We can't draw structures on the board but you can on a sheet of paper.
1. Draw a parallelogram on a sheet of paper.
2. From the center of the parallelogram, draw vertical line above and below.
3. The coordination number of the Co compound you have drawn is 6. Notice that you have 6 "corners" on your structure. H2O goes on 5 of them and NCS (or SCN) on the other one.
4. The isomer is SCN on the other one. The only difference is in the Co-NCS bond versus the Co-SCN bond.
Two isomers have the same formula but different structures: Co(H2O)5SCN, CO(H2O)5NCS. The 5's mean 5-H20's. Draw each of these structures and explain the difference between the two? I don't even know where to begin. I appreciate any kind of help at this point. Thanks!
4 answers
Thanks Dr. Bob! Along those same lines. Which of the following compounds have more than one isomer? PtCl5NH3, Co(NH3)4Cl2, Pt(CO)2Cl2, Rb3[Cr(CN)6] Thanks!
Two kinds of diagrams you need.
One is the parallelogram with vertical lines above and below for coordination number of 6. The other is a parallelogram with no lines (which has four corners) for a coordination of 4.
For PtCl5NH3, coordination number (which I will abbreviate as C.N.) of 6 so you want the octahedral (the 6 cornered one). There is only one isomer.
For Co(NH3)4Cl2 C.N. = 6, the octahedral complex is what you want, and it can have two isomers. One is the Cl at the top and Cl at the bottom (this is the trans isomer); the other one is Cl at top and the other Cl at any of the adjacent corners and this is the cis isomer.
For PtCOCl2, this is the square planar one, C.N. = 4 and it can have two isomers, the cis and trans. I'll leave that for you to draw. It's easy to see since the square planar complex is two dimensional and so is the sheet of paper on which you've made your drawing.
Don't let the Rb3[Cr(CN)6] confuse you. That is a compound that consists of the Rb^+ and the hexacyanochromate(III) anion(3- charge). The only part that can have isomers is the anion and since all of the ligands are the same there will be no isomers.
One is the parallelogram with vertical lines above and below for coordination number of 6. The other is a parallelogram with no lines (which has four corners) for a coordination of 4.
For PtCl5NH3, coordination number (which I will abbreviate as C.N.) of 6 so you want the octahedral (the 6 cornered one). There is only one isomer.
For Co(NH3)4Cl2 C.N. = 6, the octahedral complex is what you want, and it can have two isomers. One is the Cl at the top and Cl at the bottom (this is the trans isomer); the other one is Cl at top and the other Cl at any of the adjacent corners and this is the cis isomer.
For PtCOCl2, this is the square planar one, C.N. = 4 and it can have two isomers, the cis and trans. I'll leave that for you to draw. It's easy to see since the square planar complex is two dimensional and so is the sheet of paper on which you've made your drawing.
Don't let the Rb3[Cr(CN)6] confuse you. That is a compound that consists of the Rb^+ and the hexacyanochromate(III) anion(3- charge). The only part that can have isomers is the anion and since all of the ligands are the same there will be no isomers.
Thanks Dr Bob! You da man!