Download 12 - einstein classes

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p-BLOCK ELEMENTS (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
CO2 + 4H2
CH4 + H2O
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.
4N2 + 2H2O
( 4 N 2  O 2 ) 2H 2
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.
Fertilizers

Nitrogen, phosphorous and potassium are three most essential plant nutrients consumed by them. Fertility
of soil is maintained by using these nutrients in the form to fertilisers.

Fertiliser used should :
—
be water soluble
—
be stable.
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—
not disturb alkaline nature (pH)
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
NH2CONH2 (urea)
NH4NO3 (ammonium nitrate)
(NH4)2SO4 (ammonium sulphate)
NH4H2PO4 (mono ammonium hydrogen phosphate)
These are either used on their own or in conjugation with other chemicals, often to make NPK fertilisers.
These are designed to release nitrogen (N), phosphorus (P) and potassium (K) into the soil. If the NPK
fertiliser is marked 4-8-2, it means it has :
4% – N2 ;
8% – P2O5 ;
2% – K2O
Urea
Urea is widely used as a nitrogenous fertilizer. It is very soluble, and hence quick acting, but it is easily
washed away. It has a very high nitrogen content (46%). It is manufactured from NH3, and the reaction
proceeds in two stages.
180 2000 C
2 NH 3  CO 2 high

 NH 2 COONH 4  NH 2 ·CO·NH 2  H 2O
pressure
ammonium
carbamate
urea
In the soil, urea slowly hydrolyses to ammonium carbonate.
NH 2 CONH 2  2H 2O  ( NH 4 ) 2 CO 3
Phosphate Fertilizers
Phosphate rocks such as fluoroapatite [3Ca3(PO4)2 . CaF2] are very insoluble, and thus are of no use to
plants. Superphosphate is made by treating phsophate rock with concentrated H2SO4. The acid salt Ca(H2PO4)2
is more soluble, and over a period of weeks the superphosphate will dissolve in the soil water.
The CaSO4 is an insoluble waste product, and is of no value of plants, but is not removed from the product
sold.
‘Triple superphosphate’ is made in a similar way, using H3PO4 to avoid the formation of the waste product
CaSO4.
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.
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6.
PH3 (Phosphine) : Preparation
(i)
Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
Ca3P2 + 6HCl  3CaCl2 + 2PH3
(ii)
P4 (white) + 3NaOH + 3H2O  PH3 + 3NaH2PO2
(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.
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Triethyl phosphate
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.
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PCl5 + 4RCOOH  4RCOCl + H3PO4 + HCl
PCl5 + 4ROH  4RCl + H3PO4 + HCl
It reacts with P4O10, forming POCl3, and with SO2, forming thionyl chloride SOCl2.
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
Practice Problems :
1.
How is N2 differ from rest of elements ?
2
Nitrogen is inert while other (e.g. phosphorous) are righly reactive in elemental state.
3.
Why are pentahalides more covalent than trihalides ?
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4.
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements ?
5.
Why is N2 less reactive at room temperature ?
6.
Mention the conditions required to maximise the yield of ammonia.
7.
How does ammonia react with a solution of Cu2+ ?
8.
What is the covalence of nitrogen in N2O5 ?
9.
Bond angle in PH4+ is higher than that in PH3. Why ?
10.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert
atmosphere of CO2 ?
11.
What happens when PCl5 is heated ?
12.
Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water.
13.
What happens when H3PO3 is heated ?
14.
Discuss the general characteristics of Group 15 elements with reference to their electronic
configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
15.
Why does the reactivity of nitrogen differ from phosphorus ?
16.
Discuss the trends in chemical reactivity of Group 15 elements.
17.
Why does NH3 form hydrogen bond but PH3 does not ?
18.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
19.
How is ammonia manufactured industrially ?
20.
Illustrate how copper metal can give different products on reaction with HNO3.
21.
Give the resonating structures of NO2 and N2O5.
22.
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why ?
23.
Why does R3P = O exist but R3N = O does not (R = alkyl group) ?
[Due to the absence of d-orbitals, nitrogen cannot from p-d multiple bonds.]
24.
Explain, why NH3 is basic while BiH3 is only feebly basic.
25.
Nitrogen exists as diatomic molecule and phosphorus as P4. Why ?
26.
Write main differences between the properties of white phosphorus and red phosphorus.
27.
Why does nitrogen show catenation properties less than phosphorus ?
28.
Give the disproportionation reaction of H3PO3.
29.
Can PCl5 act as an oxidising as well as a reducing agent ? Justify.
30.
What are the oxidation states of phosphorus in the following :
(i)
H3PO3
(v)
POF3
(ii)
PCl3
(iii)
Ca3P2
(iv)
Na3PO4
[Answers (1) (i) N2 is gas while other elements (e.g. P4....) solid (ii)
Nitrogen is diatomic while
other elements polyatomic (P4) (2) because of small size and high EN nitrogen has tendency to form
p-p multiple bond and form triple bond due to which bond dissociation energy is very high whereas
others are more reactive due to presence of single bond p-p and exist as tetrahedral form (P4) in
which four p-atoms is bonded by three p-atoms by single p-p bonds and therefore exist as solid (4) As
we move down the Group 15, the size of the element increases and, therefore, the length of the bond
with hydrogen increases and its strength decreases. In other words, as we move down the Group, the
bond with hydrogen can break more easily to evolve H2 gas which acts as the reducing agent. Thus,
Bi — H bond is the weakest amongst the hydrides of elements of Group 15, and hence, BiH3 is the
strongest reducing agent. (5) Due to the presence of a triple bond between the two nitrogen atoms,
the bond dissociation energy of N2 (941.4 kJ mol–1) is very high. Therefore, N2 is less reactive at room
temperature. However, nitrogen reacts with metals at higher temperature to form nitrides.
(6) Ammonia is prepared by the Haber’s process. Optimum contions for the manufacture of
ammonia are as under :
. In accordance
with Le Chatelier’s principle, to maximise the yield, a high pressure of 200 × 105 Pa is used. To
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increase the rate of the reaction, a temperature of around 700 K is used and iron oxide mixed with
some K2O and Al2O3 is used as a catalyst. Sometimes, Mo is also used as a promoter to increase the
efficiency of the Fe catalyst. (7) Cu2+ ion reacts with excess of ammonia to form a deep blue coloured
complex,
according
to
the
following
reaction
:
Cu 2 (aq)  4NH 4 OH(aq) 
 [Cu(NH 3 ) 4 ]2  4H 2 O (8) 4
(9) P in PH3 is sp3-hybridised. It
Tetrammine
copper (II) ion
(Deep blue)
has three bond pairs and one lone pair around P. Due to stronger lone pair-bond pair repulsions
than bond pair-bond pair repulsions, the tetrahedral angle decreases from 109028’ to 93.60; As a
result, PH3 is pyramidal. However, when it reacts with a proton, it forms PH4+ ion which has four
bonds pairs and no lone pair. Now, there are no lone pair-bond pair repulsions. Only four identical
bond pair-bond pair interactions exist. PH4+, therefore assumes tetrahedral geometry with a bond
angle of 109028’. This explains why the bond angle in PH4+ is higher than in PH3. (10) White
phosphorus reacts with NaOH to form phosphine, as per the following equation :
Heat
P4
Phosphorou s
 3NaOH  3H 2 O CO

  PH 3  3NaHPO 2 (11) PCl 5 has three equatorial
atmosphere
2
Phosphine
Sodium
hypophosph ite
(202 pm long) and two axial (240 pm long). Axial bonds are weaker than equatorial bonds due to
greater repulsions as they are comparatively nearer. When PCl5 is heated, the less stable axial bonds
Heat
 PCl 3  Cl 2 (12) It reacts with heavy water to form phosphorus
break to form PCl3. PCl 5 
oxychloride (POCl3) and deuterium chloride (DCl). POCl3 reacts further with heavy water to form
D3PO4.
(13) The oxidation state of P in H3PO3 is +3. This value,
intermediate between the highest (+5) and lowest (–3) oxidation states of P, therefore, on heating,
H3PO4, it undergoes disproportionation to form PH3 and H3PO4 with oxidation states of –3 and +5,
3
3
Heat
respectively. 4H 3 PO 3  PH 3 
Phosphorus
acid
Phosphine
5
H 3 PO 4
(17) The electronegativity of N(3.0) is
Orthophosp horic
acid
appreciably higher than that of H(2.1). As a result, N — H bond is quite polar and hence NH3
exhibits intermolecular H-bonding.
. In contrast, P and
H bond have an electronegativity of 2.1. Therefore, P — H bond is not polar and hence PH2 does not
exhibit H-bonding. (18) by heating an equimolar aqueous solution of ammonium chloride and
sodium nitrite. (19) Ammonia is prepared on industrial scale by Haber’s process. (20) (i)
Heat
3Cu  8HNO 3 (dil .)  3Cu( NO 3 ) 2  4H 2 O  2NO
Nitric oxide
(ii) Cu  4HNO 3 (conc.) Heat
 Cu( NO 3 ) 2  2H 2 O  2NO 2
(22) In all the hydrides of Group 15
Nitric dioxide
elements, the central atom is sp3-hybridised. Three of the four sp3-orbitals from three E — H
(E stands for element of Group 15), -bonds while the fourth contains the, lone pair of electrons as
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shown
. Since the lone pair-bond pair repulsions are stronger than the bond pair-bond
pair repulsions, therefore, the bond angle decreases from 109028’ to 107.80 in NH3. As we move from
N to P to As to Sb, the atomic size goes on increasing. As a result, bond pairs of electrons, lie away and
away from the central atom. In other words, force of repulsion between the adjacent bond pair goes
on decreasing and, consequently, the bond angles keep on decreasing from NH3 to SbH3. In other
words, HNH bond angle is maximum (107.80) followed by HPH (93.60), followed by HAsH (91.80)
and HSbH (91.30). (27) The property of catenation depends upon the strength of the elementelement bond. Weaker the bond, smaller is the possibility of catenation. Since the N — N
(159 kJ mol–1) bond strength is much weaker than P — P (213 kJ mol–1) bond strength, therefore,
nitrogen shows less catenation properties than phosphorus. (29) Phosphorus has maximum
oxidation state of +5 in PCl5. It cannot increase its oxidation state further and therefore, PCl5 cannot
act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower
value, therefore, PCl5 acts as an oxidising agent.]
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
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
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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,
Po : Radioactive (highly toxic)
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.
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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
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.
SO2 + H2O  H2SO3
SO3 + H2O  H2SO4
P4O10 + H2O  H3PO4
N2O3 + H2O  HNO2
P4O6 + H2O  H3PO4
N2O5 + H2O  HNO3
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(ii)
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
2ZnS + 3O2  2ZnO + 2SO2
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
7.
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.
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(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 )
9.
Oxoacids of sulphur :
(a)
H2SO4
(b)
H2SO3
most important acid
(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)
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(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)
(iii)
Strong acidic character :
H2SO4  H+(aq) + HSO4–(aq);
K1 = very large
HSO4–(aq)  H+(aq) + SO42–(aq);
K2 = 1.2 × 10–2
Strong affinity for water (dehydrating agent) :
SO 4
C12H22O11 H
2 
 12C + 11H2O
(charring)
(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
Practice Problems :
1.
List the important sources of sulphur.
2.
Write the order of thermal stability of the hydrides of Group 16 elements.
3.
Why is H2O a liquid and H2S a gas ?
4.
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe
5.
Complete the following reactions :
(i)
C2H4 + O2 
(ii)
4Al + 3O2 
6.
Why does O3 act as a powerful oxidising agent ?
7.
How is O3 estimated quantitatively ?
8.
What happens when sulphur dioxide is passes through an aqueous solution of Fe(III) salt ?
9.
Comment on the nature of two S – O bonds formed in SO2 molecule. Are the two S – O bonds in this
molecule equal ?
10.
How is the presence of SO2 detected ?
11.
Mention three areas in which H2SO4 plays an important role.
12.
Write the conditions to maximise the yield of H2SO4 by Contact process.
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13.
Why is K a 2  K a1 for H2SO4 in water ?
14.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of
electronic configuration, oxidation, state and hydride formation.
15.
Knowing the electron gain enthalpy values of O  O– and O  O2– as –141 and 702 kJ mol–1
respectively, how can you account for the formation of a large number of oxides having O2– species
and not O– ?
16.
Which aerosols deplete ozone ?
17.
Describe the manufacture of H2SO4 by contact process.
18.
How is SO2 an air pollutant ?
19.
Write balanced equations for the following :
(i)
NaCl is heated with sulphuric acid in presence of MnO2.
(ii)
Chlorine gas is passed into a solution of NaI in water.
[Answers : (1) Sulphur occurs in the earth’s crust in the combined state primary in form of sulphates
and sulphides. As sulphates : Gypsum, CaSO4·2H2O; Epsom salt, MgSO4·7H2O; Baryte, BaSO4, etc.
As sulphides : Galena, PbS; Zinc blende, ZnS; Copper pyrites, CuFeS2; Iron pyrites, FeS2, etc. Traces
of sulphur occur as H2S and in organic matter such at eggs, proteins, garlic, onion, mustard hair and
wool. (2) As the size of the element increases down the group, the E — H (E stands for element of
Group 16) bond dissociation energy decreases and hence E — H bond breaks more easily. Thus, the
thermal stability of the hydrides of Group 16 elements decreases down the Group :
H2O > H2S > H2Se > H2Te > H2Po. (3) Due to greater electronegativity of O than S, H2O undergoes
extensive intermolecular H-bonding. As a result, H2O exists as an associated, H2O is a liquid at room
temperature.
   H — O   H — O   H — O   
|
H
|
H
|
H
. In contrast, H2S does not undergo H-bonding as the
electronegativity difference between H and S is not appreciable. It exists as discrete molecules which
are held together by weak van der Waals’ forces of attraction. To break these forces of attraction,
only a small amount of energy is required. Therefore, H2S is a gas at room temperature. (4) Platinum
is a noble metal. Its ionisation enthalpy is very large. Therefore, it does not react with oxygen
directly. In contrast, Zn, Ti and Fe are active metals and hence, directly react with oxygen to form
their respective oxides. Zn + O2  ZnO2, Ti + O2  TiO2, 4Fe + 3O2  2Fe2O3 (5) (i) C2H4 undergoes
Heat
 2CO 2  2H 2 O . (ii) Al combines with O2 to
combustion to form CO2 and H2O. C 2 H 4  3O 2 
Heat
form alumina. 4Al  3O 2  2 Al 2 O 3 (6) O3 is an endothermic compound i.e., it possesses greater
internal energy than oxygen. On heating, it readily decomposes to give dioxygen and nascent
Heat
oxygen. O 3  O 2  O (nascent oxygen). (7) When O3 is treated with excess of KI solution
buffered with borate buffer (pH 9.2), I2 is liberated quantitatively, according to the following
equation :
2I–(aq) + H2O (l) + O3 (g)  2OH–(aq) + I2(s) + O2(g). The I2 thus liberated is titrated against a
standard solution of sodium thiosulphate using starch as an indicator.
2NaS2O3 + I2  Na2S4O6 + 2NaI (8) SO2 acts as a reducing agent and reduces an aqueous solution of
Fe(III)
salt
to
Fe(II)
salt,
given
by
the
following
equation
:
(9) In the structure of SO2, S is sp2-hybridised. Two of
the three sp2-orbitals form two -bonds while the third contains the lone pair of electrons (10) SO2 is
a pungent smelling gas. Following two sets tests can be performed to detect SO2. (i) SO2 turns the
pink violet colour of KMnO4 solution to colourless due to reduction of MnO4– to Mn2+ ions.
2MnO 4  5SO 2  2H 2 O 
 2Mn 2  5SO 42   4H  ]. (ii) SO2 turns acidified K2Cr2O7 solution
( pink violet )
Einstein Classes,
( colourless )
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green due to reduction of Cr2O72– to Cr3+ ions. Cr2 O 72   3SO 2  2H  
 2Cr 3  3SO 42   H 2O
( green )
( orange )
(11) (i) It is used as an electrolyte in storage batteries. (ii) It is used in petroleum refining, detergent
industry and in the manufacture of paints, pigments and dyes. (iii) H2SO4 is used in the manufacture
of fertilizers such as ammonium sulphate, calcium superphosphate. (12) The main step in the
preparation of H2SO4 is the oxidation of SO2 to SO3.
SO2(g) + O2(g)
2SO3(g); fH0 = –196.6 kJ mol–1. The reaction is exothermic and reversible. The
forward reaction proceeds with decrease in volume. Therefore, in accordance with Le Chatelier’s
principle, to maximise the yield of SO3, a low temperature (720 K), a high pressure (2 bar) are
required. V2O5 is used as a catalyst. (13) This is because the negatively charged HSO4– ion has much
less tendency to donate a proton to H2O as compared to neutral H2SO4. (15) Consider the reaction of
a divalent metal (M) with oxygen. The formation of M2O and MO involves the following steps :
H
H
i 1
i 2
M(g ) 
 M  (g ) 
 M 2  (g ) . iH1 and iH2 are first and second ionisation enthalpies of
 H
 H
eg 1
eg 2
the metal M. O(g ) 
 O  (g ) 
 O 2 – (g ) . egH1 and egH2 are first and second electron
Lattice energy
2

Lattice energy
gain enthalpies. 2M  (g )  O  (g )     M 2 O(s ) , M (g )  O 2 (g )     MO(s ) .
Although iH2 is much more than iH1 and egH2 is much higher than egH1, yet the lattice energy of
formation of MO (s) due to higher charges is much more than that of M2O (s). In other words,
formation MO is energetically more favourable than M2O. It is due to this reason that oxygen forms
preferably oxides having the O2– species and not O–. (16) Aerosols such as chlorofluorocarbons (CFC’s),
for example, freon (CCl2F2) depletes the O3 layer by supplying Cl free radicals which convert O3 to
O 2 in
the
following
sequence
of
reactions
:
h
Cl 2 CF2 ( g )  Cl ( g )  CClF2 ( g ) ,
Freon
 Cl(g )  O 3 (g ) 
 ClO  (g )  O 2 (g ) ,
ClO  (g ) 
 Cl  (g )  O(g ) , ClO  (g )  O(g ) 
 Cl(g )  O 2 (g ) (18) (i) SO 2 is strongly
irritating to the respiratory tract. SO2 at a concentration of 5 ppm causes throat and eye irritation
causing cough, tears and redness in eyes. If causes breathlessness and affects larynx, i.e., voice box.
(ii) Even at low concentration, it has a damaging effect on the plants. If exposed for a long time, it
slows down the formation of chlorophyll resulting in injury to the leaves including loss of green
colour. This is called chlorosis. (iii) SO2 dissolves in rainwater and produces acid rain which
damages building materials especially marble (CaCO3). CaCO3 + H2SO3  CaSO3 + H2O + CO2.
(iv) It corrodes metals particularly iron and steel. It also deteriorates fabrices, leather, paper, etc.,
and effects the colour of paints.]
C3A Group 17 (Halogen family) (Salt producing)
General properties
ns2p 5
F – [He]2s22p5
F2(g) Pale yellow
Cl – [Ne]3s23p5
10
Cl2(g) greenish yellow
2
Br – [Ar]3d 4s 4p
10
2
I – [Kr]4f 5s 5p
5
5
Br2(l) reddish brown
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.
Einstein Classes,
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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.
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.
aq
1
 e
X 2 (g ) 
 X(g ) 
 X  (g ) 

 X  (aq )
H
hydration   ve
2
;
1
Hdissociation = +ve
2
Hegain enthalpy = –ve
E0 =
1
Hdissociation + Hegain enthalpy + Hhdydration
2
F2  B.D.E. (
1
Hdissociation ) is less thyan of Cl2
2
F2  Heg
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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F2  Hhydration is higher than Cl2 due to small size and high charge density of F– ion. Thus the net energy
1
1
F (g)  F–(aq) is –752 kJ whereas Cl2(g)  Cl–(aq) is –708 kJ which accounts for higher
2 2
2
oxidising power of F2.
released for
F2 + 2e–  2F–
E0 = +2.87 V
Cl2 + 2e–  2Cl–
E0 = +1.36 V
Br2 + 2e–  2Br–
E0 = +1.08 V
I2 + 2e–  2I–
E0 = +0.53 V
F2 + 2X–  X2 + 2F–
(X = Cl, Br, I)
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
Hydrides :
H – F, H– Cl, H – Br, H – I
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.
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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.
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
573K
Cl 2  3F2  2ClF3
( excess)
I 2  Cl 2 
 ICl
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
(Hypohalous acid)
Halic (III) acid
–
(Perhalic acid)
Einstein Classes,
HOI
HOClO
–
–
HOBrO2
HOIO2
HOBrO3
HOIO3
(Chlorous acid)
–
(Halic acid)
Holic (VII) acid
HOBr
(Hypochlorus acid)
(Halous acid)
Halic (V) acid
HOCl
HOClO2
(Chloric acid)
–
HOClO3
(Perchloric acid)
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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) :
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)
 H 2O
6NaOH  3Cl 2 
 5NaCl  NaClO 3  3H 2 O
( hot & conc .)
(iii)
NaOCl
sodium hypochlori de
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
Einstein Classes,
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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.
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)
Lab Preparation :
420 K
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
Practice Problems :
1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and
hydration enthalpy, compare the oxidising power of F2 and Cl2.
2.
Give two examples to show the anomalous behaviour of fluorine.
3.
Sea is the greatest source of some halogens. Comment.
4.
Give the reason for bleaching action of Cl2.
5.
Name two poisonous gases which can be prepared from chloride gas.
6.
Why is ICl more reactive than I2 ?
7.
Why are halogens strong oxidising agents ?
8.
Explain why fluorine forms only one oxoacid, HOF.
9.
Explain why inspite or nearly the same electronegativity, oxygen forms hydrogen bonding while
chlorine does not.
10.
Write two uses of ClO2.
Einstein Classes,
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11.
Why are halogens coloured ?
12.
Write the reactions of F2 and Cl2 with water.
13.
How can you prepare Cl2 from HCl and HCl from Cl2 ? Write reactions only.
14.
With what neutral molecule is ClO– isoelectronic ? Is that molecule a Lewis base ?
15.
Arrange the following in the order of property indicated for each set :
16.
(i)
F2, Cl2, Br2, I2, – increasing bond dissociation enthalpy.
(ii)
HF, HCl, HBr, HI – increasing acid strength.
(iii)
NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.
Give the formula and describe the structure of a noble gas species which is isostructural with :
(i)
ICl4–
(ii)
IBr2–
(iii)
BrO3–
[Answers : (2) (i) Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5
and +7 oxidation states also. The m.p and b.p of hydrogen halides follow the order : HF > HCl < HBr
< HI. (ii) The anomalous behaviour of fluorine is due to its (i) small size, (ii) highest electronegativity.
(3) Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and
calcium, but the main constituents is sodium chloride (2.5% by mass). Dried up sea beds contain
sodium chloride and carnallite, KCl.MgCl2.6H2O. Certain seaweeds contain up to 0.5% of iodine as
sodium iodide and chile salpetre (NaNO3) contains up to 0.2% of sodium iodate. Thus, sea is the
greatest source of halogens. (4) In presence of moisture or in aqueous solution, Ci2 liberates nascent
 2HCl 
oxygen. Cl 2  H 2O 
[O]
. This nascent oxygen brings about the oxidation of
Nascent oxygen
coloured substances present in vegetable and organic matter to convert them to colourless
substances. Coloured substance + O  Colourless substance. Thus, the bleaching action of Cl2 is due
to oxidation. (5) Two poisonous gases which can be prepared from Cl2 are : (i) Phosgene (ii) Mustard
 charcol
gas. These are prepared as follows : (i) CO  Cl 2 h
 COCl 2 (ii) Cl2 is passed through
Phosgene
boiling S, when S 2 Cl 2 is formed. This on reaction with ethene gives mustard gas.
(6) ICl is more reactive than I2 because I — Cl bond is weaker
than I — I bond. A bond between two different atoms is always weaker than that between identical
atoms. Consequently, ICl breaks easily to form halogen atoms which readily bring about the
reactions. (7) Due to low bond dissociation enthalpy, high electronegativity and large negative
electron gain enthalpy, halogens have a strong tendency to accept electrons and thus get reduced. X
+ e–  X– . In other words, halogens act as strong oxidising agents. (8) This is due to limitation of
having only one oxidation state that it forms only one oxo acid. Chlorine, bromine and iodine form
four series of oxoacids of the general formula HOX, HOXO, HOXO2 and HOXO3 in which the
oxidation states of halogen (X = Cl, Br or I) is +1, +3, +5 and +7, respectively. However, due to high
electronegativity and small size, fluorine forms only HOF with oxidation state +1. (9) Although O
and Cl have about the same electronegativity, yet their atomic size (covalent radii) are much
different with radius of oxygen being 66 pm and that of chlorine being 99 pm. As a result, electron
density per unit volume on oxygen atom is much higher than that on chlorine atom. Therefore,
oxygen forms hydrogen bonds while chlorine does not even though both have approximate, the
same electronegativity. (10) (i) It is an excellent bleaching agent. Its bleaching power is about 30
times higher than that of Cl2. It is used for bleaching flour to make white bread. (ii) ClO2 is a
powerful oxidising agent and chlorinating agent. Large quantities of ClO2 are used for bleaching
wood pulp and cellulose and for purifying drinking water. (14) ClO– has 17 + 8 + 1 = 26 electrons. A
neutral molecule with 26 electrons is OF2 (8 + 2 × 9 = 26 electrons). OF2 is a Lewis base due to the
presence of two lone pairs of electrons.]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 27
C4
Group - 18 (Noble Gases)
He2 , Ne
, Ar
,
2
6
2
6
1s
2s 2p
3 s 3p
Kr
,
Xe
,
Rn
3d 10 4s 2 4p 6 4d 10 5s 2 6s 2 6p 6 4 f 14 5d 10 6 s 2 6p 6
( Radioactiv e)
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.
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,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 28
Practice Problems :
1.
Why is helium used in diving apparatus ?
2.
Balance the following equation : XeF6 + H2O  XeO2F2 + HF.
3.
Why has it been difficult to study the chemistry of radon ?
4.
What inspired N. Bartlett for carrying out reaction between Xe and PtF6 ?
5.
How are xenon fluorides XeF2, XeF4 and XeF6 obtained ?
6.
How are XeO3 and XeOF4 prepared ?
7.
Which one of the following does not exist ?
(i)
XeOF4
(ii)
NeF2
(iii)
XeF4
(iv)
XeF6
[Answers : (1) Because of its low solubility (as compared to N2) in blood, a mixture of oxygen and
helium is used in diving kit used by deep sea divers. (2) XeF6 + 2H2O  XeO2F2 + 4HF (3) This is
because radon is a radioactive element with a short life of 3.82 days. This makes the study of
chemistry of radon difficult. (4) Neil Bartlett observed that PtF6 reacts with O2 to yield an ionic solid,
O2+, FtF–6. O2(g) + PtF6(g)  O2+[PtF6]–. In this reaction, O2 gets oxidised to O2+ by PtF6. He thought
since the first ionisation enthalpy Xe to Xe+. This inspired Bartlett to carry out the reaction between
Xe and PtF6. When Xe and PtF6 were mixed, actually a rapid reaction took place and a red solid with
278 K
a formula, Xe+PtF6– was obtained. Xe  PtF6  Xe  [PtF6 ] (7) The ionisation enthalpy of Ne
is much higher than that of Xe. As a result, F2 can oxidize Xe to Xe2+ but cannot oxidise Ne to Ne2+.
Thus, NeF2 does not exist. In contrast, all the xenon fluorides (XeF4 and XeF6) and xenon oxyfluoride
(XeOF4) do exist. This is because ionisation enthalpy of Xe is not very high. It is compensated by
electron gain enthalpy of F.]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 29
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
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 30
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
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 31
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
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CPB – 32
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
CPB – 33
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.
I

134.
135.
II
Borax 
 H 3 BO 3  B 2 O 3  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
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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 – 39
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 – 40
EXCERCISE BASED ON NEW PATTERN
1.
MATCHING TYPE PROBLEMS
Matching-1
Column - A
STATEMENT-2 : SiCl 4 is ionic and CCl 4 is
covalent.
Column - B
(A)
Asbestos
(P)
Paramagnetic
(B)
Lithium metal
(Q)
Silicates of Ca and Mg
(C)
Nitric oxide
(R)
Reducing agent
(D)
CCl4 hydrolysis
(S)
no reaction
2.
3+
Column - B
+
(A)
Bi  (BiO)
(B)
[AlO2]–  Al(OH)3
(C)
(D)
4–
4
SiO
 Si2O
(P)
6–
7
STATEMENT-1 : Boron always forms covalent
bond.
STATEMENT-2 : The small size of B3+ favours
formation of covalent bond.
3.
Matching-2
Column - A
STATEMENT-1 : Between SiCl4 and CCl4, only
SiCl4 reacts with water.
STATEMENT-1 : In water, orthoboric acid
behaves as a weak monobasic acid.
STATEMENT-2 : In water, orthoboric acid as a
proton donor.
Heat
(Q)
Hydrolysis
(R)
Acidification
2–
7
(B4O )  [B(OH)3]
(S)
Dilution by water
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
(Answers) EXCERCISE BASED ON NEW PATTERN
MATCHING TYPE PROBLEMS
1.
[A-Q; B-R; C-P; D-S]
2.
[A-Q, S; B-S, C-R; D-Q, R]
3.
C
ASSERTION-REASON TYPE
1.
C
2.
Einstein Classes,
A
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111