Which of the following is a cationic complex?
(A) Hexa amino platinum chloride
(B) Potassium ferrocyanide
(C) Sodium argento cyanide
(D) Nickel carbonyl
Answer
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Hint: A cationic complex is the one in which the complex carries a net positive charge. To determine a cationic complex first we need to write the molecular formula for the given coordination complexes. After that we will determine the oxidation number of the central metal ion. By that we can determine the cationic complex.
Complete step-by-step answer:
(A) Hexa amino platinum chloride:So, the molecular formula will be $[Pt{(N{H_3})_6}]C{l_2}$.The ligand present here is a neutral ligand so the oxidation state of $N{H_3}$ will be $0$So, the oxidation state of platinum will be:Let the oxidation state of platinum be $x$.So,
$
x + 0 + ( - 2) = 0 \\
\Rightarrow x - 2 = 0 \\
\Rightarrow x = + 2 \\ $
Therefore, the oxidation state of platinum will be $ + 2$
Here as we know after the dissociation of $[Pt{(N{H_3})_6}]C{l_2}$ it will give:
$[Pt{(N{H_3})_6}]C{l_2} \rightleftharpoons {[Pt{(N{H_3})_6}]^{ + 2}} + 2C{l^ - }$
So, this will be a cationic complex
(B) Potassium ferrocyanide:Molecular formula is ${K_4}[Fe{(CN)_6}]$.The oxidation state of ligand i.e., cyano is $ - 1$.So, the oxidation state of iron will be: Let the oxidation state of iron be $x$.
$
+ 1(4) + x + ( - 1 \times 6) = 0 \\
\Rightarrow + 4 + x - 6 = 0 \\
\Rightarrow x - 2 = 0 \\
\Rightarrow x = + 2 \\ $
Oxidation state of iron is $ + 2$
So now we know that the dissociation reaction will be:
${K_4}[Fe{(CN)_6}] \rightleftharpoons {[Fe{(CN)_6}]^{ - 2}} + 4{K^{ + 1}}$
Thus, this will be an anionic complex.
(C) Sodium argento cyanide:Molecular formula is$Na[Ag(C{N_2})]$ .The oxidation state of $Ag$ will be:Let the oxidation state of $Ag$ be x
$
+ 1 + x + ( - 1 \times 2) = 0 \\
\Rightarrow + 1 + x - 2 = 0 \\
\Rightarrow x - 1 = 0 \\
\Rightarrow x = + 1 \\ $
The oxidation state of $Ag$ is $ + 1$
The dissociation reaction will be:
$Na[Ag(C{N_2})] \rightleftharpoons {[Ag(C{N_2})]^{ - 1}} + N{a^{ + 1}}$
Hence this is also an anionic complex.
(D) Nickel carbonyl:Molecular formula is $[Ni(C{O_4})]$. Since carbonyl is a neutral ligand hence its oxidation state will be $0$.Therefore the oxidation state of nickel will also be $0$.Hence this compound will be a neutral compound.
So, the correct answer is Option (A).
Note: If the coordination compound will be accompanied by a counter ion, then the name of the cationic entity must be written before the name of the anionic entity. The coordination complexes are not completely ionizable in aqueous solution which leads to the formation of cationic complexes or anionic complexes.
Complete step-by-step answer:
(A) Hexa amino platinum chloride:So, the molecular formula will be $[Pt{(N{H_3})_6}]C{l_2}$.The ligand present here is a neutral ligand so the oxidation state of $N{H_3}$ will be $0$So, the oxidation state of platinum will be:Let the oxidation state of platinum be $x$.So,
$
x + 0 + ( - 2) = 0 \\
\Rightarrow x - 2 = 0 \\
\Rightarrow x = + 2 \\ $
Therefore, the oxidation state of platinum will be $ + 2$
Here as we know after the dissociation of $[Pt{(N{H_3})_6}]C{l_2}$ it will give:
$[Pt{(N{H_3})_6}]C{l_2} \rightleftharpoons {[Pt{(N{H_3})_6}]^{ + 2}} + 2C{l^ - }$
So, this will be a cationic complex
(B) Potassium ferrocyanide:Molecular formula is ${K_4}[Fe{(CN)_6}]$.The oxidation state of ligand i.e., cyano is $ - 1$.So, the oxidation state of iron will be: Let the oxidation state of iron be $x$.
$
+ 1(4) + x + ( - 1 \times 6) = 0 \\
\Rightarrow + 4 + x - 6 = 0 \\
\Rightarrow x - 2 = 0 \\
\Rightarrow x = + 2 \\ $
Oxidation state of iron is $ + 2$
So now we know that the dissociation reaction will be:
${K_4}[Fe{(CN)_6}] \rightleftharpoons {[Fe{(CN)_6}]^{ - 2}} + 4{K^{ + 1}}$
Thus, this will be an anionic complex.
(C) Sodium argento cyanide:Molecular formula is$Na[Ag(C{N_2})]$ .The oxidation state of $Ag$ will be:Let the oxidation state of $Ag$ be x
$
+ 1 + x + ( - 1 \times 2) = 0 \\
\Rightarrow + 1 + x - 2 = 0 \\
\Rightarrow x - 1 = 0 \\
\Rightarrow x = + 1 \\ $
The oxidation state of $Ag$ is $ + 1$
The dissociation reaction will be:
$Na[Ag(C{N_2})] \rightleftharpoons {[Ag(C{N_2})]^{ - 1}} + N{a^{ + 1}}$
Hence this is also an anionic complex.
(D) Nickel carbonyl:Molecular formula is $[Ni(C{O_4})]$. Since carbonyl is a neutral ligand hence its oxidation state will be $0$.Therefore the oxidation state of nickel will also be $0$.Hence this compound will be a neutral compound.
So, the correct answer is Option (A).
Note: If the coordination compound will be accompanied by a counter ion, then the name of the cationic entity must be written before the name of the anionic entity. The coordination complexes are not completely ionizable in aqueous solution which leads to the formation of cationic complexes or anionic complexes.
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