REDOX AND EQUIVALENT CONCEPT QUIZ-20

REDOX AND EQUIVALENT CONCEPT QUIZ-20

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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. In which of the following compounds, the oxidation number of iodine is fractional?

•  IF₃
•  IF₅
•  I₃^-
•  IF₇

Solution
Oxidation number of iodine in given species is as follows O.N. of iodine in IF₃=+3 O.N. of iodine in I₃^-=-1/3 O.N. of iodine in IF₅=+5 O.N. of iodine in IF₇=+7

Q2.How many electrons are involved in oxidation of KMnO₄ in basic medium?

•  1
•  2
•  5
•  3

Solution

Q3.  Which reaction indicates the oxidising behavior of H₂ SO₄?

•   2PCl₅+H₂ SO₄ ⟶2POCl₃+2HCl+SO₂ Cl₂
•  2NaOH+H₂ SO₄ ⟶Na₂ SO₄+2H₂ O
•  NaCl+H₂ SO₄ ⟶NaHSO₄+HCl
•  2HI+H₂ SO₄ ⟶I₂+SO₂+2H₂ O

Solution
2e+S⁽⁶⁺⁾ ⟶S⁽⁴⁺⁾ S of H₂ SO₄ is reduced.

Q4. The best oxidising agent of the oxygen family is:

•  Tellurium
•  Selenium
•  Sulphur
•  Oxygen

Solution
Oxygen has highest electron affinity in its family.

Q5.In sodium hydride, oxidation state of sodium is :

•  Zero
•  +1
•  -1
•  +2

Solution
 In ionic hydrides, H has -1 ox.no.

Q6. Oxidation number of ‘N’ in N₃ H(hydrazoic acid) is

•  -1/3
•  +3
• 0
•  -3

Solution
N₃ H (hydrazoic acid) +3(x)+1=0 3x+1=0 x=-1/3

Q7. The eq. wt. of Fe₃ O₄ in , Fe₃ O₄ + KMnO₄ ⟶ Fe₂ O₃ + MnO₂ is:

•  M/6
•  M
•  2M
•  M/3

Solution
2e+2Fe₃^(8/3)+)⟶ 3Fe₂⁽³⁺⁾ ∴E_{(Fe3 O4} =M/(No.of electrons lost or gained by one molecule) = M/1

Q8.The oxidation number of iodine in IF₅ is :

•  +5
•  -5
•  -1
•  +1

Solution
a+5 × (-1)=0 ∴a= +5

Q9.The eq. wt. of FeC₂ O₄ in , FeC2 O₄ ⟶Fe⁽³⁺⁾+2CO₂ is∶

•  its mol. wt.
•  mol. wt./3
•  mol. wt./4
•  None of these

Solution
Fe⁽²⁺⁾ ⟶Fe⁽³⁺⁾+e ( (C⁽³⁺⁾ )₂⟶2C⁽⁴⁺⁾+2e)/(Fe⁽²⁺⁾+C₂ O₄^(2-)⟶CO₂+Fe⁽³⁺⁾+3e)_ ∴E=M/3

Q10. Which compound has oxidation number of carbon equal to zero?

•  C₆ H₆
•  CH₃
•  C₂ H₄
• C₆ H₁₂ O₆

Solution
6 × a+12 × 1+6 × (-2)= 0 ∴a=0