Chemistry Aptitude Quiz
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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1. 
Solution
(d) All options are correct
(d) All options are correct
Q2.Which of the following represents a redox reaction?
Solution
(d) (a) and (b) are neutralisation reactions c. The oxidation state of Cu is +2 in both reactant and product and SO_4^(2-) ion does not change. So it is none reaction d. It is a redox reaction Zn⟶Zn^(2+)+2e^- (Oxidation) 2H^⊕+2e^-⟶H_2 (Reduction)
(d) (a) and (b) are neutralisation reactions c. The oxidation state of Cu is +2 in both reactant and product and SO_4^(2-) ion does not change. So it is none reaction d. It is a redox reaction Zn⟶Zn^(2+)+2e^- (Oxidation) 2H^⊕+2e^-⟶H_2 (Reduction)
Q3. The compound Me_3 C- NH_2 on oxidation with acidic KMnO_4 gives:
Solution
(d) Acidic KMnO_4 oxidises (-NH_2) group to (-NO_2) group. So the answer is (c)
(d) Acidic KMnO_4 oxidises (-NH_2) group to (-NO_2) group. So the answer is (c)
Q4. 
Solution
Q5.For decolourisation of 1 mol of KMnO_4, the moles ofH_2 O_2 required is
Solution
(c)
1/5 mol =1/2 mol
1 mol of MnO_4^( ⊖)=5/2 mol of H_2 O_2
(c)
1/5 mol =1/2 mol
1 mol of MnO_4^( ⊖)=5/2 mol of H_2 O_2
Q6. 
Solution
(b) Lindlar's catalyst reduces (C≡C) to(C=C) in syn (cis) addition and imultaneously reduces (-COCl) group to (-CHO) group. So the answer is (b)
(b) Lindlar's catalyst reduces (C≡C) to(C=C) in syn (cis) addition and imultaneously reduces (-COCl) group to (-CHO) group. So the answer is (b)
Q7.Starch iodide is used to test for the presence of
Solution
(c) Statement is self explanatory
(c) Statement is self explanatory
Q8.
Solution
(c) Acidic KMnO_4 breaks the double bond and also oxidises 2°alcohol to ketone, whereas PCC only oxidises 2° alcohol to ketone. So the answer is (c)
(c) Acidic KMnO_4 breaks the double bond and also oxidises 2°alcohol to ketone, whereas PCC only oxidises 2° alcohol to ketone. So the answer is (c)
Q9.Which of the following statements is not correct?v
Solution
(c) ∴Oxidation state of S=+6 (Since S_2 O_8^( 2-)has one peroxide bond) Oxidation state of Os=+8 Oxidation state of S in H_2 SO_5=+6 (Since it has one peroxide bond) K^(1+) O_2^(1-), oxidation state of O=-1/2
(c) ∴Oxidation state of S=+6 (Since S_2 O_8^( 2-)has one peroxide bond) Oxidation state of Os=+8 Oxidation state of S in H_2 SO_5=+6 (Since it has one peroxide bond) K^(1+) O_2^(1-), oxidation state of O=-1/2
Q10. The oxidation number of Cl in CaOCl_2 is
Solution
(a) Ca^(2+) (OCl^(1-) ) Cl^(1-) OCl^⊖:-2+x=-1⇒x=+1 Cl^⊖:-1 Oxidation states of Cl are -1 and +1
(a) Ca^(2+) (OCl^(1-) ) Cl^(1-) OCl^⊖:-2+x=-1⇒x=+1 Cl^⊖:-1 Oxidation states of Cl are -1 and +1
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