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What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
A)$C{{r}^{3+}}$ and $C{{r}_{2}}{{O}_{7}}^{2-}$ are formed
B) $C{{r}_{2}}{{O}_{7}}^{2-}$ and ${{H}_{2}}O$
C) $C{{r}_{2}}{{O}_{4}}^{2-}$ is reduced to +3 state of Cr
D) $C{{r}_{2}}{{O}_{4}}^{2-}$ oxidised to +7 state of Cr

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint: The answer to this question is based on the fact that nitrogen here has the maximum oxidation state present in this molecule and thus this fact gives the correct answer.

Complete step by step solution:
-In our lower classes we have come across the concept about the oxidation states and also about the transition elements.
-Now, let us see what would happen when potassium dichromate is treated with excess dilute nitric acid.
-To start with, in the compound that is potassium chromate given by the molecular formula ${{K}_{2}}Cr{{O}_{4}}$ where the oxidation state of chromium present in it has ‘+6’.
-This compound that is potassium dichromate is treated with nitric acid which has the molecular formula $HN{{O}_{3}}$ where the oxidation state of nitrogen present in this is +5. This oxidation state is the maximum oxidation state that a nitrogen atom can bear.
-Therefore, it is because of this reason that nitric acid cannot be oxidised by potassium dichromate because nitrogen cannot be further oxidised as it is already present in its maximum oxidation state.
-Regarding all these facts the products thus formed will be as \[C{{r}_{2}}{{O}_{7}}^{2-}or{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] and water that is ${{H}_{2}}O$ and thus the oxidation state of chromium in this product will also be +6 and therefore this says that it has not oxidised nitric acid nor have undergone any oxidation changes.

Thus, the correct answer is option (B)

Note: Note that it is not just potassium dichromate but also any other oxidising agent cannot oxidise any compound which is already present in its higher oxidation state which is the maximum state of that particular compound.
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