Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

How many significant figures are present in 0.0000135?

  • 7

  • 8

  • 4

  • 3


D.

3

Significant numbers in 0.0000135 are 3. It is because zeros to the left of the first non-zero digitin a number are not significant.


2.

Assertion : NF3 is a weaker ligand than N(CH3)3.

Reason : NF3 ionizes to give F- ions in aqueous solution.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


C.

If assertion is true but reason is false.

Due to high electronegativity and electron withdrawing nature of F-atom, the lone pair of N-atom in NF3 molecule is not easily available for donation. However, in N(CH3)3, -CHgroup is an electron releasing group, thus lone pair of N-atom in N(CH3)3 molecule can be ligated easily. Except for nitrogen fluoride, all other trihalides hydrolyse in water.


3.

Assertion : Molecular nitrogen is less reactive than molecular oxygen.

Reason: The bond length of N2 is shorter than that of oxygen.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


A.

If both assertion and reason are true and reason is the correct explanation of assertion.

Bond order of N2 = 3, bond order of O2 = 2. Higher the bond order, shorter is the bond length and higher is the bond dissociation energy, i.e., higher stability or lesser reactivity. Thus, N2 is less reactive than O2.


4.

Which of the following arrangement is possible?

  • n l m s
    5 2 2 +1/2
  • 2 2 0 -1/2
  • 3 -2 1 +1/2
  • 0 0 1 +1/2

A.

n l m s
5 2 2 +1/2

For a given value of nl = 0 to n - 1;

For a given value of lm = -l to +l including zero;

For a given value of ms+12, -12

Therefore, when = 5, then l = 0, 1, 2, 3, 4;

when l = 2, then = -2, -1, 0, +1, +2

The value of s can be ±12

Hence, the arrangement, n = 5, = 2, m = 2, s = +12 is possible for an electron.


5.

The amount of BaSO4 precipitated on mixing BaCl2 (0.5 M) with H2SO4 (1 M) will correspond to

  • 0.5 M

  • 1.0 M

  • 1.5 M

  • 2.0 M


A.

0.5 M

[Ba2+] = 0.5 mol L-1

[SO42-] = 1mol L-1

Ba2+ (aq) + SO42- (aq)  BaSO4 (s)

0.5 mol of Ba2+ would react with 0.5 mol of SO42- ions to form 0.5 mol of BaSO4.


6.

In a decay process XZA changes into YZ-1A. Which process is this?

  • β-decay

  • β+ decay

  • α- decay

  • γ- decay.


B.

β+ decay

In a decay process, when XZA changes into YZ-1A, the process is known as β+ decay. It can be explained as-

XZA  YZ-1A + e+10 (or β+)


7.

The pair of amphoteric hydroxides is

  • Be(OH)2, Al(OH)3

  • Al(OH)3, LiOH

  • B(OH)3, Be(OH)2

  • Be(OH)2, Mg(OH)2


A.

Be(OH)2, Al(OH)3

Al and Be show similar properties due to their diagonal relationship.


8.

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, U and w correspond to

  • U < 0, w = 0

  • U < 0, w < 0

  • U > 0, w = 0

  • U > 0, w > 0


A.

U < 0, w = 0

Bomb calorimeter commonly used to find the heat of combustion of organic substances, consists of a sealed combustion chamber, called a bomb.

The reaction in a bomb calorimeter proceeds at constant volume so, w = 0 and U = q.

Hence, U < 0, w = 0


9.

One mole of SO3 was placed in a vessel of 1 L capacity at a certain temperature and the following equilibrium was established.

2SO3 (g)  2SO2 (g) + O2 (g)

If 0.6 moles of SO2 were formed at equilibrium, Kc for the reaction will be

  • 0.036

  • 0.45

  • 0.54

  • 0.675


D.

0.675

2SO3 (g)11-0.6= 0.4   2SO2 (g)00.6 + O2 (g)00.3   initial molesat equilibrium moles

Kc[SO2]2 O2[SO3]2 = (0.6)2 0.320.42 = 0.675


10.

α-particles can be detected by using

  • thin aluminium sheet

  • barium sulphate

  • zinc sulphide screen

  • gold foil


C.

zinc sulphide screen

α particles can be detected by using zinc sulphide screen. It emits a flash of light upon  α-particle. It was first used by Rutherford.