The number of geometrical isomers of a planar complex $M_{abcd}$ is

Total number of geometrical isomers possible for the complexes $[NiCl_4]^{2-}$,$[CoCl_2(NH_3)_4]^{+}$,$[Co(NH_3)_3(NO_2)_3]$ and $[Co(NH_3)_5Cl]^{2+}$ is

Consider the complexes:
$I$. $[Pd(NH_3)_2 ClBr]$
$II$. $[Pd(NH_3)_2 Cl_2]$
$III$. $[Pd(en) Cl_2]$
$IV$. $[Pd(en) ClBr]$
$V$. $[Pd(en)_2 Cl_2]$
(en = ethylenediamine)
The total number of geometrical isomers of $(I)$ is the same as the total number of geometrical isomers of:

The pair in which both the species have the same magnetic moment (spin only) is :

Consider the following complex ions,$P$,$Q$ and $R$.
$P = [FeF_6]^{3-}$,$Q = [V(H_2O)_6]^{2+}$ and $R = [Fe(H_2O)_6]^{2+}$.
The correct order of the complex ions,according to their spin-only magnetic moment values (in $B.M.$) is

Match List $I$ with List $II$ :

List $I$ (Complex)	List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$	$I$. Optical
$B$. $[Co(en)_3]^{3+}$	$II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$	$III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$	$IV$. Linkage

Choose the correct answer from the options given below :