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.

Can you please help me with this?

  • $\begingroup$ NO isn't bidentate ligand, is it? Neither ambidentate... $\endgroup$
    – Zenix
    Commented Apr 23, 2020 at 8:11
  • $\begingroup$ @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? $\endgroup$ Commented Apr 23, 2020 at 9:32
  • $\begingroup$ I would write that as (ON)... $\endgroup$
    – Zhe
    Commented Apr 23, 2020 at 23:42
  • $\begingroup$ @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. $\endgroup$
    – hBy2Py
    Commented May 4, 2020 at 12:00

2 Answers 2


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

enter image description here

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).

  • $\begingroup$ if a and b are drawn on the vertical lines then there is no plane of symmetry so isnt it optically active? $\endgroup$ Commented Apr 23, 2020 at 9:41
  • $\begingroup$ 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. $\endgroup$ Commented Apr 23, 2020 at 10:07
  • $\begingroup$ @Oscar Lanzi Right, my bad. I'll change it. $\endgroup$ Commented Apr 23, 2020 at 11:09
  • $\begingroup$ 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. $\endgroup$ Commented Apr 24, 2020 at 9:17
  • $\begingroup$ @Oscar Lanzi can you explain what you mean by two fold rotation? i was not able to superimpose the mirror image on the molecule. $\endgroup$ Commented 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 $\ce{Cl}$,

enter image description here

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

enter image description here

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.


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