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

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.