Subject

Chemistry

Class

NEET Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

In the electronic structure of H2SO4 , the total number of unshared electrons is

  • 20

  • 16

  • 12

  • 8


2.

The correct order towards bond angle is

  • sp3 < sp2 < sp

  • sp < sp2 < sp3

  • sp < sp3 < sp2

  • sp2 < sp3 < sp


3.

Which of the following relation is correct ?

  • Ist IE of C > Ist IE of B

  • Ist IE of C < Ist IE of B

  • IInd IE of C > IInd IE of B

  • Both (b) and (c)


4.

Which of the following configurations corresponds to element of highest ionisation energy?

  • 1s2, 2s1

  • 1s2, 2s2, 2p3

  • 1s2, 2s2, 2p2

  • 1s2, 2s2, 2p6, 3s1


Advertisement
5.

Which of the following has a bond formed by overlap of sp-sp3 hybrid orbitals?

  • CH3 - C  C - H

  • CH3 - CH = CH - CH3

  • CH2 = CH - CH = CH2

  • HC CH


6.

The density of a gas is found to be 1.56 g/L at 745 mm pressure and 65°C. What is the moleculer mass of the gas?

  • 44.2 u

  • 4.42 u

  • 2.24 u

  • 22.4 u


7.

Considering H2O as weak field ligand, the number of unpaired electrons in [Mn(H2O)6] will be (Atomic number of Mn = 25)

  • five

  • two

  • four

  • three


8.

The quantum numbers +12 and -12 for the electron spin represent

  • rotation of the electron in clockwise and anticlockwise direction respectively

  • rotation of the electron in anticlockwise and clockwise direction respectively

  • magnetic moment of the electron pointing up and down respectively

  • two quantum mechanical spin states which have no classical analogues


Advertisement
Advertisement

9.

Which of the following has the highest bond order?

  • N2

  • O2

  • He2

  • H2


A.

N2

N2( 7 + 7 = 14e-) = KK*σ2s2σ*2s2π2px2

                            π2py2 , σ2pz2

                   BO    = 12(8 - 2) = 3

O(8 + 8 =16e-) = KK*σ2s2σ*2s2σ2pz2

                             π2px2  π2py2 , π* 2px1  π* 2py1

                   BO   = 12 (8 - 4) = 2

He( 2 + 2 = 4e-) = σ1s2σ*1s2

                       BO = 12(2 - 2) = 0

H2 (1 + 1 = 2e-) = σ1s2

                  BO   = 12(2 - 0) = 1

Hence, N2 has the highest bond order.


Advertisement
10.

In long form of Periodic Table, the properties of the elements are a periodic function of their

  • atomic size

  • ionisation energy

  • atomic mass

  • atomic number


Advertisement