In the cyanide ion,the formal negative charge | vedclass

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(A) The Lewis structure of the cyanide ion $(CN^-)$ consists of a triple bond between carbon and nitrogen $(C \equiv N)$.In this structure,the carbon

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$C$

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(A) The Lewis structure of the cyanide ion $(CN^-)$ consists of a triple bond between carbon and nitrogen $(C \equiv N)$.In this structure,the carbon atom has $4$ valence electrons and the nitrogen atom has $5$ valence electrons.For the $CN^-$ ion,the formal charge on an atom is calculated as: $	ext{Formal Charge} = 	ext{Valence electrons} - 	ext{Non-bonding electrons} - \frac{1}{2} 	imes 	ext{Bonding electrons}$.For Carbon: $4 - 2 - \frac{1}{2}(6) = 4 - 2 - 3 = -1$.For Nitrogen: $5 - 2 - \frac{1}{2}(6) = 5 - 2 - 3 = 0$.Wait,re-evaluating: The structure is $[:C \equiv N:]^-$. Carbon has $4$ valence electrons,$2$ non-bonding,and $6$ bonding. Formal charge on $C = 4 - 2 - 3 = -1$. Nitrogen has $5$ valence electrons,$2$ non-bonding,and $6$ bonding. Formal charge on $N = 5 - 2 - 3 = 0$.Therefore,the formal negative charge is on the carbon atom $(C)$.

In the cyanide ion,the formal negative charge is on

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