Question

If the ammonia solution of Na were not completely converted to NaNH2, how would this affect...

If the ammonia solution of Na were not completely converted to NaNH2, how would this affect the subsequent preparation of [Cr(NH3)6](NO3)3?

If a large excess of Fe(NO3)39H2O were added to the NH3 solution of Na, how would this affect the yield of [Cr(NH3)6](NO3)3 and why?

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Answer #1

Preparation of [Cr(NH3)6](NO3)3 involves two main reactions:

CrCl3 + 6 NH3 [Cr(NH3)6]Cl3   .....(1)

[Cr(NH3)6]Cl3 + 3 HNO3 [Cr(NH3)6](NO3)3 + 3 HCl .....(2)

In the first step NaNH2 is added for the formation of [Cr(NH3)6]Cl3. Without NaNH2 formation of [Cr(NH3)5 Cl]Cl3 takes place. NaNH2 releases NH2- in the reaction medium. Since NH2- is a better nucleophile than NH3, it facilitates the removal of Cl directly bound to Cr from its chromium complex thus forming [Cr(NH3)5(NH2)]2+ which adds a proton to form [Cr(NH3)6]Cl3. So if the Na solution of liquid ammonia is not completely converted to NaNH2 we will have less NH2- in the solution and thus some amount of [Cr(NH3)5Cl]Cl3 will also be formed. This will decrease the yield of our product.

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