Question

What is the correct “stock” name for $SnB{{r}_{4}}$? (meaning include roman numerals).

Answer 1

Hint: Every compound be it inorganic or organic requires a specific name and to name a compound there are particular rules which should be followed to keep the use of it simple worldwide. Like IUPAC is used to name the organic compounds, similarly, the stock naming system is generally used for inorganic compounds.

Complete answer:
Many elements show more than one oxidation state like both tin and lead shows +2, +4 therefore to know which compound is considering which oxidation state, we use the stock naming system.
Firstly to solve the above questions let us see an example very similar to it i.e $SnB{{r}_{4}}$ and $SnB{{r}_{2}}$, now there should be a way to differentiate between both of them, so the first step is to know which compound contains Tin(II) and which one has Tin(IV). So, as we know that bromide possesses a single negative charge therefore calculating the oxidation state, it comes out that in$SnB{{r}_{4}}$, tin is in +4 oxidation state whereas in$SnB{{r}_{2}}$, tin is in +2 oxidation state. So, if we write both of their names using the stock name system, we can easily differentiate between them:
$SnB{{r}_{2}}$- Tin(II) bromide
$SnB{{r}_{4}}$- Tin(IV) bromide

Note:
The basic rule followed to write a stock name is to write the element’s name and its oxidation state in the roman numeral closed between the brackets with no spacing in between. If the oxidation state is 1, it is not mandatory to mention the roman numeral in brackets because it is self-understood like NaCl named as sodium chloride and not sodium(I) chloride.