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Redox And Equivalent Concept Quiz-10

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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. Which is not a redox reaction?
  •  H2+Br2 ⟶2HBr
  •  NH4 Cl ⟶NH3+HCl
  •  NH3 NO3 ⟶N2 O+2H2 O
  •  Fe+S ⟶FeS
Solution
(b) No change in ox.no. of any species.  

Q2.The halogen that shows same oxidation state in all its compounds with other elements is:
  •  I2
  •  F2
  •  Cl2
  •  Br2
Solution
(b) F2 shows only −1 ox.no.

Q3. Oxidation states of X,Y,Z are +2, +5 and -2 respectively. Formula of the compound formed by these wii be  
  •  X2 YZ6
  •  XY2 Z6
  •  XY5
  •  X3 YZ4
Solution
(b) The sum of the oxidation states is always zero in neutral compound. 
 The oxidation state of X,Y, and Z are +2, +5 and -2 respectively.
 In X2 YZ6 2×2+5+6(-2)≠0 In XY2 Z6 2+5×2+6(-2)=0 
 In XY5 2+5×5≠0 In X3 YZ4 3×2+5+4(-2)≠0 
 Hence, the formula of the compound is XY2 Z6.

Q4. In acid medium Zn reduces nitrate ion to NH4+ ion according to the reaction Zn+NO3 Zn2++NH4++H2 O (unbalanced) How many moles of HCl are required to teduce half a mole of NaNO3 completely? Assume the availability of sufficient Zn.
  •  5
  •  4
  •  3
  •  2
Solution
(a) ∴4Zn+NO3-+10H+⟶4Zn2++NH4++3H2 O(Net equation)
 4Zn+NO3-+10HCl⟶4Zn2++NH4++5Cl2+3H2
 ∵1 mole of NO3- (0r NaNO3) is reduced by =10 moles of HCl 
 ∴1/2 mole of No3- will be reduced by = 10×1/2 moles of HCl = 5 moles of HCl

Q5.Which one of the compound does not decolourised an acidified solution of KMnO4?
  •  SO2
  •  FeCl3
  •  H2 O2
  •  FeSO4
Solution
  (b) FeCl3 cannot be oxidised because Fe has highest oxidation state.  

Q6. In the aluminothermic process, aluminium acts as :
  •  An oxidising agent
  •  A flux
  •  A reducing agent
  •  A solder
Solution
(c) Cr2 O3+2Al ⟶Al2 O3+2Cr.  

Q7.An element A in a compound ABD has oxidation number An-. It is oxidized by Cr2 O72- in acidic medium. In the experiment 1.68 × 10-3 mole of K2 Cr2 O7 were used for 3.26 × 10-3 mole of ABD. The new oxidation number of A after oxidation is :
  •  3
  •  3-n
  •  n-3
  •  +n
Solution
(b) An-⟶Aa++(a+n)e Cr26++6e ⟶2Cr+3 
 Also, Meq of A=Meq.of K2 Cr2 O7 3.26 × 10-3 (a+n)=1.68 × 10-3 × 6 Or a+n=3 
 ∴ a=3-n

Q8.CrO5 reacts with H2 SO4 to give Cr2 (SO4 )3,H2 O and O2. Moles of O2 liberated by 1 mole of CrO5 in this reaction are :
  •  2.5
  •  1.25
  •  4.5
  •  1.75
Solution
(d) 4CrO5+6H2 SO4 ⟶2Cr2 (SO4 )3+6H2 O+7O2  

Q9.Reaction of Br2 with Na2 CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is
  •  1
  •  3
  •  5
  •  7
Solution
(c) Br2 is disproportionated in basic medium as 3Br2+3Na2 CO3⟶5NaBr+NaBrO3+3CO2

Q10. MnO4- is a good oxidising agent in different medium changing to MnO4- ⟶Mn2+ ⟶ MnO42- ⟶ MnO2 ⟶Mn2 O3 Changes in oxidation number respectively are
  •  1,3,4,5
  •  5,4,3,2
  •  5,1,3,4
  •  2,6,4,3
Solution
(c) MnO4-=Mn=+7 
 MnO42-=Mn=+6 
 MnO2=Mn=+4 
 Mn2 O3=Mn=+3 
 Hence, changes in oxidation number are 5,1,3,4.


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