Which is Lewis acid and base?

Question

Which is Lewis acid and base?

1.) Determine
a)CN- (is it Lewis acid or base)
b)MgBr2 (is it Lewis acid or base)

Write acid-base reaction for both CN- and MgBr2 and show how electrons shifted.

It would be a great help if you could just tell me if I'm right assuming that CN- is a base and MgBr2 an acid, I'm not sure how to write the reactions...

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Answer 1 · 2017-01-02T16:34:32

A $Lewis acid$ $"Lewis acid"$ is an electron pair acceptor, and a.........

Explanation:

.............a $Lewis base$ $"Lewis base"$ is an electron-pair donor.

If we look at the standard reaction for a Bronsted acid, $H^{+}$ $H^+$ , with a $Bronsted base$ $"Bronsted base"$ , $^{-} O H$ $""^(-)OH$ it is easy to see which is the $Lewis acid$ $"Lewis acid"$ , and which is the $Lewis base$ $"Lewis base"$ .

$H^{+} +^{-} O H \to H_{2} O$ $H^(+) + ""^(-)OH rarr H_2O$

So which species has donated the electrons, and which species has accepted the electron to form the $H - O$ $H-O$ bond?

Now you have given us $cyanide anion$ $"cyanide anion"$ , $^{-} C \equiv N$ $""^(-)C-=N$ , which is moderately basic (and also of course highly poisonous):

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

Given what I have said, clearly cyanide anion is the electron pair donor, and therefore is the Lewis base.

The designation of $M g B r_{2}$ $MgBr_2$ is much less clear cut. In aqueous solution, magnesium bromide speciates to $M g^{2 +}$ $Mg^(2+)$ and $B r^{-}$ $Br^-$ species, as the aquated ions:

$M g B r_{2} (s) \to M g^{2 +} + 2 B r^{-}$ $MgBr_2(s) rarr Mg^(2+) +2Br^-$

The magnesium ion is solvated by several water molecules, so that we could reasonably represent this as ${[M {g (O H_{2})}_{6}]}_{6}^{2 +}$ $[Mg(OH_2)_6]^(2+)$ . In other words, the magnesium cation has accepted electron density, i.e. donation, from the water ligands, and here has acted as a Lewis acid. Which is the Lewis base in this scenario?

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Which is Lewis acid and base?

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