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# NEET Chemistry Solved Question Paper 2014

#### Multiple Choice Questions

1.

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

2.

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.

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

3.

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.

# 4.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 of6.6 × 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 =$\frac{6.6×{10}^{-27}}{100×100}=6.6×{10}^{-31}$

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

• $\frac{\mathrm{X}}{2}$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 Vvs 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

6.

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

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 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.

8.

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$$\frac{1}{\sqrt{\mathrm{M}}}=\frac{\mathrm{V}}{\mathrm{t}}$

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

Thus, for H2 gas,

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

For O2 gas,

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

From equation (i) and (ii)

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

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

9.

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

10.

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.

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