Question

Nucleophilicity order is correctly represented by:

• A. ${CH}_3^{-}<{NH}_2^{-}<{HO}^{-}<F^{-}$
• B. ${CH}_3^{-}={NH}_2^{-}>{HO}^{-}=F^{-}$
• C. ${CH}_3^{-}>{NH}_2^{-}>{HO}^{-}>F^{-}$
• D. ${NH}_2^{-}>F^{-}>{HO}^{-}>{CH}_3^{-}$

Answer 1

The correct option is C ${CH}_3^{-}>{NH}_2^{-}>{HO}^{-}>F^{-}$

Nucleophilicity is the tendency of nucleophile to donate its pair of electron to electrophile or electron deficient species to form covalent bond.As electronegativity of central atom decreses, its nucleophilicity increases. Here Carbon has least electronegativity, so $C{H}_3^{-}$ will have highest nucleophilicity.therfore the correct order of nucleophilcity is $C{H}_3^{-}$ $>$ $N{H}_2^{-}$ $>$$H{O}^{-}$ $>$ $F^{-}$.so the answer is option C.