Arrange the following substances in the incre | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The oxidation numbers of $Mn$ in the given compounds are calculated as follows:$MnCl_2$: $x + 2(-1) = 0 \Rightarrow x = +2$$MnO_2$: $x + 2(-2) = 0

Vedclass · Accepted Answer

$MnCl_2 < Mn(OH)_3 < MnO_2 < KMnO_4$

Vedclass · Answer

(D) The oxidation numbers of $Mn$ in the given compounds are calculated as follows:$MnCl_2$: $x + 2(-1) = 0 \Rightarrow x = +2$$MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$$Mn(OH)_3$: $x + 3(-1) = 0 \Rightarrow x = +3$$KMnO_4$: $1(+1) + x + 4(-2) = 0 \Rightarrow x = +7$Thus,the increasing order of the oxidation number of $Mn$ is: $MnCl_2 (+2) < Mn(OH)_3 (+3) < MnO_2 (+4) < KMnO_4 (+7)$.

Arrange the following substances in the increasing order of the oxidation number of $Mn$: $MnCl_2, MnO_2, Mn(OH)_3, KMnO_4$.

Similar Questions

What is the oxidation state of chlorine atom in chloric acid?

Determine the oxidation state of $Cr$ in $K_2Cr_2O_7$.

What is the oxidation number of $S$ in $SO_4^{2-}$?

What are the oxidation states of the most electronegative element in the products of the reaction between $BaO_2$ and $H_2SO_4$?

The oxidation number of chlorine in $HOCl$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module