Covalent and Ionic Bonds


 
 
Concept Explanation
 

Covalent and Ionic Bonds

Covalent Bonds Formation :-

The bonds which are formed by the sharing of an electron pair between the atoms either same or different atoms are known as covalent bonds.

Formation of Carbon Compounds :-

Atomic number of carbon (C) is 6. So, its electronic configuration= 2, 4. Thus, there are 4 electrons in its outermost shell and its octet can be completed by the following two ways:

(i) It could lose 4 electrons and form large C^4^+ cation. But a large amount of energy required to remove 4 electrons leaving behind a carbon cation with 6 protons in  nucleus holding on just two electrons together, which is not possible.

(ii) It could gain 4 electrons and can form large C^4^-anion. But for a nucleus have 6 protons, it would be difficult to hold on 10 electrons, that is  4 extra electrons.

Therefore, in order to overcome this problem, carbon shares its valence electrons with other atoms of carbon or with atoms of other elements. These shared electrons belong to the outermost shells of both the atoms and in this way, both atoms attain the noble gas configuration. This type of bonding is called as Covalent bonding.

Some examples depicting covalent bonding:-

Formation of Hydrogen Molecule large (H_2) :

Atomic number of H = 1  ,Electronic configuration = 1.

It has 1 electron in its K-shell and needs 1 one more electron  to fill the K-shell completely. Two H-atoms share each of their electrons  to form an large H_2 molecule and thus, each H-atom attains the nearest noble gas configuration, i.e. configuration of Helium gas (having two electrons in its K-shell). Valence electrons are depicted by using crosses. The shared pair of electrons constitutes a single bond , which is represented by a line between the two H atoms.

Formation of Chlorine Molecule large (Cl_2) :

Atomic number of Cl =17,   Electronic configuration = 2, 8, 7.

It has 7 electrons in its outermost shell and thus requires 1 more electron to fulfill its outermost shell configuration. This is achieved by sharing 1 electron with another , forming a chlorine diatomic molecule large (Cl_2).

Formation of Oxygen Molecule large (O_2)

Atomic number of O = 8,    Electronic configuration= 2, 6.

It has 6 electrons in its outermost shell thus require 2 electrons to complete of its octet for attaining noble gas configuration. This is achieved by sharing 2 electrons with another oxygen atom. The two electrons contributed by each oxygen atom give rise to two shared pairs of electrons.

Here, a double bond is formed between two oxygen atoms there by forming an oxygen molecule. The above figure represents the sharing of 4 electrons.

Formation of Nitrogen Molecule large (N_2) :

Atomic number of N =  7,        Electronic configuration = 2, 5.

It needs 3 more electrons to attain noble gas configuration. Thus, 2 nitrogen atoms share 3 each of their electron, forming a triple bond between two nitrogen atoms thereby forming nitrogen molecule.

Formation of Methane large (CH_4) : Atomic number of C= 6     Electronic configuration = 2,4

Atomic number of H= 1     Electronic configuration =1

In the formation of a methane molecule, one carbon atom shares its 4 electrons with four hydrogen atoms (one electron of each hydrogen atom). It shows carbon is tetravalent because it possesses 4 valence electrons and hydrogen is monovalent because it has only 1 valence electron.

Note: Methane is a carbon compound which is also called marsh gas. It is used as a fuel and a major component of CNG (Compressed Natural Gas) and biogas. It is one of the simplest compounds formed by carbon.

Formation of Ammonia large (NH_3) :

Atomic number of N = 7,         Electronic configuration = 2, 5.          

Atomic number of H = 1         Electronic configuration = 1. 

To attain the electronic configuration of the nearest noble gas, nitrogen needs 3 electrons and hydrogen needs 1 electron. When a molecule of ammonia is to be formed, one atom of nitrogen shares its three electrons, one with each of the three atoms of hydrogen.

Note : Ammonia gas large (NH_3) can be used as refrigerant.

Formation of Water large (H_2O)

Atomic number of O = 8        Electronic configuration = 2, 6. 

Atomic number of H = 1       Electronic configuration = 1. 

To attain the stable electronic configuration of the nearest noble gas, hydrogen needs 1 electron and oxygen needs 2 electrons. In case of a water molecule, each of the two hydrogen atoms share an electron pair with the oxygen atom such that hydrogen acquires a duplet configuration and oxygen an octet, resulting in the formation of two single covalent bonds.

Formation of Carbon Dioxide large (CO_2) :

Atomic Number of C = 6       Electronic configuration =2, 4.

Atomic number of O = 8       Electronic configuration 2, 6. 

To attain the stable electronic configuration, carbon needs 4 electrons, while oxygen needs 2 electrons. So, inlarge (CO_2) , each of oxygen atom share two electrons from carbon. Thus, oxygen and carbon both attain octet.

Formation of Sulphur Molecule large (S_8):

Atomic number of S = 16     Electronic configuration = 2, 8, 6.

To attain the electronic configuration of the nearest noble each sulphur needs 2 electrons.

Ionic Bond Formation :-

Ionic bond is formed by transfer of electrons from one atom to another.

In this one atoms can donate electrons to achieve the inert gas electron configuration and the other atom need electrons to achieve the inert gas configuration.

Metals having 1, 2, 3 electrons in their shell donate electrons.

Metals having 4, 5, 6, 7 electrons in their outer shell accept electrons.

Formation of Sodium Chloride (NaCl) :-

Sodium chloride is formed when sodium atoms interact with chlorine atoms. This makes sodium slightly  positive and chlorine slightly negative.

Sodium becomes Na^+and Chlorine becomes Cl^-

As, it can be seen that positive charge of the sodium is due to the donation of electron or we can say that sodium get reduced in the process.

Chlorine, on the other hand gain electron pair which is denoted  negative sign. As there is gain of electrons therefore we can say that chlorine get oxidized in the process.

Formation of Lithium Chloride (LiCl) :-

Lithium chloride is formed when Lithium atom interacts with chlorine atom. Lithium becomes Li^+ by losing 1 electron, and chlorine becomes Cl^- by accepting electrons. So, the ionic compound, LiCl is formed.Hence, the ionic bond formation of lithium chloride involves transfer of electrons form lithium to chlorine and forming Li^+ and Cl^-in LiCl.

Formation of Sodium Fluoride (NaF) :- Sodium Fluoride is formed of Sodium and fluorine atom. In this reaction, the sodium atom loses its single valence electron to the fluorine atom. So, we can say that in formation of Sodium Fluoride Sodium get reduced and fluorine get oxidized.
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Sample Questions
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Question : 1

Carbon(C) atom forms which type of bonds ?

Right Option : B
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Question : 2

In the formation of NaF, the atom which forms cation is _________________ .

Right Option : A
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Explanation
Question : 3

The property of catenation is shown by which of the following atom?

Right Option : A
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