Question:
The correct stability order of the following resonance
structure is
(I) $H_2C = _{N}^{+} = _{N}^{-} \, \, \, \, $ (II) $ H_2 \, _{C}^{+} - N = _{N}^{-}$
(c) $H_2 \, _{C}^{-} - _{N}^{+} = N \, \, \, \, $ (IV) $H_2 {C}^{-} - N = N $
The correct stability order of the following resonance
structure is
(I) $H_2C = _{N}^{+} = _{N}^{-} \, \, \, \, $ (II) $ H_2 \, _{C}^{+} - N = _{N}^{-}$
(c) $H_2 \, _{C}^{-} - _{N}^{+} = N \, \, \, \, $ (IV) $H_2 {C}^{-} - N = N $
Updated On: Jun 14, 2022
- $ (I) > (II) > (IV) > (III) $
- $(I) > (III) > (II) > (IV)$
- $(II) > (I) > (III) > (IV)$
- $(III) > (I) > (IV) > (II)$
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
I is most stable because it has more covalent bonds and negative
charge on electronegative nitrogen. Ill is more stable than II and
IV due to greater number of covalent bonds. Between II and IV,
II is more stable since, it has negative charge on electronegative
atom and positive charge on electropositive atom. Hence,
overall stability order is
I > III > II > IV
charge on electronegative nitrogen. Ill is more stable than II and
IV due to greater number of covalent bonds. Between II and IV,
II is more stable since, it has negative charge on electronegative
atom and positive charge on electropositive atom. Hence,
overall stability order is
I > III > II > IV
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