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Tuesday, 6 October 2015

FSc Notes Chemistry Part 1 Chapter 6 Chemical Bonding Lecture 1

FSc Notes Chemistry Part 1 Chapter 6 Chemical Bonding Lecture 1


Chemical Bond:


The force of attraction between two atoms or ions that hold them together in a unit is called chemical bond. Or
The force which keeps the atoms of a compound together is known as chemical bond.
Actually chemical bond is the main factor, which makes molecules and compounds. By the interaction of outer electrons, greater forces of attraction are developed b/w two atoms. This force of attraction is known as chemical bond. The process of formation of chemical bond is known as chemical bonding.

Why Atoms Form Chemical Bond?
Except H and He, all other atoms tend to get eight electrons in their outer most (valence) shell. This is known as “octet rule ". As H and He have only K shell, which need only two electrons to complete the valence shell (K). This is called duplet rule. When H and He get a electrons in their valence shell and all other atoms get 8 electrons in their valence shell, they get stabilized. From the above discussion it is clear, that the atoms either transfer or share their electrons to get 8 electrons (except H and He) in their valence shell. Due to the transference or sharing of electrons, chemical bonds are formed b/w the atoms. So we can say that atoms form bond for gaining stability.

Energetic of Bond formation:
Bond formation b/w the atoms, leads t a decrease in energy due to which both bonding atoms get stabilized. Whenever a chemical bond forms B/W two atoms, then two types of forces operate. One is the forces of attractions, which brings the binding atoms close to each other and thus decrease the potential energy of the system. The other force is the repulsion which tends to push the atoms apart and thus increase the potential energy of the system. If the amount of potential energy decreased during attraction is greater than the amount of the P.E increased repulsions, then chemical bond will be formed b/w the atom, otherwise no bond will be formed.

Periodicity & Periodic Properties:
The repetition of a property after equal interval is known as periodicity and such a property which repeats itself after equal interval is known as periodic property. Some important periodic properties are,
Atomic size (atomic radii, ionic radii and covalent radii)

  1. Ionization potential
  2. Electronaffinity
  3. Electronegativity
  4. Electropositivity

They are explained as:

Atomic size
Atomic size is expressed in terms of atomic radius, ionic radius and covalent radius.

Atomic radius
Half of the distance between the nuclei of two adjustment atoms, is known as atomic radius. Thus in simple words we can say that, atomic radius is actually the distance between the nucleus and valence shall of an atom. The SI units for atomic radius are: nano meter and picometre.


In Group: From top to bottom in any group of the periodic table, the atomic radius of atoms increases.
Reason: As we go down the group, an extra shell is constitutively added in the atoms. Thus the valence shell moves away from the nucleus and we say that atomic radius increases.
In Period: from left to right in any period of the periodic table, atomic radius of the atoms decreases.
Reason:
From left to right in any period of the periodic table, the atomic radius decreases because an electron is successively added in the same shell. It means that in a particular period of the periodic table, same valence shell is possessed by all the elements of that period. From left to right the no. of electrons increases that valence shell; hence no. of protons in the nucleus also increases. Thus the valence shell is attracted by the nucleus due to increased nuclear charge and thus the atomic radius decreases

 

Ionic radius:

The distance between the valence electrons and nucleus of an ion is known as ionic radius. As we have two types of ions i.e. caption ( + ve ion ) and anion ( -ve ion ). Thus we can say that, ionic radius may be a cationic radius or anionic radius.

For example, Na atom has one electron in its valence shell. To get eight electrons in its valence shell ( octet rule ), Na atom loses its single valence shell thus producing a cation ( ion ) i.e. Na+
Na (atom ) ------------------ Na+ ( ion i.e. cation )
K=2 l= b m=1 k=2 L=8

Ionic radius of a cation i.e. cationic radius is always smaller than the atomic radius of its corresponding atom. It is because of with ejection of valence electrons, the valence shell is lost and thus the size.



Shielding Effect:

The repulsion offered by the core electrons upon the valance electrons, away from the nucleus, is
Known as shielding effect.

Core electrons are those which le beneath the valence electrons, decreases and hence we say that the ionic radius of caution is less than atomic radius of its corresponding atom. Due to increased nuclear charge, shielding effect decreases and the 2nd last shell of the atoms, which is the valence shell of cation, comes close to the nucleus and thus we say that the ionic radius of a cation is less than the at radius of the corresponding atom.

Similarly chlorine atom has of electrons in its valence shell. It needs only one electron to complete its valence shell. Thus it gains the electron and forms a negative ion (anion) i.e. 17 Cl (atom) + ve- ------------------Cl- (ion i.e. anion)
K=2 L=8 M=f K=2 L=8 M=8

The anionic radius of an anion is always larger than the atomic radius of its corresponding atom. It is because of the fact that when the atom gains one or more electrons in its valence shell to form an anion, the nuclear charge doesn’t increase but remains the same. Thus the repulsive force of the core electrons upon the valence electrons increases. I.e. shielding effect increases and thus the valence shell of an anion moves away from the nucleus. That is why the ionic radius of an anion is always greater than the atomic radius of its corresponding atom.

Covalent Radius:

Covalent radius of an element can be defined as:
Half of the single bond length between two similar covalently bonded atoms in a molecules. In simple words we can say that if two similar atoms form a covalent bond among them, then the covalent radius of that element ( whose atoms have ) bonded together will be equal to half of the bond length of the molecule.

Bond length is actually the distance b/w the nuclei of two bonded (covalently) atoms. Let’s consider the example of H2 molecule. In H2 molecule, two hydrogen atoms (H) are bonded together by means of a single covalent bond.

The bond length of H2 molecule is 75.4 nm, ie Therefore, the covalent radius of the hydrogen element will be: 75.4/2====== 37.7nm

The covalent radius of an atom is bonded together by means of a single covalent.
It can be explained by considering an example,

Let’s consider CH3CL (Methyl Chloride molecular). Here in CH3-CL, the experimentally determined bond length of C- CL bond is 176.7 pm

If we know the covalent radius of either "C" or “CL ", them the covalent radius of another atom can be calculated easily.

As we know that the covalent radius of Cl is 99.4 pm (Picometre), then the covalent radius of "C" can be calculated by subtracting the covalent radius of Cl ( i.e. 99.4 Pm ) from the C-Cl bond length ( ie 176.7 Pm ) i.e. Covalent Radius of C = 176.7-99.4= 77.3 to find the bond length of the molecule in which two atoms ( different similar ) are bond together by means a single covalent bond.

Lets explain it with the help of an example.

Silicon Carbide or Carborundum ( Si – C ) is a molecule having silicon ( Si) and carbon ( C ) atoms bonded together by means of a single covalent bond. If we know the covalent radii of both C & Si then by adding them we can get the bond length of Si – C.

As in C-C bond distance or length is 0.154nm, so the covalent radius of each C atom will be:  0.154 = 0.077nm.

On the other hand, in Si-Si, bond distance is 0.236nm, so the covalent radius of Si atom will be : 0.236/2 = 0.118nm.

Thus the bond length or bond distance of Si-C (carborundum) will be: 0.077 + 0.118 = 0.195nm which is in close agreement with the experimentally determined Si- C bond length i.e. 0.1993nm.

It is important to note that the above mentioned methods is not true for all molecules, i.e. in most of the molecule, the bond length b/w unlike atoms is not equal to the sum of the covalent radii of the bonded atoms.
There are several factors for this fact for example some factors are: multiple bond ( double or triple covalent Bond) ionic charade ( polarity ), geometry of molecule etc.

Ionization Energy or Ionization Potential:

The minimum amount of energy required to eject an electron from the valence shell of a gaseous atom is known as I.E or I.P of the element of that atom.
It is expressed in the unit of KJ/ mole.

Types:
Depending upon the no. of electrons present in the valence shell of an atom of an element, its ionization energy may be 1st I.E, and so on.

1st Ionization Energy:
The minimum amount of energy required to eject the 1st electron from the valence shell of a neutral atom is known as its 1st I.E

2nd Ionization Energy:
The minimum amount of energy required to eject a second electron from the valence shell of a uni-positive ion (cation) is known as its 2nd I.E. Lets consider magnesium element the atomic No. of Mg is 12. Hence its electronic configuration will be:
12Mg = 1S2, 2S2, 2P6, 3S2
When one mole of Mg atoms eject one electron each, from their valence shell, 738 Kj of energy is absorbed:
So 1st Ionization Energy of Mg is 738 Kj/ mole i.e.
Mg----------------Mg+1 + 1e- ( I.E )1st = 738 KJ / mole
K=2 L=8 M=2 K=2 L=8 M=1
When these uni-positive magnesium ions ( Mg+1 ) eject their 2nd valence electron 1451 KJ / mole of energy is absorbed.
Thus the 2nd I.E of Mg is 1451 Kj / mole i.ee
Mg+1----------Mg+2 + 1e- ( I.E ) 2nd = 1451KJ / Mole
K=2 L=8 M=1 K=2 L=8

2nd Ionization Energy is always greater than 1st I.E.
It is because of the fact that, in case of neutral atom, the atomic size is large and hence the valence electrons are away from the nucleus. Thus the removal of 1st electron requires less energy. But after valence shell of a neutral atom, the –ve charge decreases and hence the nuclear charge dominates thus bringing the remaining valence electron or electrons closer to the nucleus.
Therefore, the removal of a 2nd electron will require greater amount of energy. Hence 2nd Ionization Energy is greater than 1st Ionization Energy. Similarly 3rd Ionization Energy will be greater then 2nd Ionization Energy and so on.

Factors Affecting Ionization Energy:
Ionization Energy of atoms is affected by the following factors.

  1. Atomic radius of atom: Ionization Energy inversely proportional to atomic radius.
  2. Nuclear charge (no of protons of the atom): Ionization Energy directly proportional to nuclear charge.
  3. Shielding effect: Ionization Energy inversely proportional to Shielding effect.
  4. Nature of orbital: for different orbital’s the order of i.e is S> P> d > f.


Trends of Ionization Energy in Periodic Table:

In group:
From top to bottom in any group of the periodic table: the I.E of elements decreases.
Reason:
As we go down the group, the atomic size of elements increases because of the successive addition of shell, thus the valence electrons get away from the nucleus; therefore less energy is required to eject them.

In period:
From left to right, in any period of the periodic table, the I.E of element increases.
Reason:
As we go from left to right in any period of the periodic table, an electron is successively added in same shell remains the same throughout the period. Hence greater no of electrons and protons will bring the valence shell closer to nucleus i.e. atomic size deceases from left to right in any of the periodic table, hence the removal of valence electron will require greater energy.

  • Elements with low Ionization Energy are metals.
  • Elements with high Ionization Energy are non-metals.
  • Elements with intermediate Ionization Energy are metalloids.


Written by: Asad Hussain

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