Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

Green chemistry means such reactions which 

  • produce colour during reactions

  • reduce the use and production of hazardous chemicals

  • are related to the depletion of the ozone layer

  • are related to the depletion of the ozone layer


B.

reduce the use and production of hazardous chemicals

Green chemistry involves such reactions which reduce the use and production of hazardous or toxic chemical to reduce pollution from the environment.

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22.

The value of KP1 and Kp2 for the reactions

X ⇌ Y +Z  ... (i)
A ⇌ 2B    ... (ii)

are in the ratio of 9:1. if the degree of dissociation of X and A be equal, then the total pressure at equilibrium (i) and (ii) are in the ratio

  • 3:1

  • 1:9

  • 36:1

  • 36:1


C.

36:1

From equation.

X     ⇌    Y  +   Z
1            0        0     initial mole
(1-α)     α         α     mole at equilibrium


K subscript p subscript 1 space end subscript space equals space fraction numerator p subscript y italic space x italic space p subscript z over denominator straight p subscript straight x end fraction
equals fraction numerator open square brackets begin display style fraction numerator straight alpha space plus straight p subscript 1 over denominator 1 plus straight alpha end fraction end style close square brackets squared over denominator open square brackets begin display style fraction numerator 1 minus straight alpha over denominator 1 plus straight alpha end fraction end style close square brackets straight p subscript 1 end fraction
K subscript p subscript 1 space end subscript space equals space fraction numerator alpha squared p subscript 1 over denominator 1 minus alpha squared end fraction space... space left parenthesis straight i right parenthesis
From space equation
straight A space space space space space space space space leftwards harpoon over rightwards harpoon space space space space space space 2 straight B
1 space space space space space space space space space space space space space space space space space space space 0 space space space space space space space Initial space mole
left parenthesis 1 minus straight alpha right parenthesis space space space space space space space space space space 2 straight alpha space space space space space space mole space at space equilibrium

straight k subscript straight p subscript 2 end subscript equals fraction numerator open square brackets begin display style fraction numerator 2 straight alpha space over denominator 1 plus straight alpha end fraction. straight p subscript 2 end style close square brackets squared over denominator open square brackets begin display style fraction numerator 1 minus straight alpha over denominator 1 plus straight alpha end fraction end style close square brackets straight p subscript 2 end fraction

K subscript p subscript 1 space end subscript space equals space fraction numerator italic 4 alpha squared p italic 2 over denominator 1 minus alpha squared end fraction space.. space left parenthesis ii right parenthesis
From space eqa space left parenthesis straight i thin space right parenthesis and space left parenthesis ii right parenthesis
straight K subscript straight p subscript 1 end subscript over straight K subscript straight p subscript 2 end subscript space equals space fraction numerator straight p subscript 1 over denominator 4 straight p subscript 2 end fraction
equals space 9 over 1 space equals space fraction numerator straight p subscript 1 over denominator 4 straight p subscript 2 end fraction
rightwards double arrow space straight p subscript 1 over straight p subscript 2 space equals space 36 over 1

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23.

On the basis of the following Eo values; the strongest oxidising agent is

[Fe(CN)6]4- → [Fe(CN)6]3-] +e- ;
Eo = -0.35 V

Fe2+ → Fe3+  +e-; E = -0.77 V

  • [Fe(CN)6]4-

  • Fe2+

  • Fe3+

  • Fe3+


C.

Fe3+

Oxidised form + ne- → Reduced Form

The substance which has lower reduction potential are stronger reducing agent while the substances which have higher reduction potential are a stronger oxidising agent.

[Fe(CN)6]3- + e- →[Fe(CN)6]4- ; Eo = 0.35 V

Fe3+ + e- → Fe2+ ;                      Eo = 0.77 V

The reduction potential of Fe3+/ Fe2+ is higher, hence, Fe3+ is a strongest oxidising agent.

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24.

The value of equilibrium constant of the reaction HI (g) ⇌ H2(g)/2 + I2/2 is 8.0
The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2 HI (g)  will be

  • 1/16

  • 1/64

  • 16

  • 16


B.

1/64

HI (g) ⇌ H2/2(g) + I2/2(g)

straight K subscript 1 space equals space fraction numerator left square bracket straight H subscript 2 right square bracket to the power of 1 divided by 2 end exponent left square bracket straight I subscript 2 right square bracket to the power of 1 divided by 2 end exponent over denominator left square bracket HI right square bracket end fraction space.. space left parenthesis straight i right parenthesis
straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight I subscript 2 space left parenthesis straight g right parenthesis space space 2 space HI space left parenthesis straight g right parenthesis
straight K subscript 2 space equals space fraction numerator left square bracket HI right square bracket squared over denominator left square bracket straight H subscript 2 right square bracket left square bracket straight I subscript 2 right square bracket end fraction space.. space left parenthesis ii right parenthesis

From space eqs space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
straight K subscript 1 superscript 2 space equals space 1 over straight K subscript 2
straight K subscript 1 space equals space 8.0
straight k subscript 2 space equals space fraction numerator 1 over denominator straight K subscript 1 superscript 2 end fraction space equals space 1 over 8 squared space equals space 1 divided by 64

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25.

The stability of carbanions in the following

1 right parenthesis space RC space identical to straight C with minus on top straight H
space 2 right parenthesis space straight C subscript 6 straight H subscript 5 to the power of minus
3 right parenthesis space straight R subscript 2 straight C space equals space straight C to the power of minus straight H
4 right parenthesis space straight R subscript 3 straight C space equals space straight C to the power of minus straight H

is in the order of 

  • (1) > (2) > (3) > (4)

  • (2) > (3) > (4) > (1)

  • (4) > (2) > (3)  > (1)

  • (4) > (2) > (3)  > (1)


A.

(1) > (2) > (3) > (4)

When s character increases, then electrons becomes more closer to the nucleus and structure is of lower energy and is more stable.

The stability order of carbanions is as:

RC space identical to straight C to the power of negative space end exponent space greater than thin space straight C subscript 6 straight H to the power of minus subscript 5 space greater than thin space straight R subscript 2 space straight C space equals space straight C to the power of minus straight H space greater than thin space straight R subscript 3 straight C space minus space straight C to the power of minus straight H subscript 2

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26.

In the hydrocarbon

straight C with 6 below straight H subscript 3 space minus straight C with 5 below straight H equals straight C with 4 below straight H minus straight C with 3 below straight H subscript 2 minus straight C with 2 below identical to straight C with 1 below straight H

The state of hybridization of carbon 1, 3, and 5 are in the following sequence

  • sp2, sp, sp3

  • sp, sp3, sp2

  • sp, sp2, sp3

  • sp, sp2, sp3


B.

sp, sp3, sp2

negative straight C minus straight C minus sp cubed
minus straight C equals straight C minus sp squared
minus straight C space identical to space straight C minus space sp
equals straight C equals straight C equals sp

straight H subscript 3 straight C with 6 below and sp cubed on top space minus space straight C with 5 below and sp squared on top straight H space equals straight C with 4 below and sp squared on top straight H space minus space straight C with 3 below and sp cubed on top straight H subscript 2 minus straight C with 2 below and sp on top identical to straight C with 1 below and sp on top straight H

Hence the state of hybridization of carbon 1,3 and 5 are sp, sp3, sp2 respectively.
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27.

Volume occipied by one molecule of water (density = 1 g cm-3) is

  • 9.0 x 10-23

  • 6.023 x 10-23 cm3

  • 3.0 x 10-23 cm3

  • 3.0 x 10-23 cm3


C.

3.0 x 10-23 cm3

6.02 x 1023 molecules of water = 1 mol
                                                  = 18 g
Therefore, Mass of one molecule of water
 fraction numerator 18 over denominator 6.023 space straight x space 10 to the power of 23 space straight g end fraction
straight d equals space straight m over straight V
therefore comma space
straight V space equals space straight m over straight d space equals space fraction numerator 18 over denominator 6.023 space straight x space 10 to the power of 23 space straight x 1 end fraction
space equals space 3 space straight x space 10 to the power of negative 23 space end exponent space cm cubed

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28.

The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrieds decreases in which of the following orders?

  • CsH > RbH > KH > NaH >LiH

  • KH > NaH> LiH > CsH >RbH

  • NaH > LiH >KH> RbH> CsH

  • NaH > LiH >KH> RbH> CsH


D.

NaH > LiH >KH> RbH> CsH

Small anion forms stable compounds with small cation.

The thermal stability of alkali metal hydrides decreases as:

LiH> NaH > KH > RbH  >CsH

because the size of cation increases as;
Li+ < Na+ < K+ < Rb+ <Cs+

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29.

Which of the following complexes exhibits the highest paramagnetic behaviour?

Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities

(At. no. ; Ti = 22, V = 23, Fe =26, Co = 27)

  • [V(gly)2 (OH)2 (NH3)2]+

  • [Fe(en)(bpy)(NH3)2]+

  • [CO(OX)2(OH)2]-

  • [CO(OX)2(OH)2]-


C.

[CO(OX)2(OH)2]-

Greater is the number of unpaired electrons, larger is the paramagnetism.
[V(gly)2(OH)2(NH3)2]+
V23 = [Ar] 4s2, 3d2

Oxidation state of V in [[V(gly)2(OH)2(NH3)2]+ is
x + (-1) x 2 + (-1) x 2 + (0) x 2 = +1
x=+5
V5+ = [Ar]3d0
[Fe(en)(bby)(NH3)2]2+
x + (0) + (0) + (0) x 2 = +2
x = +2

Fe2+ = [Ar] 3d6

But en , bby and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons.
[Co(OX)2 (OH)2]- is 
Oxidation state of Co in [Co(OX)2(OH)2]- is 
x + (-2) x 2 + (-1) x 2 = -1
x-6 = -1
x = +5
Co5+ = [Ar]3d4
[Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.

Hence, [Co(OX)2(OH)2]- has highest paramagnetic behaviour.
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30.

Kohlrausch's law states that at

  • finite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

  • infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.

  • infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever be the nature of the other ions of the electrolyte.

  • infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever be the nature of the other ions of the electrolyte.


D.

infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever be the nature of the other ions of the electrolyte.

According to Kohlrausch's law "at infinite dilution when the dissociation is complete, each ion makes a definite contribution towards equivalent condctivity of the electrolyte irrespective of the nature of the order ion with which it is associated.

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