The atomic number of hydrogen is 1. The atomic number of chlorine is 17. The oribitals corresponding to the third energy level could hold u to 8 electrons.
1.What kind of bond is formed between hydrogen and chlorine atoms?
2. Describe the informatin of this bond and the total number of electrons in the oribitals of each energy level.
The atomic number of hydrogen is 1. Its lone electron is in the first (K) shell.
The atomic number of chlorine is 17. So, its electons are distributed as:
K rarr 2, L rarr 8, M rarr 7
The oribital corresponding to the third energy level could hold up to 8 electrons. From the octet rule, it follows that during bond formation with hydrogen, chlorine would like to have 8 electrons in its outermost (M) shell. This can be achieved in two ways:
1. By gaining an electron from H, to form Cl^-, H becomes H^+, -electrovalent bond results.
2. By sharing an electron with hydrogen, hydrogen is also stabilised through attainment of next inert gas (helium)-like electronic configuration. In this case, a covalent bond results.
Again, the Pauling scale electronegativity values of hydrogen and chlorine are 2.1 and 3.1 respectively.
Difference in electronegativity: 3.1-2.1=1.0.
Typically, an ionic bond forms when the difference between the electronegativity values is 1.7 or higher.
Therefore, a covalent bond formation via path 2 as described above would be the most probable way of bond formation.
However, appreciable diference in electronegativity value implies that the bond would not be absolutely covalent in nature. There will be some ionic nature associated with it. In actual fact the bond is ~~17% ionic and ~~83% covalent.