NEET Exam  >  NEET Test  >  Chemistry Class 11  >  Test: Oxidation Number in Redox Reaction - NEET MCQ

Oxidation Number in Redox Reaction - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Oxidation Number in Redox Reaction (18 Questions)

You can prepare effectively for NEET Chemistry Class 11 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Oxidation Number in Redox Reaction". These 18 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 18 minutes
  • - Number of Questions: 18

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Oxidation Number in Redox Reaction - Question 1

In which case change of oxidation number of V is maximum? 

Detailed Solution: Question 1



Test: Oxidation Number in Redox Reaction - Question 2

In which of the following set of compounds oxidation number of oxygen is not (- 2)?

Detailed Solution: Question 2


OF2 Fluorine is most electronegative atom with oxidation number = - 1
x - 2 = 0
x = + 2
H2O2   2 + 2x = 0 ⇒ x = - 1 

Test: Oxidation Number in Redox Reaction - Question 3

Which is chlorate (I) ion?

Detailed Solution: Question 3

  • ClO3: A very reactive inorganic anion.
  • The term chlorate can also be used to describe any compound containing the chlorate ion, normally chlorate salts. 
  • Example: Potassium chlorate, KClO3

Test: Oxidation Number in Redox Reaction - Question 4

Which of the following is not an example of redox reaction?

Detailed Solution: Question 4

a) BaCl2 + H2SO4 → BaSO4 + 2HCl is not a redox reaction, as there is no change in the oxidation state of any element.
It is an example of double displacement reactions.

Test: Oxidation Number in Redox Reaction - Question 5

Which of the following atom has been assigned only single oxidation number?

Detailed Solution: Question 5

  • Fluorine is the most electronegative element. It can gain one electron.​
    Thus, Oxidation number = -1
    F+ e- → F-
  • Due to very high (IE), it cannot lose electrons.

(a) H- (-1), (H+ (+1)
(b) O -2 in oxide, -1 in peroxide, and +2 in OF2
(c) +3 in N2O3, -3 in NH3, +3 in HNO2, +5 in HNO3, + 1 in N2O

Test: Oxidation Number in Redox Reaction - Question 6

The oxidation number of P in Ba(H2PO2)2, Ba(H2PO3)2 and Ba(H2PO4)2 are respectively

Detailed Solution: Question 6



Test: Oxidation Number in Redox Reaction - Question 7

the oxidation number of sulphur in S8,S2F2,H2S respectively are

Detailed Solution: Question 7

(i) Oxidation state of element in its free state is zero.

(ii) Sum of oxidation states of all atoms in compound is zero.

O.N of S in S8=0;    O.N of S in S2F2=+1

O.N of S in H2S=-2

Test: Oxidation Number in Redox Reaction - Question 8

In which of the following reactions oxidation number of chromium has been affected?

Detailed Solution: Question 8


Oxidation number of Cr changes from +6 to +3.

Test: Oxidation Number in Redox Reaction - Question 9

Prussian blue is represented by KFex[Fe(CN)6]. Value of x is

Detailed Solution: Question 9

Test: Oxidation Number in Redox Reaction - Question 10

In the following reaction, , oxidation number of

Detailed Solution: Question 10

In NH4NO2, the oxidation number of N in NH4+ is -3, and of N in NO2- is +3. 

Test: Oxidation Number in Redox Reaction - Question 11

Select the set of compounds having fractional oxidation number in one or more atoms.

Detailed Solution: Question 11

*Multiple options can be correct
Test: Oxidation Number in Redox Reaction - Question 12

Direction (Q. Nos. 13 and  14) This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. Which of the following species has/have oxidation number of the metal as + 6 ?

Detailed Solution: Question 12




Test: Oxidation Number in Redox Reaction - Question 13

Oxidation numbers of P in PO4−3, of S in SO42− and that of Cr in Cr2O72− are respectively,

Detailed Solution: Question 13

The correct answer is option A
(I) xPO43−​ ⇒ x + 4 × (−2) = −3
                ⇒x = −3 + 8 = +5
                ⇒x = +5
Oxidation number of P = +5
(II) xSO42−​ ⇒ x + 4 × (−2) = −2
                ⇒x = −2 + 8
                  ⇒x = +6
Oxidation number of S=+6
(III) xCr2​O72− ​    ⇒2x + 7 × (−2) = −2
                        ⇒2x =−2+14
                        ⇒2x=12
                        ⇒x= 12/2​ = +6

*Multiple options can be correct
Test: Oxidation Number in Redox Reaction - Question 14

Direction (Q. Nos. 15-16) This section contains  a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)

The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO(nitrosyl).

Q.Oxidation number of the Fe in the ring is

Detailed Solution: Question 14

Let, oxidation number of Fe = x

*Multiple options can be correct
Test: Oxidation Number in Redox Reaction - Question 15

The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO(nitrosyl).

Q. Magnetic moment  of Fe in the ring is 

Detailed Solution: Question 15

Fe2+ is added asFeSO4
Fe+ is formed by charge transfer from NO to Fe2+ 



Fe+ has three unpaired electrons (N).

Test: Oxidation Number in Redox Reaction - Question 16

Direction (Q. Nos. 17) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.

Match the compounds/ions having underlined atoms of different oxidation number (in Column I) with values (in Column II).



Detailed Solution: Question 16

(i) CaOCI2 has



Oxidation number = + 6
Oxidation number = - 2
(ii) → (p,u)


(iii) → (q.v)


*Answer can only contain numeric values
Test: Oxidation Number in Redox Reaction - Question 17

Direction (Q. Nos. 18 and 19) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

Q.   S2O32- has two types of sulphur atoms. What is the difference in the oxidation states of two types of sulphur atoms?


Detailed Solution: Question 17

 The two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation.  

So difference in the oxidation states of two types of sulphur atoms will be

= (+6)- (-2) 

= +8

                                              

Test: Oxidation Number in Redox Reaction - Question 18

The difference in the oxidation number of the two types of sulphur atoms in Na2S4O6 is _________. [HTJEE2011]

Detailed Solution: Question 18

the difference of oxidation states of two S is 5−0 = 5

98 videos|243 docs|71 tests
Information about Test: Oxidation Number in Redox Reaction Page
In this test you can find the Exam questions for Test: Oxidation Number in Redox Reaction solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Oxidation Number in Redox Reaction, EduRev gives you an ample number of Online tests for practice
98 videos|243 docs|71 tests
Download as PDF