Question

${K_2}C{r_2}{O_7}$ react with hydrazine to form a product. The oxidation state of ${\text{Cr}}$ in the product will be:
A) ${\text{ + 4}}$
B) ${\text{ + 3}}$
C) ${\text{ + 5}}$
D) ${\text{ + 2}}$

Answer 1

Hint: The chemical formula for the hydrazine is ${N_2}{H_4}$. One can write the schematic representation of the reaction and find out the product. While calculating the oxidation state of an atom one should calculate the surrounding atoms oxidation state which shows the central atom oxidation state which is the opposite.

Complete step by step answer:
1) First of all let's learn the concept of oxidation state where it shows the degree of oxidation of an atom in the chemical compound.
2) Now let's see the chemical reaction scheme for the reaction of ${K_2}C{r_2}{O_7}$ with hydrazine as follows,
${K_2}C{r_2}{O_7} + {N_2}{H_4} + {H_2}S{O_4} \to C{r_2}{(S{O_4})_3} + {N_2} + {K_2}S{O_4} + {H_2}O$
3) As we got the product of the reaction as $C{r_2}{(S{O_4})_3}$ and let's calculate the oxidation state of the chromium atom in the formula.
4) The chemical formula $C{r_2}{(S{O_4})_3}$ has negatively charged oxygen atoms and positively charged sulfur atoms. The oxygen atom has an $ - 2$ oxidation state, the sulfur has $ + 6$ and we need the value of the oxidation state of chromium which we are considering as $x$. Let's calculate the oxidation state as follows,
Chemical formula, $C{r_2}{(S{O_4})_3}$
Let's put the values of each element in the equation to find out the oxidation state,
$2x + \left[ {\left( { + 6} \right) \times \left( 3 \right) + \left( { - 2} \right) \times \left( 4 \right) \times \left( 3 \right)} \right] = 0$
Lets first do the sum of all the numerical values in the square bracket,
$2x + ( + 18) + ( - 24) = 0$
$\Rightarrow 2x + ( - 6) = 0$
Now, let's take the value of $2x$ on one side,
$2x = + 6$
The negative sign changes to positive after changing the side,
$x = \dfrac{{ + 6}}{2}$
$\Rightarrow x = + 3$
Oxidation state $ = + 3$
Therefore, the oxidation state of the chromium atom in the product $C{r_2}{(S{O_4})_3}$ is $ + 3$ which shows option B as the correct choice.

Note:
The oxidation state of an element is the hypothetical charge that an atom would possess if all the bonds to atoms of different elements were completely ionic. The oxidation state of an element is generally represented by an integer value which can be positive, negative, or zero. The sum of the oxidation state of all the elements present in the chemical formula is equal to the charge present on that chemical formula. If there is no charge then it is taken as zero.