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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.) enter image description here I'm looking forward to seeing your answer.

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    $\begingroup$ Yes, mer, that's him all right. $\endgroup$ – Ivan Neretin May 7 '20 at 17:00
  • $\begingroup$ @IvanNeretin Is there not some energy issue? $\endgroup$ – Woojin K May 7 '20 at 17:03
  • $\begingroup$ Why would that be? $\endgroup$ – Ivan Neretin May 7 '20 at 17:04
  • $\begingroup$ @IvanNeretin like stablity $\endgroup$ – Woojin K May 7 '20 at 17:04
  • $\begingroup$ Please be more specific. $\endgroup$ – Ivan Neretin May 7 '20 at 17:05
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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., $\ce{[Co(NH3)3(NO2)3]}$):

Geometrical facial and meridonal isomers

  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.

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