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.

 Multiple Choice QuestionsMultiple Choice Questions

1.

The valence shell electronic configuration of Cr2+ ion is :

  • 4s03d4

  • 3p64s2

  • 4s33d2

  • 4s23d0


A.

4s03d4

The valence shell electronic configuration of Cr+ ion is :

= 1s2 , 2s2 2p6 , 3s2 3p6 3d4 , 4s0


2.

The structure of IF5 can be best described as :

  • None of these


D.

None of these

Number of hybrid orbitals = number of bp + number of lp

= 5 + 1 = 6

Thus , hybridisation is sp3d2 but geometry , due to the presence of one lone pair , is square pyramidal , i.e. ,


3.

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


C.

0.502

Mole fraction of H2O = number of moles of H2Ototal number of moles of all components

Let the total mass of solution = 100 g

Mass of H2O = 25 g

Mass of C2H5OH = 25g

Mass of CH3COOH = 50 g

Moles of H2O = 2518 = 1.388

( Molar mass of H2O = 18)

Moles of C2H5OH =2546 = 0.543

( Molar mass of C2H5OH = 46)

Moles of CH3COOH = 5060 = 0.833

( Molar mass of CH3COOH = 60)

Total no. of moles = 1.388 + 0.543 + 0.833  = 2.764

 Mole fraction of H2O = 1.3882.764 = 0.502


4.

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


B.

1 x 10-3

PbCl2 completely ionised in the solution as 

PbCl2  Pb2+ + 2Cl-

i.e , 1 mole of PbCl2 in the solution gives 1 mole of Pb2+ ion and 2 moles of Cl- ions .

Now , as the solubility of PbCl2

= 6.3 x 10-2 mol/L

 [Pb2+] = 6.3 x 10-2 mol/L

and [Cl-] = 2 X 6.3 X 10-2 mol/L

= 12.6 x 10-2 mol/L

Ksp for PbCl2 = [Pb2+][Cl-]2

= (6.3 x 10-2) x (12.6 x 10-2)2

= 1 x 10-3


5.

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

  • 2 electrons

  • 1 electron

  • 4 electrons

  • All electrons


C.

4 electrons

A double bond connecting two atoms , thus there is a sharing of 4 electrons .We can explain it by taking example of C-C double bond .


6.

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

  • 25

  • 33

  • 50

  • 75


D.

75

Each P atom in P4 is sp3 hybridised , hence the percentage ofs and p character is 25% and 75% .


7.

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


8.

Ionisation potential is lowest for :

  • alkali metals

  • inert gases

  • halogens

  • alkaline earth metals


A.

alkali metals

The atomic radii of alkali metals are the largest in their respective periods due to which the ionisation enthalpy of the alkali metals are the lowest as compared to the elements in other groups .


9.

Which of the following has largest atomic radii ?

  • Al

  • Al+

  • Al2+

  • Al3+


A.

Al

Atomic radii varies inversely with nuclear charge , ie ,

Atomic radii  1nuclear charge


10.

pH of a solution can be expressed as :

  • -loge [H+]

  • -log10 [H+]

  • loge [H+]

  • log10[H+]


B.

-log10 [H+]

pH is defined as the negative logarithm of the hydrogen ion concentration in solution .
pH = -log10 [H+]