Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# In the following reaction: $Cr{(OH)_3} + O{H^ - } + I{O_3}^ - \to Cr{O_4}^{2 - } + {H_2}O + {I^ - }$:A. $I{O_3}^ -$ is oxidising agentB. $Cr{(OH)_3}$ is oxidisedC. $6{e^ - }$ are being taken per iodine atomD. None of these

Last updated date: 20th Jun 2024
Total views: 405.3k
Views today: 9.05k
Verified
405.3k+ views
Hint: We know that the transition elements can show more than one type of valency so here we have to use charge balance to calculate the charge on any transition elements. And for oxide the valency is $- 1$.

Transition elements: Those elements which are in groups from three to eleven. They are called a transition because they are in between the s- block elements and p-block elements. For example: Scandium, iron, zinc, etc. They have fully or at least one electron in their d-orbits. For example: scandium has one d-electron, zinc has ten d-electrons. Due to the different number of electrons present in their d-shells they show different valencies. For example: Scandium having one electron can show three valencies as one, two and three. Because the atomic number of scandium is $21$ and its electronic configuration is $Ar3{d^1}4{s^2}$. By losing only one s-electron it will attain one valency and by losing both the s-electrons it will attain valency two. And by losing two s-electrons and one d-electron it will achieve three valencies which are stable states of scandium. Because after losing three electrons it will attain electronic configuration of noble gas argon.
In the given reaction, on the reactant side the oxidation number of chromium is three and on the product side the oxidation number of chromium is six. So chromium hydroxide is oxidized. And for the iodide on the reactant side the oxidation number of iodide is $+ 5$ and on the product side the oxidation number of iodide is $- 1$. So iodide is reduced. Hence $I{O_3}^ -$ is an oxidising agent in the reaction. And the electron gain for the iodine is $5 - ( - 1) = 6$.