00:01
So in this question we find out the oxidation state and name of the following complexes and also draw the crystal field energy level diagram and then find out the number of unpaired electron in the following complexes.
00:14
So our first complex is chromium complex that is nh4 cr h2o6 so4 twice.
00:25
So the name of this complex is ammonium hexa so we can write ammonium hexa quackromium hexa hexa aqua chromium three sulfate chromium chromium three sulfate so that means the oxidation state of chromium here is plus three and crystal -filled energy level diagram for this octahedral complexes so here one two and here one two three so now we can fill the electron in these orbitals so here one 2, 3.
01:08
So that means there is 3 unpaired electron is present in this complex.
01:18
Now part b of the question is similarly, part b of the question is the complex is given is malibidinum carbonyl 6.
01:26
So the name of this complex is hexa carbonyl, hexacarbonile, molybidum and oxidation state here is 0.
01:43
That means molybidum oxidation state is 0.
01:46
So we can make a crystal field energy level diagram for this octahedral complexes.
01:54
So here 1, 2, 3, 4, 5, 6.
01:57
So that means this is the low spin complex and there is no unpaired electron.
02:05
There is no unpaired electrons are present in this complex because we can see all the electrons are paired in this complex.
02:14
So there is no unpaired electron present in this complex.
02:19
Now next element is, next complex is we can see this is nickel complex.
02:27
So that is nickel, nh3, 4, water twice.
02:35
And then this is no3 twice.
02:41
So the name of this complex is tetraamine.
02:44
So we can write tetraamine di aqua di aqua nickel to nitrate nickel to nitrate because here oxidation state of nickel is plus two so that that mean nickel in plus two...