Explain the formation of the following molecules using valence bond theory. The molecules are in N2 and O2
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Answered by
6
Answer:
total no of electrons in N2 is 14
total no of electron in O2 is 16
Explanation:
N2 is diamagnetic all orbitals are filled while O2 is paramagnetic two unpaired electrons are present
Answered by
8
Answer:
a) Formation of N _{ 2 } molecules by valence bond theory :
The electronic configuration of ‘N’ atom is 2, 5 and to have Octet in the valence shell it requires three more electrons.
When two nitrogen atoms approach each other, each atom contributes 3 electrons for bonding.
There are six electrons shared between two nitrogen atoms in the form of three pairs.
Therefore, there is a triple bond between two nitrogen atoms in N _{ 2 } molecules.
b) Formation of O _{ 2 } molecule by valence bond theory :
The electronic configuration of { }_{ 8 }{ O } is 2, 6.
Oxygen atom has six electrons in its valence shell.
It requires two more electrons to get Octet in its valence shell.
Therefore oxygen atoms come close and each oxygen atom contributes two electrons for bonding.
Thus there exist two covalent bonds between two oxygen atoms in O _{ 2 } molecule as there are two pairs of electrons shared between them.
Two pairs of electrons are distributed between two oxygen atoms.
So, we can say that a double bond is formed between two oxygen atoms in O _{ 2 } molecule.
By viewing the following diagram, both the oxygen atoms have Octet in the valence shell.
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