Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

Which of the following statement is correct for a nucleophile? 

  • Nucleophile is s Lewis acid

  • Ammonia is a nucleophile

  • Nucleophiles attack low electrons density sites

  • Nucleophiles attack low electrons density sites


A.

Nucleophile is s Lewis acid

Nucleophiles are electron rich species. Hence, act as a Lewis base but not Lewis acid.

822 Views

12.

An aqueous solution of which of the following compounds is the best conductor of electric current?

  • Acetic acid C2H4O2

  • Hydrochloric acid, HCl

  • Ammonia, NH3

  • Ammonia, NH3


B.

Hydrochloric acid, HCl

HCl is strong acid and dissociates completely. Hence, it conducts electricity best in its aqueous solution.

952 Views

13.

If the equilibrium constant for 

straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 space NO space left parenthesis straight g right parenthesis space is space straight K comma space the space equilibrium space constant space for
1 half straight N subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space NO space left parenthesis straight g right parenthesis space will space be

  • K1/2

  • K/2


A.

K1/2

As we can see the reaction for which we have to find out equilibrium constant is different only in stoichiometric coefficient as compared to the given reaction. Hence, we can find equilibrium constant as,

straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 space NO space left parenthesis straight g right parenthesis thin space is space straight K
straight i. straight e. space straight K space equals 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 space space space.. space left parenthesis straight i right parenthesis
Let space equilibrium space constant space for space the space reaction comma
1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 NO space left parenthesis straight g right parenthesis space is space straight K apostrophe
straight i. straight e space space straight K apostrophe space equals space fraction numerator left square bracket NO right square bracket over denominator left square bracket straight N subscript 2 right square bracket to the power of 1 divided by 2 end exponent left square bracket straight O subscript 2 right square bracket to the power of 1 divided by 2 end exponent end fraction
On space squaring space both space sides
straight K apostrophe squared space equals 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 space space.. space left parenthesis ii right parenthesis
On space comparing space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
straight K equals space straight K apostrophe squared
or
straight K to the power of apostrophe space equals space square root of straight K

1553 Views

14.

In which of the following pairs, both the species are not isostructural?

  • SiCl4.PCl4+

  • Diamond, carbide

  • NH3, PH3

  • NH3, PH3


D.

NH3, PH3

a) SiCl4.PCl4: Both are isostructural because their central atom is sp3 hybridised and both have a tetrahedral arrangement.

b) Diamond and silicon carbide (SiC): Both are isostructural because their central atom is sp3 hybridised and both have a  tetrahedral arrangement.

c) NH3 and PH3 have sp3 geometry.

d) XeF4 has sp3d2 hybridizations while XeO4 has sp3 hybridizations.Hence, XeF4 and XeO4 are not isostructural.



1205 Views

15.

The number of water molecules is maximum in 

  • 18 molecules of water

  • 1.8 g of water

  • 18 g of water

  • 18 g of water


D.

18 g of water

Mole is the biggest unit to measure the number of molecules/atom/ions.

therefore, 

1 mole of water contains molecules = 6.02 x 1023

therefore, 18 moles of water contain molecules


= 18 x 6.02 x 1023 molecules 

Now, 1 mole of water = 18 g of water

Hence, number of molecules in 18 g of water
 = 6.02 x1023

and 1.8 g of water contains  = 6.02 x 1022 molecules.
1478 Views

16.

Two possible stereo-structures of CH3CHOH.COOH, which are optically active, are called

  • diastereomers

  • atropisomers

  • enantiomers

  • enantiomers


C.

enantiomers


Both are enantiomers.
1397 Views

17.

On heating which of the following releases CO2 most easily?

  • K2CO3

  • Na2CO3

  • MgCO3

  • MgCO3


C.

MgCO3

Order of thermal stability is 

K2CO3 > Na2CO3 > CaCO3 > MgCO3
Thus, MgCO3 releases CO2 most easily

MgCO3 + heat ---> MgO + CO2

1639 Views

18.

Gadolinium belongs to 4f series. Its atomic number is 64. which of the following is the correct electronic configuration of gadolinium?

  • [Xe]4f8 6d2

  • [Xe]4f95s1

  • [Xe]4f7 5d16s2

  • [Xe]4f7 5d16s2


C.

[Xe]4f7 5d16s2

Atomic number of Gadolinium = 64

electronic configuration of Gd,

[Xe]4f7 5d16s2

743 Views

19.

What is the mole fraction of the solute in a 1.00 m aqueous solution? 

  • 0.177

  • 1.770

  • 0.0354

  • 0.0354


D.

0.0354

Molality space left parenthesis straight m right parenthesis space equals space fraction numerator 1000 space straight x space straight n over denominator straight N space straight x space straight M end fraction

where comma space straight n space equals space number space of space moles space of space solute
straight N space equals space number space of space moles space of space solvent
straight M space equals space molar space mass space of space solvent
Given comma space straight m space equals space 1
1 space equals space fraction numerator 1000 space straight x space straight n over denominator straight N space straight x space 18 end fraction space rightwards double arrow space straight n over straight N space equals space 18 over 1000

Or space fraction numerator straight n over denominator straight n space plus space straight N end fraction space equals space 18 over 1018 space equals space 0.0177

Alternat space method space
straight N space equals space 1000 over 18 space equals 55.5 space mol
left square bracket therefore space 1 space straight m space solution space implies space that space 1 space mole space of space solute space is space present space in space 1 space kg space of space 1000 space straight g space water right square bracket
therefore comma
Mole space fraction space of space solute
space equals space fraction numerator straight n over denominator straight n space plus straight N end fraction space equals space fraction numerator 1 over denominator 1 plus 55.5 end fraction space equals space 0.0177
863 Views

20.

The number of structural isomers possible from the molecular formula C3H9N is 

  • 4

  • 5

  • 2

  • 2


A.

4

Structural isomers of C3H9N are 


1543 Views