Question

[Pt(Py)4] [Ptcl4] I.U.P.AC. name ​

Answer 1

Answer:

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Answer 2

Explanation:

[Pt(py)_{4}] [Pt(Cl)_{4}]

IUPAC Name :-

Tetrapyridineplatinum(II) Tetrachloridoplatinate(II)

Explaination :-

In the process of writing IUPAC Name in inorganic chemistry Some rules to be followed

• For IUPAC Name :- For cationic complex = Ligan number + Ligan name + Metal name (In English) + O.N. (oxidation number)
• For anionic complex = Ligan number + Ligan name + Metal name (Latin) + -ate + O.N.

Here,

[Pt(py)_{4}] [Pt(Cl)_{4}]

First complex is cationic complex. So, we write name of first complex with Cationic rule :-

Here,

ligan number = 4

ligan name = pyridine (py)

metal name (English) = platinum

So,

IUPAC name of Cationic complex is

Tetrapyridineplatium()

• Give a bracket for entering O.N.
• we write O.N. in last.

Now,

In Anionic complex :-

ligan number = 4

ligan name = chlorido (CL)

metal name (Latin + -ate) = platinate

So,

IUPAC name of Anionic complex is

Tetrachloridoplatinate()

Calculation of O.N. :-

Let the oxidation number of Pt in cationic complex be x

And

The oxidation number of Pt in anionic complex be y

Now,

→ x + 0×4 + y - 1× 4 = 0 [we know that sum of O.N. in all atoms of a molecule is zero]

→ x + y = 4

Now,

Divide 4 in two equal part

It means 2 and 2.

Now,

Put 2 as oxidation number of Pt in cationic complex.

And

Put 2 as oxidation number of Pt in anionic complex.

In Cationic complex :-

Tetrapyridineplatinum(II)

In Anionic complex :-

Tetrachloridoplatinate(II)

Note :-

The oxidation number is written in Roman number.