From elementary molecular orbital theory we can deduce the electr

Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The structure of IF5 can be best described as :

  • None of these


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2.

From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion as :

  • σ1s2σ*1s2σ2s2σ*2s2π2p4σ2p1

  • σ1s2σ*1s2σ2s2σ*2s2σ2p2π2p3

  • σ1s2σ*1s2σ2s2σ*2s2σ2p3 , π2p2

  • σ1s2σ*1s2σ2s2σ*2s2σ2p2π2p4


A.

σ1s2σ*1s2σ2s2σ*2s2π2p4σ2p1

N2+ = (7 + 7 - 1 = 13)

σ1s2σ*1s2σ2s2σ*2s2π2px2  π2py2 , σ2pz1


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3.

Which of the following has largest atomic radii ?

  • Al

  • Al+

  • Al2+

  • Al3+


4.

Ionisation potential is lowest for :

  • alkali metals

  • inert gases

  • halogens

  • alkaline earth metals


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5.

The percentage of p-character in the orbitals forming P-P bond in P4 is :

  • 25

  • 33

  • 50

  • 75


6.

The valence shell electronic configuration of Cr2+ ion is :

  • 4s03d4

  • 3p64s2

  • 4s33d2

  • 4s23d0


7.

A solution contains 25% H2O , 25% C2H5OH and 50% CH3COOH by mass .The mole fraction of H2O would be :

  • 0.25

  • 2.5

  • 0.502

  • 5.03


8.

If the solubility of PbCl2 at 25°C is 6.3 x 10-2 mol/L , its solubility product is :

  • 1 x 10-6

  • 1 x 10-3

  • 1.1 x 10-6

  • 1.1 x 10-5


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9.

A double bond connecting two atoms .There is a sharing of :

  • 2 electrons

  • 1 electron

  • 4 electrons

  • All electrons


10.

pH of a solution can be expressed as :

  • -loge [H+]

  • -log10 [H+]

  • loge [H+]

  • log10[H+]


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