Question:

Solution of $Na_2CrO_4$ in water is coloured. Oxidation state of $Cr$ in $Na_2CrO_4$ is $+6$.

Updated On: Jul 28, 2022
  • If both assertion and reason are true and the reason is the correct explanation of the assertion
  • If both assertion and reason are true and the reason is not the correct explanation of the assertion
  • If assertion is true but reason is false
  • If both assertion and reason are false
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The Correct Option is B

Solution and Explanation

$Na _{2} CrO _{4}$ solution in water is yellow in colour, when acidified changes to orange colour (due to the formation of $Cr _{2} O _{2}{ }^{2-}$ ions). $Na _{2} CrO _{4}+ H _{2} SO _{4} \longrightarrow Na _{2} Cr _{2} O _{7}+ Na _{2} SO _{4}+ H _{2} O$ Oxidation state : $Na _{2} CrO _{4}$ Let oxidation state of $Cr$ be $x$. $\Rightarrow 2(+1)+x+4(-2)=0 \Rightarrow x=+6$.
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Concepts Used:

D and F Block Elements

The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements. 

They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.

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