Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The emission spectrum of hydrogen discovered first and the region of the electromagnetic spectrum in which it belongs, respectively are

  • Lyman, ultraviolet

  • Lyman, visible

  • Balmer, ultraviolet

  • Balmer, visible


D.

Balmer, visible

Balmer explained the wave numbers of spectral lines present in the visible region in hydrogen spectrum. These are given by-

v(cm-1) = 109677122-1n2

Here, n = 3, 4, 5, ....

Thus, Balmer spectrum of hydrogen was discovered first and it lies in the visible region.


2.

At a certain temperature the time required for the complete diffusion of 200-mL of H2 gas is 30 min. The time required for the complete diffusion of 50 mL of O2 gas at the same temperature will be

  • 60 min

  • 30 min

  • 45 min

  • 15 min


B.

30 min

According to Graham's law of diffusion,

Rate of diffusion, r1M=Vt

Here, M = molecular mass; V = volume and t = time

Thus, for H2 gas,

20030=12 ...(i)

For O2 gas,

50t=132 ...(ii)

From equation (i) and (ii)

200×t30×50=322

t =16×30×50200=4×304= 30 min


3.

Four gases P, Q, R and S have almost same values of 'b' but their 'a' values (a, b are van der Waals' constants) are in the order Q < R < S < P. At a particular temperature, among the four gases, the most easily liquefiable one is

  • P

  • Q

  • R

  • S


A.

P

Higher the value of 'a', more will be the tendency to get liquefy. Since, value of 'a' is highest for P1 thus, P is the most liquefiable gas among the given


4.

The correct order of decreasing length of the bond as indicated by the arrow in the following structures is

  • I > II > III

  • II > I > III

  • III > II > I

  • I > III > II


C.

III > II > I

The length of carbon-carbon single bond is always greater than that of the carbon-carbon double bond.

In III, positive charge is not in conjugation of double bond, so the bond does not acquire a double bond character. However, chances of acquiring double bond character (of the indicated bond) is much more in I than that in case of II. Thus, the correct order of decreasing bond length is


5.

For one mole of an ideal gas,the slope of V vsT curve at constant pressure of 2 atm is X L mol-1 K-1.The value of the ideal universal gas constant 'R' in terms of X is

  • X L atm mol-1 K-1

  • X2L atm mol-1 K-1

  • 2X L atm mol-1 K-1

  • 2 X atm L-1 mol-1 K-1


C.

2X L atm mol-1 K-1

Ideal Gas Equation -

pV = nRT

or, V =nRTp+C

or, V =RTp+0 (for 1 mol)

When Vvs T is plotted, a straight line is obtained, slope of ehich is given by R/p. Thus,

Slope =Rp

X =R2

or R = 2 X L atm mol-1 K-1


6.

β- emission is always accompained by

  • formation of antineutrino and α-particle

  • emission of α-particle and γ-ray

  • formation of antineutrino and γ-ray

  • formation of antineutrino and positron


C.

formation of antineutrino and γ-ray

During the β-emission, a neutron gets converted into a proton along with the formation of antineutrino and electron.

n10H11Proton+e0-1βparticle+vAnti-neutrino+γ-ray


7.

As per de-Broglie's formula a macroscopic particle of mass 100 g and moving at a velocity of 100 cm s-1 will have a wavelength of

  • 6.6 × 10-29 cm

  • 6.6 × 10-30 cm

  • 6.6 × 10-31 cm

  • 6.6 × 10-32 cm


C.

6.6 × 10-31 cm

According to de-Broglie wavelength,λ =hmv

Given, m = 100 gm; v = 100 cm s-1

h = 6.6× 10-34 J-s = 6.6× 10-27 erg-s

On substituting values, we get

λ =6.6×10-27100×100=6.6×10-31


8.

The compound that will have a permanent dipole moment among the following is

  • I

  • II

  • III

  • IV


A.

I

Symmetrical molecules and symmetricaltrans-alkene have a net dipole moment zero. CH2Cl2 is not a symmetrical molecule, thus it will have a permanent dipole.

μ1234 = 0 ;μ3 andμ41 andμ2. Thusμnet≠ 0

Symmetricaltrans-alkeneμnet≠ 0

μ1234net = 0

μnet = 0


9.

During the emission of a positron from a nucleus, the mass number of the daughter element remains the same but the atomic number

  • is decreased by 1 unit

  • is decreased by 2 units·

  • is increased by 1 unit

  • remains unchanged


A.

is decreased by 1 unit

When a positron is emitted, a proton is converted into neutron and neutrino. Thus, the number of protons and hence the atomic number decreases by one unit.

H11Protonn01Neutron+e01Positron+vNeutrino


10.

The correct order of decreasing H-C-H angle in the following molecule is

  • I > II > III

  • II > I > III

  • III > II > I

  • I > III > II


B.

II > I > III

The correct order of decreasing H-C-H bond angle is in the order II > I > III.

It is because H-C-H has a bond angle of about 120°, when C is bonded through a double bond. But in case of CH4 the H-C-H bond angle is only 109° 28'.

Cyclic compounds have H-C-H bond angle greater than 109° 28' due to hindrance.