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 Multiple Choice QuestionsMultiple Choice Questions

1.

Among the following, the maximum covalent character is shown by the compound

  • FeCl2

  • SnCl2

  • AlCl3

  • MgCl2


C.

AlCl3

The covalent character in ionic compounds is governed by Fazanís Rule.
(i) larger the charge on the ions.
(ii) smaller the size of anions.
(iii) larger the size of anion.
(iv) larger the polarizing power larger the covalent character.

AlCl3 will show Maximum covalent character on account of the higher polarising power of Al3+ because of its having higher positive charge and smaller size.

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2.

The species in which the N atom is in a state of sp hybridization is:

  • NO2-

  • NO3-

  • NO2

  • NO2+


D.

NO2+

765 Views

3.

Using MO theory predict which of the following species has the shortest bond length?

  • O22+

  • O2+

  • O2

  • O22−


A.

O22+

Bond space length space proportional to fraction numerator 1 over denominator bond space order end fraction
straight B. straight O space equals space 1 half space equals space left square bracket straight N subscript straight b minus space straight N subscript straight a right square bracket
space equals space 1 half space left square bracket 10 minus 4 right square bracket
Bond space orders space of space straight O subscript 2 superscript plus space comma space straight O subscript 2 superscript minus comma space straight O subscript 2 superscript 2 minus end superscript space and space straight O subscript 2 superscript 2 plus end superscript space are space
2.5 comma space 1.5 space and space 3
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4.

The ionic radii (in Å) of N3–, O2– and F are respectively:

  • 1.36, 1.40 and 1.71

  • 1.36, 1.71 and 1.40

  • 1.71, 1.40 and 1.36

  • 1.71, 1.36 and 1.40


C.

1.71, 1.40 and 1.36

Number of electrons in N3- = 7+3 = 10
Number of electrons in O2- = 8+2 = 10
Number of electrons in F- = 9+1 = 10
Since, all the three species have each 10 electrons hence they are isoelectronic species.
It is considered that, in case of isoelectronic species as the negative charge increase, ionic radii increase and therefore the value of ionic radii are

N3- = 1.71 (highest among the three)
O2- = 1.40
F- = 1.36 (lowest among the three)

N3- > O2-> F-



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5.

Which of the following exists as covalent crystals in the solid state?

  • Iodine

  • Silicon

  • Sulphur

  • Phosphorus


B.

Silicon

D.

Phosphorus

Silicon (Si) – covalent solid
Sulphur (S8) – molecular solid
Phosphorous (P4) – Molecular solid
Iodine (I2) – Molecular solid

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6.

Stability of the species Li2, Li2 and Li2+ increases in the order of:

  • Li2 < Li2+ < Li2-

  • Li2 < Li2+ < Li2 

  • Li2< Li2 < Li2+

  •  Li2 <Li2< Li2+


B.

Li2 < Li2+ < Li2 

Li subscript 2 left parenthesis 6 right parenthesis space equals space straight sigma 1 straight s squared space straight sigma asterisk times 1 straight s squared straight sigma 2 straight s squared
straight B. straight O space equals space fraction numerator 4 minus 2 over denominator 2 end fraction space equals space 1
Li subscript 2 superscript plus space left parenthesis 5 right parenthesis space equals space straight sigma 1 straight s squared space straight sigma asterisk times 1 straight s squared straight sigma 2 straight s to the power of 1
straight B. straight O space equals space fraction numerator 3 minus 2 over denominator 2 end fraction space equals 0.5
Li subscript 2 superscript minus space left parenthesis 7 right parenthesis space space equals space space straight sigma 1 straight s squared space straight sigma asterisk times 1 straight s squared straight sigma 2 straight s squared straight sigma asterisk times 2 straight s to the power of 1
straight B. straight O space equals space fraction numerator 4 minus 3 over denominator 2 end fraction space equals space 0.5
Li2+ is more stable than Li2 because Li2 has more numbers of antibonding electrons.
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7.

The hybridization of orbitals of N atom in NO3-, NO2+ and NH4+ are respectively

  • sp, sp2,sp3

  • sp2 , sp, sp

  • sp, sp3 , sp2

  • Sp2, sp3, sp


B.

sp2 , sp, sp

NO2+
Number of electron pairs = 2
Number of bond pairs = 2
Number of lone pair = 0
So, the species is linear with sp hybridisation.
NO3-
Number of electron pairs = 3
Number of bond pairs = 3
Number of lone pair = 0
So, the species is trigonal planar with sp2 hybridisation

NH4+
Number of electron pairs = 4
Number of bond pairs = 4
Number of lone pair = 0
So, the species is tetrahedral with sp3 hybridisation.

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8.

In which of the following pairs the two species are not isostructural?

  • CO32- and NO3-

  • PCl+4 and SiCl4

  • PF5 and BrF5

  • AlF63- and SF6


C.

PF5 and BrF5

i) CO32- and NO3- = Triangular Planar (sp2)
ii) PCl+4 and SiCl4  = Tetrahedral (sp3)
iii) PF5 and BrF5   = Tetrahedral (sp3)
iv) AlF63- and SF6 = sp3d2 hybridized, octahedral

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9.

The group having isoelectronic species is

  • O2– , F, Na+ , Mg2+

  • O– , F, Na, Mg+

  • O2–, F, Na ,Mg2+

  • O , F, Na+ , Mg2+


A.

O2– , F, Na+ , Mg2+

ions O2-
F Na+ Mg+2
Atomic number 8 9 11 12
No of electron 10 10 10 10
429 Views

10.

In which of the following pairs of molecules/ions, both the species are not likely exist?

  • straight H subscript 2 superscript plus comma space He subscript 2 superscript 2 minus end superscript
  • straight H subscript 2 superscript minus comma space He subscript 2 superscript 2 minus end superscript
  • straight H subscript 2 superscript 2 plus end superscript comma He subscript 2
  • straight H subscript 2 superscript minus comma space He subscript 2 superscript 2 plus end superscript

C.

straight H subscript 2 superscript 2 plus end superscript comma He subscript 2

species which have zero or negative bond order does not exist.

362 Views