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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which of the following compounds is not coloured yellow?

  • Zn2[Fe(CN)6]

  • K3[Co(NO2)6]

  • (NH4)3[As(Mo3O10)4]

  • BaCrO4


A.

Zn2[Fe(CN)6]

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2.

The Colour of KMnO4 is due to 

  • M → L charge transfer transition

  • d-d transition

  • L →M charge transfer transition

  • σ →σ*


C.

L →M charge transfer transition

KMnO4 → K+ + MnO-4
Therefore,
In MnO4-, Mn has +7 oxidation state having no electron in d- orbitals.
It is considered that higher the oxidation state of metal, greater is the tendency to occur L →M charge transfer because ligand is able to donate the electrons into the vacant d- orbital of metal.
Since, charge transfer is Laporte as well as spin allowed, therefore, it shows colour.

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3.

Which of the following compounds will exhibit geometrical isomerism?

  • 1-phenyl-2-butene

  • 3-phenyl-1-butene

  • 2-phenyl-1-butene

  • 1,1-diphenyl-1-propane


A.

1-phenyl-2-butene

Alkene in which different groups are attached with the double bonded carbon atoms, exhibit geometrical isomerism.

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4.

The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine)

  • 2

  • 3

  • 4

  • 6


B.

3

[Pt (Cl) (py) (NH3) (NH2OH)]+  is square planar complex.
The structures are formed by fixing a group and then arranging all the groups. Hence, this complex shows three geometrical isomers.

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5.

Which of the following properties are not shown by NO?

  • It is diamagnetic in gaseous state

  • It is a neutral oxide

  • It combines with oxygen to form nitrogen dioxide

  • Its bond order 2.5


A.

It is diamagnetic in gaseous state

D.

Its bond order 2.5

NO is paramagnetic in the gaseous state because, in a gaseous state, It has one unpaired electron.
Total number of electron present = 7+8 = 15e-
Hence, there must be the presence of unpaired electron in a gaseous state while in a liquid state, it dimerises due to the unpaired electron.

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6.

The octahedral complex of a metal ion M3+ with four Monodentate Ligands, L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue respectively. The increasing order of ligand strength of the four ligands is

  • L4< L3<L2<L1

  • L1<L3<L2<L4

  • L3<L2<L4<L1

  • L1<L2<L4<L3


B.

L1<L3<L2<L4

Ligand space field space strength
space space space space space space space space space space space space space space space space space space space space space space space proportional to Energy space of space light space absorbed
space space space space space space space space space space space space space space space space space space space space space space space proportional to space fraction numerator 1 over denominator Wavelength space of space light space absorbed end fraction

λ L1 L2 L3 L4
Absorbed light Red Green Yellow Blue
the wavelength of absorbed light decreases

Therefore, increasing order of energy of wavelengths abosrbed reflects greater extent of crystal - field splitting, hence higher field strength of the ligand.
Energy: Blue(L4)> Green (L2)> Yellow (L3) > Red (L1)
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7.

Which of the following complex species is not expected to exhibit optical isomerism? 

  • [Co(en)3]3+

  • [Co(en)2Cl2]+

  • [Co(NH3)3Cl3]

  • [Co(NH3)3Cl3]


C.

[Co(NH3)3Cl3]

Complexes of the type [MA2(AA2), [M(AA3)] exhibit optical isomerism.

Optical Isomerism is shown by only those complexes which lack elements of symmetry.
In the given complexes, [Co(NH3)3Cl3] shows facial as well as meridional isomerism. But both of the forms contain a plane of symmetry. Thus, only [Co(NH3)3Cl3] complex does not show optical isomerism.

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8.

The pair having the same magnetic moment is:

[At. No. : Cr=24, Mn=25, Fe=26, Co=27]

  • [Cr(H2O)6 ]2+ and [Fe(H2O)6 ]2+

  • [Mn(H2O)6 ]2+ and [Cr(H2O)6 ]2+

  • [CoCl4 ]2− and [Fe(H2O)6 ]2+

  • [Cr(H2O)6 ]2+ and [CoCl4 ]2−


A.

[Cr(H2O)6 ]2+ and [Fe(H2O)6 ]2+

Complex ion Electronic configuration metal ion Number of unpaired electrons (n)
[Cr(H2O)6 ]2+  Cr2+ ; [ar] 3d4
[Fe(H2O)6 ]2+ Fe2+ ; [Ar]3d6
[Mn(H2O)6 ]2+  Mn2+ ; [Ar] 3d5
[CoCl4 ]2−  Co2+ ; [Ar]3d7
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9.

The number of unpaired electrons in[NiCl4]2-, Ni(CO)4 and (Cu(NH3)4]2+ respectively are

  • 0, 2, 1

  • 2, 0, 1

  • 0, 2, 1

  • 2, 2, 0


B.

2, 0, 1

Configuration for metal (M) in various complex.

Ni2+ has 2 unpaired electrons.

Ni0 has zero unpaired electrons with sp3-hybridisation.

Cu2+ has one unpaired electron with dsp2-hybridisation.

Therefore, number of unpaired electrons is (2, 0, 1).


10.

Which one of the following complexes shows optical isomerism?

  • cis[Co(en)2Cl2]Cl

  • trans[Co(en)2Cl2]Cl

  • [Co(NH3)4Cl2 ]Cl

  • [Co(NH3)3Cl3 ]


A.

cis[Co(en)2Cl2]Cl

The optically active compound capable of rotating the plane-polarized light to the right or left.

cis[Co(en)2Cl2]Cl is optically active compound.

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