The electrons in a bond between two iodine atoms (I2) are shared
2 answers:
Answer : The electrons in a bond between two iodine atoms are shared equally and non-polar.
Explanation :
Non-polar covalent bond : It is a type of bond that is present between the two similar and different atoms. This type of bond is formed when the electrons are shared equally between the atoms.
This type of bond is termed as non-polar because of the electronegativity difference between the atoms is negligible.
The given molecule is formed by the equal sharing of electrons and the bond formed between the iodine-iodine is non-polar due to negligible electronegativity difference.
Therefore, the electrons in a bond between two iodine atoms are shared equally and non-polar.
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Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,