Question

The IUPAC name of $ [Pt{{(N{{H}_{3}})}_{4}}(N{{O}_{2}})Cl]S{{O}_{4}}$ is :
(A) tetraamine chloro nitro platinum (II) sulphate
(B) tetraamine chloro nitro platinum (IV) sulphate
(C) chlor tetrammine nitro platinum (IV) sulphate
(D) chlor tetrammine nitro platinum (II) sulphate

Answer 1

Hint :use the nomenclature for coordinate compounds and don’t include the tetra, di, tri in considering the alphabetic order for writing the names.

Complete Step By Step Answer:
Firstly let us look at the structure of our compound which is $ [Pt{{(N{{H}_{3}})}_{4}}(N{{O}_{2}})Cl]S{{O}_{4}}$ now here we can see that the whole compound is neutral as whole but if remove the $ S{{O}_{4}}$ it will get $ +2$ charge on whole like this $ {{[Pt{{(N{{H}_{3}})}_{4}}(N{{O}_{2}})Cl]}^{2+}}$ as the $ S{{O}_{4}}$ have $ -2$ charge on it , now we will look inside the structure of bracket here we have $ 4$ amines and one chlorine and one nitro group so according to the nomenclature we have to write the naming according to alphabetical order , here the first will be ammine as ‘a’ comes first in alphabetical order, here keep note that tetra is not included in considering which alphabet will come first so firstly we will write $ tetrammine$ now to next part after ‘a’ from ammine it comes ‘c’ from chlorine so its name will be $ chloro$ so our name will become $ tetraamine chloro$ after ‘c’ from chlorine it come ‘n’ from nitro so our name will become $ tetraamine chloro nitro$ after this there is no group left except our central atom which is platinum now our name will become $ tertramminechloronitro platinum$ but after this we also have to write in which oxidation state the platinum is in so the nitro will have $ -1$ charge and $ chloro$ will also have $ -1$ charge, and ammine doesn’t contribute any charge so its charge will be zero and let the oxidation number of platinum is $ x$ so using the charges in the formula $ {{[Pt{{(N{{H}_{3}})}_{4}}(N{{O}_{2}})Cl]}^{2+}}$ we will get the value of $ x$
$ x-1-1=+2 \\
 so\text{ x = +4} $
Hence platinum is in this oxidation state and our formula becomes $ tetraamine chloro nitro platinum\left( IV \right)$
Now the $ sulphate$ is also there so it will come in the end so our name will become $ tetraamine chloro nitro platinum\left( IV \right)sulphate$ which is our B option.

Note :
Always check the charge on the compound which is in the bracket that if it is neutral or not like in this question it is not neutral and has a charge on it. And always check the oxidation number on the central atom.