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D and F Block Elements question

2002 · Shift 0 · Q45
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D and F Block Elements question

2002 · Shift 0 · Q45

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
When KMnO4KMnO_4KMnO4​ acts as an oxidising agent and ultimately forms [MnO4]2−[MnO_4]^{2-}[MnO4​]2−, MnO2MnO_2MnO2​, Mn2O3Mn_2O_3Mn2​O3​, Mn2+Mn^{2+}Mn2+ then the number of electrons transferred in each case respectively is :
  1. A
    4, 3, 1, 5
  2. B
    1, 5, 3, 7
  3. C
    1, 3, 4, 5
  4. D
    3, 5, 7, 1
View written solutionFree

Correct answer: C

  1. Find oxidation state of Mn in KMnO4KMnO_4KMnO4​

In MnO4−MnO_4^-MnO4−​: x+4(−2)=−1x + 4(-2) = -1x+4(−2)=−1 x−8=−1x - 8 = -1x−8=−1 x=+7x = +7x=+7 So, Mn is initially in oxidation state +7+7+7.

  1. Find electrons gained in each product

Since KMnO4KMnO_4KMnO4​ acts as an oxidising agent, Mn gets reduced from +7+7+7 to lower oxidation states.

(i) Formation of [MnO4]2−[MnO_4]^{2-}[MnO4​]2−

Let oxidation state of Mn be xxx: x+4(−2)=−2x + 4(-2) = -2x+4(−2)=−2 x−8=−2x - 8 = -2x−8=−2 x=+6x = +6x=+6 Change in oxidation state: +7→+6+7 \to +6+7→+6 So electrons gained =1= 1=1.

(ii) Formation of MnO2MnO_2MnO2​

Let oxidation state of Mn be xxx: x+2(−2)=0x + 2(-2) = 0x+2(−2)=0 x−4=0x - 4 = 0x−4=0 x=+4x = +4x=+4 Change in oxidation state: +7→+4+7 \to +4+7→+4 So electrons gained =3= 3=3.

(iii) Formation of Mn2O3Mn_2O_3Mn2​O3​

Let oxidation state of Mn be xxx: 2x+3(−2)=02x + 3(-2) = 02x+3(−2)=0 2x−6=02x - 6 = 02x−6=0 2x=62x = 62x=6 x=+3x = +3x=+3 Change in oxidation state: +7→+3+7 \to +3+7→+3 So electrons gained per Mn =4= 4=4.

(iv) Formation of Mn2+Mn^{2+}Mn2+

Change in oxidation state: +7→+2+7 \to +2+7→+2 So electrons gained =5= 5=5.

  1. Collect the electron transfers respectively

1,3,4,51, 3, 4, 51,3,4,5

  1. Match with options

This corresponds to Option C.

  1. Comparison with stored correct answer

Stored correct answer is C, which matches the derived answer.

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