
For inorganic nomenclature, on what does the stock system rely?
Answer
486.3k+ views
Hint: Stock system is a type of nomenclature system which is particularly used for naming the inorganic compounds. The stock nomenclature system was first described by Alfred Stock. This system indicates the name as well as oxidation number of the element in roman digit.
Complete answer:
A particular nomenclature is required in the field of chemistry to name the compounds and avoid confusion which is created by their local name. The name which is assigned by the stock system is accepted by IUPAC.
The stock system is required to encode the particular structure and composition of the compound to give unambiguous reference for the compound. All these systems make absolute clarity in the name of inorganic compounds.
There are some rules while naming any inorganic compound by stock nomenclature system-
A suffix ($ - ide$) is added to the end of the name of any anion.
Example: $\left( {{O^ - }} \right)$as oxide, $\left( {{H^ - }} \right)$ as hydride
Compound which is formed by the cation then the name of the compound is cation’s name which precedes the name of anion.
Example: $KCl$ as potassium chloride, $NaCl$ as sodium chloride
If the cation contains more than one charge in magnitude then the numerical digit in roman system is added in the name of the compound.
Example: $\left( {C{u^{ + 2}}} \right)$ a roman digit $\left( {II} \right)$ is added, $\left( {A{l^{ + 3}}} \right)$ a roman digit $\left( {III} \right)$ is added.
Suffixes (-ite) and (-ate) are added in the case of polyoxoanions.
Now take the example of compound $\left[ {KMn{O_4}} \right]$: we have to calculate the oxidation number of the manganese atom which is mentioned in roman number in the name.
As we know oxidation number of oxygen is $\left( { - 2} \right)$ and potassium is $\left( { + 1} \right)$, oxidation number of $Mn$ is-
$1 + x + 4 \times \left( { - 2} \right) = 0$
After solving the above equation, we get
$x = 7$
Hence according to the stock system name of $\left[ {KMn{O_4}} \right]$ should be potassium manganate $\left( {VII} \right)$.
$ \Rightarrow $ From the above discussion we find that the stock system for inorganic compounds rely on oxidation number of the element which is represented in roman digit.
Note:
Red book is used as the literature for nomenclature of inorganic compounds. According to the stock system there should be no space left between the name of compound and its oxidation number in roman. Other examples may include $Ca{F_2}$ as calcium$\left( {II} \right)$ fluoride.
Complete answer:
A particular nomenclature is required in the field of chemistry to name the compounds and avoid confusion which is created by their local name. The name which is assigned by the stock system is accepted by IUPAC.
The stock system is required to encode the particular structure and composition of the compound to give unambiguous reference for the compound. All these systems make absolute clarity in the name of inorganic compounds.
There are some rules while naming any inorganic compound by stock nomenclature system-
A suffix ($ - ide$) is added to the end of the name of any anion.
Example: $\left( {{O^ - }} \right)$as oxide, $\left( {{H^ - }} \right)$ as hydride
Compound which is formed by the cation then the name of the compound is cation’s name which precedes the name of anion.
Example: $KCl$ as potassium chloride, $NaCl$ as sodium chloride
If the cation contains more than one charge in magnitude then the numerical digit in roman system is added in the name of the compound.
Example: $\left( {C{u^{ + 2}}} \right)$ a roman digit $\left( {II} \right)$ is added, $\left( {A{l^{ + 3}}} \right)$ a roman digit $\left( {III} \right)$ is added.
Suffixes (-ite) and (-ate) are added in the case of polyoxoanions.
Now take the example of compound $\left[ {KMn{O_4}} \right]$: we have to calculate the oxidation number of the manganese atom which is mentioned in roman number in the name.
As we know oxidation number of oxygen is $\left( { - 2} \right)$ and potassium is $\left( { + 1} \right)$, oxidation number of $Mn$ is-
$1 + x + 4 \times \left( { - 2} \right) = 0$
After solving the above equation, we get
$x = 7$
Hence according to the stock system name of $\left[ {KMn{O_4}} \right]$ should be potassium manganate $\left( {VII} \right)$.
$ \Rightarrow $ From the above discussion we find that the stock system for inorganic compounds rely on oxidation number of the element which is represented in roman digit.
Note:
Red book is used as the literature for nomenclature of inorganic compounds. According to the stock system there should be no space left between the name of compound and its oxidation number in roman. Other examples may include $Ca{F_2}$ as calcium$\left( {II} \right)$ fluoride.
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