redox-and-equivalent-concept-quiz-6

Redox And Equivalent Concept Quiz-6

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..

Q1. What is the oxidation number of chlorine in ClO₃^-?

•  +5
•  +3
•  +4
•  +2

Solution
(a) Oxidation number of Cl in ClO₃^-. ClO₃=-1 x+3(-2)=-1 x=+6-1 x=+5  

Q2.Equivalent mass of oxidizing agent in the reaction is. 

 SO₂+2H₂ S ⟶3S+2H₂ O

•  32
•  64
•  16
•  8

Solution

(c) 4e+S⁴⁺⟶S⁰ 

 ∴ E^{(SO₂ )} =64/4=16

Q3. Oxidation number of P in P₂ O₇^4- is : 

•  +3
•  +4
•  +5
•  +6

Solution
(c) 2 × a+7 × (-2)= -4 ∴ a= + 5

Q4. In CH₂ Cl₂, the oxidation number of C is :

•  -4
•  +2
•  Zero
•  +4

Solution
(c) a+2 × 1+2 × (-1)=0 ∴a=0

Q5.Bleaching action of chlorine in presence of moisture is :

•  Reduction
•  Oxidation
•  Hydrolysis
•  substitution

Solution
  (b) Cl₂+H₂ O ⟶2HCl+O; thus, matter is oxidised by liberated oxygen.  

Q6. How many milliliter of 0.5 N SnCl₂ solution will reduce 600 mL of 0.1 N HgCl₂ to Hg₂ Cl₂?

•  120 mL
•  60 mL
•  30 mL
•  240 mL

Solution
(a) Meq.if SnCl₂=Meq.of HgCl₂ 0.5 × V=600 × 0.1 ∴V=120 mL  

Q7.K₃ Fe(CN)₆ is used as ……. Indicator for FeSO₄ vs.K₂ Cr₂ O₇ titrations.

•  Self
•  External
•  Internal
•  Not an

Solution

(b) Since, K₃ Fe(CN)₆ reacts with FeSO₄ (if added internally) to give blue colour of iron complex.

Q8.Oxidation number of chromium in K₂ Cr₂ O₇ is :

•  +2
•  +3
•  +6
•  -4

Solution
(c) 2 × 1+2 × a+7 × (-2)=0 ∴a= +6  

Q9.In which transfer of five electrons takes place?

•  MnO₄^- ⟶Mn²⁺
•  CrO₄^2- ⟶Cr³⁺
•  MNO₄^- ⟶MnO₂
•  Cr₂ O₇^2- ⟶2Cr³⁺

Solution

(a) Mn⁷⁺+5e ⟶Mn²⁺.

Q10. In the equation, H₂ S+2HNO₃ ⟶2H₂ O+2NO₂+S. The equivalent weight of hydrogen sulphide is :

•  17
•  34
•  68
•  18

Solution

(a) S^2- ⟶S⁰+2e ∴E=M/2=34/2=17