# Number of optically active isomers of [Co(NH3)2Cl2(NO)(OH)]?

My Attempt:

Since Compound is of the form M[a2b2cd]

I got three optically active isomers of the form M[(ab)(ab)(cd)] , M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)] and since NO is a bidentate ligand, total number of optically active isomers should be multiplied by 2 due to linkage isomerism.

Hence i was getting the answer as 2*3=6 but answer given was 4.

• NO isn't bidentate ligand, is it? Neither ambidentate... Apr 23, 2020 at 8:11
• @Zenix then i am still getting only 3 isomers but according to answer it is 4. Also doesnt NO donate through both oxygen and nitrogen? Apr 23, 2020 at 9:32
• I would write that as (ON)...
– Zhe
Apr 23, 2020 at 23:42
• @RajeshKoothrapalli NO might be able in general to donate through both N and O, but it's sterically incapable of donating from both N and O to the same metal ion center. May 4, 2020 at 12:00

Try drawing out [M(ab)(ab)(cd)]: you'll see that it's actually optically inactive!

So, there are 2 optically active isomers, which along with their respective enantiomers adds up to 4 (M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)], as you've mentioned).

• if a and b are drawn on the vertical lines then there is no plane of symmetry so isnt it optically active? Apr 23, 2020 at 9:41
• The plane of symmetry should pass through every type of ligand that occurs an odd number of times, here that means is must be vertical passing through the c and d ligands. It exists in this isomer but you drew it wrong. Also to get a superposition you need to combine refkection with a twofold rotation, which is allowed for an achiral molecule. Apr 23, 2020 at 10:07
• @Oscar Lanzi Right, my bad. I'll change it. Apr 23, 2020 at 11:09
• You have two choices really. (1) use the plane of tge paper and add the twofold ritation. (2) use a plane perpendicular to tge paper and passing through c and d, bisecting the a-M-a and b-M-b angles; this does not need the additional rotation. Apr 24, 2020 at 9:17
• @Oscar Lanzi can you explain what you mean by two fold rotation? i was not able to superimpose the mirror image on the molecule. May 1, 2020 at 6:42

Since there are two of $$\ce{NH3}$$ and $$\ce{Cl}$$, therefore in trans-form, we can always draw a plane of symmetry (in figure) to which the trans-line (in figure) is normal as depicted below for $$\ce{Cl}$$,

So, only that configuration will be optically active which has both the groups in cis-form as follows,

which gives us two optical isomers (this and its mirror image). The other two optical isomers can be drawn by swapping the positions of $$\ce{NO}$$ and $$\ce{OH}$$ in the above figure.

Hence, we get a total of four.