Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve! - Redox Reaction and Electrochemistry - JEE Main-2

When KMnO4 acts as an oxidising agent and ultimately forms \left [ MnO_{4} \right ]^{-1},MnO_{2},Mn_{2}O_{3},Mn^{2+} then the number of electrons transferred in each case respectively is

  • Option 1)

    4, 3, 1, 5

  • Option 2)

    1, 5, 3, 7

  • Option 3)

    1, 3, 4, 5

  • Option 4)

    3, 5, 7, 1

 

Answers (1)

As we learnt in

Oxidation Number -

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to more electronegative elements.

-

 

 Let us take the constituents one by one and state their oxidation states.

 

K\overset{+7}{Mn}O_{4},\: O.S. = +7

[\overset{+6}{Mn}O_{4}]^{2}, \:O.S.= +6

MnO_{2},\:O.S.= +4

Mn_{2}O_{3},\:O.S.= +3

Mn^{2+},\:O.S.= +2

\therefore The number of electrons transferred in each case 

= 7-6, 7-4, 7-3, 7-2 respectively.

number of electrons transferred is 1,3,4,5

 


Option 1)

4, 3, 1, 5

This option is incorrect.

Option 2)

1, 5, 3, 7

This option is incorrect.

Option 3)

1, 3, 4, 5

This option is correct.

Option 4)

3, 5, 7, 1

This option is incorrect.

Posted by

Sabhrant Ambastha

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question