Question

The compound $N{a_2}IrC{l_6}$ reacts with triphenylphosphine in diethylene glycol in an atmosphere of $CO$ to give $[IrCl(CO){(PP{h_3})_2}]$ known as ‘Vaska’s compound’. What is the number of unpaired electrons and oxidation state of iridium in the complex respectively?
[Atomic number of $Ir = 77$]
A. $0,0$
B. $0,1$
C. $1,1$
D. $1,2$

Answer 1

As we know that number of electrons can be determined by knowing the number of moles multiplied with the Avogadro’s number. And the atomic number of iridium is $77$ which means its electronic configuration will be given as $5{d^8}$ having eight electrons in the outermost valence shell.

Complete answer:
As we are given that compound $N{a_2}IrC{l_6}$ reacts with triphenylphosphine in the solution of diethylene glycol in an atmosphere of $CO$ and results in the formation of a complex compound $[IrCl(CO){(PP{h_3})_2}]$ which is commonly known as ‘Vaska’s compound’. We can represent this reaction in the form of an unbalanced chemical equation as:
$N{a_2}IrC{l_6} + PP{h_3} + CO \to [IrCl(CO){(PP{h_3})_2}]$
So we can see that the complex which is formed does not have any unpaired electrons as every molecule is attached with another one thereby forming a complex. So the number of unpaired will be automatically zero for this complex.
Now we know that the oxidation number is the degree of an atom in a given chemical compound. So we can calculate the oxidation number of iridium in the complex. Let it be $x$, and we know that the oxidation number of carbonyl ligand is zero because it is a neutral ligand, similarly the oxidation state of triphenylphosphine will be zero. So the oxidation state of iridium can be:
$x + ( - 1) + 0 + 2 \times 0 = 0$
$x = + 1$
Thus the oxidation state of iridium is found to be $+ 1$.

**Therefore the correct answer is (B).

Note:**
Always remember the neutral chemical species or ligand will always have zero charge and thus zero oxidation number or oxidation state, an element in it native state will also have zero charge and zero oxidation number, monoatomic molecules will have $+ 1$ oxidation state like that of hydrogen, diatomic molecules will have $+ 2$ oxidation state. And a complex compound may always have zero unpaired electrons.