Practice Questions

Q49.The pair that contains two P −H bonds in each of the oxoacids is: (1) H3 PO3 and H3 PO2 (2) H4P2O5 and H3 PO3 (3) H4P2O5 and H4P2O6 (4) H3 PO2 and H4P2O5

Q49.Consider the given plots for a reaction obeying Arrhenius equation (0°C < T < 300°C) : ( K and Ea are rate constant and activetion energy, respectively ) (I) (II) (1) I is wrong but II is right (2) Both I and II are wrong (3) I is right but II is wrong (4) Both I and II are correct

Q49.Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of log x vs logp is shown in the given graph. x is proportional to, m m (1) p3 (2) p2 (3) p2 / 3 (4) p3 / 2

Q49.The pair that does not require calcination is (1) ZnO and Fe2 O3. xH2 O (2) ZnO and MgO (3) ZnCO3 and CaO (4) Fe2 O3 and CaCO3. MgCO3

Q50.The correct statements among I to III are: (I) Valence bond theory cannot explain the color exhibited by transition metal complexes. (II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes. (III) Valence bond theory cannot distinguish ligands as weak and strong field ones. (1) (I) and (II) only (2) (II) and (III) only (3) (I) and (III) only (4) (I), (II) and (III)

Q50.The transition elements that has the lowest enthalpy of atomisation is: (1) V (2) Fe (3) Zn (4) Cu

Q50.The correct statement is: (1) Pig iron is obtained from cast iron. (2) The Hall-Heroult process is used for the production of aluminium and iron. (3) leaching of bauxite using concentrated NaOH (4) the blistered appearance of copper during the solution gives sodium aluminate and sodium metallurgical process is due to the evolution of silicate. CO2.

Q50.Chlorine on reaction with hot and concentrated sodium hydroxide gives. (1) Cl− and ClO−2 (2) Cl− and ClO−3 (3) Cl− and ClO− (4) ClO−3 and ClO−2

Q50. A ⟶ 2 B + 2H2O B ⟶4HCl 2C + MnO2 + 2H2O 2C H2O,KI⟶2 A + KOH + D In the above (Green) ( Purple ) sequence of reactions, A and D, respectively, are: (1) KI and KMnO4 (2) MnO2 and KIO3 (3) KIO3 and MnO2 (4) KI and K2MnO4

Q50.Consider the hydrated ions of Ti2+, V2+, Ti3+ and Sc3+ . The correct order of their spin-only magnetic moments is: JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper (1) Sc3+ < Ti3+ < V2+ < Ti2+ (2) Ti3+ < Ti2+ < Sc3+ < V2+ (3) Sc3+ < Ti3+ < Ti2+ < V2+ (4) V2+ < Ti2+ < Ti3+ < Sc3+

Q50.The correct option among the following is. (1) Brownian motion in colloidal solution is faster if (2) Colloidal particles in lyophobic sols can be the viscosity of the solution is very high. precipitated by electrophoresis. (3) Colloidal medicines are more effective because (4) Addition of alum to water makes it unfit for they have small surface area. drinking.

Q50.The correct statement about ICl5 and ICl4- is: (1) ICl5 is trigonal bipyramidal and ICl4- is (2) Both are isostructural. tetrahedral. (3) ICl5 is square pyramidal and ICl4- is tetrahedral. (4) ICl5 is square pyramidal and ICl4- is square planar.

Q50.Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y , is: (1) 1 (2) 5 (3) 3 (4) 7

Q50.Match the catalysts (Column I) with products (Column II) Column I Column II Catalyst Product A. V2O5 P. Polyethylene B. TiCl4/Al(Me)3 Q. Ethanal C. PdCl2 R. H2SO4 D. Iron Oxide S. NH3 (1) (A) −(S); (B) −(R); (C) −(Q); (D) −(P) (2) (A) −(R); (B) −(S); (C) −(P); (D) −(Q) (3) (A) −(Q); (B) −(R); (C) −(P); (D) −(S) (4) (A) −(R); (B) −(P); (C) −(Q); (D) −(S)

Q51.The chloride that CANNOT get hydrolysed is: (1) PbCl4 (2) CCl4 (3) SnCl4 (4) SiCl2

Q51.The magnetic moment of an octahedral homoleptic Mn(II) complex is 5 .9 B. M. . The suitable ligand for this complex is: (1) CN − (2) CO (3) Ethylenediamine (4) NCS −

Q51.The lanthanide ion that would show colour is: (1) Lu3 + (2) La3 + (3) Gd3 + (4) Sm3 +

Q51.A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet-coloured) and B (green-coloured). A can show optical activity, but, B is optically inactive. What type of isomers do A and B represent? JEE Main 2019 (10 Jan Shift 2) JEE Main Previous Year Paper (1) Coordination isomers (2) Linkage isomers (3) Ionisation isomers (4) Geometrical isomers

Q51.Homoleptic octahedral complexes of a metal ion ′M 3+′ with three monodentate ligands L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is: JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper (1) L3 > L1 > L2 . (2) L1 > L2 > L3 . (3) L2 > L1 > L3 . (4) L3 > L2 > L1 .

Q51.The species that can have a trans-isomer is: (en = ethane −1, 2 −diamine, ox = oxalate) (1) [Cr(en)2(ox)]+ (2) [Pt(en)Cl2] (3) [Zn(en)Cl2] (4) [Pt(en)2Cl2]2+

Q51.The pair that has similar atomic radii is: (1) Mo and W (2) Sc and Ni (3) Mn and Re (4) Ti and Hf

Q51.Increasing order of reactivity of the following compounds for SN1 substitution is: (1) B < C < D < A (2) B < A < D < C (3) A < B < D < C (4) B < C < A < D

Q51.The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2 , is: (1) Co2+ and Fe2+ (2) V 2+ and Fe2+ (3) Cr2+ and Mn2+ (4) V 2+ and Co2+

Q51.The correct statement is: (1) Aniline is a froth stabilizer. (2) Zone refining process is used for the refining of titanium. (3) Zincite is a carbonate ore. (4) Sodium cyanide cannot be used in the metallurgy of silver. JEE Main 2019 (10 Apr Shift 2) JEE Main Previous Year Paper

Q52.The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is: (1) Fe2+ (2) Co2+ (3) Ni2+ (4) Mn2+