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Chapter TWO
Introduction
Chemical formula
Chemical compound : It is a chemical substance that is formed by the combination
of the atoms of two or more elements and it can be represented by a
chemical
formula
The ionic compound is formed by the combination of +ve and –ve ions , while
the covalent compound is formed by the combination of atoms of elements with each
other.
NaCl
H 2O
Ionic
covalent
Chemical (molecular) formula : It is a simple symbolic formula shows the type and
the number of atoms or ions in the molecule of a compound .
Emperical formula : It is the formula which represents the simplest composition
percent of elements in a compound .
C6H12O6
CH2O
Molecular
Emperical
How can one write a chemical formula ?
1) Chemical symbols of elements :
Chemists use symbols which easily express the elements .
Element
Symbol
Element
Symbol
Element
Symbol
Hydrogen
H
Lithium
Li
Aluminium
Al
Carbon
C
Calcium
Ca
Zinc
Zn
Nitrogen
N
Sodium*
Na
Iron *
Fe
Oxygen
O
Chlorine
Cl
Copper
Cu
Sulphur
S
Magnesium
Mg
Silicon
Si
Phosphorous
P
Lead *
Pb
Silver *
Ag
Potassium *
K
Bromine
Br
Gold *
Au
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2) Valency of the elements : It is the number of electrons lost , gained or shared
by the atom in a chemical reaction .
Atoms in free state are unstable so they combine to form stable element or
stable compound .
Element : Combination of two or more similar atoms
Hydrogen H2 , Oxygen O2 , Chlorine Cl2 , ……………………..
Compound : Combination of two or more different atoms
Water H2O , Carbon dioxide CO2 , …………………………..
Note : Noble gases as Helium He , Neon Ne , Argon Ar , ……. exist as free atoms
because they are stable so do not lose , gain or share electrons , i.e do not
make chemical reactions .
* Elements are classified into :
H ,, Li
Li ,, Cl
Cl ,, Br
Br , I , Na , K , ……………………………………
Monovalent : H
Divalent :: Mg
Mg ,, Ca
Ca ,, O
O ,, SS ,, Cu
Cu ,, Zn
Zn ,, Fe
Fe ,, ……………………………….
……………………………….
Divalent
Trivalent :: Al
Al ,, N
N ,, PP ,, Fe
Fe ,, ……………………………………
……………………………………
Trivalent
Tetravalent : C , Si , ……………………
Tetravalent : C , Si , ………………………………….
3) Radicals : Group of atoms combined together and behave like single atom .
They carry either + ve or – ve charge so can not exist in free state .
Monovalent : Hydroxide OH-- , Nitrite NO2 -- , Nitrate NO3 -- , Bicarbonate
HCO3 -- , Ammonium ion NH4 + , ……………
Divalent : Carbonate CO3 –2 , Sulphate SO4 –2 , ……………………..
Trivalent : Phosphate PO4 –3
On writing a chemical formula of a compound , the valency of the two radicals
are exchanged putting these numbers below the symbol of the radical to the
right .
Examples
1) The chemical formula of Calcium phosphate :
2
Calcium
Ca
2
Formula
Valency
The formula is
Ca3(PO4)2
2)The chemical formula of Sodium carbonate :
Sodium
Formula
Na
Valency
1
The formula is
Phosphate
PO4
3
Carbonate
CO3
2
Na2CO3
Apply the same steps on the following chemical compounds :
Sodium chloride
Calcium chloride
Potassium sulphate
Sodium hydroxide
Sodium bicarbonate
Aluminium sulphate
NaCl
CaCl2
K2SO4
NaOH
NaHCO3
Al2(SO4)3
Classification of elements
The long form periodic table
The long form periodic table
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Elements are arranged in an increasing order according
to their atomic numbers , where each element has one electron more than the
element before it .
This is based on Aufbau principle .
The f-block is placed below the table so that the long form periodic table not
be too wide and be easy to study .
Description of the long form periodic table :
1-
It consists of 7 horizontal periods and 18 vertical groups .
2-
The elements are arranged in increasing order according to their atomic no .
Example : 1H 2He 3Li
4Be
5B
6C
7N
8O
9F
10Ne
3-
Each element in the same period increases by one electron than that before
it till the last element in the period ( noble gas )
4-
Each new period begins by filling a new energy level of larger (n) value
by
one electron .
5-
Elements of the same vertical group have the same number of electrons
in
the last energy sublevel , so they have similar properties .



S-block elements
They are placed in the left hand block .
The outermost electrons occupy the s-sublevel
Consists of two groups : ( s-sublevel is filled with 2 e- )
- group IA has the configuration ns1 ( 11Na )
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- group IIA has the configuration ns2 ( 20Ca )



-



-
Elements of s- and pBlocks are known as
Representative ( main
group) elements .
p-block elements
They are placed in the right hand block
The outermost electrons occupy the p-sublevel
Consists of six groups : ( p-sublevel is filled with 6 e- )
group IIIA has the configuration ns2 np1 ( 13Al )
group IVA has the configuration ns2 np2 ( 6C )
group VA , VIA , VIIA ( 7N , 8O , 9F )
group Zero has the configuration ns2 np6 ( 10Ne )
d-block elements
They are placed in the middle block
The outermost electrons occupy the d-sublevel
Consists of 10 vertical columns: ( d-sublevel is
Elements of d-block are
filled with 10 e- )
Known as Transition
7 columns belong to B-groups ( IB,IIB,IIIB,…..) elements
3 columns belong to group VIII
The d-block is divided into 3 horizontal series :





1st transition series
Includes elements in which sublevel 3d is filled successively ,
and they are placed in the 4th period .
2nd transition series
Includes elements in which sublevel 4d is filled successively ,
and they are placed in the 5th period .
3rd transition series
Includes elements in which sublevel 5d is filled successively ,
and they are placed in the 6th period .
f-block elements
The outermost electrons occupy the f-sublevel
Consists of 14 vertical columns : ( f-sublevel is
filled with 14 e- )
Elements of f-block are
Known as Inner
transition elements
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The f-block is divided into 2 horizontal series :

Lanthanide series
Includes elements in which sublevel 4f is filled successively ,
and they are placed in the 6th period .
They have the electronic configuration 54[Xe] 6s2 5d1 4f1-14
They are oxides and known as “Rare earths”

Actinide series
Includes elements in which sublevel 5f is filled successively ,
and they are placed in the 7th period .
They have the electronic configuration 86[Rn] 7s2 6d1 5f1-14
They are radioactive elements .
Position of elements
1)
2)
3)
To determine the position of an element , you have to define :
Its block : The outermost sublevel ( s , p , d or f )
Its period : The n value of the outermost energy level
Its group : The number of electrons in the outermost energy level
19X
: 1s2 2s2 2p6 3s2 3p6 4s1
s-block
period 4 ( highest n = 4 )
group 1A ( the last electron number = 1 )
16Y
: 1s2 2s2 2p6 3s2 3p4
p-block
period 3
group 6A
31Z
: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1
p-block
period 4
group 3A
Trends and periodicity of properties in
the long form periodic table
1.
The atomic radius
“ It’s half the distance between the nuclei of two similar atoms
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in a diatomic molecule “
Diatomic molecules of similar atoms as H2 , Cl2 , O2 , Br2 , F2 , ……………………
The bond length
“ It’s the distance between the nuclei of two bonded similar or
different atoms “
Bond length
Bond length
atomic radius
Atomic radius = Bond length / 2 ( Bonded atoms are similar )
G.R for : It is incorrect to define the atomic radius as the distance from the
nucleus to the farthest electron
*Because it is impossible to determine the exact position of any electron
around the nucleus .
Example : If the bond length in the chlorine molecule Cl—Cl is 1.98 Ao and the
bond length between carbon and chlorine atoms C—Cl is 1.76 Ao .
Calculate the atomic radius of carbon .
* atomic radius of Cl atom = 1.98 / 2 = 0.99 Ao
atomic radius of carbon atom = 1.76 – 0.99 = 0.77 Ao
Example : If the bond length in the chlorine molecule Cl2 is 1.98 Ao and the bond
length in hydrogen chloride molecule HCl is 1.29 Ao . Calculate the
bond length in hydrogen molecule H2 .
* atomic radius of Cl atom = 1.98 / 2 = 0.99 Ao
atomic radius of hydrogen atom = 1.29 – 0.99 = 0.30 Ao
bond length of H2 molecule = 2 X 0.30 = 0.6 Ao
Trend of atomic radius in the periodic table
The atomic radius decreases through the period
(from left to right) and increases through the
group (from up to down) .
period
Atomic radius
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*In the period (from left to right), the atomic
radius decreases due to increase in nuclear
group
+
charge (number of ve protons) which attract the outermost electrons more strongly.
Example : atomic radius of 12Mg is greater than that of 17Cl
*In the group (from up to down), the atomic radius increases due to :
1increased number energy levels
2increased number of electrons which cause repulsion and thus the atom
takes
larger volume in space .
Example : atomic radius of 19K is greater than that of 11Na
The cationic radius is always smaller than the atomic radius of Metal atoms
due to increased nuclear charge in the cation which attract the remaining
electrons more strongly and decrease the radius .
Example : 11Na ( 2,8,1 ) lose one electron 11Na+ ( 2,8 ) cation
protons = 11
electrons = 11
(greater size)
protons = 11
electrons = 10
(smaller size)
The anionic radius is always greater than the atomic radius of Nonmetal
atoms due to increased number of electrons without increase in the nuclear
charge , so repulsion increases between electrons and atomic radius increases.
Example : 17Cl ( 2,8,7 ) gain one electron 17Cl- ( 2,8,8 ) anion
protons = 17
electrons = 17
(smaller size)
2.
protons = 17
electrons = 18
(greater size)
The ionization energy
“ It is the amount of energy required to remove the most
loosely bound electron completely from an atom “
M+ + e-
1st ionization energy
M+ lose one electron M++ + e-
2nd ionization energy
M
lose one electron
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Trend of ionization energy in the periodic table
The ionization energy increases through the
period (from left to right) and decreases through
the group (from up to down)
period
Ionization energy
*In the period , the ionization energy increases
group
from left to right because the atomic radius decreases and the valence electrons
become more attracted to the nucleus so more energy is required to remove the
electron .
*In the group , the ionization energy decreases from up to down because the
atomic radius increases and the outermost electrons become less bound to the
nucleus and need less energy to be removed from the nucleus .
The ionization energy is inversely proportional to atomic radius
Give reason for : In noble gases , the first ionization energy is much greater than
the second ionization energy .
Because it is difficult to remove an electron from completely filled energy level
which cause the stability of noble gases .
10Ne
1st
10Ne
2,8
+
2nd
10Ne
2,7
++
2,6
completely filled
(stable)
1 st > 2nd
Give reason for : The 1st , 2nd and 3rd ionization energies for magnesium metal
12Mg are 737 , 1450 and 7730 kj respectively .
Mg
1st
Mg+
2,8,2
737 kj
2,8,1
2nd
1450 kj
Mg++
2,8
3rd
7730 kj
Mg+++
2,7

The 2nd ionization energy is higher than the 1st because when the first
electron
is removed , the remaining electrons become more strongly bound to the nucleus
and need more energy .

The 3rd ionization energy is much higher than the 2nd because the electron
is
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removed from completely filled energy level .
3.
Electronegativity
“ It is the tendency of an atom to attract the electrons of the
chemical bond to itself “
period
*In the period (from left to right),
the electronegativity increases
as the atomic radius decreases
because the nuclear charge inc.
and the attraction of bonded eincreases .
Electonegativity
group
Fluorine 9F is the highest electronegative element .

In the group (from up to down), the electronegativity decreases because
the
nuclear charge decreases and the attraction of bonded electrons decreases .
Cesium Cs is the lowest electronegative element .
The electronegativity is inversely proportional to atomic radius
Each element has a certain electrongativity value as Cl=3 , Na=0.9 , H=2.1
and so on . The difference in electronegativity between the bonded atoms may
determine the type of bond between them .
Electronegativity difference > 1.7
Ionic bond
~
~
< 1.7
Polar covalent bond
~
~
= zero
Pure covalent bond
Example : NaCl
Cl2
HCl
E.D = 3 - 0.9 = 2.1
E.D = 3 – 3 = zero
E.D = 3 – 2.1 = 0.9
Ionic
pure covalent
polar covalent
Oxidation number
“It is the number that refers to the electric charge ( +ve or –ve )
that an atom or an ion would have in a covalent or ionic compound “
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The oxidation number is the valency of the element
preceeded by +ve or -ve charge
1A
H
Li
Na
K
+1
2A
Be
Mg
Ca
+2
3A
4A
5A
6A
7A
zero
B
Al
C
Si
N
P
O
S
F
Cl
He
Ne
Ar
+3
+4
-3
-2
-1
Zero
Exceptions
1- Oxidation number of H is –1 in metal hydrides NaH
2- Oxidation number of O is –1 in peroxides
Na2O2 , H2O2
Oxidation number of O is –1/2 in superoxides KO2
To calculate the oxidation number :
*The oxidation number of all elements equals Zero
Al , Fe, Cu , Cl2 O2, O3 , P4 , S8 , …………………………………
*The sum of oxidation numbers of all atoms in a compound equal Zero
Na2CO3
2Na + C + 3O = zero
(+1x2) + (+4) + (-2x3) = zero
*The sum of oxidation numbers of all atoms in a radical equal te charge of the
radical
[NH4]+1
N + 4H = +1
-3 + (+1x4) = +1
Examples
1- Calculate the oxidation number of Mn in KMnO4
K + Mn + 4O = zero
+1 + Mn + (-2x4) = zero
so, Mn = +7
2-
Calculate the oxidation number of N in NO3 –1
N + 3O = -1
N + (-2x3) = -1
so, N = +5
Oxidation : “ It is the process of losing electrons and increase in +ve charge”
Fe+2
oxidation
Fe+3 + e11
Reduction : “ It is the process of gaining electrons and decrease in +ve charge”
Cl2 + 2e-
reduction
2Cl-1
In the following reaction :
K2Cr2O7 + FeCl2 + HCl
KCl + CrCl3 + FeCl3 + H2O
Which is oxidized and which is reduced : Cr or Fe ?
K2Cr2O7
Cr = +6
CrCl3
FeCl2
FeCl3
Cr = +3
Fe = +2
Fe = +3
Reduction
Oxidation
Cr is called Oxidizing agent
Fe is called Reducing agent
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