the size of the central atom increases and thereby its tendency to form
stable covalent bond with hydrogen decreases as a result the M-H bond strengh decreases and thermal
stability decreases.
(v)
Reducing Character : H2O < H2S < H2Te < H2PO due to decrease in thermal stability.
2.
Preparation of Sulphur and its properties
(a)
Hydrogen sulphide is oxidized to sulphur by passing oxygen at low temperature
(b)
H 2S 
1
low temp .
O 2   S  H 2O
2
H 2S 
3
O2 
 SO 2  H 2 O
2
3
Fe O catalyst
SO 2  2H 2 S 2 3  
 S 8  2H 2 O
673 K
8
Uses of Sulphur and other Elements :
(i)
S : used as a manufacture of H2SO4 & other industrially compounds.
(ii)
Se as photoconductor in photocopying (xerox) machines.
(iii)
major use is as a decolouriser of glass.
Te : Alloy,
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Allotropy of S and Se :
S exists 3 main allotropic forms
(i)
Rhombic (orthorhombic)  : yellow, stable at room temperature
(ii)
Monoclinic  : which is formed by heating rhombic sulphur to about 368 K.
Both ,  are soluble in CS2, exist as S8 molecules with puckered ring structures and has a crown shape.
(iii)
Plastic Sulphur : It is amorphous form of sulphur when molten sulphur heated to about 625 K is poured into
cold water, a soft rubber like mass is obtained called plastic sulphur. It is soft and elastic in begining but
hardens on standing and gradually changes to rhombic sulphur. It is also regarded as supercooled liquid.
(iv)
Cyclo-S6 : chair form
(v)
Catena-Sn : chain polymer
(vi)
S2 is paramagnetic
S in vapour state exhibit paramagnetic behaviour because in vapour state S partly exists as S2 molecule and
S 2 molecule like O 2 has 2 unpaired electrons in the antibonding * orbital and hence exhibit
paramagnetism.
3.
Halides of Group 16 : EX2, EX4, EX6
S
Se, Te
S2X2(X = F, Cl, Br, I)
Se2Cl2, Se2Br2
SX2 (X = F or Cl)
SeCl2, TeCl2, SeBr2, TeBr2
SF4
SeX4, TeX4 (X = F, Cl, Br), TeI4
SF6
SeF6, TeF6
The stability of the halides decreases in the following order because of bond length increases
F– > Cl– > Br– > I–. The highest O.S. is in the fluorides only with iodine only TeI4 is known.
Prepration : Chalcogens halides prepared by direct combination.
(i)
1
S 8 (s )  3F2 (g ) 
 SF6 (g )
8
(ii)
1
S 8 ( l )  Cl 2 (g ) 
 S 2 Cl 2 ( l )
4
(iii)
Te(s )  2Cl 2 (g ) 
 TeCl 4 (g )
(iv)
Te(s )  I 2 (s ) 
 TeI 4 (s )
(v)
3SCl 2  4NaF      S 2Cl 2  SF4  4NaCl
MeCN( acetonitri le )
350 K
SF4 undergoes hydrolysis while SF6 not because SF6 is inert in nature. Its inertness is due to the presence of
sterically protected sulphur atom which does not allow thermodynamically favourable reactions like
hydrolysis while SF4 and SeF4 are less sterically hindered and undergo hydrolysis readily.
Because of its inertrness and good dielectric properties SF6 is used as a gaseous insulator in high voltage
generator. Both SF4 and SF6 are used as a fluorinating agents for conversion of –COOH into CF3 and
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C = O, P = O groups into CF2, PF2 groups.
SeCl4, SeBr4, TeCl4, TeBr4, TeI4 exist as tetramers. While TeF4  polymeric structure.
+6 O.S. stability decreases down the group and +4 O.S. increases because of inert pair effect
4.
Oxides : EO2, EO3 (SO2, SO3, SeO2, SeO3)
SO2 is a gas and SeO2 is solid :
(i)
SeO2 is solid because it has a chain polymeric structure whereas SO2 forms discrete units.
(ii)
Reducing power of dioxides decreases down the group.
(iii)
SO3 gas while SeO3, TeO3 solids.
(iv)
Dioxides and trioxides both are acidic in nature
5.
Oxides are of four types :
(i)
Acidic : Oxides that combine with water to give acid. They are non-metallic oxides e.g. SO2, NO2 etc.
(ii)
SO2 + H2O  H2SO3
SO3 + H2O  H2SO4
P4O10 + H2O  H3PO4
N2O3 + H2O  HNO2
P4O6 + H2O  H3PO4
N2O5 + H2O  HNO3
Basic oxides : Metallic oxides and form bases when dissolved in water Na2O, K2O, CaO
CaO + H2O  Ca(OH)2, K2O + H2O
(iii)
KOH
Amphoteric Oxides : Reacts with acids as well as base
Al2O3 + 6HCl + 9H2O  2[Al(H2O)6]3+ + 6Cl–
Al2O3 + 6NaOH + 3H2O  2Na3 [Al(OH)6]
(iv)
Neutral oxides : CO, NO, N2O.
6.
Oxygen (O2)
Preparation :

(i)
2KClO 3 MnO
 2KCl  3O 2
2
(ii)
By thermal decomposition of oxides :
2Ag2O(s) 
 4Ag(s) + O2(g)
2HgO(s) 
 2Hg(l) + O2(g)
2Pb3O4(s) 
 6PbO(s) + O2(g)
2PbO2(s) 
 2PbO(s) + O2(g)
(iii)
divided
2H2O2(aq) singly
  
 2H2O + O2(g)
metal
Properties :
(i)
Paramagnetic nature
(ii)
2Ca + O2  2CaO
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7.
4Al + 3O2  2Al2O3
CH4 + O2  CO2 + 2H2O
P4 + SO2  P4O10
2O 5
2SO2 + O2 V
 2SO3
C + O2  CO2
4HCl + O2 CuCl
2  2Cl2 + 2H2O
Ozone (O3) :
Very reactive
Preparation : When a slow dry steam of oxygen is passed through a silent electrical discharge, oxygen gets
converted to ozone. This is called ozonised oxygen (10%).
3O2  2O3, H = +x (endothermic process)
Properties :
(i)
Pale blue gas, dark blue liquid, violet black solid.
(ii)
Its decomposition to oxygen liberate heat and increase entropy  G has a large negative value  ozone
is thermodynamically unstable. Also high concentrations of ozone can be dangerously explosive.
(iii)
It is a powerful oxidising agent because it liberates atoms of nascent oxygen O3  O2 + O
PbS(s) + 4O3(g)  PbSO4(s) + 4O2(g)
4I–(aq) + H2O(l) + O3(g)  2OH–(aq) + I2 + O2
Estimation of Ozone is done by reacting ozone with KI solution buffered with a barate buffer (pH 9.2) I2 is
liberated which can be liberated against a standard solution of sodium thiosulphate
(iv)
Depletion of ozone layer : Is caused by
(a)
use of freons i.e.,CFC’s used in aerosol sprays and refrigerants.
(b)
Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes
NO + O3  NO2 + O2
8.
SO2 (Sulphur dioxide) :
Preparation :
(i)
S + O2  SO2
(ii)
SO32– + 2H+  H2O + SO2 (lab preparation)
i.e.,
(iii)
Na2SO3 + H2SO4(dil.)  Na2SO4 + SO2 + H2O
4FeS2 + 11O2  2Fe2O3 + 8SO2 (industrial preparation)
Properties :
(i)
SO2 + H2O
H2SO3 (aq)
(ii)
2NaOH + SO2  Na2SO3 + H2O
Na2SO3 + H2O + SO2  2NaHSO3
(iii)
( catalyst )
SO2 + Cl2 charcoal

 SO2Cl2
2O 5
2SO2 + O2 V
 2SO3
(iv)
Reducing agent :
(a)
2Fe3+ + SO2 + 2H2O  2Fe2+ + SO42– + 4H+
(b)
5SO2 + 2MnO4– + 2H2O  5SO42– + 4H+ + 2Mn2+
(KMnO4 colour decolourize)
(v)
SO2 can be detected by passing through the line water
SO 2  Ca(OH ) 2 
 CaSO 3  H 2 O
( milky )
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9.
Oxoacids of sulphur :
(a)
H2SO4
most important acid
(b)
H2SO3
(c)
H2SO3 and H2S2O3 are unstable and cannot be insoluble. They are only known in aqueous solutions or in
the form of their salts.
(d)
(e)
(f)
10.
H2SO4 manufacture by contact process
There are three stages :
(i)
S + O2  SO2
FeS + O2  SO2 + Fe2O3
(ii)
2O 5
2SO2 + O2 V
 2SO3;
(iii)
SO3 + H2SO4  H2S2O7
H = –196.6 kJ/mol
(SO3 observed by H2SO4 to form oleum)
H2S2O7 + H2O  2H2SO4 (96-98%)
Properties of H2SO4 :
The concentration H2SO4 acid must be added slowly into water with constant stirring during the
manufacture of dil. H2SO4 form conc. as there is evolution of large quantity of heat (exo.)
(i)
Low Volatility : because of low volatility can be used to manufacture more volatile acids from their
corresponding salts
2MX + H2SO4  2HX + M2SO4 (X = F–, Cl–, NO3–)
as 2NaNO3 + H2SO4  Na2SO4 + 2HNO3
(ii)
Strong acidic character :
H2SO4  H+(aq) + HSO4–(aq);
–
4
+
2–
4
HSO (aq)  H (aq) + SO (aq);
(iii)
K1 = very large
K2 = 1.2 × 10–2
Strong affinity for water (dehydrating agent) :
SO 4
C12H22O11 H
2 
 12C + 11H2O
(charring)
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(iv)
Strong oxidizing agent : reduce itself to SO2. It oxidizes both metals and non-metals.
C + H2SO4  CO2 + 2SO2 + 2H2O
Cu + 2H2SO4  CuSO4 + SO2 + 2H2O
C3A Group 17 (Halogen family) (Salt producing)
General properties
ns2p 5
F – [He]2s22p5
2
F2(g) Pale yellow
Cl – [Ne]3s 3p
5
Cl2(g) greenish yellow
Br – [Ar]3d104s24p5
10
2
I – [Kr]4f 5s 5p
Br2(l) reddish brown
5
I2(s) violet vapours
At  Radioactive element
Physical State : Diatomic F2, Cl2 are gases, Br2 is liquid and I2 is solid because of increase of atomic size
and increase of vander waal’s interaction.
Colour : The colour of the halogen is due to absorption of visible light molecules resulting in the excitation
of outer electron to higher energy levels. F has smaller size so high excitation energy light (violet light) is
required to excite the e– and shows reflected complementary pale yellow light while I due to large size low
excitation energy yellow light absorbs and  appears violet black. In between the colour of Cl is
yellowish-green and Br is reddish brown i.e., in the group the size increases low excitation energy is
required and hence dark colour. Thus the colour deepens down the group.
Bond dissociation energy : Cl2 > Br2 > F2 > I2
F2 bond dissociation energy is smaller than Cl2 because of large electronic repulsions of lone pair - lone pair
e–s and i.e. due to small size of F  e–s are much closer to each other. B.D.E. decrease from Br2 to I2 due to
large bond length.
Atomic Size : Halogens have the smallest atomic radii in their respective periods because of maximum
effective nuclear charge.
Electron gain enthalpy : Halogens have maximum EGE because they have one electrons less than stable
noble gas configuration i.e. they have strong tendency accept the electron.
EGE  down the group because of increase of atomic size
F : Heg is lower than Cl because of small size e– – e– repulsion occur in 2p-compact subshell and the added
e– does not feel much attracting resulting in low value of Heg.
EN : Halogens are highly EN and EN decreases down the group because of decrease of effective nuclear
charge.
Melting and Boiling Point : Increases downing the group because of  of atomic size and hence  of
vander waal’s interaction.
Oxidation State : F
–1
Cl
–1, +1, +3, +5, +7
Br
–1, +1, +3, +5, +7
I
–1, +1, +3, +5, +7
(i)
F only shows –1 o.s. not positive o.s. because F is most EN.
(ii)
F not shows higher o.s. because of absence of d-orbitals.
(iii)
–1 o.s. is common for all elements because of high EN.
(iv)
Cl to Br all show +1 to +7 o.s. because of vacant d-orbitals.
C3B
Nature of bonds with other elements : Halogens combine with metal to form predominantly ionic
halides. This halides ionic character decreased down the group while covalent character increases because
of decrease of EN and increase of halogen size and hence polarizability increases and covalent character
increases. e.g.
(i)
Increasing order of ionic character of metal halide : M – I < M – Br < M – Cl < M – F (M = Na, K etc.)
(ii)
AlF3 is ionic while AlCl3, AlBr3 are covalent.
(iii)
Higher o.s. of metal and non-metals halides are covalent but polar and this polarity  down the group.
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Oxidising Power : Halogens act as strong oxidising agent because of high electron affinity that they have
strong tendency to accept the e–s. The O.A. tendency decreases down the group from F to I. i.e.,
F2 > Cl2 > Br2 > I2.
F2 is a better oxidising agent then Cl2 although its electron affinity is less than that of Cl2 because of high
reduction electrode potential value of F2 and Cl2.
Reactivity : All the halogens are very reactive but F2 is most reactive because of high Electronegativity
decreases moving down in the group, the reactivity decreases because decrease of EN. F2 > Cl2 > Br2 > I2
(reactivity order)
F2 + 2X–  X2 + 2F–
–
(X = Cl, Br, I)
–
Cl2 + 2X  X2 + 2Cl
–
(X = Br, I)
–
Br2 + 2I  2Br + I2
H – F, H– Cl, H – Br, H – I
Hydrides :
Thermal Stability : H – F > H – Cl > H – Br > H – I
Reducing Character : H – F < H – Cl < H – Br < H – I
Acid Strength : HF < HCl < HBr < HI
because of increase of X size and increases M – H bond length and decrease of B.D.E. and hence easily
gives H+ ion.
B.P. : HF > HI > HBr > HCl [HF : intermolecular H-bond, HI > HBr > HCl : Vander waal’s forces]
% ionic character : HF > HCl > HBr > HI
Dipole moment : HF > HCl > HBr > HI [HF : electronegative]
Oxides : Halogens form many oxides with oxygen but most of them are unstable.
(i)
F2 forms two oxides OF2 and O2F2 in which OF2 is thermally stable at room temperature. These are called
oxygen fluorides because of the higher electronegativity of F then oxygen. Both are strong fluorinating
agents.
(ii)
Cl, Br, I form oxides of o.s. from +1 to +7. The kinetics and thermodynamic stability oxides formed by
halogen order is I > Br > Cl. The higher oxides are more stable than lower.
(iii)
Cl oxides : Cl2O, ClO2, Cl2O6, Cl2O7
(iv)
Br oxides : Br2O, BrO2, BrO3 least stable halogen oxides and exist only at low temperature and powerful
oxidising agent.
(v)
I oxides : I2O4, I2O5, I2O7 are insoluble solid and decompose on heating. I2O5 is a very good oxidising agents
and used for estimation of CO.
Reactivity with Metals : Halogens react with metals to form metal halides Mg + Br2  MgBr2
Ionic Character : MF > MCl > MBr > MI
Covalent Character : Halides in higher oxidation state will be more covalent than lower o.s.
e.g., SnCl4, PbCl4, SbCl5, UF6 more covalent than SnCl2, PbCl2, SbCl3, UF4.
Halides : Interhalogens compounds : Binary compounds of two different halogen of general formula XXn
are called interhalogen compounds (where n = 1, 3, 5, 7)
– The larger halogen atom is the central atom. All are covalent
– Interhalogens compounds are more reactive than halogens because X  X is more polar bond and weaker
than X – X bonds.
Shape :
XX(sp 3 )
XX3 (sp 3d )
XX5 (sp 3d 2 )
XX7 (sp 3d 3 )
ClF, BrF,
ClF3, BrF3
BrF5, ClF5
IF7
IF, BrCl, ICl
IF3, ICl3
IF5
Linear
T-shape
Square pyramidal
Pentagonal bipyramidal
X = large size and more electropositive halogen
X  = smaller size and more electronegative halogen.
Einstein Classes,
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–
I only shows higher halogens fluorides i.e. IF7. All are covalent, diamagnetic in nature. They are volatile
solids or liquids except ClF which is gas. Melting point higher than halogens.
–
All these undergoes hydrolysis giving halide ion derived from the smaller halogen and a hypohalite ( XX ),
halite ( XX3 ), halate ( XX5 ) and perhalate ( XX7 ).
XX  H 2 O 
 HX  HOX
Hypohalite
XX3  2H 2O 
 3HX  HOXO (or HXO 2 ) Halite
XX5  3H 2 O 
 5HX  HOXO 2 (or HXO 3 )
Halate
XX7  4H 2 O 
 7HX  HOXO 3 ( HXO 4 )
perhalate
Shapes : on the basis of VSEPR theory
Preparation of bond interhalogens compounds can be prepared by the direct combination of halogen
473 K
Cl 2  F2  ClF
I 2  Cl 2 
 ICl
573K
Cl 2  3F2  2ClF3
( excess)
I 2  3Cl 2 
 ICl 3
( excess)
Br2  3F2 
 2BrF3
Br2  5F2 
 2BrF5
( excess)
Oxoacids :
Due to high EN and small size, F forms only one oxoacids i.e. HOF.
Halic (I) acid
HOF
HOCl
(Hypohalous acid)
HOBr
HOI
–
–
HOBrO2
HOIO2
HOBrO3
HOIO3
(Hypochlorus acid)
Halic (III) acid
–
HOClO
(Halous acid)
(Chlorous acid)
Halic (V) acid
–
HOClO2
(Halic acid)
(Chloric acid)
Holic (VII) acid
–
HOClO3
(Perhalic acid)
(Perchloric acid)
Acidic Strength Order :
(a)
due to EN
1
(b)
3
5
7
HOCl  HOClO  HOClO 2  HOClO 3 (due to increase stabilization of conjugate base)
Chlorine (Cl2) :
Einstein Classes,
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Preparation :
(1)
(2)
Lab preparation :
(a)
MnO 2  4HCl 
 MnCl 2  Cl 2  2H 2 O
(b)
4NaCl  MnO 2  4H 2 SO 4 
 MnCl 2  4NaHSO 4  2H 2 O  Cl 2
(c)
2KMnO 4  16HCl 
 2KCl  2MnCl 2  8H 2O  5Cl 2
Deacon’s Process :
CuCl
4HCl  O 2 2  2Cl 2  2H 2 O
(3)
Electrolytic Process :
electrolysis
NaCl   
 Cl 2 
( Brine )
( At Anode )
Properties :
It is a greenish yellow gas with pungent and suffocating odour.
(a)
(b)
Reaction with metals and non-metals : to form chlorides
2Al + 3Cl2  2AlCl3
P4 + 6Cl2  PCl3
2Fe + 3Cl2  2FeCl3
S8 + 4Cl2  4S2Cl2
Reacts with hydrogen :
H2 + Cl2  2HCl
C10H16 + 8Cl2  16HCl + 10C
H2S + Cl2  2HCl + S
(c)
Reaction with Ammonia :
8NH 3  3Cl 2 
 6NH 4 Cl  N 2
( excess)
NH 3  3Cl 2 

( excess )
(d)
NCl 3  3HCl
Nitrogen trichlorid e (exp losive )
Reaction with alkali :
(i)
2NaOH  Cl 2 
 NaCl 
( cold & dil .)
(ii)
NaOCl
sodium hypochlori de
6NaOH  3Cl 2 
 5NaCl  NaClO 3  3H 2 O
( hot & conc .)
(iii)
 H 2O
2Ca(OH) 2 
(dry slaked lime)
( sodium clorate )
2Cl 2 
 Ca(OCl) 2  CaCl 2  2H 2 O
(Bleaching powder)
The composition of bleaching powder is Ca(OCl)2 · CaCl2 · Ca(OH)2 · 2H2O
(iv)
with hydrocarbons :
UV
CH 4  Cl 2  CH 3 Cl  HCl
Room
C 2 H 4  Cl 2 
temp .
(v)
C 2 H 4 Cl 2
1, 2  Dichloroet hane
with water :
H 2O  Cl 2 
 HCl  HOCl 
 HCl 
[O]
nascant oxygen
due to nascant oxygen it is responsible for oxidising and bleaching properties of chlorine.
Cl2 is a powerful bleaching agent; bleaching action is due to oxidation.
Einstein Classes,
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Coloured substance + O  Colourless substance
Bleaching effect of Cl2 is permanent.
Oxidising Agent :
Na2SO3 + Cl2 + H2O  Na2SO4 + 2HCl
SO2 + 2H2O + Cl2  H2SO4 + 2HCl
I2 + 6H2O + 5Cl2  2HIO3 + 10HCl
2FeSO4 + H2SO4 + Cl2  Fe2(SO4)3 + 2HCl
HCl (Hydrogen Chloride)
420 K
Lab Preparation :
NaCl  H 2 SO 4  NaHSO 4  HCl
823 K
NaHSO 4  NaCl  Na 2 SO 4  HCl
HCl gas can be dried by passing through concentrated H2SO4.
Properties :
(i)
It is colourless, pungent smelling gas and easily liquified to colourless liquid and extremely soluble in
water and ionizes as HCl + H2O  H3O+ + Cl– [Ka = 107]
(ii)
It’s aqueous solution is called hydrochloric acid. High value of dissociation constant (Ka) indicates that it is
a strong acid.
(iii)
NH3 + HCl  NH4Cl (white fumes)
(iv)
Aqua regia  3 parts of HCl + 1 part of HNO3 and used for dissolving noble metals like Au, Pt.
Au + 4H+ + NO3– + 4Cl–  AuCl4– + NO + 2H2O
3Pt + 16H+ + 4NO3– + 18Cl–  3PtCl62– + 4NO + 8H2O
(v)
Na2CO3 + 2HCl  2NaCl + H2O + CO2
NaHCO3 + HCl  NaCl + H2O + CO2
Na2SO3 + 2HCl  2NaCl + H2O + SO2
(vi)
Fe + 2HCl  FeCl2 + H2
C4
Group - 18 (Noble Gases)
He2 , Ne
, Ar
,
2
6
2
6
1s
2s 2p
3 s 3p
Kr
,
Xe
,
Rn
3d10 4s 2 4p 6 4d10 5s 2 6s 2 6p 6 4 f 14 5d10 6 s 2 6p 6
( Radioactive)
He, Ne, Ar  no compound known
Kr  only few compounds known (KrF2)
Xe  many compounds
Rn  identical but not isolated
(i)
The elements of group 18 known as noble gases because their valence shell orbitals completely filled and
 react with a few elements only under certain conditions. Therefore they are known as noble gases.
General Electronic Configuration : ns2np6 except He = 1s2
(ii)
I.E. : (a) Due to stable (fully filled) configuration the I.E. very high. (b) down the group I.E. decreases
because of increase of atomic size.
(iii)
Atomic radii : down the group increases because  in atomic number of shells.
(iv)
E.G.E. : They have large positive EGE values because of stable electronic configuration, they have no
tendency to accept the electron.
(v)
m.p. and b.p. : low because of weak dispersion forces.
B.P. are low of noble gases because noble gases being monoatomic have no inter atomic forces except weak
dispersion forces and hence can be liquified at very low temp. and have low B.P.
Ist compound of noble gas is Xe+ [PtF6]– be neil Bartlett.
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Xenon-fluorine compounds :
Xe  F2 
 XeF2
( excess )
Xe  2F2 
 XeF4
(1:5 )
Xe
(1: 20 ratio )
 3F2 
 XeF6
Properties :
(i)
Xe-Fluorine are powerful fluorinating agent.
(ii)
Hydrolysis :
2XeF2 + 2H2O  2Xe + 4HF + O2
6XeF4 + 12H2O  4Xe + 2XeO3 + 24HF + 3O2
XeF6 + H2O  XeOF4 + 2HF
XeF6 + 2H2O  XeO2F2 + 4HF
XeF6 + 3H2O  XeO3 + 6HF
Complete Hydrolysis :
XeF2 + PF5  [XeF]+ [PF6]–
XeF4 + SbF5  [XeF3]+ [SbF6]–
XeF6 + MF  M+ [XeF7]–
Structure of Fluorides :
Einstein Classes,
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E X E R C I S E (OBJECTIVE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
From which of the following we cannot extract
boron ?
(a)
orthoboric acid (b)
borax
(c)
cryolite
(d)
kernite
Ist ionization enthalpy of boron family follows ?
(a)
B > Al > Ga > In
(b)
B < Al < Ga < In
(c)
Al > B > Ga > In
(d)
B > Ga > Al > In
The relative stability of +1 oxidation state for
heavier elements :
(a)
Al < Ga < In < Tl
(b)
Al > Ga > In > Tl
(c)
Ga < Al < In < Tl
(d)
Al < Ga < Tl < In
AlCl3 achieves stability by forming a dimer. In
dimer it exists in
(a)
planar
(b)
tetrahedral
(c)
octahedral
(d)
linear
Which of the following element does not react
(a)
B
(b)
Al
(c)
Ga
(d)
In
Which of the following is incorrect ?
(a)
Boron trifluoride easily reacts with Lewis
bases.
(b)
Due to the absence of d orbital that the
maximum covalence of B is 3.
(c)
AlCl3 dimerised through halogen
bridging.
(d)
Trichlorides of Boron covalent in nature
The formula of Borax is
(a)
Na2B2O7.10H2O
(b)
Na2B4O7.10H2O
(c)
Na2B2O4.10H2O
(d)
Na2B4O4.10H2O
If orthoboric acid get strongly heated then it convert into
(a)
H2BO3
(b)
BO
(c)
BO2
(d)
B2 O 3
Diborane cannot be prepare by the reaction
(a)
BF3 + LiAlH4 
(b)
NaBH4 + I2 
(c)
Na2B4O7 + HCl + H2O 
(d)
BF3 + NaH 
In diborane each B atom hybridises of
(a)
sp
(b)
sp2
(c)
sp3
(d)
dsp2
Einstein Classes,
11.
12.
13.
14.
15.
16.
In the following oxides, which is amphoteric in
nature ?
(a)
CO2
(b)
SiO2
(c)
GeO2
(d)
SnO2
Which of the following element reacted with steam
water ?
(a)
C
(b)
Si
(c)
Ge
(d)
Sn
Which of the following tetrachlorides are not
easily hydrolysed ?
(a)
CCl4
(b)
SiCl4
(c)
GeCl4
(d)
SnCl4
The difference between water gas and producer gas
is
(a)
Water gas is naturally prepared whereas
producer gas commercially prepared.
(b)
C + H2O  CO (water gas)
C + O2 + N2  CO (producer gas)
(c)
C + H2O  CO (producer gas)
C + O2 + N2  CO (water gas)
(d)
no difference
SiO2 + HF  products, one of the products is
(a)
SiF
(b)
SiF4
(c)
H2
(d)
O2
Silocones is
(a)
17.
18.
19.
Si
(b)
(c)
both
(d)
None
Which of the following is incorrect regarding the
nitrogen family
(a)
The tendency to show –3 oxidation state
decreases down the group
(b)
The stability of +5 oxidation state
decreases down the group
(c)
+3 oxidation state in arsenic, stable as
compare to disporportionation
(d)
Nitrogen have maximum covalency of 3.
Arsenic can form bond with transition metal
(a)
p – p
(b)
p – d
(c)
d – p
(d)
d – d
NH3 have higher b.p. because of
(a)
Size
(b)
Non-availibility of d-orbital
(c)
Hydrogen bonding
(d)
All of these
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CPB – 44
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
In which of the following reaction we can prepare
dinitrogen ?
(a)
NH 4Cl (aq)  NaNO2 (aq) 

(b)
( NH 4 ) 2 Cr2 O 7 
(c)
Ba( N 3 ) 2 
31.

32.

(d)
By all
In Haber process the catalyst used is
(a)
V2O5
(b)
Al2O3
(c)
SO2
(d)
All of these
When AgCl(s) reacts with ammonia, it form colour
that is
(a)
deep blue
(b)
white
(c)
red
(d)
colourless
HNO3 exist with the structure of
(a)
tetrahedral
(b)
planar
(c)
linear
(d)
octahedral
When copper reacted with concentrated nitric acid
it give
(a)
NO
(b)
NO2
(c)
N2O3
(d)
N2O5
Which of the following reaction is incorrect ?
(a)
I2 + HNO3  HIO3
(b)
C + HNO3  CO
(c)
S8 + HNO3  H2SO4
(d)
P4 + HNO3  H3PO4
Which of the following allotrops of phosphorus is
more reactive as compare to others ?
(a)
white
(b)
red
(c)
black
(d)
all equal
Which of the following reaction is incorrect ?
(a)
Ca3P2 + H2O  PH3
(b)
Ca3P2 + NaOH  PH3
(c)
P4 + NaOH + H2O  PH3
(d)
PH4 + KOH  PH3
HgCl2 + PH3  products, one of the product is
(a)
H2
(b)
Hg2P3
(c)
HCl
(d)
None
Number of P–OH bond in H4P2O7 is
(a)
2
(b)
3
(c)
4
(d)
5
Which of the following statements is incorrect ?
(a)
The stability of –2 oxidation state
decrease down the group in group 16.
(b)
Electronegativity of oxygen is high.
(c)
The stability of +4 oxidation state
increases down the group in group 16
(d)
Oxygen shows only –2 oxidation state
Einstein Classes,
33.
34.
35.
36.
37.
38.
39.
Correct order regarding b.p. is
(a)
H2O > H2S > H2Se > H2Te
(b)
H2O > H2Se > H2Te > H2S
(c)
H2O > H2Te > H2Se > H2S
(d)
H2O > H2Se > H2S > H2Te
Which of the compound after heat does not give
the oxygen ?
(a)
KClO3
(b)
H2O 2
(c)
HgO
(d)
PbO
Which of the following is basic oxide ?
(a)
Mn2O7
(b)
Cl2O7
(c)
CO2
(d)
BaO
S + H2SO4(conc.)  products, one of the product is
(a)
O2
(b)
H2
(c)
SO2
(d)
H2O 2
Out of which we cannot extract Fluorine ?
(a)
fluorspar
(b)
cryolite
(c)
fluoroapatite
(d)
carnallite
Which have the higher bond dissociation energy ?
(a)
F2
(b)
Cl2
(c)
Br2
(d)
I2
Which of the chlorine oxides is used as a bleaching
agent ?
(a)
Cl2O
(b)
Cl2O6
(c)
Cl2O7
(d)
ClO2
By which reaction we can prepare chlorine ?
(a)
MnO2 + HCl 
(b)
NaCl + MnO2 + H2SO4 
(c)
KMnO4 + HCl 
(d)
by all
NH 3  Cl 2 
 products , one of the product is
( excess)
(a)
(c)
40.
41.
42.
N2
NCl3
(b)
(d)
NH4Cl
H2
Out of halogens, which form only one oxoacid ?
(a)
F
(b)
Cl
(c)
Br
(d)
I
Out of inter halogens, which does not exist ?
(a)
IF7
(b)
ICl3
(c)
ClF3
(d)
ICl5
Which of the following statements is incorrect ?
(a)
Interhalogen compound are covalent in
nature
(b)
Interhalogen are paramagnetic in nature
(c)
IF5 have square pyramidal structure
(d)
ICl3 have T shape structure
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 45
43.
44.
45.
46.
47.
48.
49.
Out of following which reaction is not correct ?
(a)
Xe + F2  XeF2
(b)
Xe + 2F2  XeF4
(c)
Xe + 3F2  XeF6
(d)
XeF4 + O2F2  XeF4
Which of the following statement is incorrect
(a)
Bi5+ salts do not exist
(b)
Pb4+ salts are good oxidising agents
(c)
Sn4+ salts are good oxidising agents
(d)
Tl3+ salts are good oxidising agents
Of the following acids :
I : hypo phosphorous acid,
II : hydrofluoric acid,
III : oxalic acid,
IV : glycine
(a)
I, II are monobasic, III dibasic acid and
IV amphoteric
(b)
II monobasic, I, III dibasic acid, IV am
photeric
(c)
I monobasic, II, III dibasic, IV
amphoteric
(d)
I, II, III dibasic, IV amphoteric
Be and Al show diagonal relationship and thus
(a)
their oxides are soluble in alkali solution
forming [Be(OH)4]2— and [Al(OH)6]3—
(b)
they form complex anion BeF2— and
AlF3—
(c)
BeCl2 is Lewis acid but AlCl3 is a Lewis
base
(d)
BeO is basic but Al2O3 is amphoteric
In the following reaction :
B(OH)3 + H2O  [B(OH)4]— + H+
(a)
B(OH)3 is a Lewis acid
(b)
B(OH)3 is a Lewis base
(c)
B(OH)3 is amphoteric
(d)
none is correct
Select correct statements :
(a)
hydrides of B and Si are volatile and catch
fire on exposure to air
(b)
oxides of B and Si (B2O3 and SiO2) are
acidic in nature
(c)
borates and silicates have tetrahedral
BO4 and SiO4 structural units
(d)
All of above
Select correct statements
(a)
catenation is maximum in carbon
(b)
carbon has pronounced ability to form
p – p multiple bonds to itself and to
other elements like O and N
(c)
both correct
(d)
none is correct
Einstein Classes,
50.
51.
52.
53.
54.
55.
56.
57.
Consider following statements
I:
In diamond, each carbon atom is linked
tetrahedrally to four other carbon atoms
by sp3 bonds.
II :
Graphite has planar hexagonal layers of
carbon atoms held together by weak
vander Walls forces
III :
Silicon exists only in diamond structure
due to its tendency to form p — p
bonds to itself.
In this :
(a)
only I and II are correct
(b)
only I is correct
(c)
only II and III are correct
(d)
all are correct statements
In the following statements, select the correct
statement(s)
(a)
N(CH3)3 has pyramidal structure
(b)
N(SiH3)3 shows planar arrangement
(c)
both correct
(d)
none is correct
Out of CO2, SiO2, GeO2, SnO2 and PbO2
(a)
CO2 and SiO2 are acidic. SnO2 is
amphoteric and PbO2 is an oxidising
agent
(b)
PbO2 is converted to Pb(NO3)2 on
reaction with HNO3
(c)
both correct
(d)
none is correct
Which is true statement about silicones ?
(a)
they are synthetic polymers containing
repeated R2SiO2 units
(b)
they are formed by hydrolysis of R2SiCl2
units
(c)
both (a) and (b) are correct
(d)
none is correct
Metalloids are
(a)
N, P
(b)
As, Sb
(c)
Bi, As
(d)
Bi, Sb
Which forms p – p multiple bonds with itself and
with C and O
(a)
P, As
(b)
N, As
(c)
N, P
(d)
N
Maximum number of covalent bonds formed by N
and P respectively are
(a)
3, 5
(b)
3, 6
(c)
4, 5
(d)
4, 6
Which of the following compounds do not exist
N4, P4, PCl5, NCl5, NCl3, P2O5, NO2, PO
(a)
N4, NCl5, NO2 (b)
N4, NCl5, PO2
(c)
PCl5, NCl5, PO2 (d)
all
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 46
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
Which of the following is/are paramagnetic
NO2, NO, N2O4, N2O2, N2O5
(a)
only NO2
(b)
NO2, NO
(c)
NO, NO2, N2O5
(d)
all are paramagenetic
Which of the following is dibasic acid showing
geometrical isomerism ?
(a)
hyponitrous acid H2N2O2
(b)
maleic acid C4H4O2
(c)
fumeric acid C4H4O2
(d)
all of above
Which of the following has peroxy linkage ?
(a)
H2SO5
(b)
H2 S 2 O 8
(c)
both
(d)
none
Which of the following can be used as dehydrating
agents ?
(a)
conc. H2SO4
(b)
POCl3
(c)
P2 O 5
(d)
all
1 mol each of H 2PO 2, H 3PO 3 and H 3PO 4 will
neutralise respectively x mol of NaOH, y mol of
Ca(OH)2 and z mol of Al(OH)3 (assuming all as
strong electrolytes). x, y, z are in the ratio of
(a)
3 : 1.5 : 1
(b)
1:2:3
(c)
3:2:1
(d)
1:1:1
Which is a set of acid salts and can react with
base ?
(a)
NaH2PO2, Na2HPO3, NaH2PO4
(b)
Na2HPO3, NaH2PO3, Na2HPO4
(c)
NaHCO3, NaH2PO3, Na2HPO4
(d)
all
Which is used as a rocket fuel
(a)
N2H4
(b)
polybudiene
(c)
both
(d)
none
Semiconductor pairs are
(a)
Se and Te
(b)
Ge and Si
(c)
both
(d)
none
In case of oxygen family (Group 16 family)
(a)
the tendency for catenation decreases
markedly as we go down the group
(b)
maximum coordination of oxygen is four
due to lack of d-orbital but tha of other
elements is six due to presence of
d-orbital
(c)
the tendency to form multiple bonds with
C, N and O decrease as going down the
group from S to Te.
(d)
all correct
Which has (S—S) bonds ?
(a)
H2 S 2 O 3
(b)
H2 S 2 O 6
(c)
both
(d)
none
Einstein Classes,
68.
69.
70.
71.
72.
73.
74.
75.
76.
Which has max. pKa value ?
(a)
H2 O
(b)
H 2S
(c)
H2Se
(d)
H2Te
The active constituents of bleaching powder is
(a)
Ca(OCl)2
(b)
Ca(OCl)Cl
(c)
Ca(ClO2)2
(d)
Ca(ClO2)Cl
Which one of the following is the weakest base as
per the Bronsted concept ?
(a)
[ClO4]—
(b)
[ClO3]—
—
(c)
[ClO2]
(d)
[ClO]—
The first ionization potentials of Na, Mg, Al and Si
and such that
(a)
Na < Mg < Al > Si
(b)
Na < Al < Mg < Si
(c)
Na > Mg > Al > Si
(d)
Na < Mg < Al > Si
Which one of the following is not a true
“per acid” ?
(a)
HClO4
(b)
H2SO5
(c)
H2 S 2 O 8
(d)
H3PO5
Which is/are true statements ?
(a)
sulphur trioxide exists as cyclic trimer in
solid state, S3O9
(b)
selenium trioxide solid is a cyclic tet
ramer, Se4O12
(c)
TeO3 is a solid with a network structure
in which TeO6 octohedral share all
vertices
(d)
all correct
Which is true statement ?
(a)
hydrofluoric acid is a dibasic acid and
attacks glass
(b)
oxidising power is in order
F2 > Cl2 > Br2 > I2
(c)
both
(d)
none
HClO4, HNO3 and HCl are all very strong acids in
aqueous solution. In glacial acetic acid medium,
their acid strength is such that
(a)
HClO4 > HNO3 > HCl
(b)
HNO3 > HClO4 > HCl
(c)
HCl > HClO4 > HNO3
(d)
HCl > HClO4 ~ HNO3
Consider the following reactions :
1.
2Na + 2H2O  2NaOH + H2
2.
2NaOH + Cl2  NaCl + NaOCl + H2O
3.
4OH—  O2 + 2H2O + 4e—
4.
2Cl—  Cl2 + 2e—
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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77.
78.
79.
80.
81.
82.
83.
In the diaphragm cell used for the electrolysis of
brine, the reactions that occur would include
(a)
1, 2, and 4
(b)
2, 3 and 4
(c)
1, 3 and 4
(d)
1, 2 and 3
Consider the following species
1.
[O2]2—
2.
[CO]+
3.
[O2]+
Among these species sigma bond alone present in
(a)
1, 2 and 3
(b)
2 alone
(c)
2 and 3
(d)
1 alone
The dipole moments of the given molecule are such
that
(a)
BF3 > NF3 > NH3
(b)
NF3 > BF3 > NH3
(c)
NH3 > NF3 > BF3
(d)
NH3 > BF3 > NF3
The electronegativities of N, C, Si and P are such
that
(a)
P < Si < C < N (b)
Si < P < N < C
(c)
Si < P < C < N (d)
P < Si < N < C
The electron affinities of N, O, S and Cl are such
that
(a)
N < O < S < Cl (b)
O < N < Cl < S
(c)
O = Cl < N = Cl (d)
O < S < Cl < N
Which is true statement ?
(a)
basic nature of X— is in order
F— > Cl— > Br— > I—
(b)
HI is strongest acid of HF, HCl, HBr and
HI
(c)
The ionic character of M-X bond
decreases in the order
M — F > M — Cl > M — Br > M — I
(d)
all of above
Select correct statement :
(a)
ClO2 and Cl2O are used as bleaching
agents for paper pulp and textiles
(b)
ClO— (hypohalites) salts are used as
sterilising agents
(c)
ClO— disproportionates in alkaline
medium
(d)
all
Which one of the following species is capable of
functioning both as a Bronsted acid and Bronsted
base ?
(a)
F—
(b)
CO32—
(c)
HS—
(d)
S2—
Einstein Classes,
84.
85.
86.
87.
88.
89.
Consider the following boron halides
1.
BF3
2.
BCl3
3.
BBr3
4.
BI3
The Lewis acid characters of these halides are such
that
(a)
1<2<3<4
(b)
1<3<2<4
(c)
4<3<2<1
(d)
4<2<3<1
Which of the following pairs are correctly
mathced ?
1.
Haber process....................Manufacture
of ammonia
2.
Bayer process....................Manufacture of
sulphuric acid
3.
Birekeland-Eyde
process................Manufacture of nitric
acid
4.
Solvay process.....................Manufacture
of sodium carbonate
Select the correct answer using the codes given
below
(a)
2, 3 and 4
(b)
1, 3 and 4
(c)
1, 2 and 4
(d)
1, 2 and 3
Which one of the following bonds has the highest
bond energy ?
(a)
C—C
(b)
Si—Si
(c)
Ge—Ge
(d)
Sn—Sn
Which of the following species have undistorted
octahedral structures ?
1.
SF6
2.
PF6—
3.
SiF62—
4.
XeF6
Select the correct answer using the codes give
below :
(a)
2, 3 and 4
(b)
1, 3 and 4
(c)
1, 2 and 3
(d)
1, 2 and 4
Which of the following compounds posses Lewis
acid character ?
1.
BF3
2.
SiF4
3.
PF5
Select the correct answer using the codes given
below :
Codes :
(a)
1 alone
(b)
1, 2 and 3
(c)
2 and 3
(d)
1 and 3
Inter-halogen compounds can be
(a)
ICI3
(b)
B2F 5
(c)
FI7
(d)
FCl3
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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90.
91.
92.
93.
94.
Select correct statement :
(a)
helium has the lowest m.p. and b.p.
(b)
helium can diffuse through rubber, PVC
and even glass
(c)
Ar, Kr and Xe form clathrate
(d)
all of above
Consider following transformations :
I
XeF6 + NaF  Na+[XeF7]—
II
2PCl5(s)  [PCl4]+[PCl6]—
III
[Al(H2O)6]3+ + H2O  [Al(H2O)5OH]2+ +
H3 O +
Possible transformations are
(a)
I, II, III
(b)
I, III
(c)
I, II
(d)
II, III
Consider the following statements. In diborane
1.
boron is approximately sp3 hybridized
2.
B—H—B angle is 1800
3.
there are two terminal B—H bonds for
each boron atom
4.
there are only 6 bonding electrons
available
Of these statements :
(a)
1, 2 and 4 are correct
(b)
1, 2 and 3 are correct
(c)
2, 3 and 4 are correct
(d)
1, 3 and 4 are correct
Which of the following statements are true of
zeolites ?
1.
They are formed by the replacement of
some of the silicon atoms of the SiO2
lattice, by say, aluminium
2.
They have a more closed structure than
feldspar
3.
They can absorb CO2, NH3 and EtOH
4.
They can separate straight chain
hydrocarbons from a mixture containing
both straight chain and branched chain
hydrocarbons.
Of these statements :
(a)
1, 2 and 3 are correct
(b)
1, 3 and 4 are correct
(c)
1, 2 and 4 are correct
(d)
2, 3 and 4 are correct
Consider the following compounds :
1.
Sulphur dioxide
2.
Hydrogen peroxide
3.
Ozone
Among these compounds, those which can act as
bleaching agents would include :
(a)
1 and 3
(b)
2 and 3
(c)
1 and 2
(d)
1, 2 and 3
Einstein Classes,
95.
96.
97.
98.
99.
100.
101.
102.
103.
104.
Molecule(s) possessing three-centre two electron
bonds and three-centre-four electron bonds would
include
(a)
B2H6 and SiF4 (b)
XeF6 and BF6
(c)
B2H6 and XeF6 (d)
B2H6 alone
Oxidation of hydrogen halide, HX affords a method
for the industrial and laboratory preparation of the
halogen, X2, in the free state in respect of all of the
following except
(a)
fluorine
(b)
chlorine
(c)
bromine
(d)
iodine
Which of the following statements are correct for
all three halogens (X), Cl, Br and I ?
(a)
they all form oxyacids that are strong acid
in aqueous solution
(b)
they all react with hot, conc. NaOH (aq.)
to give XO4— ions
(c)
they all need to gain one electron acquire
stable configuration
(d)
all of above
H3BO3 and HBO2 not differ in
(a)
oxidation number
(b)
basicity
(c)
m.p.
(d)
structure
A solution of sodium metal in liquid NH3 is
(a)
strongly reducing
(b)
blue in colour
(c)
good conductor
(d)
all of above
Which one of the following properties of the
elements of Group 14th of the Periodic Table does
not show an increase with increasing relative atomic
mass ?
(a)
the first ionisation energy
(b)
the basic character of the oxides
(c)
the ionic character of the compounds
(d)
the stability of the +2 oxidation state
Lead as impurity in the extraction of silver is
removed by
(a)
Parke’s process (b)
Solvay process
(c)
cyanide process (d)
none
Which involves breaking of covalent bonds ?
(a)
boiling of H2S (b)
melting of SiO2
(c)
melting of KCN (d)
boiling of CCl4
Glacial phosphoric acid is
(a)
HPO3
(b)
H3PO3
(c)
H3PO4
(d)
H4 P2 O 7
Which is silane ?
(a)
SiO2
(b)
SiO3
(c)
SiH4
(d)
Si
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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105.
106.
107.
108.
109.
110.
111.
112.
113.
114.
115.
116.
117.
118.
119.
Which of the following is not a silicate ?
(a)
feldspar
(b)
water glass
(c)
calgon
(d)
zeolite
Cl2 can be used in the preparation of all but not in
(a)
bleaching power (b)
PVC
(c)
chlorophyll
(d)
lindane
Phosgene can be obtained when
(a)
white phosphorus reacts with alkali
(b)
calcium pohsphide reacts with water
(c)
chloroform reacts with air
(d)
bone comes in contact with water
Carborundum is
(a)
CaC2
(b)
Fe3C
(c)
CaCO3
(d)
SiC
Lithophone is a mixture of
(a)
ZnSO4 and CuS
(b)
PbS and BaSO4
(c)
ZnS and BaSO4
(d)
ZnSO4 and CuS
Gas that strikes in thundering of light is
(a)
CO
(b)
NO
(c)
CO2
(d)
H2
Which is soluble in H2O
(a)
AgF
(b)
CaF2
(c)
AgCl
(d)
AgBr
Which can be extracted into ether ?
(a)
LiCl
(b)
AlCl3
(c)
SiCl4
(d)
all
Inorganic benzene is
(a)
B3 N 3 H 6
(b)
B 3 O3 H 6
(c)
B2 H 6
(d)
(BH3)3
Oil of vitriol is
(a)
HNO3
(b)
HCl
(c)
H2SO4
(d)
H3PO4
Baryta water is
(a)
Ca(OH)2
(b)
Ba(OH)2
(c)
B(OH)3
(d)
AgOH
Following metals are soluble in aqua regia
(a)
Pt
(b)
Au
(c)
Ag
(d)
all
Dry ice is
(a)
H2O(s)
(b)
NH3(g)
(c)
CO2(s)
(d)
PH3(g)
Tincture of iodine is
(a)
CHI3 is alcohol (b)
I2 in alcohol
(c)
I2 in KI
(d)
CHI3 in KI
Artificial gem used for cutting glass is
(a)
graphite
(b)
diamond
(c)
SiC
(d)
CaCN2
Einstein Classes,
120.
121.
122.
123.
124.
Hypervalent ion is
(a)
SO32—
(b)
PO43—
(c)
SO42—
(d)
all
Match List I with List II and select the correct
answer using the codes given below the list
List I
List II
(Manufacturing Process)
(Catalyst used)
A.
Deacon’s process
1.
Finely divided
iron with
molybdenum
for chlorine
as promoter
B.
Hydrogenation of
2.
Copper (II)
chloride
vegetable oils
C.
Ostwald’s process
3.
Finely divided
nickel powder
for nitric acid
D.
Haber’s process for
ammonia
4.
Platinum
gauze
Codes :
A
B
C
D
(a)
4
1
3
2
(b)
2
3
1
4
(c)
1
4
2
3
(d)
2
3
4
1
Red lead used as primer for iron to prevent it from
rusting is
(a)
Pb3O4
(b)
PbO
(c)
PbO2
(d)
Pb2O
Ionic hydrides are usually
(a)
liquids at room temperature
(b)
good reducing agents
(c)
good electrical conductors when solid
(d)
easily reduced
For the molecule PF4CH3, which of the following
structures is the most stable, considering that
CH3— is more electropositive than F— ?
(a)
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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132.
Na3AlF6 is added to Al2O3 to
(a)
improve the electrical conductivity of the
cell
(b)
lower the m.p. of the mixture
(c)
both
(d)
none
133.
 NaBO2 + A + H2O
Na2B4O7.10H2O 
(b)
(c)

(d)
125.
126.
127.
128.
A considerable part of the harmful UV rays of the
sun does not reach the surface of the earth. This is
because high above the earth atmosphere, there is
a layer of
(a)
O3
(b)
CO2
(c)
SO2
(d)
NO
Polymeric fluorocarbons tend to be
(a)
good oxidizing agents
(b)
good reducing agents
(c)
corrosive to the skin
(d)
very unreactive
Most abundant metal is
(a)
oxygen
(b)
silicon
(c)
aluminium
(d)
iron
Borax is converted into B by steps.
134.
135.
I

II
Borax 

H 3 BO 3 

B2 O3 

B
129.
130.
131.
I and II reagents are :
(a)
acid, Al
(b)
acid, C
(c)
acid, Fe
(d)
acid, Mg
Boron carbide is used
(a)
in nuclear reactor to absorb neutrons
(b)
as an abrasive
(c)
both correct
(d)
none is correct
Borax is used
(a)
as a flux is brazing and in silver
soldering
(b)
in making enamel
(c)
in leather tanning
(d)
all correct
Soda free glass fibres are made of
(a)
H3BO3, B2O3 and Ca3(BO3)2
(b)
H3BO3, B2O3
(c)
H3BO3
(d)
B2 O 3
Einstein Classes,

136.
137.
138.
139.
 B
A + MnO 
A and B are
(a)
Na3BO3, Mn3(BO3)2
(b)
Na2(BO2)2, Mn(BO2)2
(c)
B2O3, Mn(BO2)2
(d)
none is correct
Match the column A (name of process) with
column B (one of the intermediate) for extraction
of Al :
A
B
I:
Baeyer
X:
AlN
II :
Hall
Y:
Al2O3
III :
Serpeck
Z:
Na3AlF6
I
II
III
(a)
X
Y
Z
(b)
Y
Z
X
(c)
Y
X
Z
(d)
Z
Y
X
Light-emitting diodes (LED), laser-diodes and
memory chips of computers are made of
(a)
gallium
(b)
aluminium
(c)
arsensic
(d)
galliumarsenide
Indium is used
(a)
to dope crystals to make p-n-p transitors
(b)
in thermistors
(c)
in low m.p. solder
(d)
all correct
When boron is fused with NaOH, products formed
are
(a)
Na2B4O7 and H2 (b)
NaBO2 and H2
(c)
Na3BO3 and H2 (d)
B2O3 and H2

H O
2
 Y 

 NH 3 X and Y
Metal X  air 
can be
(a)
Mg
(b)
Al
(c)
both
(d)
none
Thermite welding uses Al because of
(a)
its low m.p.
(b)
its lightness
(c)
its greater affinity for oxygen
(d)
all correct
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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140.
141.
142.
143.
144.
145.
146.
147.
Consider following reactions :
I:
ZnO + C  Zn + CO
II :
Cr2O3 + 2Al  Al2O3 + 2Cr
III :
Al2O3 + 2Cr  Cr2O3 + 2Al
In this, possible reactions are :
(a)
I, III
(b)
II,III
(c)
I, II
(d)
I, II, III
Al2(SO4)3 is used in the following but not
(a)
as a coagulant and precipitant in
treating drinking water and sewage
(b)
in paper industry
(c)
as a mordant in dyeing
(d)
in plastic industry
General formula of aluminium alums is
(a)
[M1(H2O)6][Al(H2O)6](SO4)2
(b)
[M1(H2O)4][Al(H2O)6](SO4)4
(c)
[M1(H2O)4][Al(H2O)4](SO4)4
(d)
[M1(H2O)6][Al(H2O)6](SO4)4
(M1 is monovalent cation)
Aqueous solution of borax reacts with two mol of
acids. This is because of
(a)
formation of 2 mol of B(OH)3 only
(b)
formation of 2 mol of [B(OH)4]— only
(c)
formation of 1 mol each of B(OH)3 and
[B(OH)4]—
(d)
formation of 2 mol each of [B(OH)4]— and
B(OH)3 of which only [B(OH)4]—
reacts with acid
Borax is used as a buffer since
(a)
its aqueous solution contains equal
amount of weak acid and its salt
(b)
it is easily available
(c)
its aqueous solution contains equal
amount of strong acid and its salt
(d)
statement that borax is a buffer, is wrong
While testing BO33—, there is green-edged flame on
heating the salt with conc. H2SO4 and CH3OH.
Green colour is of
(a)
(CH3)3B
(b)
(CH3O)3B
(c)
B2 O 3
(d)
H3BO3
Which is not soluble in excess of NH4OH ?
(a)
Al(OH)3
(b)
Cu(OH)2
(c)
Zn(OH)2
(d)
Cd(OH)2
Diborane is a Lewis acid forming addition
compound B2H6.2NH3 with NH3, a Lewis base. This
(a)
is ionic and exists as [BH2(NH3)2]+ and
[BH4]— ions
(b)
on heating, is converted into borazine,
B3N3H6 (called inorganic benzene)
(c)
both correct
(d)
none correct
Einstein Classes,
148.
149.
150.
151.
152.
153.
154.
Which is true statement ?
(a)
diamond is unaffected by conc. acid but
graphite reacts with hot conc. HNO3
forming mellitic acid C6(COOH)6
(b)
CO is toxic because it forms a complex
with haemoglobin in the blood
(c)
C3O2, carbon suboxide, is a foul-smelling
gas
(d)
all
Which is true statement about glass ?
(a)
glass is a solid solution - a supercooled
liquid
(b)
the glass would be water soluble if only
Na2O or K2O were used
(c)
glass of high refractive index has PbO
and opal glass has CaF2
(d)
all
Which is/are true statement(s) about hydrazine
(a)
it is a reducing as well as oxidising agents
(b)
it forms hydrazone with carbonyl
compounds
(c)
it is used as a rocket fuel
(d)
all correct
Choose correct statements :
(a)
superphosphate is [3Ca(H2PO4)2 +
7CaSO4]
(b)
it forms hydrazone with carbonyl
compounds
(c)
it is used as a rocket fuel
(d)
all correct
N2O (laughting gas) finds use in the following
except
(a)
as a propellant for whipped ice-cream
(b)
as an anaesthetic
(c)
for the preparation of N3H
(d)
as fuel for rockets
following are neutral oxides except
(a)
NO
(b)
N 2O
(c)
CO
(d)
NO2
Liquid N 2O 4 is used as non-aqueous solvent
except :
(a)
itself ionises as NO+, NO3—
(b)
substance containing NO+ is said to be
acid and that containing NO3— is said to
be base
(c)
N2O4 is paramagnetic
(d)
NO2 dimerises to N2O4 with
disappearance in paramagnetism
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 52
155.
156.
157.
158.
159.
160.
161.
162.
163.
Acid rain may cause
(a)
rusting easier
(b)
stone-cancer in Taj Mahal
(c)
non-fertility of soil
(d)
all correct
Select correct statement :
(a)
mixture of NH4Cl and NaNO3 on heating
gives N2 gas
(b)
CFC is used as refrigerating fluid and as
propellant in aerosols
(c)
phosgene is formed when P4 reacts with
NaOH
(d)
Phosgene dissolves in water forming P2O5
Glowing boards (for advertisement) use gas
(a)
N2
(b)
He
(c)
Ne
(d)
Ar
Force keeping atoms of Ar in solid state is
(a)
H-bond
(b)
van der Waal
(c)
cohesive
(d)
gravity
Pseudo alum does not have
(a)
Na+, K+
(b)
NH4+, K+
(c)
Cu2+, Mg2+
(d)
Fe3+, Cr3+
Lapis Leziul is a blue coloured mineral used as
valuable artificial gemstone. It is
(a)
cryolite
(b)
sodium aluminosilicate
(c)
basic copper acetate
(d)
basic chromium carbonate
Mixture of Al(OH)3 and Fe(OH)3 can be separated
by
(a)
HCl
(b)
NH4OH
(c)
NaOH
(d)
HNO3
N2O is formed when
(a)
moist Fe reacts with NO
(b)
Sn2+ reacts with conc. HNO3 in presence
of HCl
(c)
cold dil. HNO3 reacts with Cu and Zn
(d)
all of above
Yellow-coloured solution FeCl3 changes to light
green colour when
(a)
SnCl2 is added
(b)
NH4CNS is added
(c)
both
(d)
none
Einstein Classes,
164.
165.
166.
167.
168.
Which of the following statements are true
(a)
cold and very dilute HNO3 forms
NH4NO3 with Zn or Sn.
(b)
conc. HNO3 forms H2SnO3 with Sn
(c)
cold and more concentrated HNO3 form
N2 with Cu
(d)
all correct
NO2 is not obtained when following is heated
(a)
Pb(NO3)2
(b)
AgNO3
(c)
LiNO3
(d)
KNO3
Which is least basic :
(a)
NF3
(b)
NCl3
(c)
NBr3
(d)
NI3
NH3 can’t be obtained by
(a)
heating of NH4NO3 or NH4NO2
(b)
heating of glycine
(c)
both
(d)
none
Red and white phosphorus will differ but not in
(a)
smell
(b)
solubility in CHCl3
(c)
exhibiting phosphorescence
(d)
reactions with HNO3
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 53
ANSWERS
1.
c
31.
b
61.
d
91.
a
121.
d
151.
c
2.
d
32.
a
62.
d
92.
d
122.
a
152.
d
3.
a
33.
d
63.
c
93.
b
123.
b
153.
d
4.
b
34.
c
64.
c
94.
c
124.
a
154.
c
5.
a
35.
d
65.
c
95.
d
125.
a
155.
d
6.
b
36.
b
66.
d
96.
a
126.
d
156.
b
7.
b
37.
d
67.
c
97.
d
127.
c
157.
c
8.
d
38.
d
68.
d
98.
a
128.
d
158.
b
9.
c
39.
c
69.
b
99.
d
129.
c
159.
a
10.
c
40.
a
70.
a
100.
a
130.
d
160.
b
11.
d
41.
d
71.
b
101.
a
131.
a
161.
c
12.
d
42.
b
72.
a
102.
b
132.
c
162.
d
13.
a
43.
d
73.
d
103.
c
133.
c
163.
c
14.
b
44.
c
74.
c
104.
c
134.
b
164.
d
15.
b
45.
c
75.
a
105.
c
135.
d
165.
d
16.
b
46.
a
76.
c
106.
c
136.
d
166.
a
17.
d
47.
a
77.
d
107.
c
137.
c
167.
c
18.
d
48.
d
78.
c
108.
d
138.
c
168.
d
19.
c
49.
c
79.
c
109.
c
139.
c
20.
d
50.
d
80.
a
110.
b
140.
c
21.
b
51.
c
81.
d
111.
a
141.
d
22.
d
52.
c
82.
d
112.
d
142.
a
23.
b
53.
c
83.
c
113.
a
143.
d
24.
b
54.
b
84.
c
114.
c
144.
a
25.
b
55.
d
85.
b
115.
b
145.
b
26.
a
56.
d
86.
a
116.
d
146.
a
27.
b
57.
b
87.
c
117.
c
147.
c
28.
c
58.
b
88.
a
118.
a
148.
d
29.
c
59.
d
89.
a
119.
c
149.
d
30.
d
60.
c
90.
d
120.
d
150.
d
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 54
TEST YOURSELF
1.
Stability of trivalent and monovalent cation of
group III A (Boron family) will be in order
(a)
(b)
2.
3.
4.
5.
7.
3+
3+
3+
3+
3+
Ga < In < Tl
Ga < In > Tl
(a)
OF2
(b)
O2F 2
(c)
ICI
(d)
all
(c)
Ga > In > Tl
(d)
Ga+ < In+ < Tl+
(a)
+
+
9.
Of the following which are known
The correct structural representation of diborane
is
+
[BH2]+[BH4]—
Select correct statements
(a)
B2H6 has odd-electron multicentres bonds
(b)
AlH3 is colourless solid and is polymeric
containing Al—H—Al bridges
(c)
LiAlH4 is ionic compound
(d)
All of above
(b)
(c)
Select correct statement :
(a)
carbon graphite has a layer structure
(b)
Silicon dioxide (silica) is a network solid
with tetrahedral coordination and is a
giant molecule
(c)
GeO2, SnO2 and PbO2 are all network
solids with octahedral coordination
(d)
all of above
(d)
10.
Match List I with List II and select the correct
answer using the codes given below the list
List I
In P4 (tetraheral)
(a)
each P is joined to four P
(b)
each P is joined to three P
(c)
each P is joined to two P
(d)
P4 does not exist
For the hydrides of nitrogen family go down the
group
List II
A.
XeF4
1.
Distorted
octahedral
B.
XeF6
2.
Tetrahedral
C.
XeO3
3.
Square planar
D.
XeO4
4.
Pyramidal
Codes :
A
B
C
D
stability decreases
(a)
1
2
3
4
(b)
reducing activity increases
(b)
3
1
4
2
(c)
bond angle HMH decreases
(c)
3
1
2
4
(d)
all of above
(d)
2
4
1
3
(a)
6.
3+
8.
The thermal stability of hydrides of oxygen family
is in order
(a)
H2Po < H2Te < H2Se < H2S < H2O
(b)
H2Po < H2O < H2Te < H2Se < H2S
(c)
H2S < H2O < H2Te < H2Se < H2Po
(d)
H2O < H2S < H2Te < H2Se < H2Po
Which of the underlined atoms in oxyacids have
sp3 hybridised atoms
(a)
HClO 4 , H 2 SO 4 , H NO 2
(b)
H 2 SO 4 , H 3 PO 4 , H NO 3
ANSWERS
1.
d
6.
a
2.
d
7.
c
3.
d
8.
d
(c)
HClO 4 , H 2 SO 4 , H 2 SO5
4.
b
9.
c
(d)
HClO 4 , H NO3 , HClO3
5.
d
10.
c
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111