Chemistry

NEET Class 12

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1.

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

I

II

III

IV

A.

I

Symmetrical molecules and symmetrical *trans- *alkene have a net dipole moment zero. CH_{2}Cl_{2} is not a symmetrical molecule, thus it will have a permanent dipole.

μ_{1} + μ_{2} - μ_{3} - μ_{4} = 0 ; μ_{3} and μ_{4} > μ_{1} and μ_{2}. Thus μ_{net} ≠ 0

Symmetrical *trans*-alkene μ_{net} ≠ 0

μ_{1} + μ_{2} = μ_{3} + μ_{4} ; μ_{net} = 0

μ_{net} = 0

2.

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 = $\frac{\mathrm{nRT}}{\mathrm{p}}+\mathrm{C}$

or, V = $\frac{\mathrm{RT}}{\mathrm{p}}+0$ (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}

3.

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.

${}_{0}\mathrm{n}^{1}\to \underset{\mathrm{Proton}}{{}_{1}\mathrm{H}^{1}}+\underset{\mathrm{\beta}\mathrm{particle}}{{}_{-1}\mathrm{e}^{0}}+\underset{\mathrm{Anti}-\mathrm{neutrino}}{\overline{)\mathrm{v}}}+\mathrm{\gamma}-\mathrm{ray}$

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.

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}cm6.6 × 10

^{-30}cm6.6 × 10

^{-31}cm6.6 × 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 = $\frac{6.6\times {10}^{-27}}{100\times 100}=6.6\times {10}^{-31}$

6.

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 $\left(\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{\mathrm{n}}^{2}}\right)$

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

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

7.

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 CH_{4} 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.

8.

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 P_{1} thus, P is the most liquefiable gas among the given

9.

At a certain temperature the time required for the complete diffusion of 200-mL of H_{2} gas is 30 min. The time required for the complete diffusion of 50 mL of O_{2} 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 $ $\frac{1}{\sqrt{\mathrm{M}}}=\frac{\mathrm{V}}{\mathrm{t}}$

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

Thus, for H_{2} gas,

$\frac{200}{30}=\frac{1}{\sqrt{2}}$ ...(i)

For O_{2} gas,

$\frac{50}{\mathrm{t}}=\frac{1}{\sqrt{32}}$ ...(ii)

From equation (i) and (ii)

$\frac{200\times \mathrm{t}}{30\times 50}=\frac{\sqrt{32}}{\sqrt{2}}$

t = $\frac{\sqrt{16}\times 30\times 50}{200}$ = $\frac{4\times 30}{4}$= 30 min

10.

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.

$\underset{\mathrm{Proton}}{{}_{1}{}^{1}\mathrm{H}}\to \underset{\mathrm{Neutron}}{{}_{0}{}^{1}\mathrm{n}}+\underset{\mathrm{Positron}}{{}_{1}\mathrm{e}^{0}}+\underset{\mathrm{Neutrino}}{\mathrm{v}}$

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