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# NEET Chemistry Solved Question Paper 2014

#### Multiple Choice Questions

1.

If a gas expands at constant temperature indicates that

• kinetic energy of molecules decreases

• pressure of the gas increases

• kinetic energy of molecules remains as same

• number of the molecules of gas increase

C.

kinetic energy of molecules remains as same

Here, the temperature of gas remains constant hence kinetic energy of molecules remains the same.

2.

If n = 6, the correct sequence for filling of electrons will be

• ns $\to$(n - 1)d $\to$(n - 2)f $\to$np

• ns $\to$(n - 2)f $\to$np $\to$(n - 1)d

• ns $\to$np $\to$(n - 1)d $\to$(n - 2)f

• ns $\to$(n - 2)f $\to$(n - 1)d $\to$np

D.

ns $\to$(n - 2)f $\to$(n - 1)d $\to$np

For n = 6 electron are filled as

6s $\to$ 4f $\to$ 5d $\to$ 6p

This is because electron first enters in that orbitals for which (n + 1) is lower. In case, if (n + 1) is same, it goes first in that orbital for which n is lower.

3.

An electron is moving in Bohr's fourth orbit. Its de-Broglie wave length is $\mathrm{\lambda }$. What is the circumference of the fourth orbit?

• $\frac{2}{\mathrm{\lambda }}$

• 2$\mathrm{\lambda }$

• 4$\mathrm{\lambda }$

• $\frac{4}{\mathrm{\lambda }}$

C.

4$\mathrm{\lambda }$

According to Bohr's concept, an electron always move in the orbit with angular momentum (mvr) equal to $\frac{\mathrm{nh}}{2\mathrm{\pi }}$

4.

For the reaction, N2(g) + O2(g) $⇌$ 2NO (g), the equilibrium constant is K1. The equilibrium constant is K2 for the reaction, 2NO(g) + O2)g) $⇌$ 2NO2(g). What is K for the reaction, NO2(g) $⇌$ ?

• $\frac{1}{4{\mathrm{K}}_{1}{\mathrm{K}}_{2}}$

• ${\left[\frac{1}{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}\right]}^{\frac{1}{2}}$

• $\frac{1}{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}$

• $\frac{1}{2{\mathrm{K}}_{1}{\mathrm{K}}_{2}}$

B.

${\left[\frac{1}{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}\right]}^{\frac{1}{2}}$

N2(g) + O2(g) $⇌$2NO(g); K1  ...(i)

2NO(g) + O2(g) $⇌$2NO2(g); K2  ...(ii)

N2(g) + 2O2(g) $⇌$2NO2(g); K = K1 $×$ K2

$\therefore$ For NO2(g)  $⇌$$\frac{1}{2}$N2(g) + O2(g); K = ${\left[\frac{1}{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}\right]}^{\frac{1}{2}}$

5.

Among the following the molecule possessing highest dipole moment is

• CO2

• BF3

• SO2

• Trans-2-butene

C.

SO2

Among the given compounds, SO2 has unsymmetrical or angular structure. Thus, it has the highest dipole moment.

6.

The angular shape of ozone molecule (O3) consists of

• 1 sigma and 2 pi bonds

• 2 sigma and 2 pi bonds

• 1 sigma and 1 pi bonds

• 2 sigma and 1 pi bonds

D.

2 sigma and 1 pi bonds

Single bond contains= 1 $\mathrm{\sigma }$ bond

Double bond contains = 1 $\mathrm{\sigma }$ bond + 1 $\mathrm{\pi }$ bonds

Triple bond contains= 1$\mathrm{\sigma }$ bond + 2 $\mathrm{\pi }$ bonds

The molecule of O3 is bent with an angle 116.8° and equal O - O distance of 128 pm. The ozone molecule consists of 2 sigma and 1 pi bonds. The angular shape of molecule is represented as

7.

The correct order of magnitude of bond angles among the compounds CH4, NHand H2O is

• CH4 > H2O < NH3

• H2O < NH3 < CH4

• NH3 < CH4 < H2O

• NH3 < H2O < CH4

B.

H2O < NH3 < CH4

Bond angles depend upon number of lone pair (s) of electrons. Higher the number of lone pairs of electrons, lesser the bond angle.

8.

The dissociation equilibrium of a gas AB2 can be represented as

2AB2 (g) $⇌$2AB (g) + B2(g)

The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant KP and total pressure p is

• $\frac{2{K}_{p}}{p}$

• ${\left(\frac{2{K}_{p}}{p}\right)}^{\frac{1}{3}}$

• ${\left(\frac{2{K}_{p}}{p}\right)}^{\frac{1}{2}}$

• $\frac{{K}_{p}}{p}$

B.

${\left(\frac{2{K}_{p}}{p}\right)}^{\frac{1}{3}}$

Initial moles           1             0           0

At equillibrium    2(1-x)         2x          x

where, x = degree of dissociation

Total moles at equilibrium = 2 - 2x + 2x + x

=  (2 + x)

So,

9.

A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of N${\mathrm{H}}_{4}^{+}$ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1. 8 x 10-5, what is the pH of this solution? (log 2. 7 =0. 43).

• 9.43

• 11.72

• 8.73

• 9.08

A.

9.43

pOH = pKb$\mathrm{log}\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

=

=

= 5 - 0.25 + (-0.176)

= 4.75 - 0.176 = 4.57

$\therefore$ pH = 14 - 4.57 = 9.43

10.

Which of the following is least likely to behave as Lewis base?

• NH3

• BF3

• OH-

• H2O

B.

BF3

Boron in BF3 is electron poor or electron deficient species. It has an empty orbital, so it can accept a pair of electrons, therefore, making it a lewis acid.