
IUPAC name of ${K_3}\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$ is:
1.Potassium aluminum oxalate
2.Potassium trioxalatoaluminate $\left( {III} \right)$
3.Potassium aluminium $\left( {III} \right)$oxalate
4.Potassium tris oxalato aluminate$\left( {VI} \right)$
Answer
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Hint: This question gives the knowledge about the IUPAC naming. It is the way of naming or representing compounds according to the rules formed by the International Union of Pure and Applied Chemistry. Naming is done in the alphabetical order.
Complete step-by-step answer: IUPAC naming is the way of naming or representing compounds according to the rules formed by the International Union of Pure and Applied Chemistry. Naming in the alphabetical order. IUPAC naming is mostly preferred for complex organic and inorganic molecules. And for simpler compounds common names are used.
Consider the given complex ${K_3}\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$, and determine the oxidation state of the central metal ion which is Aluminum $Al$.
The charge on potassium ion $K$ is $ + 1$ , but here three ions of potassium are present. So, the total charge on the potassium will be $ + 3$ .
Therefore, the overall charge on $\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$ will be $ - 3$. The oxalate ion ${C_2}{O_4}$ carries a charge of $ - 2$ . In total there are three oxalate ions present in the complex. So, the total charge on the oxalate ion will be $ - 6$.
Now, calculate the oxidation state of aluminum $\left( x \right)$ using these charges.
$ \Rightarrow x - 6 = - 3$
On simplifying, we get
$ \Rightarrow x = + 3$
Therefore, the oxidation state of aluminum is $ + 3$.
Now, according to IUPAC naming, the name of ${K_3}\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$ is Potassium trioxalatoaluminate $\left( {III} \right)$.
Therefore, option $2$ is correct.
Note: Oxidation state is the sum of all the charges of the individual atoms present in the chemical compound. And IUPAC naming is the process of naming or representing compounds according to the rules formed by the International Union of Pure and Applied Chemistry.
Complete step-by-step answer: IUPAC naming is the way of naming or representing compounds according to the rules formed by the International Union of Pure and Applied Chemistry. Naming in the alphabetical order. IUPAC naming is mostly preferred for complex organic and inorganic molecules. And for simpler compounds common names are used.
Consider the given complex ${K_3}\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$, and determine the oxidation state of the central metal ion which is Aluminum $Al$.
The charge on potassium ion $K$ is $ + 1$ , but here three ions of potassium are present. So, the total charge on the potassium will be $ + 3$ .
Therefore, the overall charge on $\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$ will be $ - 3$. The oxalate ion ${C_2}{O_4}$ carries a charge of $ - 2$ . In total there are three oxalate ions present in the complex. So, the total charge on the oxalate ion will be $ - 6$.
Now, calculate the oxidation state of aluminum $\left( x \right)$ using these charges.
$ \Rightarrow x - 6 = - 3$
On simplifying, we get
$ \Rightarrow x = + 3$
Therefore, the oxidation state of aluminum is $ + 3$.
Now, according to IUPAC naming, the name of ${K_3}\left[ {Al{{\left( {{C_2}{O_4}} \right)}_3}} \right]$ is Potassium trioxalatoaluminate $\left( {III} \right)$.
Therefore, option $2$ is correct.
Note: Oxidation state is the sum of all the charges of the individual atoms present in the chemical compound. And IUPAC naming is the process of naming or representing compounds according to the rules formed by the International Union of Pure and Applied Chemistry.
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