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. 


Can you please help me with this?




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


[![enter image description here](https://i.stack.imgur.com/Qd7JC.png)](https://i.stack.imgur.com/Qd7JC.png)


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). 


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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](https://i.stack.imgur.com/ysXAj.png)](https://i.stack.imgur.com/ysXAj.png)


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


[![enter image description here](https://i.stack.imgur.com/eMkZD.jpg)](https://i.stack.imgur.com/eMkZD.jpg)


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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