Question

When acidified \[{K_2}C{r_2}{O_7}\] solution is added to \[S{n^{2 + }}\] salts, then \[S{n^{2 + }}\] changes to:
A. \[Sn\]
B. \[S{n^{3 + }}\]
C. \[S{n^{4 + }}\]
D. \[S{n^ + }\]

Answer 1

Hint:As we are well aware with the fact that potassium dichromate is a strong oxidising agent and tin has an atomic number of \[50\] and electronic configuration as $[Kr]4{d^{10}}5{s^2}5{p^2}$ having two valence electrons to lose and four to gain.

Complete answer:
As we know that Tin is a group fourteen element with a symbol \[\;Sn\] stands for stannum and it can exist in two oxidation states which are $ + 2$ and $ + 4$. Tin having an atomic number of \[50\] and electronic configuration as $[Kr]4{d^{10}}5{s^2}5{p^2}$ suggests that it possess two valence electrons so it has the ability to lose two electrons or gain four electrons to complete their octet.
We know that \[{K_2}C{r_2}{O_7}\] is a strong oxidising agent, so under acidic conditions, it oxidizes the dipositive tin \[S{n^{2 + }}\] to tetrapositive \[S{n^{4 + }}\] and free elements. We can show this with the help of an equation as:
$S{n^{2 + }} + C{r_2}O_7^{2 - } \to S{n^{4 + }} + C{r^{3 + }}$
Here, tin was present in $ + 2$ oxidation state and chromium was in $ + 6$ oxidation state, under acidic conditions redox reaction took place and \[S{n^{2 + }}\] was oxidised to \[S{n^{4 + }}\] and $C{r^{6 + }}$ was reduced to $C{r^{3 + }}$.
In option A, that is, tin is present in its elemental form thus it has an oxidation state of zero which is the reduced form of $S{n^{2 + }}$, so option ‘a’ is incorrect and in the case of option d, that is \[S{n^ + }\] which is present in reduced form already so it is also incorrect. Similar is the case with \[S{n^{3 + }}\] as it cannot be obtained by strong oxidizing agents like potassium dichromate from \[S{n^{2 + }}\], so option ‘b’ is also incorrect.
Therefore from the above explanation we can say that the correct answer is \[S{n^{4 + }}\].

Hence the right option is (C). \[S{n^{4 + }}\]

Note:
When potassium dichromate is added with \[S{n^{2 + }}\], the natural orange color of potassium dichromate changes to green color and after acidified reaction tin in \[S{n^{2 + }}\] oxidation state oxidizes to its highest oxidation state of \[S{n^{4 + }}\]resulting in the blue color of the solution.