# MA3B3 isomers - is this same as meridional one? [closed]

I'm new to studying stereo chemistry. (Now I'm studying coordination complexes.)

In my class material and other some datas(from google), it says that there is two, fac and mer.
I want to know is whether the image is meridional one. By the explanation, two A-M-A bond angles are 90 degrees, and the other is 180 degrees - so I guess it is the mer one. Or not, I want to know why it is not possible, or not stable.

(I'm guessing my picture is the less stable version.) I'm looking forward to seeing your answer.

• Yes, mer, that's him all right. – Ivan Neretin May 7 '20 at 17:00
• @IvanNeretin Is there not some energy issue? – Woojin K May 7 '20 at 17:03
• Why would that be? – Ivan Neretin May 7 '20 at 17:04
• @IvanNeretin like stablity – Woojin K May 7 '20 at 17:04
• Please be more specific. – Ivan Neretin May 7 '20 at 17:05

Your description of: In my class material and other some data (from google), it says that there is two, fac and mer, is somewhat misleading. These two type of isomers are only possible in $$\ce{MA3B3}$$ type of complexes (e.g., $$\ce{[Co(NH3)3(NO2)3]}$$):
1. In facial isomer (fac-), all three same type of ligands have $$90^\circ$$ $$\ce{A-M-A}$$ angles (same is for all three $$\ce{B-M-B}$$ angles). As shown in following diagram, all three of same types of ligands are clustered in one side of the octahedron so that they looks like the base of a triogonal pyramid (shown as red triangle on $$\ce{NO2}$$), head of which is the metal itself. The three other kind of ligands are on the opposite side of the octahedron, arranged in similar manner.
2. In meridional isomer (mer-), all the three ligands of same type are arranged in a T-shape in same plane including the metal such that there are two $$90^\circ$$ $$\ce{A-M-A}$$ angles and one $$180^\circ$$ $$\ce{A-M-A}$$ angles, making it 'T' with metal in the middle (same is true for $$\ce{B-M-B}$$ angles, though on the opposite side). Also keep in mind that $$\ce{A-M(A)-A}$$ plane is perpendicular to $$\ce{B-M(B)-B}$$ plane.
Also note that neither is optically active since both have plane of symmetry. For mer-isomer, both $$\ce{A-M(A)-A}$$ and $$\ce{B-M(B)-B}$$ planes are planes of symmetry. For fac-isomer, for example, the plane (indicated by dotted line in the figure) going through trans-$$(\ce{A-M-B})$$ axis, dividing the octahedron to two is a plane of symmetry.
According to above description, your drown structure is mer-isomer (see red dotted lines for T-shape with $$\ce{A}$$ and blue dotted lines for T-shape with $$\ce{B}$$), which you have realized after discussion with Ivan Neretin.