Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

If the value of an equilibrium constant ofr particular reaction is 1.6 x1012 then at equilibrium the system will contains

  • all reactants

  • mostly reactants

  • mostly products

  • similar amounts of reactants and products


C.

mostly products

For a reaction,

stack straight A space with Reactant below leftwards harpoon over rightwards harpoon space straight B with product below

straight K space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket space subscript eq end fraction space equals space 1.6 space straight x space 10 to the power of 12 space equals space fraction numerator left square bracket straight B right square bracket subscript eq over denominator left square bracket straight A right square bracket subscript eq end fraction

therefore space left square bracket straight B right square bracket subscript eq space greater than greater than space left square bracket straight A right square bracket subscript eq
So, mostly the product will be present in the equilibrium mixture.

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2.

Magnetic moment 2.84 BM is given by (At. no. Ni= 28, Ti= 22, Cr=24, Co = 27)

  • Ni2+

  • Ti3+

  • Cr3+

  • Co2+


A.

Ni2+

Magnetic space moment space left parenthesis straight mu right parenthesis space equals space square root of straight n space left parenthesis straight n plus 2 right parenthesis space end root space BM space where comma space
straight n equals space number space of space unpaired space electrons
straight mu space equals space 2.84 space left parenthesis given right parenthesis
284 space equals space space square root of straight n left parenthesis straight n plus 2 right parenthesis space end root space straight B. straight M
left parenthesis 2.84 right parenthesis squared space equals space straight n space left parenthesis straight n plus 2 right parenthesis
8 equals straight n squared plus 2 straight n
straight n squared space plus 2 straight n space minus 8 space equals space 0
solving space the space equation
straight n squared space plus 4 straight n minus 2 straight n minus 8 space equals 0
straight n left parenthesis straight n plus 4 right parenthesis minus 2 left parenthesis straight n plus 4 right parenthesis space equals space 0
straight n equals 2

The electronic configuration of the given ions,
Ni2+ = [Ar]3d8 4s0 (two unpaired electron)
Ti3+ = [Ar] 3d1 4s0 ( one unpaired electrons)
Cr3+ = [Ar] 3d3 ( three unpaired electrons)
Co2+ = [Ar],3d7, 4s0 (three unpaired electrons)
So, only Ni2+ has 2 unpaired of electrons.
2504 Views

3.

The angular momentum of electrons in d orbital is equal to

  • square root of 6 space h
  • square root of 2 straight h
  • 2 square root of 3 straight h
  • 0 h


A.

square root of 6 space h

Angular momentum of electrons in d- orbital is 

square root of straight l space left parenthesis straight l plus 1 right parenthesis end root space fraction numerator h over denominator 2 pi end fraction semicolon space f o r space d space minus o r b i t a l space space l space equals 2
equals space square root of 2 space left parenthesis 2 plus 1 right parenthesis end root space equals space h space equals space square root of 6 h end root space space space space open parentheses therefore space h equals space fraction numerator h over denominator 2 pi end fraction close parentheses space

827 Views

4.

Which of the following statements is correct for a reversible process in a state of equilibrium?

  • ΔG = -2.303RT log K
  • ΔG = 2.303RT log K

  • ΔG0 = -2.303RT log K

  • ΔG0 = 2.303RT log K


C.

ΔG0 = -2.303RT log K

 ΔG =ΔG0+2.303RT log K log Q
At equilibrium when Δ G = 0 and Q = K
 then ΔG = ΔG0 +2.303 RT log K = 0

ΔG0 = -2.303 RT log K

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5.

Solubility of the alkaline earth's metal sulphates in water decreases in the sequence

  • Mg>Ca>Sr>Ba

  • Ca>Sr>Ba>Mg>

  • Sr>Ca>Mg>Ba

  • Ba>Mg>Sr>Ca


A.

Mg>Ca>Sr>Ba

The solubility of the sulphates. The sulphates becomes less soluble as  you fo down the group i.e

Mg> Ca>Sr> Ba

The magnitude of the lattice energy remains almost constant as the size of the sulphate ion is so big that small increase in the size of the cation from Be to Ba does not make any difference. However, the hydration energy decreases from Be2+ to Ba2+ appreciably as the size of the cation increases down the group. The significantly high solubility of MgSO4 is due to high enthalpy of solvation on the smaller Mg2+ ions.

734 Views

6.

Which of the following species contains an equal number of σ and Π bond? 

  • HCO3-

  • XeO4

  • (CN)2

  • CH2(CN)2


B.

XeO4

Structure

σ and Π 

 

 σ bond -4
 Π bond-1 

 

 σ bond -4
 Π bond-4

 straight N identical to straight C minus straight C identical to straight N

 σ bond -3
 Π bond-4

 

 σ bond -6
 Π bond-4 

2307 Views

7.

Which of the following pairs of ions are isoelectronic and isostructural? 

  • CO subscript 3 superscript 2 minus end superscript comma space SO subscript 3 superscript 2 minus end superscript
  • ClO subscript 3 superscript minus comma space CO subscript 3 superscript 2 minus end superscript
  • SO subscript 3 superscript 2 minus end superscript comma NO subscript 3 superscript minus
  • ClO subscript 3 superscript minus comma space SO subscript 3 superscript 2 minus end superscript

D.

ClO subscript 3 superscript minus comma space SO subscript 3 superscript 2 minus end superscript

Isostructural chemical compounds have similar chemical structures.
Isoelectronic terms refer to two atom, ions or molecules that have the same number of valence electrons.


ClO3- = SO32-
 Number of electrons
CO32- = 6+2+24 =32
SO32- = 16+2+24 = 32
ClO32- = 17+24+1 = 42
NO32- = 7+2+24 = 33
Hence ClO32- and SO32-  are isoelectronic and are pyramidal in shape.

2058 Views

8.

The number of d- electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which one of the following?

  • S- electronic in Mg (Z=12)

  • p-electrons in Cl( Z=17)

  • d- electrons in Fe (Z=26)

  • p-electrons in Ne (Z=10 )


B.

p-electrons in Cl( Z=17)

Electronic configuration of Fe2+ is [Ar]3d6 4s0 
therefore, Number of electrons  = 6
Mg- 1s2 2s2 2p6  (6s electrons)
It matches with the 6d electrons of Fe2+ 
Cl- 1s2 2s2 2p6 3s2 3p5 (11 p electrons)
It does not match with the 6d  electrons of Fe2+
Fe- [Ar] 3d6 4s2 (6d electrons)

It does not match with the 6d electrons of Fe2+
Ne- 1s2 2s2 2p6 3s2 3p 6 (6p electrons)
It matches with the 6d electrons of Fe2+.
Hence, Cl has 11 p electrons which do not match in number with 6d electrons of Fe2+

984 Views

9.

The Ksp of Ag2CrO4, AgCl, AgBr and AgI are respectively, 1.1 x 10-12, 1.8 x 10-11, 8.3 x 10-17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4

  • AgI

  • AgCl

  • AgBr

  • Ag2CrO4


D.

Ag2CrO4

Ag subscript 2 CrO subscript 4 space rightwards harpoon over leftwards harpoon space 2 Ag to the power of plus space plus CrO subscript 4 superscript 2 minus end superscript
solubility space product
straight K subscript sp space equals space left parenthesis 2 straight s right parenthesis squared space straight x space straight S space equals space 4 straight s cubed
straight K subscript sp space equals space left parenthesis 1.1 space straight x space 10 to the power of negative 12 end exponent right parenthesis
straight S space equals space 3 square root of straight K subscript sp over 4 end root space equals space 0.65 space straight x space 10 to the power of negative 4 end exponent
AgCl space rightwards harpoon over leftwards harpoon space Ag to the power of plus space plus Cl to the power of minus
straight K subscript sp space equals space straight S space xS space space space space space space space space space left parenthesis straight K subscript sp space equals space 1.8 space straight x space 10 to the power of negative 10 end exponent right parenthesis
straight S space equals space square root of straight K subscript sp end root space equals space 1.34 space straight x space 10 to the power of negative 5 end exponent

AgBr space rightwards harpoon over leftwards harpoon Ag to the power of plus space plus Br to the power of minus space
straight K subscript sp space equals space straight S space straight x space straight S space space space space space left parenthesis straight K subscript sp space equals space 5 space straight x space 10 to the power of negative 13 end exponent
straight S space equals space square root of straight K subscript sp end root space equals space 0.71 space straight x space 10 to the power of negative 6 end exponent

AgI space rightwards harpoon over leftwards harpoon space Ag to the power of plus plus space straight I to the power of minus
Ksp space equals space straight S space xS space space space space space space space straight K subscript sp space equals space 8.3 space straight x space 10 to the power of negative 17 end exponent right parenthesis
straight S space equals space square root of straight K subscript sp space end root space equals space 0.9 space straight x space 10 to the power of negative 8 end exponent space

therefore space comma space solubility space of space Ag subscript 2 CrO subscript 4 space is space highest space space so space it space will space precipitate space last.
4272 Views

10.

The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase?

  • Ar < K+< Ca2+

  • Ca2+ < Ar < K+

  • Ca2+ < K+ < Ar

  • K+< Ar < Ca2+


B.

Ca2+ < Ar < K+

Ca2+ < K+ < Ar
 Ar, K+ and Ca2+ are isoelectronic i.e with the same number of electrons, 18. For isoelectronic species, ionic radii decrease with increases in effective (relative) positive charge. Also Ar, K and Ca belong to the same period.

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