Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which of the following ions has the highest value of ionic radius:

  • O2-

  • F-

  • Li+

  • B3+


A.

O2-

O2- has the highest value of ionic radius and this can be explained on the basis of ze (nuclear charge / number of electrons) ratio. when ze ratio increases the size decreases and when ze ratio decreases , the size increases.

ze  1r


2.

A cylinder of compressed gas contains nitrogen and oxygen in the ratio of 3:1 by mole. The total mass of the gas mixture if the cylinder is known to contain 2.5 x 10g of oxygen will be:

  • 8 x 103 g

  • 9 x 10g

  • 9 x 103 g

  • 8 x 102 g


B.

9 x 10g

Number of moles of oxygen in the cylinder

=mass in grammolecular mass in gram

=2.5 × 10432 = 781.25

 Number of moles of N2 = 3 x 781.25

=2343.75

Mass of nitrogen in cylinder 

=6.5625 x 104 g

Hence , the total mass of gas cylinder 

= 2.5 x 104 + 6.56259 x 104

= 9.0625 x 104 g


3.

On adding A to the reaction at equilibrium, AB (s)  A(g) + B(g) the new equilibrium concentration of A becomes double, the equilibrium concentration of B would become:

  • 1/2 of its original value

  • 1/4 of its original value

  • 1/8 of its original value

  • Twice its original value


A.

1/2 of its original value

K[A] [B][AB]

Now , if [A] = 2 × [A] , B should be 12 [B].

So , that Kc remains constant.


4.

The entropy will be positive in which of the following reaction change.

  • H2(g) + I2 (g)  2HI (g)

  • HCl(g) + NH3 (g)  N2O(g) + 2H2O(g)

  • NH4NO3(s)N2O(g) + 2H2O(g)

  • MgO(s) + H2(g)  Mg(s) + H2O(l)


C.

NH4NO3(s)N2O(g) + 2H2O(g)

Entropy is a measure of randomness of the system, so entropy of substance in different physical states is in the order Gas > Liquid > Solid

ΔS=Sproduct -Sreactant

In option (c), solid is converted into gaseous state, so ΔS will be positive.


5.

The vapour pressure of pure water is 25.0 mm Hg and that of an aqueous dilute solution of urea in 20mm Hg at 25°C. The molarity of the solution is:

  • 12.6

  • 14.9

  • 13.8

  • 9.6


C.

13.8

The given values are:

PA=25mmHg , Ps=22mmHg , m =?

Using Raoult's law

PA-PSPA=XB

Substituting the values:

XB=25-2025=0.2 XB=1-XB

=1-0.2=0.8


6.

Assuming that Hund's rule in violated,the bond order and magnetic nature of the  diatomic molecule B2 is:

  • 1 and diamagnetic

  • 0 and diamagnetic

  • 1 and paramagnetic

  • O and paramagnetic


A.

1 and diamagnetic

B2 : Total electrons = 10

Configuration : σ1s2 , σ* 1s2 ,  σ2s2 , σ* 2s2 ,  π2px1  π2py1

If Hund's rule is violated, then

σ1s2 , σ* 1s2 ,  σ2s2 , σ* 2s2 ,  π2px2  π2py0

So, diamagnetic bond order = 6-42 = 1


7.

Wave number of a spectral line for a given transition is xcm-1 for He+ then its value for Be+ (iso electronic of H+) for same transition is:

  • x cm-1

  • 4x cm-1

  • x4cm-1

  • 2x cm-1


B.

4x cm-1

v- =(wave number)

=RHZ21n12-1n22

ν1(He+ , z=2)=RH(22)1n12-1n22

v2(Be3+ , z=4)=RH(42)1n12-1n22

v2v1=4

 v2=4v1=4x


8.

A solution contains 10 mL of 0.1 N NaOH and 10mL of 0.05 NH2SO4.pH of this solution is:

  • zero

  • less than 7

  • greater than 7

  • 7


C.

greater than 7

Number of milliequivalents of NaOH = 0.1 x 10 = 1

Number of milliequivalents of H2SO4
= 10 x 0.05 = 0.5

Milliequivalent of H2SO4 will neutralise NaOH so that (1-0.5) = 0.5 milliequivalent of NaOH is left unneutralised.

Therefore, pH of the resulting solution will be greater than seven i.e., pH > 7,


9.

The correct sets of quantum numbers which represents the highest energy level in an atomis:

  • n=4 , l=0 , m=0 , s = + 12

  • n=3 , l=0 , m=0 , s = + 12

  • n=3 , l=1 , m=1 , s = + 12

  • n=3 , l=2 , m= -1 , s = + 12


D.

n=3 , l=2 , m= -1 , s = + 12

The order of energy of orbitals can be calculated from (n + 1) rule. The lower the value of (n +l) For an orbital, lower is its energy. If two orbitals have same (n+J) value, the orbital with lower value of n has lower energy.

(a) n+l = 4  (b) n+l = 3

(c) n+l = 4   (d) n+l = 4

 

 The correct set which represents the highest energy level is n = 3 , l = 2 , m = -2 and s = + 12


10.

The hybridisation and shape of IF5 are respectively:

  • sp3d , trigonal bipyramide

  • sp3d , see saw

  • sp3d , square pyramidal

  • sp3d , pentagonal pyramidal


C.

sp3d , square pyramidal

IF5 has 5 bond pair and 1 lone pair

The geometry is square pyramidal with sp3d hybridisation