The oxoanion in which the oxidation state of | vedclass

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(D) To find the oxidation state of the central atom in each oxoanion,we use the rule that the sum of oxidation states equals the charge on the ion.$1$

Vedclass · Accepted Answer

$Cr_2O_7^{2-}$

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(D) To find the oxidation state of the central atom in each oxoanion,we use the rule that the sum of oxidation states equals the charge on the ion.$1$. For $SO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$. Sulfur is in Group $16$,so oxidation state $+6 
eq 16$.$2$. For $VO_2^+$: $x + 2(-2) = +1 \Rightarrow x = +5$. Vanadium is in Group $5$,so oxidation state $+5 = 5$.$3$. For $MnO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$. Manganese is in Group $7$,so oxidation state $+6 
eq 7$.$4$. For $Cr_2O_7^{2-}$: $2x + 7(-2) = -2$ $\Rightarrow 2x = +12$ $\Rightarrow x = +6$. Chromium is in Group $6$,so oxidation state $+6 = 6$.Note: Both $VO_2^+$ and $Cr_2O_7^{2-}$ satisfy the condition. However,in standard competitive chemistry contexts,$Cr_2O_7^{2-}$ is the classic example of an oxoanion where the oxidation state equals the group number $(6)$. Given the options,$D$ is the most appropriate answer.

The oxoanion in which the oxidation state of the central atom is same as its group number in the periodic table is

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