Question

The E A N of Platinum in Potassium hexachloroplatinate ($IV$) is:
(A) $46$
(B) $86$
(C) $36$
(D) $84$

Answer 1

Hint: E A N stands for Effective atomic number. We first know what the effective atomic number is. Then analyte the structure of platinum in potassium hexachloroplatinate($iv$).

Complete Step by Step solution:
Effective atomic number(E A N), a number that represents the total number of electrons surrounding the nucleus of a metal atom in a metal’s complex. It is composed of the metal atom’s electrons and the bonding electrons from the surrounding electrons-donating atoms and molecules.
The mathematical expression of effective atomic number (E A N) is,
\[EAN = [atomic{\text{ }}number{\text{ }}of{\text{ }}metal{\text{ }}atom - oxidation{\text{ }}number{\text{ }}of{\text{ }}metal{\text{ }}atom] \times 2 \times number{\text{ }}of{\text{ }}ligands\]
Calculate effective atomic number of Coordinate compounds:-
  Effective atomic Number (E A N)
 The sum of the number of electrons, donated by all ligands and those present on the central metal ion on atoms in the ion complex is called effective atomic number (E A N).
Generally E A N of central metal ions will be equal to the number of electrons in the nearest noble gas.
If the E A N of the central metal is equal to the number of electrons in the nearest noble gas then the complex possesses greater stability.

$EAN = [\left( {atomic{\text{ }}number{\text{ }}of{\text{ }}central{\text{ }}metal} \right) - \left( {the{\text{ }}oxidation{\text{ }}state{\text{ }}of{\text{ }}the{\text{ }}metal} \right) + \left( {the{\text{ }}number{\text{ }}of{\text{ }}electrons{\text{ }}gained{\text{ }}by{\text{ }}the{\text{ }}metal{\text{ }}from{\text{ }}ligands{\text{ }}through{\text{ }}Co - ordination} \right)]$$EAN = [7metal - \left( {oxidation{\text{ }}state{\text{ }}of{\text{ }}the{\text{ }}metal} \right) + 2\left( {coordination{\text{ }}number{\text{ }}of{\text{ }}the{\text{ }}metal} \right)] - \left( 1 \right)$
Hence ,
Potassium Hexachloroplatinate $ \to {K_2}[PtC{l_6}]$
${K_2}[PtC{l_6}] \to 2{K^ + } + {[P{t^{ + 4}}C{l_6}^{ - 1}]^{\left( { - 2} \right)}}$
Hence, Platinum in Potassium Hexachloroplatinate exists as a $ + 4$ oxidation state.
We know that the atomic number of Platinum is $78.$
Hence Coordination number of Platinum in $[PtC{l_6}]$ is $6$
$EAN = [78 - 4 + \left( {2 \times 6} \right)]$ [According to the $(1)$ equation]
$ = [78 - 4 + 12]$
$ = 86$
Hence, The E A N of Platinum in potassium hexachloroplatinate $\left( {IV} \right)$ is (B) $86$.

Hence, the correct answer is option B.

Note: The English Nevil V.Sidgwick made the observation, since known as E A N rule, that in a number of metal complexes the metal atom tends to surround itself with sufficient ligands that the resulting effective atomic number is numerically equal to the atomic number of noble gas element found in the same period in which the metal is situated. This rule seems to hold for most of the metal complexes with carbon monoxide, the metal carbonyls as well as many organometallic compounds.