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# NEET Chemistry Solved Question Paper 2014

#### Multiple 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-

$\overline{)\mathrm{v}}$(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.

2.

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.

3.

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

# 4.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 be60 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

5.

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.

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

6.

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 =$\frac{6.6×{10}^{-27}}{100×100}=6.6×{10}^{-31}$

7.

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

8.

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

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.

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

10.

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