Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

Name the gas that can readily decolourises acidified KMnO4 solution

  • CO2

  • SO2

  • NO2

  • NO2


B.

SO2

SO2 is readily decolourises acidified KMnO4.

1440 Views

22.

An example of a sigma bonded organometallic compound is :

  • Ruthenocene

  • Grignard's reagent

  • Ferrocene

  • Ferrocene


B.

Grignard's reagent

Grignard's reagent i.e., RMgX is σ-bonded organometallic compound.

3978 Views

23.

The correct order of the stoichiometries of AgCl formed when AgNO3 in excess is treated with the complexes: CoCl3.6NH3, CoCl3.5NH3, CoCl3.4NH3 respectively is

  • 1 AgCl, 3 AgCl, 2 AgCl

  • 3 AgCl, 1 AgCl, 2 AgCl

  • 3 AgCl, 2 AgCl, 1 AgCl

  • 3 AgCl, 2 AgCl, 1 AgCl


C.

3 AgCl, 2 AgCl, 1 AgCl

Complexes are respectively [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2and [Co(NH3)4Cl2]Cl

1445 Views

24.

Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co3+ is

  • [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+

  • [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+

  • [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+

  • [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+


A.

[Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+

The order of the ligand in the spectrochemical series
H2O < NH3< en
Hence, the wavelength of the light observed will be in the order

[Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+
Thus, wavelength absorbed will be in the opposite order
[Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+

2206 Views

25.

A first-order reaction has a specific reaction rate of 10–2 s–1. How much time will it take for 20 g of the reactant to reduce to 5 g?

  • 238.6 second

  • 138.6 second

  • 346.5 second

  • 346.5 second


B.

138.6 second

straight t subscript 1 divided by 2 end subscript space equals space fraction numerator 0.693 over denominator 10 to the power of negative 2 end exponent end fraction
For the reduction of 20 g of reactant to 5 g, two t1/2 is required.
therefore space straight t space equals space 2 space straight x space fraction numerator 0.693 over denominator 10 to the power of negative 2 end exponent end fraction space second
space equals space 138.6 space second
1056 Views

26.

Which of the following is dependent on temperature?

  • Molality

  • Molarity

  • Mole fraction

  • Mole fraction


B.

Molarity

Molarity includes volume of solution which can change with change in temperature.

962 Views

27.

In which pair of ions both the species contain S – S bond?

  • S2O72–, S2O32–

  • S4O62–, S2O32–

  • S2O72–, S2O82–

  • S2O72–, S2O82–


B.

S4O62–, S2O32–

1229 Views

28.

In the electrochemical cell :
Zn|ZnSO4(0.01M)||CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the following, which one is the relationship between E1 and E2?
(Given, RT/F= 0.059)

  • E1= E2

  • E1< E2

  • E1> E2

  • E1> E2


C.

E1> E2

Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu

therefore space straight E subscript 1 space equals space straight E subscript cell superscript 0 minus fraction numerator 2.303 space RT over denominator 2 space straight x space straight F end fraction space straight x space log space fraction numerator left parenthesis 0.01 right parenthesis over denominator 1 end fraction
when space concentrations space are space changed
therefore space straight E subscript 2 space equals space straight E subscript cell superscript 0 space minus space fraction numerator 2.303 over denominator 2 straight F end fraction space straight x space log space fraction numerator 1 over denominator 0.01 end fraction
space straight i. straight e. comma space straight E subscript 1 space greater than space straight E subscript 2

1855 Views

29.

Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?

  • Na

  • K

  • Rb

  • Rb


D.

Rb

Li+ being smallest has maximum charge density
∴ Li+ is most heavily hydrated among all alkali metal ions. The effective size of Li+ in aq solution is therefore, largest.
∴ Moves slowest under the electric field.

1352 Views

30.

The equilibrium constants of the following are.

straight N subscript 2 space plus space 3 straight H subscript 2 space rightwards harpoon over leftwards harpoon with space on top space 2 NH subscript 3 space space straight K subscript 1
straight N subscript 2 space plus space straight O subscript 2 space rightwards harpoon over leftwards harpoon with space on top space 2 NO space space space space straight K subscript 2
straight H subscript 2 space plus space 1 half straight O subscript 2 space rightwards arrow with space on top space straight H subscript 2 straight O space straight K subscript 3

The equilibrium constant (K) of the reaction:

2 NH subscript 3 space plus space 5 over 2 straight O subscript 2 space rightwards harpoon over leftwards harpoon with straight K on top space 2 NO space plus space straight H subscript 2 straight O comma space will space be

  • straight K subscript 1 straight K subscript 3 superscript 3 divided by straight K subscript 2
  • straight K subscript 2 straight K subscript 3 superscript 3 space divided by space straight K subscript 1
  • straight K subscript 2 straight K subscript 3 superscript 3 space divided by space straight K
  • straight K subscript 2 straight K subscript 3 superscript 3 space divided by space straight K

B.

straight K subscript 2 straight K subscript 3 superscript 3 space divided by space straight K subscript 1
straight N subscript 2 space plus 3 straight H subscript 2 space rightwards harpoon over leftwards harpoon with space on top space 2 NH subscript 3 space semicolon space straight K subscript 1 space fraction numerator left square bracket NH subscript 3 right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight H subscript 2 right square bracket cubed end fraction
left parenthesis II right parenthesis space straight N subscript 2 space plus space straight O subscript 2 space rightwards harpoon over leftwards harpoon with space on top space 2 NO space semicolon space straight K subscript 2 space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction
left parenthesis III right parenthesis space straight H subscript 2 space plus 1 half straight O subscript 2 space rightwards arrow with space on top space straight H subscript 2 straight O semicolon space straight K subscript 3 space equals space fraction numerator left square bracket straight H subscript 2 straight O right square bracket over denominator left square bracket straight H subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction
left parenthesis II space plus space 3 space straight x space III minus II right parenthesis space will space space give
2 NH subscript 3 space plus space 5 over 2 straight O subscript 2 space rightwards harpoon over leftwards harpoon with space straight K on top space 2 NO space plus space 3 straight H subscript 2 straight O
therefore space straight K space equals space straight K subscript 2 space straight x space straight K subscript 3 superscript 3 divided by straight K subscript 1
1094 Views