Write the resonance structures for , and .
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There are seven resonance structures for SO3.
Explanation
When you draw the Lewis structure, you first get the three structures at the top.
In each of them, S has a formal charge of +2 and two of the O atoms have formal charges of -1.
In each of the three structures in the middle, Shas a formal charge of +1 and one of the Oatoms has a formal charge of -1.
In the bottom structure, all atoms have a formal charge of zero.
From the point of view of formal charges, the bottom structure is the most stable structure.
For NO2, we have 5 valence electrons with the Nitrogen; plus 6 for the Oxygen, we have 2 Oxygens. So 5 plus 12 equals 17 valence electrons. Nitrogen is the least electronegative, so we'll put that in the center, and let's put an Oxygen on either side.
NO3– There are 3 possibleresonance structures that satisfy the octet rule. In each resonance structure, the formal charge on N = +1; the formal charge on the N = O oxygen = 0; and the formal charge on each of the N – O oxygens = –1. The actualstructure is an equal mixture of the 3resonance structures.
Explanation
When you draw the Lewis structure, you first get the three structures at the top.
In each of them, S has a formal charge of +2 and two of the O atoms have formal charges of -1.
In each of the three structures in the middle, Shas a formal charge of +1 and one of the Oatoms has a formal charge of -1.
In the bottom structure, all atoms have a formal charge of zero.
From the point of view of formal charges, the bottom structure is the most stable structure.
For NO2, we have 5 valence electrons with the Nitrogen; plus 6 for the Oxygen, we have 2 Oxygens. So 5 plus 12 equals 17 valence electrons. Nitrogen is the least electronegative, so we'll put that in the center, and let's put an Oxygen on either side.
NO3– There are 3 possibleresonance structures that satisfy the octet rule. In each resonance structure, the formal charge on N = +1; the formal charge on the N = O oxygen = 0; and the formal charge on each of the N – O oxygens = –1. The actualstructure is an equal mixture of the 3resonance structures.
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