Practice Questions

Q33.The potassium ferrocyanide solution gives a Prussian blue colour, when added to : (1) CoCl3 (2) CoCl2 (3) FeCl2 (4) FeCl3

Q33.The oxide that gives H2O2 most readily on treatment with H2O is : (1) Na2 O2 (2) BaO2. 8H2O (3) SnO2 (4) PbO2 JEE Main 2021 (27 Aug Shift 2) JEE Main Previous Year Paper

Q33.(A) HOCl + H2O2 →H3O+ + Cl- + O2 (B) I2 + H2O2 + 2OH- →2I- + 2H2O + O2 Choose the correct option. (1) H2O2 act as oxidizing and reducing agent (2) H2O2 acts as oxidising agent in equations ( A ) respectively in equations ( A ) and ( B ) . and ( B ) . (3) H2O2 acts as reducing and oxidising agent (4) H2O2 acts as reducing agent in equations ( A ) respectively in equations ( A ) and ( B ) . and ( B ) .

Q33.Which one of the following methods is most suitable for preparing deionized water? (1) Synthetic resin method (2) Calgon's method (3) Clark's method (4) Permutit method

Q33.Given below are two statements: One is labelled as Assertion A and the other labelled as reason R Assertion A : During the boiling of water having temporary hardness, Mg (HCO3)2 is converted to MgCO3 Reason R : The solubility product of Mg(OH)2 is greater than that of MgCO3 . In the light of the above statements, choose the most appropriate answer from the options given below: (1) Both A and R are true but R is not the correct (2) A is true but R is false explanation of A (3) Both A and R are true and R is the correct (4) A and R both are false. explanation of A

Q33.In the following the correct bond order sequence is: (1) O2−2 > O+2 > O−2 > O2 (2) O+2 > O−2 > O2−2 > O2 (3) O+2 > O2 > O−2 > O2−2 (4) O2 > O−2 > O2−2 > O+2

Q33.The correct shape and I −I −I bond angles respectively in I−3 ion are: (1) T-shaped; 180° and 90° (2) Distorted trigonal planar; 135° and 90° (3) Trigonal planar; 120° (4) Linear; 180°

Q33.Match List-I with List-II List - I List - II (Elements) (Properties) (a) Ba (i) Organic solvent soluble compounds (b) Ca (ii) Outer electronic configuration 6 s2 (c) Li (iii) Oxalate insoluble in water (d) Na (iv) Formation of very strong monoacidic base Choose the correct answer from the options given below: (1) (a)-(ii), (b)-(iii), (c)-(i) and (d)-(iv) (2) (a)-(iv), (b)-(i), (c)-(ii) and (d)-(iii) (3) (a)-(iii), (b)-(ii), (c)-(iv) and (d)-(i) (4) (a)-(i), (b)-(iv), (c)-(ii) and (d)-(iii)

Q33.The oxidation states of nitrogen in NO, NO2, N2O and NO−3 are in the order of : (1) NO−3 > NO2 > NO > N2O (2) NO2 > NO−3 > NO > N2O (3) N2O > NO2 > NO > NO−3 (4) NO > NO2 > N2O > NO−3

Q33.Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : In TlI3, isomorphous to CsI3 , the metal is present in +1 oxidation state. Reason R : Tl metal has fourteen f electrons in its electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below: JEE Main 2021 (26 Feb Shift 2) JEE Main Previous Year Paper (1) A is correct but R is not correct (2) Both A and R are correct and R is the correct explanation of A (3) A is not correct but R is correct (4) Both A and R are correct but R is NOT the correct explanation of A

Q33.A group 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is : (1) Sb (2) P (3) As (4) Bi

Q33.Deuterium resembles hydrogen in properties but: JEE Main 2021 (27 Aug Shift 1) JEE Main Previous Year Paper (1) reacts vigorously than hydrogen (2) reacts just as hydrogen (3) emits β+ particles (4) reacts slower than hydrogen

Q33.Water does not produce CO on reacting with : (1) CO2 (2) CH4 (3) C3H8 (4) C JEE Main 2021 (25 Feb Shift 2) JEE Main Previous Year Paper

Q33.Which of the following compound CANNOT act as a Lewis base? (1) NF3 (2) PCl5 (3) SF4 (4) ClF3

Q33.Given below are two statements: Statement I : o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding. Statement II : o-Nitrophenol has high melting due to hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below: (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Statement I is false but Statement II is true (4) Both Statement I and Statement II are true

Q33.The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is :- (1) 1 and 1 (2) 3 and 1 (3) 2 and 1 (4) 2 and 2

Q33.According to molecular orbital theory, the species among the following that does not exist is: (1) He−2 (2) O2−2 (3) He+2 (4) Be2

Q33.The species given below that does NOT show disproportionation reaction is: (1) BrO−4 (2) BrO− (3) BrO−2 (4) BrO−3 JEE Main 2021 (20 Jul Shift 1) JEE Main Previous Year Paper

Q33.The bond order and magnetic behaviour of O−2 ion are, respectively : (1) 1. 5 and diamagnetic. (2) 1. 5 and paramagnetic. (3) 2 and diamagnetic. (4) 1 and paramagnetic

Q34.The oxidation states of ′P′ in H4P2O7, H4P2O5 and H4P2O6, respectively, are : (1) 7, 5 and 6 (2) 5, 4 and 3 (3) 5, 3 and 4 (4) 6, 4 and 5

Q34.Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Lithium salts are hydrated. Reason (R) : Lithium has higher polarising power than other alkali metal group members. In the light of the above statements, choose the most appropriate answer from the options given below : (1) (A) is not correct but (R) is correct. (2) (A) is correct but (R) is not correct. (3) Both (A) and (R) are correct but (R) is NOT the (4) Both (A) and (R) are correct and (R) is the correct explanation of (A). correct explanation of (A).

Q34.The major component/ingredient of Portland Cement is : (1) dicalcium silicate (2) tricalcium aluminate (3) dicalcium aluminate (4) tricalcium silicate

Q34.What are the products formed in sequence when excess of CO2 is passed in slaked lime? (1) CaO, CaCO3 (2) Ca (HCO3)2′ CaCO3 (3) CaCO3, Ca (HCO3)2 (4) CaO, Ca (HCO3)2

Q34.The set of elements that differ in mutual relationship from those of the other sets is: (1) Li −Mg (2) B −Si (3) Be −Al (4) Li −Na

Q34.Al2O3 was leached with alkali to get X . The solution of X on passing of gas Y, forms Z . X, Y and Z respectively are (1) X = AlOH3, Y = CO2', Z = Al2O3 (2) 𝑋= NaAlOH4, Y = SO2, Z = Al2O3 (3) X = AlOH3, Y = SO2, Z = Al2O3 · xH2O (4) 𝑋= NaAlOH4, Y = CO2, Z = Al2O3, 𝑥H2O