Is ${CN}^{-}$ $"CN"^(-)$ a Lewis base? Why?

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Is ${CN}^{-}$ $"CN"^(-)$ a Lewis base? Why?

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Answer 1 · 2018-05-09T14:54:47

My word....

Explanation:

By definition, a $Lewis base$ $"Lewis base"$ is an electron pair donor..and certainly cyanide anion acts as a ligand in coordination chemistry. Cyanide is isolectronic with $C O$ $CO$ , and while not as $π - acidic$ $pi-"acidic"$ as the carbonyl can accept some electron density from the metal centre.

As a Bronsted base, cyanide salts (and they will not allow undergrads to play with them) will give a slightly basic solution:

$^{-} C \equiv N + H_{2} O (l) ⇌ H C \equiv N (a q) + H O^{-}$ $""^(-)C-=N+H_2O(l)rightleftharpoonsHC-=N(aq) + HO^(-)$

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Is ${CN}^{-}$ $"CN"^(-)$ a Lewis base? Why?

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