00:01
We have to name the coordination compounds here and first of all we have co, nh3, oll 6, cl2.
00:23
So now to name this, first of all we will write the lygand name followed by the metal and iron name.
00:32
So here we have 6 amine lysans.
00:35
So for 6 we will use the prefix hexa.
00:38
This will be hexamile.
00:42
Then we have cobalt here so we will write cobald okay the metal atom or the metal ion we have that is cobald now i have we have to calculate here the oxidation number so this amine is a neutralized we have cl here so that carries minus one charge they are two in number so that's why the oxidation number will be second okay the oxidation number is written in the roman method, roman numeral.
01:16
And then i have cl, so for it, when it is present outside the coordination sphere, so it is written as chloride, okay? lysence present inside the coordination sphere.
01:31
So they will be written like very different from the license present outside the coordination sphere you will see in the further parts so now we have co h2o whole 6 inside the coordination is fair i3 so here we have lysand which is present inside the coordination sphere so we will write the lysent name first and then the meta line name so they are six in numbers so the prefix will be hexa and now for h2o we we write aqua then we have cobalt here so cobald will be written as it is then the oxidation state so i have i3 here okay so that carries minus 1 chart so the overall oxidation number oxidation state is 3 plus 3 that is written in the roman numeral form as third and now i have iodide here okay because it is present outside the coordination sphere so the license are like we have c -l -i or we are halogens okay so they are named as adding i'd okay we will add the suffix i'd to them so this will be named as iodide okay now c part k -2 pt c -l -4 so here we have pt cl4 present inside the coordination sphere and k2 that is outside the coordination sphere.
03:19
Whenever this happens, whenever we have different metal which is present outside the coordination sphere then it is named as it is.
03:28
So this will be potassium here.
03:34
This case potassium here.
03:41
Then we have here this cl is present inside the coordination system.
03:49
For this reason it will be named as chloro.
03:54
Okay, remember when the cliz was present outside the coordination sphere, it was named as chloride, okay? adding ired suffix to it.
04:05
But whenever it is present inside the coordination sphere, it is named as chloro, okay, adding o suffix.
04:12
So now then i have this platinum here.
04:17
So see, this coordination is where we have.
04:20
So this carries minus 1 charge here okay or you can see that minus 2 is present here okay since we have 2 potassium ions present here so of course it carries plus 2 charge so since this is a complex here overall so this should carry minus 2 charge so now since it carries minus charge negative charge so that's why for this reason this metal iron present that is named after adding 8 suffix.
04:55
So this will be platinate.
04:59
Okay? this platinum will be written like platinade.
05:06
So now we will write the oxidation state.
05:19
So oxidation state is calculated.
05:26
So i have overall minus 2 charge.
05:28
Okay, if i am considering this two potassium ions.
05:31
So this means that for complex, this coordination is fair it carries minus two charge this c l has minus one so if i consider for this platinum as x oxidation number so this can be written as minus four okay and is equal to minus two so the oxidation number will be plus two right so that's why in roman number it is written as second now we have d part k4 p t cl6.
06:16
So here we have this metal potassium present outside the coordination sphere so it is named as it is potassium...