00:01
Hello, here we have to solve the following problem.
00:03
So here we have to determine the number of de -elections associated with essential metal ion.
00:10
And then we have to draw the electron configuration and determine whether the complex is diamagnetic or paramagnetic.
00:28
So let's do this.
00:36
So let's start with a.
00:45
That is iron c l6 3 minus so here the oxidation state of iron is 3 plus therefore that is iron configuration is d5 so that is an octahedral field that's why its configuration is like it looks like this so that is e g g 2g so and that is that's going to be high spin so that is paramagnetic now let's move on to b which is low spin that is low spin iron n .02 6 3 minus so let's let's first determine the oxidation state of iron here so here iron oxidation state is 3 plus therefore here here it carries so all five electrons are on the eg on t2g level so still that is paramagnetic so now let's move on to question c that is malibdenum and h3 -62 plus in a ligand field in so in a weak ligand field oh and here in the previous one i forgot to mention configuration of iron that's d5 sorry now let's take care of malibdum so malibdum is from chromium family that's why if it's 2 plus s here then its configuration is d4 so that is the weak field that's why for electrons will look like yeah that will be high spin complex and that is part now let's look at d which is a cobalt aquacomplex of cobalt, cobalt, h206, 3 plus in a strong ligand field.
04:56
So let's determine configuration of cobalt.
05:01
Here configuration of cobalt is d6.
05:12
And now let's show the electrons.
05:14
So that is a strong field that's why that's why that is diamagnetic.
05:33
Okay, that's done.
05:35
What is left? next one is cadmium sulfate.
05:50
So, the configuration of, yeah, let's determine configuration of cadmium.
05:56
That is d10.
06:06
So that's why all the electrons are filled.
06:19
And that is diamagnetic...