p- BLOCK ELEMENTS
The general valence shell electronic configuration of p-block elements is ns2np1-6
GROUP 15 ELEMENTS
Group 15 elements: N, P, As, Sb & Bi constitute 15th group of the periodic table.
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Their general electronic configuration is ns2np3
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Occurrence: molecular nitrogen comprises 78% by volume of the atmosphere.
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In earth crust N occurs as NaNO3(called Chilesaltpetre) &KNO3(Indian salt petre)
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Phosphorous occurs in minerals of the apatite family main component of phosphate
rocks. It is also present as an essential constituent in animal & plant matter.
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As, Sb & Bi are mainly found as sulphide minerals.
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o
o
o
o
o
PHYSICAL PROPERTIES:
Dinitrogen is a diatomic gas while all others are solids.
Metallic character increases down the group.
N & P are non-metals. As & Sb metalloids & Bi is a metal .This is due to decrease in
ionisation enthalpy & increase in atomic size.
Electro negativity decreases down the group.
CHEMICAL PROPERTIES:
Group 5 elements exhibit -3, +3 & +5 state.
Nitrogen exhibits a wide range of oxidation states.
The negative oxidation states N2 are due to higher electro negativity
Due to inert effect bismuth do not exhibit +5 state.
In case of nitrogen all oxidation states from +1 to +4 tend to disproportionate in acid
solution for ex: 3HNO2 HNO3 + H2O + 2NO
ANAMALOUS PROPERTIES OF NITROGEN:
These properties are due to:

Smaller size

High electro negativity
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High ionization enthalpy &

Non-available of d-orbitals

N2 has unique ability to form pπ-pπ multiple bonds with itself & with other elements
like C & O.

Heavier members of this group do not form pπ-pπ bond because their atomic orbitals
are so large & diffuse that they cannot have effective overlapping.

Nitrogen exists as diatomic molecule with triple bond between two atoms N triple
bond N consequently its bond enthalpy is high.

Where as other elements form single bonds in elemental state

N cannot form dπ-dπ bond where as other elements can.
1
REACTIVITY TOWARDS HYDROGEN:
All the elements form hydrides of the type EH3
(E=N, P, As, Sb, or Bi)
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Stability of the hydrides decreases from NH3 to BiH3
Bond dissociation enthalpy decreases from NH3 to BiH3.
Reducing character increases from NH3 to BiH3.
Basic character decreases from NH3 to BiH3.
NH3>PH3>AsH3>SbH3>=BiH3.
REACTIVITY TOWARDS OXYGEN:
These elements form two types of oxides (1) E2O3 (2) E2O5 the oxide in high
oxidation state of the element is more acidic than that of lower oxidation state .
The acidic character decreases down the group.
E2O3 of N & P are purely acidic.
E2O3 of As & Sb are amphoteric Bi2O3 is basic.
REACTIVITY TOWARDS HALOGEN
These elements form two series of halides EX3 & EX5.
Nitrogen does not form pentahalides due to absence of ‘d’ orbitals.
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Pentahalides are more covalent than Trihalides.
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All Trihalides except N2 are stable.
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NF3 is known to be stable.
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Trihalides are covalent except BiF3.
REACTIVITY TOWARDS METALS:
All these metals react with metals to form binary compounds
Showing -3 oxidation states Ex: Ca3N2,
DINITROGEN (N2)
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PREPARATION:Dinitrogen is produced commercially by the liquification & fractional distillation of
air. Liquid N2 (bb 77.2k) distils out first leaving behind liquid O2 (bp 90 k).
In the laboratory it is prepared by treating on aqueous solution of NH4Cl with sodium
nitrate.
NH4Cl (aq) +NaNO2 (aq) N2 (g) +2H2O (l) +NaCl (aq)
Thermal decomposition of ammonium dichromate also gives N2.
(NH4)2Cr2O7N2+4H2O+Cr2O3
Thermal decomposition of barium or sodium azide.
2
PROPERTIES:
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It is a colourless non toxic gas.
It has 2 stable isotopes 14N & 15N.
Dinitrogen is inert at room temperature because of the high bond enthalpy of N triple
bond N.
At high temperature it combines with metals to form ionic nitride (Mg3N2), with non
metals covalent nitrates.
USES:
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In the manufacture of ammonia by (haber process) CaCN2 calcium cyanamide.
To creating inert atmosphere
Refrigerant
In cryosurgery.
AMMONIA:
PREPARATION:
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In solid & air NH3 is formed by the decomposition of organic matter. ex: urea
NH2CONH2+ 2H2O (NH4)2CO32NH3+H2O+O2
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In laboratory it is prepared by heating ammonium salt with NaOH or lime
2NH4Cl+ Ca (OH) 22NH3+ 2H2O +CaCl2
(NH4)2SO4+ 2NaOH2NH3+2H2O+Na2SO4
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In large scale it is manufactured by Haber’s process
N2 (g) +3H2 = 2NH3 (g)
∆H0= -46.1 chi/mol
According to Lechatlier’s principle the favourable conditions for the manufacture of
NH3 are
Optimum temperature : 700k
High pressure : 200atm
Catalyst : iron oxide
Promoter : K2O or Al2O3
PROPERTIES:
Ammonia is a colourless gas with pungent odour.
Highly soluble in water.
In solid & liquid states it exists as an associated molecule due to hydrogen bonding
which accounts for high melting & boiling points of NH3.
Trigonal pyramidal shape NH3 molecule.
3
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Aqueous solution of ammonia is weakly basic due to the formation of OH- ions.
It precipitates the hydroxides of many metals from their salt solution Ex:
ZnSO4+2NH4OHZn(OH)2+(NH4)2SO4
Ammonia can form coordinate bonds by donating its lone pair on nitrogen, ammonia
forms complexes.
CuSO4+4NH3 [Cu (NH3)4] SO4
AgCl+2NH3 [Ag (NH3)2] Cl
Ammonia is used in the manufacture of fertilizers such as (NH4)2SO4, urea, NH4NO3
etc.
Used as a refrigerant
Nitric acid by Ostwald process.
OXIDES OF NITROGEN:
Nitrogen forms 5 types of oxides in different oxidation state
NAME
Nitrous oxide or Laughing
gas
Nitric oxide
Dinitrogen trioxide
Dinitrogen tetra oxide
FORMULA
N2O
OXIDATION
STATE
+1
NO
+2
N2O3
+3
N2O4 or
+4
NO2
Dinitrogen pentoxide
N2O5
+5
Note: As the oxidation state of N2 increases acidity increases.
NITRIC ACID:
CHEMICAL
NATURE
Neutral
Neutral
Acidic
Acidic
Acidic
Nitrogen forms oxo acids such as hypo nitrous acid (H2N2O2), nitrous acid (HNO2) & nitric
acid (HNO3). Amongst them HNO3 is the most important.
4
1.
2.
1)
2)
3)
PREPARATION OF HNO3:
In the laboratory it is prepared by heating KNO3 or NaNO3 and concn.H2SO4
NaNO3+ H2SO4NaHSO4+HNO3
On large scale it is prepared mainly by Ostwald’s process. It is based upon catalytic
oxidation of ammonia by atmospheric oxidation. The main steps involved are
4NH3+5O2-pt500k, 9 bar --4NO+6H2O
2NO+O22NO2
3NO+H2O2HNO3+NO
NO thus formed is recycled & the aq HNO3 can be concentrated to get 98%
H2SO4

Concn. HNO3 is a strong oxidizing agent & attacks most metals except noble metals
gold & Pt.

Cr & Al do not dissolve in conc. HNO3 because of the formation of a positive film of
oxide on the surface.

It oxidises non metals like I2 to HIO3 ,C to CO2,S to H2SO4

Brown ring test is used to detect the presence of NO3- ion .This test is based on the fact
that Fe2+ ions can reduce nitrates to NO , which reacts with Fe2+ ions to form a brown
coloured complex [Fe(H2O)5NO].
PHOSPHOROUS:
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Allotropy: The acceptance of an element in different physical forms with the same
chemical properties is known as allotropy.
Phosphorous exists in a variety of forms called white, red, α block & β block etc.
The common forms are white & red.
White phosphorous is more reactive than red phosphorous because white P exists as
discrete P4 molecules. In red P several P4 molecules are linked to formed polymeric
chain.
It is prepared in laboratory by heating white P with conc. NAOH solution in an air
inert atmosphere of CO2 P4+3NaOH+3H2OPH3+3NaH2PO2
Phosphorous forms two types of halides PX3 & PX5 (X=F, l, Br)
Trihalides have pyramidal shape & penta halides have a trigonal bipyramidal
structure.
5
OXOACIDS OF PHOSPHROUS:

The acids in +3 oxidation state disproportionate to higher & lower oxidation.
4H3PO33H3PO4+PH3
Acids which contains P—H bond have strong reducing properties. EX:-H3PO2
Hydrogen atoms which are attached with oxygen in P—OH form
are ionisable & acuse the bascity.

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Ex:- H3PO3 & H3PO4 are di & tri basic
STRUCTURES OF SOME IMPORTANT OXOACIDS OF PHOSPHOROUS
1:
H3PO4
2:
H4P2O7
6
3:
H3PO3
4:
H3PO2
TRY THESE INTEXT QUESTIONS
Q1.Why does PCl3 fume in moisture?
Solution: In the presence of (H2O), PCl3 undergoes hydrolysis giving fumes of HCl
PCl3 + 3H2O
H3PO3 + 3HCl
Q2.Why ammonia has higher boiling point than phosphine?
Solution: NH3 forms H-bonds but PH3 does not and hence the boiling point of NH3 is
higher than that of PH3.
Q3.Why is white phosphorous kept under water?
7
Solution: Ignition temperature of white phosphorous is very low (303 K). Therefore on
exposure to air, it spontaneously catches fire forming P4O10. Therefore to protect it
from air, it is kept under water.
Q4.What is laughing gas? How is it prepared?
Solution: N2O is laughing gas and it is prepared by heating NH4NO3.
Q5.Name 3 allotropes of phosphorous? Which of the three is the most reactive?
Solution: White, Red and black phosphorous. White phosphorous is the most reactive.
GROUP-16 ELEMENTS
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Oxygen, S, Selenium, Tellurium& Polonium constitute group-16 of the periodic table.
They are also known as chalcogens (ore farming)
Oxygen is a gas other elements are solids.
The general electronic configuration of this group ns2 np4.
Atomic radius increases from oxygen to polonium.
Ionization enthalpy decreases from oxygen to polonium.
Oxygen atom has less negative electron gain enthalpy than S because of the compact
nature of the oxygen atom. However from the S onwards the value again becomes less
negative upto polonium.
Eletronegativity gradually decreases from oxygen to polonium, metallic character
increases from oxygen to polonium.
Oxygen & S are non-metals, selenium & tellurium are metalloids. Po is a radioactive
metal.
Oxygen is diatomic gas while S, Se & Te are octa atomic S8, Se8 & Te8 molecules
which has puckered ‘ring’ structure.
All these elements exhibit allotropy.
All these elements except oxygen exhibit -2, 2, 4 and 6 oxidation states.
Since oxygen is second most electro negative element next to fluorine, oxygen never
exhibits positive oxidation states except in the compounds of fluorine.
Oxygen exhibits +1 state in O2F2, + 2 in OF2.
In other compounds it exhibits only -2.
Oxygen cannot exhibit oxidation number beyond +2 due to the absence of d-orbitals in
the valency shell.
8
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From S to Te the stability of +6 oxidation state decreases and stability of +4 oxidation
state increase due to inert pair effect.
Anamalous behavior of oxygen i.e., due to its small size, high electro negativity and
absence of d-orbitals.
All the elements of group-16 forms hydrides of the type H2E
Acidic character of hydrides increases from water to H2Te.
Reducing character increases from H2S to H2Te
All these elements form oxides of EO2 &EO3 types (E=S, Se, Te or Po) O3 &SO3 are
gases while SeO2 is solid. Reducing property of dioxides decreases from SO2 to TeO2.
Form a large number of halides of the type EX6,EX4 & EX2 (E=16th group elements
X=halogen)
The stability of the halides decreases in the order F>Cl>Br>I
Hexa fluorides are gaseous in nature, octahedral structure SF6 in stable for steric
reasons
The allotrope of oxygen is ozone.
S forms various allotropes of which the yellow rhombic (α-sulphur) and monoclinic
(β-sulphur) forms are the most important.
Rhombic sulphur is the most stable form at room temperature.
In vapour state S partly exists as S2 molecules which has 2 unpaired electrons in the
antibonding orbitals like oxygen & hence exhibits paramagnetism.
DIOXYGEN:
Dioxygen can be obtained by oxygen containing salts like chlorites, nitrates
2KClO3-heat------ 2KCl+3O2
MnO2
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Oxygen does not react with metals like Au, Pt &some noble gases.
Its reaction with other elements is always exothermic
HYDRIDES:
Form hydrides of the type H2E.
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Stability decreases from water to PoH2
Acidic nature gradually increases
Hydrogen bonding decreases
Water is liquid & other hydrites are gases
OXIDES: -
Form oxides of the types EO2 &EO3 with 0.state, +4 &+6 respectively. Sulphur forms stable
SO2 & SO3. Se & Te forms SeO2, SeO3, and TeO2&TeO3.

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Basic nature of oxides gradually increases
SO2 & SO3 –acidic
9
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SeO2,SeO3,TeO2&TeO3—amphoteric
PoO—basic
HALIDES:
Forms di, tetra & hexa halides

DI HALIDES:-sp 3 hybridisation but angular structure.
Stability increases from oxygen to Po


TETRA HALIDES:-sp3d hybridization –distorted trigonal bipyramid
HEXA HALIDES:-sp3 d2 ,octahedral SF6
OXIDES:-
A binary compound of oxygen with another element is called oxide. Oxides can be
classified on the basis of nature
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o
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Acidic oxides: - Non metallic oxides. Aqueous solutions are acids. Neutralize bases to
form salts. Ex: SO2, Co2, N2O5 etc.
Basic oxides: - metallic oxides. Aqueous solutions are alkalis. Neutralize acids to form
salts.Ex:Na2O, K2O, Mgo etc.
Amphorteric oxides:- some metallic oxides exhibit a dual behavior. Neutralize both
acids & bases to form salts.
Ex:- AlO3, Sb2O3, SnO, PbO2 etc………..
OZONE:Ozone is prepared by subjecting cold, dry oxygen to silent electric discharge
3O2= 2O3
∆H=142kJ/mol
It is a pale blue gas
Liquid ozone is deep blue in colour
Solid ozone is violet black crystals
It is a powerful oxidizing agent
It oxidizes black PbS into white PbSO4
PbS+4O3 PbSO4+ 4O2
Moist I- to I2
2KI+H2O+O32KOH+I2+O2
Ozone used as a germicide, disinfectant & for sterilising water
Bleaching oils, Ivory, Starch etc……
10
SO2:- (SULPHUR DIOXIDE)
Preparation:
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Burning of S in air
S+O2SO2
Roasting of sulphide minerals gives SO
(Iron pyrites) 4FeS2+11O22Fe2O3+8SO2
(Zinc blend) 2ZnS+3O22ZnO+2SO2
PROPERTIES:-
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It is a colourless gas with pungent smell
Highly soluble in water to form solution of sulphurous water
SO2+H2OH2SO3
SO2 reacts with Cl2 to from sulphuryl chloride
SO2+Cl2SO2Cl2
It reacts with oxygen to form SO3 in presence of V2O5 catalyst
2SO2+O22SO3
Moist SO2 behaves as a reducing agent .It converts Fe (III) ions Fe (II) ions &
delocalises acidified potassium permanganate (VII) solution.
2Fe3+ +SO2+2H2O2Fe2++SO24—+4H+
5SO2+2MnO4+2H2O5SO24-+4H++2Mn2+
SO2 molecule is angular acid resonance hybrid of the following forms:
SO2 uses:
SO2 is used as good bleaching agent. Its bleaching action is due to reduction. It
bleaches silk& wool. as an anti-chlor, disinfectant& preservative.
In refining petroleum & sugar.
OXOACIDS OF S:
It forms a number of oxoacids such as
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H2SO3 sulphurous acid
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H2SO4 sulphuric acid
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H2S2O7 pyrosulphuric(oleum)acid

H2S2O8 peroxydisulphuric acid
1.
2.
11
H2SO3
H2SO4
3.
H2S2O8
4.
H2S2O7
SULPHURIC ACID:
It is manufactured by contact process which involves 3 steps
1.
burning of S or sulphide ores in air to generate SO2
2.
conversion of SO2 to SO3 in presence of V2O5 catalyst
3.
Absorption of SO2 in H2SO3 to give oleum.


PROPERTIES:
It is colourless oily dense liquid.
While making dilute solution concn acid must be added slowly in to water with
constant stirring.
12
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The chemical reactions are due to its
1.
Low volatility
2.
Strong acidic character
3.
Strong affinity for water
4.
Ability to act as an oxidising agent
In aqueous solution it ionizes into 2 steps
H2SO4+H2OH3O++HSO4HSO4-+H2OH3O++SO24It is a strong dehydrating agent Ex: cleansing action of sugar
C12H22O11-H2SO4---- 12C+11H2O
USES:
It is used in
o
Petroleum refining
o
Manufacture of pigments, paints& dyestuff intermediates.
o
Storage batteries.
TRY THESE INTEXT QUESTIONS
Q1.Why are group 16 elements called chalcogens?
Solution: Chalcogens means ore forming. The elements of group 16 are called chalcogens
because many metals are found as oxides and sulphides and a few as selenides and tellurides.
Q2.What are the allotropes of oxygen?
Solution: The allotropes of oxygen are Dioxygen (O2) and Ozone (O3).
Q3.Why H2S is acidic and H2O is neutral?
Solution: The S---H bond is weaker than O---H bond because the size of S atom is bigger
than that of O atom. Hence H2S can dissociate to give H+ ions in aqueous solution.
Q4.Why hydride of oxygen is a liquid whereas hydride of sulphur is a gas?
Solution: The hydride of oxygen i.e H2O undergoes molecular association due to hydrogen
bonding and hence exists as a liquid. But H2S does not undergo H-bonding. As a result, it
exist as discrete molecules and hence in a gas.
Q5.Why does sulphur exhibit greater tendency for catenation than selenium?
Solution: As we move from sulphur to selenium, the atomic size increases and hence the
strength of E---E bond decreases. As a result S---S bond is much stronger than Se---Se bond
and consequently S shows greater tendency for catenation than selenium.
13
GROUP 17 ELEMENTS
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Fluorine, chlorine, bromine, iodine & astatine constitute 17th group of the periodic
table.
These are also known as halogens (salt producers).
The outer electronic configuration of halogens is ns2 np5.
Atomic and ionic radii increase from fluorine to iodine due to increase of number of
shells decrease in effective nuclear charge.
Ionisation enthalpy gradually decreases from fluorine to iodine due to increase in
atomic size.
Electron gain enthalpy of fluorine is less than that of chlorine. It is due to the small
size of fluorine & repulsion between newly added electron & electrons already present
in its small 2p orbital.
The decreasing order is: chlorine  fluorine  bromine  iodine.
Electronegativity decreases from fluorine to iodine.
Fluorine  chlorine  bromine  iodine.
Fluorine is the most electronegative element in the periodic table.
Fluorine is light yellow in colour, chlorine is light green, bromine is reddish brown &
iodine is violet solid.
The colour of halogens is due to absorption of visible light for excitation of outermost
electrons into higher energy levels.
Bond dissociation enthalpy of fluorine is smaller than that of chlorine is due to
electron-electron repulsion among the lone pair in fluorine molecules where they are
much to each other than in case of chlorine. The trend: Cl -Cl  Br-Br  I-I.
Except fluorine other halogens exhibit +1, +3, +5 &+7 oxidation states.
Halogens exhibit –1 or +1 in the ground state, +3 in the first excited state, +5 in the
second excited & +7 in the third excitation state.
All the halogens are highly reactive.
Fluorine is the strongest oxidizing halogen due to
 Low enthalpy of dissociation.
 High enthalpy of hydration due to small size
And the oxidising ability gradually decreases.
Compounds formed by halogens with electropositive metals are ionic.
Compounds of halogens with non-metals are covalent. Metals with low oxidation
states are ionic & high oxidation states are covalent.
Reactivity of halogens with hydrogen decrease from fluorine to iodine & the order of
stability is HF  HCl  HBr  HI.
The oxides of fluorine (O2F2&OF2) are called fluorides of oxygen because fluorine is
more electronegative than oxygen.
Chlorine, bromine & iodine form oxides in which oxidation states of halogens range
from +1 to +7.
14
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Chlorine oxides Cl2O,ClO2,Cl2O2 & Cl2O7 are highly reactive. Acidity increases as the
oxidation state of chlorine increases & strong oxidizing agents.
Bromine oxides Br2O,BrO2,BrO3 are least stable but powerful oxidizing agents.
Iodine oxides I2O4,I2O5 &I2O3 are insoluble solids.I2O5 is a good oxidizing agent and
is used in estimation of CO.
Halogens combine amongst themselves to form a number of compounds known as
inter halogens of the type XX, XX3, XX5 & XX7(where X=large size halogen &
X=smaller size halogen).
CHLORINE:
It is prepared by following methods:
Lab methods:
1.
MnO2 + 4HCl
MnCl2 + Cl2 + 2H2O
2.
4NaCl + MnO2 + 4H2SO4
MnCl2 + 2NaHSO4 + 2H2O + Cl2
3.
KMnO4 + 16HCl
2KCl + 2MnCl2 + 8H2O + 5Cl2
DEACONS PROCESS:
4HCl + O2 CuCl 2Cl2 + 2H2O

Electrolytic process: It is obtained by electrolysis of brine solution. Cl 2 is obtained at
anode.

Cl2 is greenish yellow gas with pungent odour.

Cl2 reacts with metals and non-metals to form chlorides.

With excess ammonia it gives N2 and NH4Cl.
8NH3 +3Cl2
6NH4Cl + N2
With excess chlorine NCl3 is (explosive) formed.
NH3 + 3Cl2
NCl3 + 3HCl

With cold and dilute alkalies Cl2 produces a mixture of chloride and Hypochlorite but
with hot and concentrated alkalies it gives chloride and chlorite.
2NaOH + Cl2
NaCl + NaOCl + H2O
6NaOH + 3Cl2
5Nacl + NaClO3 + 3H2O

With dry slaked lime it gives bleaching powder.
2Ca(OH)2 + 2Cl2
Ca(OH)2 + CaCl2 + 2H2O

It oxidises ferrous to ferric, sulphur dioxide to H2SO4 and iodine to iodic acid.

It is a powerful bleaching agent; bleaching action is due to oxidation.
Cl2 + H2O
2HCl + (O)
Coloured substance + (O)
colourless substance

By the action of concentrated H2SO4 and NaCl give HCl gas.
NaCl + H2SO4 420k
NaHSO4 + HCl

3: 1 ratio of concentrated HCl and HNO3 is known as aquaregia used for dissolving
noble metals like Au and Pt.
OXOACIDS OF HALOGENS:
15
Halic (I) acid
(Hypohalous
acid)
Halic (III)
acid
(Halous acid)
Halic (V) acid
(Halic acid)
Halic (VII)
acid
(Perhalic acid)
HOF
(Hypofluorous
acid)
-
HOCl
(Hypochlorous
acid)
HOClO
(chlorous acid)
HOBr
(Hypobromous
acid)
HOI
(Hypoiodous
acid)
-
-
-
HOIO2
(Iodic acid)
-
HOClO2
(Chlorous acid)
-
HOClO3
HOBrO3
HOIO3
(perchloric acid) (perbromic acid) (Periodic acid)
STRUCTURES OF OXOACIDS OF CHLORINE
Hypochlorous acid
Chlorous acid
16
Chloric acid
Perchloric acid
INTER HALOGEN COMPOUNDS:
Inter halogen compounds are prepared by the direct combination. Ex:ClF,ClF3,BrF5,IF7.

They are more reactive than halogens because X-X weaker than X-X bonds in
halogens (except F-F).

XX3 has bent ‘T’ shape XX5 compounds. Square pyramidal & IF7 has pentagonal
bipyramidal structure.
BrF3
17
TRY THESE INTEXT QUESTIONS
Q1.Name the halogen which does not exhibit positive oxidation state.
Solution: Fluorine being the most electronegative element does not show positive oxidation
states.
Q2.Iodine forms I3- but F2 does not form F3- ions. Why?
Solution: Due to the presence of vacant d-orbitals, I2 accepts electrons from I- ions to form I-3
ions, but F2 because of the absence of d-orbital does not accept electrons from F- ions to
form F-3 ions.
Q3. Arrange HClO4, HClO3, HClO2, HClO in order of acidic strength give reasons?
Solution: Acidic strength: HClO4>HClO3>HClO2>HCLO
Reason: stability of ClO-4>ClO-3>ClO-2>ClOQ4.Why interhalogens more reactive than halogens?
Solution: This is because the bond in the interhalogen (X-X') is weaker than X-X and X'-X'
bond in the halogens. This is due to less effective overlapping between orbitals of dissimilar
atoms than those between similar atoms.
Q5.Why HF acid is stored in wax coated glass bottles?
Solution: This is because HF does not attack wax but reacts with glass. It dissolves SiO2
present in glass forming hydrofluorosilicic acid.
SiO2 + 6HF
H2SiF6 + 2H2O
Q6.. Why Halogens are coloured?
Ans: They absorb visible portion of light and emit complementary
colours.





GROUP 18 ELEMENTS
He, Ne, Ar, Kr, & Xe constitute 18th group of the periodic table.
They have ns2np6 valence shell electronic configuration except He which has 1s2
Due to complete octet of outermost shell they have less tendency to form compound
.The commercial source of He i.e. natural gas
The first noble compound prepared by Neil Bartlett was XePtF6 by mixing PtF6 &
xenon
Xenon form compounds only with F2 &O2 due to their high
Electronegativity
Main commercial source of He is natural gas. He is second most abundant element in
the universe.
18
88Ra








226
86 Rn222 +2 He4
Xe + F2 673k, 1bar XeF2
Xe(g) + 3F2(g) 573k,60 – 70bar XeF6(s)
XeF6 + MF  M+ [XeF7]XeF6 + H2O  XeO3 + 6HF
Xe(g) + 2F2(g) 873k, 7bar XeF4(s)
XeF2 + PF5  [XeF+] [PF6]XeF6 + 3H2O  XeO3 + 6HF
XeF6 + H2O  XeO2 F2 + 4HF
TRY THESE INTEXT QUESTIONS
Q1.Noble gases have very low boiling points. Why?
Solution: Noble gases are monoatomic. Their atoms are held together by weak dispersion
forces and hence can be liquefied at very low temperatures. Therefore, they have low boiling
points.
Q2.Why are elements of group 18 known as noble gases?
Solution: The elements of group 18 have their valance shell orbitals completely filled. As a
results they react with only few elements (oxygen and fluorine) only under certain
conditions. Therefore they are called noble gases.
Q3.He and Ne do not form compounds with fluorine. Why?
Solution: He and Ne do not contain d-orbitals in their respective valance shells and hence
their electrons cannot be promoted to higher energy levels like that in He to form bonds.
Therefore He and Ne do not form compounds with fluorine.
Q4.Neon is generally used for warning signals. Why?
Solution: Neon lights are visible from long distances even in fog and mist and hence neon is
generally used for warning signals.
QUESTIONS AND ANSWERS
1) Give the common oxidation states of V-A group elements. Bismuth mainly exhibits +3
oxidation state but not +5 state .why?
Ans: the common oxidation state of V-A group elements are -3, +3 and +5 .Bismuth does
not exhibits +5 state due to inert pair effect.
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ELEMENT OXIDATION STATES WITH EXAMPLE:
NITROGEN
PHOSPHORUS
ARSENIC
ANTIMONY
BISMUTH
+3
NH3,
PH3
Ca3P2
AsH3
SbH3
,
+3
NF2,
N2O3
PCl3
+5
N2O5
AsCl3
SbCl3
BiF3
AsCl5
SbCl5
PCl5
2) Nitrogen is chemically inert, though it has high electronegativity, while
Phosphorus is highly reactive?
Ans: Nitrogen exists as N2 molecules .There is a triple bond between nitrogen
atoms (N
N).The bond Dissociation energy is very high (945.4 kJ
/mol) .So N2 is chemically inactive.
Phosphorus exists as P4 molecules, P4 molecule is tetrahedral in shape, but the
bond angle is only 60 degree. So there is a lot of strain within P4
molecule and the p-p bonds are easily cleaved (broken).
3) How do i) melting and boiling points
ii) Catenation vary in V-A group elements?
Ans: (i)Generally melting and boiling point increases from Nitrogen to
Antimony due to inert pair effect. (ii)The catenation ability decreases
from Nitrogen to Arsenic due to decrease in bond dissociation energy.
4) What is allotropy? Explain the phenomenon of allotropy in V-A group
elements.
Ans: Allotropy: the existence of an element in different physical forms with
same chemical properties is known as allotropy.
1.
Nitrogen exists in 2 allotropic crystalline states a)alpha nitrogen b)beta nitrogen
2.
Phosphorus exists in a variety of forms called white, red, scarlet, alpha black,
beta black, and violet etc,
The common forms are white and red phosphorus, white phosphorus is
more reactive than red phosphorus because white phosphorus exists as
discrete P4 molecules. In red phosphorus several p4molecules are linked
to form polymeric form
3. Red phosphorus is obtained by heating white phosphorus in absence of air at
523 k in presence of iodine
4. Arsenic exists in three allotropic forms they are grey, yellow and black forms
20
5. Antimony exists in three allotropic forms they are metallic, yellow and explosive
forms.
6. Bismuth does not exhibit allotropy
5) Explain why and how nitrogen differs from the rest of the elements in nitrogen
family?
Ans: nitrogen differs from the rest of the elements in nitrogen family due to its
1. smaller atomic size
2. higher electronegativity
3. lack of d-orbitals
4. ease of bond formation
1.
2.
3.
4.
5.
PROPERTIES OF NITROGEN:
nitrogen is a gas while others are solids
nitrogen is a diatomic molecule while others are tetratomic (Bismuth is
monoatomic).
nitrogen exhibits a large number of oxidation states ranging from -3 to +5,
while others exhibit -3 to +3 and +5 oxidation states.
nitrogen cannot form penta halides while others form
nitrogen can form hydrogen bonds ,while others cannot form
HYDRIDES:
6) Write briefly about the hydrides of V-A group elements?
Ans:
I. These elements form RH3 type covalent hydrides (R=N,P,As,Sb,.Bi).ammonia=NH3
phosphine-PH3,arsine-AsH3, stibine-SbH3, bismuthine-BiH3
II. Stability decreases and reducing power from NH3 to BiH3
III. They have pyramidal structures and bond angles decreases from NH3 to BiH3
IV. All are Lewis bases .basic nature decreases from PH3 to BiH3.
V. Volatility decreases from PH3 to BiH3.
7) Give the names of different hydrides of nitrogen with their formulae?
Ans: Hydrazoic acid (HN3)-acidic hydride
Ammonia (NH3), hydrazine (N2H4)—basic hydrides
8) Explain why ammonia has higher boiling point than
phosphine? Mention the shape of ammonia molecule?
21
Ans: Ammonia has higher boiling point than phosphine, because intermolecular
hydrogen bonds exist in ammonia.
The shape of ammonia molecule is pyramidal.
9)Give the shapes of Ncl3 and Pcl3 is pyramidal.
Ans: the shape of the Ncl3 or Pcl3 is pyramidal. The central atom undergoes sp3
hybridization.
10) Phosphorus forms pcl5 but nitrogen cannot form Ncl5 .why?
Ans: The electronic configuration of nitrogen in ground state is 1s22s2 2p3 .it has
no vacant D-sub shell nitrogen cannot form NCl5.
N=7
1s2
2s2
2p3
The electronic configuration of exited phosphorus atom is 1s2 2s2 2p6 3s1 3Pz1 3d1
Phosphorus undergoes sp3d hybridization and forms Pcl5 with five chlorine atoms.
The shape of Pcl5 molecule is trigonal bi-pyramidal.
P in ground state
P in exited state
3s1
3p3
3d1
11)What is laughing gas? Why is it so called? How is it prepared?
Ans: Nitrous oxide (N2O) is called laughing gas, because when inhaled it produces
hysterical laughter. It is prepared by gently heating ammonium nitrate.
NH4NO3
N2O + 2H2O.
12) What are:
a) Nitration mixture
b) Aquaregia?
Ans: a) A mixture of concentrated HNO3 and concentrated H2SO4 in 1:1 ratio is called
nitration mixture.
b) A mixture of concentrated HNO3 and concentrated HCL in 1: 3 ratio is called
aquaregia a good solvent for gold.
22
13) What are chalcogens? Why are they called so?
Ans: The first four elements (O, S, Se and Te) of VI-A group elements are called chalcogens,
because they are ore forming elements.
14) What is the atomicity of sulphur and ozone molecules?
Ans: the atomicity of sulphur is eight (S8) .the atomicity of ozone is three (O3).
15). Draw the shape of S8 and S6 allotropes of sulphur.
Ans:
S6 allotrope
S8 Puckered ring
16). Like O2, S2 is paramagnetic. Why?
Ans: Due to presence of unpaired electrons in antibonding molecular orbital. S 2 has
bonding similar to O2 in gaseous phase.
17) What structures of the rhombic sulphur and plastic sulphur?
Ans: Rhombic sulphur shows puckered ring or crown structure .plastic sulphur shows broken
chains sulphur molecules.
18) What is meant by transition temperature? What is transition temperature of sulphur?
Ans: The temperature where 2 crystalline allotropic forms exist in a state of equilibrium is
known as transition temperature. The transition temperature of sulphur is 95.5’C
19) Oxygen cannot form more than two bonds, where as sulphur can
form six bonds .why?
Ans: Oxygen has no‘d’ orbitals .It has only two unpaired electrons, while sulphur has
vacant‘d’ orbitals hence it can expand its octet.
20) Write briefly about the hydrides of VI-A group elements.
Ans: i) These elements form H2M type of covalent hydrides (M=O, S, Se, Te) H2O, H2Shydrogen sulphide, H2Se-Hydrogen selenide, H2Te-Hydrogen Telluride.
ii) Except H2O, the others are poisonous gases at room temperature. H2O is a liquid.
iii) Thermal stability decreases from H2O to H2Te.
iv) They have bent structures and bond angles decrease from H2O to H2Te.
v) Volatility decreases from H2S to H2Te as molecular weight increases.
vi) Reducing property increases from H2O to H2Po.
vii) They are weak acids and acidic strength increases from H2O to H2Te.
23
21) H2S is a gas at room temperature, while H2O is a liquid”’ explain.
Ans: There is hydrogen bonding between the molecules of H2O, so H2O is an associated
liquid.H2S is a gas at room temperature because there is no hydrogen bonding among H2S
molecules. The large size and lower electronegativity of sulphur atom does not permit
hydrogen bonding in H2S.
22) Which hydride of VI group elements is liquid at room temperature and why?
Ans: Water or hydrogen oxide is a liquid at room temperature. It exists as associated
molecules due to intermolecular hydrogen bonding.
23) Why H2S is more acidic than H2O and why it exists as gas?
Ans: Because H-S bond is weaker than H-O bond (conjugate base is HS- is weaker than OH-)
It does not exhibit inter molecular hydrogen bonding.
24) What is the hybridisation of sulphur in SO2 and in SO3 molecules? What are their bond
angles?
Ans: The hybridisation in both molecules is SP2. The bond angles are 119’ and 120’.
25) Give the order of strength of ‘ous’ and ‘ic’ acids of sulphur, selenium, and tellurium.
Ans: H2SO4> H2SeO3>H2TeO3.
H2SO4> H2SeO4> H2TeO4.
26) What is meant by tailing of mercury?
Ans: In the presence of ozone mercury sticks to the walls of glass tube due to the formation
of Hg2O.This is called tailing of mercury.
26) How does ozone act as a bleaching agent?
Ans: Ozone bleaches by oxidation
O3
O2 + [O]
Coloured vegetable organic matter + [o]
colourless vegetable organic matter.
Thus vegetable colours get decolourised by the action of ozone gas.
27) Give uses of ozone.
Ans: 1) It is used to detect double or triple bonds in organic compounds.
2) It is used for dry bleaching of oils, starch and wax.
3) Mixture of O3 and Cynogen (C2N2) is used as rocket fuel.
4) It is used in the sterilising of drinking water.
28) Which elements have high electron affinities in the periodic table? Compare the electron
affinities of F2 and Cl2.
Ans: Halogen elements have high electron affinities. (Order of electron affinity: F 2< Cl2).
Reason: Fluorine is very small in atomic size. So the electron repulsion is very high.
29) What is order of bond dissociation energy of halogens?
24
Ans: Cl2>Br2>F2>I2.
30) Noble gases have very low boiling point. Why?
Ans: They have no interatomic forces except weak dispersion forces, hence they are
liquefied at very low temperatures & have low boiling point.
31) Name the binary fluorides of Xe & how are they prepared?
Ans: XeF2, XeF4 &XeF6
They are prepared by the direct reaction of Xe with F2 under appropriate experimental
conditions.
673K, 1bar
Xe+F2------------XeF2
873K, 7bar
Xe+2F2----------XeF4
573K, 60-70bar
Xe+3F2----------XeF6
32) Complete the following
XeF2+PF5[XeF] + [PF6]XeF4+SbF5[XeF3] + [SbF6]XeF6+MFM+ [XeF7] (M=Na, K, Rb)
33) How are XeO3 & XeOF4 are prepared?
Ans: (a) Hydrolysis of XeF6 with water gives XeO3
XeF6+3H2OXeO3+6HF
(b)Partial hydrolysis of XeF6 gives oxyfluoride XeOF4
XeF6+H2OXeOF4+2HF
34) Which reaction prompted Bartlett to prepare first noble gas compound? Which
was the compound?
Ans: The following reaction prompted Bartlett to prepare first noble gas compound.
O2 (g) + PtF6(g)  O2+ [PtF6]-. The first I.E. of Xe and O2 and their size are
comparable. The compd was Xe[PtF6]
35) Write the steps involved in the preparation of H2SO4 by contact process.
Ans: Stages: 1) Burning of sulphur or sulphide ore in air. S + O2  SO2
ii) Conversion of SO2 to SO3
V2O5
2SO2(g) + O2 (g)  2SO3(g) + 196.6 KJ / mol
This reaction is reversible and highly exothermic. According to Le-Chatlier principle
low temperature is favoured. In practice an optimum temperature of 720K is use
25
and a pressure of 2 bar is employed.
iii) Absorption of SO3 in H2SO4 to give oleum.
SO3 + H2SO4  H2S2O7
oleum on dilution gives sulphuric acid
H2S2O7 +
H2SO4  2H2SO4.
36) Draw the structures of the following
(a) XeO3 (b) XeF4 & (c) XeF6
Ans:
MULTIPLE CHOICE QUESTIONS
Q1.What is the maximum covalency of sulphur?
(A) 2
(b) 4
(c) 6
(d) 8
Q2.Which of the following shows only negative oxidation states?
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Fluorine
Reason: Fluorine has no d orbitals in its valance shell so it cannot expand its octate.
Q3.Which of the following hydrogen halides is most volatile?
(a) HF
(b) HCl
(c) HBr
(d) HI
26
Q4.Put in Correct of acidic strength
(a) Cl2O7>SO2>P4O10
(b) CO2>N2O5>SO3
(c) Na2O>MgO>Al2O3
(d) K2O>CaO>MgO
Q5.Which of the following is planar?
(a) XeO4
(b) XeO3F
(c) XeO2F2
(d) XeF4
Q6.Which of the following is a typical metal?
(a) Phosphorous (b) Antimony
(c) Arsenic
(d) Bismuth
Q7.Which of the following is a diatomic gas?
(a) Bismuth
(b) Antimony
(c) Nitrogen
(d) Phosphorous
Q8.Which is the strongest acid?
(a) H2SO4
(b) HCl
(c) HClO4
(d) HNO3
Q9.Which of the following is the most basic oxide?
(a) SeO2
(b) Al2O3
(c) Sb2O3
(d) Bi2O3
Q10.Name the compound which has molecular nature in gas phase but ionic in solid
state?
(a) PCl5
(b) CCl4
(c) PCl3
(d) POCl3
Q11.In which of the following molecules all the bonds are not equal?
(a) NF3
(b)ClF3
(c) BF3
(d) AlF3
Q12.Which of the following oxides of nitrogen is thermally more stable?
(a) N2O5
(b) N2O
(c) NO
(d) N2O3
Q13.How many hydrogen atoms are attached to phosphorous atom in hypo
phosphorous acid?
(a) 0
(b) 2
(c) 1
(d) 3
27
Q14.Number of sigma bonds in P4O10 is:
(a) 6
(b) 16
(c) 20
(d) 7
Q15.Which of the following halides is most acidic?
(a) PCl3
(b) SbCl3
(c) BiCl3
(d) CCl4
Q16.Which of the following has pπ-dπ bonding?
(a)SO32(b) SO42(c)S2O42(d) HSO4Q17.Water is oxidized to oxygen by:
(a)ClO2
(b) KMnO4
(c)H2O2
(d) Fluorine
Q18.Which of the following are in correct order of electron gain enthalpy (with negative
sign) of given atomic species?
(a)F<Cl<O<S
(b) S<O<Cl<F
(c)O<S<F<Cl
(d) Cl<F<S<O
Q19.How many types of F-S-F bonds are present in SF4?
(a)2
(b) 4
(c)3
(d) 5
Q20.Which of the following oxides is most acidic?
(a)N2O5
(b) P2O5
(c)As2O5
(d) Sb2O5
ANSWERS
1. C
2. D
3. B
4. A
5. D
6. D
7. C
8. C
9. D
10. A
11. B
12. C
13. B
28
14.
15.
16.
17.
18.
19.
20.
B
A
B
D
D
A
A
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