1. Complete and balance the following as net ionic equations. Assume the Cr and

Question

1. Complete and balance the following as net ionic equations. Assume the Cr and Fe complexes will react completely with Cl- (or SCN-) to form an octahedral complex while the Cu complex will react to form a tetrahedral complex. If there was no discernable change within the observation period, write "no discernable reaction."

a) [Cr(H2O)6]3+ + Cl- ---->

b) [Cu(H2O)6]2+  + Cl-  ---->

c) [Fe(C2O4)3]3- + SCN- ---->

2. assign the previous complexes as inert or labile

3. does c have a high spin or low spin electron configuration?

Explanation / Answer

a) [Cr(H2O)6]3+ + 2Cl- ---->[ Cr(H2O)4Cl2]+ + 2H2O [common complex formed,green colored solution]

[Cr(H2O)6]3+ + 3Cl- ----> CrCl3 + 6H2O[anhydrous chromium chloride is green colored]

b) [Cu(H2O)6]2+  + 4Cl-  ----> [CuCl4]2-+6H2O

pale blue to yellow colored solution

octahedral hexaaquacopper(II) complex changes to tetrahedral tetrachlorocopper(II) complex ion.

Cl-is larger ligand than H2O and is charged,so tetrahedral geometry preferred as less repulsion between ligands takes place.

c) [Fe(C2O4)3]3- + SCN- ----> "no discernable reaction." as C2O42-(oxalate ) ions are chelating ligands that forms extra stable complex ,that do not undergo ligand exchange

2)[Cr(H2O)6]3+ is labile ,exchanges ligands

[Cu(H2O)6]2+ is labile

[Fe(C2O4)3]3- is inert ,stable,no exchange of ligand or geometry takes place

3)Fe3+ in octahedral complexes is low spin ,d5,t2g5eg0,one unpaired electron

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