The question asks for the total number of possible isomers for the square-planar complex $[Pt(Cl)(NO_2)(NO_3)(SCN)]^{2–}$. This is a square-planar complex of the type MA_B_C_D_, where all four ligands ($Cl$, $NO_2$, $NO_3$, $SCN$) are different.
For a square-planar complex with four different ligands (MA_B_C_D_), we consider the relative positions of the ligands. Let's fix one ligand, $Cl$, at a specific position.
Therefore, the number of geometric isomers is calculated as:
Number of choices for the opposite ligand $\times$ Number of arrangements for the remaining two ligands
Which is $3 \times 2 = 6$ geometric isomers.
The complex contains two ligands known to exhibit linkage isomerism:
Linkage isomerism leads to different isomers for the same connectivity.
The total number of isomers is the product of the number of geometric isomers and the number of possible linkage forms. The provided options suggest that only one of the potentially isomerizing ligands contributes to the total count, doubling the number of isomers.
This calculation assumes that either $NO_2$ or $SCN$ exhibits linkage isomerism, leading to the answer 12.
Based on the calculation considering both geometric and linkage isomerism (assuming one ligand exhibits linkage isomerism), the total number of possible isomers is 12.
The crystal field splitting energy of $[\text{Co(oxalate)}_3]^{3-}$ complex is '$n$' times that of the $[\text{Cr(oxalate)}_3]^{3-}$ complex. Here '$n$' is ___________. (Assume $\Delta_o >> P$)
The correct order of spin-only magnetic moments among the following is :
(Atomic number : Mn = 25, Co = 27, Ni= 28, Zn = 30)
The crystal field splitting energy of $[\text{Co(oxalate)}_3]^{3-}$ complex is '$n$' times that of the $[\text{Cr(oxalate)}_3]^{3-}$ complex. Here '$n$' is ___________. (Assume $\Delta_o >> P$)