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# How many lone pairs are present in $${N_2}{O_4}$$ ?

Last updated date: 10th Aug 2024
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The Lewis structure of $${N_2}{O_4}$$is given below,
In the above structure we can see that one nitrogen atom is bonded to another nitrogen atom by a single bond. Both of the nitrogen atoms form N-O bonds. Two of the four N-O bonds are single bonded and other two are double bonded. The total number of valence electrons by nitrogen atoms are 10. The valence electrons given by four oxygen atoms are 24. So the total valence electron will be 34. The total valence electron pair will be 17. After bonding we can see in the structure that 10 pairs of electrons are not able to take part in the bonding so they are unshared. These remaining 10 pairs of valence electrons are called lone pairs. The Lewis structure is said to be stable when the charges on the atom are least. But in the above structure all oxygen atoms have one negative charge and nitrogen atoms have two positive charges. So it is not the most stable structure of $${N_2}{O_4}$$.The structure with the stable one with less charge is the Lewis structure below. The oxygen atom in this structure has one negative charge each except the double bonded oxygen and the one nitrogen atom attached to double bonded oxygen has one positive charge and the other has two positive charges.