Solution

P. $$[Cr(NH_{3})_{4}Cl_{2}]Cl$$ is Tetraamminedichlorochromium (III) Chloride.
Charge present on each ion is $$Cl = -1$$, $$Cr =x$$, $$NH_{3}=0$$.
Since chloride is outside the complex and it has a negative charge. So, the complex gets +1 charge.
Now, the addition of each ion inside = charge of overall complex
$$x-2=1 \Rightarrow x=3$$
$$Cr=Z=24$$; Electronic Configuration = $$3d^{5}4s^{1}$$
$$Cr^{3+}=3d^{3}$$
Therefore, hybridisation is $$d^{2}sp^{3}$$ and as there is no pairing of electron, it is left with 3 unpaired electrons.
So, it is paramagnetic. Also, it exhibits cis-trans isomerism (geometrical) as it is a octahedral complex of the form $$MA_{4}B_{2}$$
Q. $$[Ti(H_{2}O)_{5}Cl](NO_{3})_{2}$$
$$Ti$$ has an electronic configuration of $$[Ar]3d^{2}4s^{2}$$.
Thus, it has unpaired electrons, hence it is paramagnetic.
Also, here the counter ion, i.e., $$NO_{3}$$ can also act as a ligand. Hence, it also shows ionisation isomerism.
R. $$[Pt(en)(NH_{3})Cl]NO_{3}$$
$$Pt$$ has an electronic configuration of $$[Xe]4f^{14}5d^{9}6s^{1}$$
It does not have any unpaired electrons, hence it is diamagnetic.
Also, here the counter ion, i.e., $$NO_{3}$$ can also act as a ligand. Hence, it also shows ionisation isomerism.
S. $$[Co(NH_{3})_{4}(NO_{3})_{2}]NO_{3}$$
$$Co$$ has electronic configuration $$[Ar]3d^{7}4s^{2}$$
It does not have any unpaired electrons, hence it is diamagnetic.
Also, it also shows cis-trans isomerism.