The oxoanion in which the oxidation state of | vedclass

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Question

(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

$Cr_2O_7^{2-}$

Vedclass · Answer

(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.

Column $A$	Column $B$
$(1) \ NiCl_2 \cdot 6H_2O$	$(a) \ \text{Pink}$
$(2) \ Co(NO_3)_2 \cdot 6H_2O$	$(b) \ \text{Colorless}$
$(3) \ FeCl_3$	$(c) \ \text{Blue}$
$(4) \ CuSO_4 \cdot 5H_2O$	$(d) \ \text{Green}$
	$(e) \ \text{Yellow}$

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

Similar Questions

Match Column $A$ with Column $B$:

Column $A$ Column $B$

$(1) \ NiCl_2 \cdot 6H_2O$ $(a) \ \text{Pink}$

$(2) \ Co(NO_3)_2 \cdot 6H_2O$ $(b) \ \text{Colorless}$

$(3) \ FeCl_3$ $(c) \ \text{Blue}$

$(4) \ CuSO_4 \cdot 5H_2O$ $(d) \ \text{Green}$

$(e) \ \text{Yellow}$

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The fusion of manganese dioxide with $KOH$ in the presence of an oxidizing agent like $KNO_3$ yields a compound. Identify the compound and its color.

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