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# 1.The compound that will have a permanent dipole moment among the following isI II III IV

A.

I

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

μ1 + μ2 - μ3 - μ4 = 0 ; μ3 and μ4 > μ1 and μ2. Thus μnet ≠ 0

Symmetrical trans-alkene μnet ≠ 0

μ1 + μ2 = μ3 + μ4 ; μnet = 0

μnet = 0

2.

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

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.

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.

5.

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.

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.

7.

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-

$\overline{)\mathrm{v}}$ (cm-1) = 109677

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

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

8.

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 = $\frac{\mathrm{h}}{\mathrm{mv}}$

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 =

9.

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, r $\propto$

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

Thus, for H2 gas,

...(i)

For O2 gas,

...(ii)

From equation (i) and (ii)

t =  = = 30 min

10.

For one mole of an ideal gas,the slope of V vs T 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

• $\frac{\mathrm{X}}{2}$L 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 =

or, V =   (for 1 mol)

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

Slope = $\frac{\mathrm{R}}{\mathrm{p}}$

X = $\frac{\mathrm{R}}{2}$

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