## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# NEET Chemistry Solved Question Paper 2016

#### Multiple Choice Questions

1.

Which of the following ions has the highest value of ionic radius:

• O2-

• F-

• Li+

• B3+

A.

O2-

O2- has the highest value of ionic radius and this can be explained on the basis of $\frac{\mathrm{z}}{\mathrm{e}}$ (nuclear charge / number of electrons) ratio. when $\frac{\mathrm{z}}{\mathrm{e}}$ ratio increases the size decreases and when $\frac{\mathrm{z}}{\mathrm{e}}$ ratio decreases , the size increases.

2.

A cylinder of compressed gas contains nitrogen and oxygen in the ratio of 3:1 by mole. The total mass of the gas mixture if the cylinder is known to contain 2.5 x 10g of oxygen will be:

• 8 x 103 g

• 9 x 10g

• 9 x 103 g

• 8 x 102 g

B.

9 x 10g

Number of moles of oxygen in the cylinder

=

= = 781.25

$\therefore$ Number of moles of N2 = 3 x 781.25

=2343.75

Mass of nitrogen in cylinder

=6.5625 x 104 g

Hence , the total mass of gas cylinder

= 2.5 x 104 + 6.56259 x 104

= 9.0625 x 104 g

3.

On adding A to the reaction at equilibrium, AB (s) $⇌$ A(g) + B(g) the new equilibrium concentration of A becomes double, the equilibrium concentration of B would become:

• 1/2 of its original value

• 1/4 of its original value

• 1/8 of its original value

• Twice its original value

A.

1/2 of its original value

K

Now , if [A] = 2 $×$ [A] , B should be $\frac{1}{2}$ [B].

So , that Kc remains constant.

4.

The entropy will be positive in which of the following reaction change.

• H2(g) + I2 (g) $⇋$ 2HI (g)

• HCl(g) + NH3 (g) $⇋$ N2O(g) + 2H2O(g)

• NH4NO3(s)$⇋$N2O(g) + 2H2O(g)

• MgO(s) + H2(g) $⇋$ Mg(s) + H2O(l)

C.

NH4NO3(s)$⇋$N2O(g) + 2H2O(g)

Entropy is a measure of randomness of the system, so entropy of substance in different physical states is in the order Gas > Liquid > Solid

$\mathrm{\Delta }$S=Sproduct -Sreactant

In option (c), solid is converted into gaseous state, so $\mathrm{\Delta }$S will be positive.

5.

The vapour pressure of pure water is 25.0 mm Hg and that of an aqueous dilute solution of urea in 20mm Hg at 25°C. The molarity of the solution is:

• 12.6

• 14.9

• 13.8

• 9.6

C.

13.8

The given values are:

Using Raoult's law

$\frac{{\mathrm{P}}_{\mathrm{A}}^{\circ }-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}_{\mathrm{A}}^{\circ }}={\mathrm{X}}_{\mathrm{B}}$

Substituting the values:

=1-0.2=0.8

6.

Assuming that Hund's rule in violated,the bond order and magnetic nature of the  diatomic molecule B2 is:

• 1 and diamagnetic

• 0 and diamagnetic

• 1 and paramagnetic

• O and paramagnetic

A.

1 and diamagnetic

B2 : Total electrons = 10

Configuration :

If Hund's rule is violated, then

So, diamagnetic bond order = $\frac{6-4}{2}$ = 1

7.

Wave number of a spectral line for a given transition is xcm-1 for He+ then its value for Be+ (iso electronic of H+) for same transition is:

• x cm-1

• 4x cm-1

• $\frac{\mathrm{x}}{4}$cm-1

• 2x cm-1

B.

4x cm-1

$\stackrel{-}{\mathrm{v}}$ =(wave number)

=RHZ2$\left[\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right]$

${\mathrm{\nu }}_{1}$(He+ , z=2)=RH(22)$\left[\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right]$

v2(${\mathrm{Be}}_{3}^{+}$ , z=4)=RH(42)$\left[\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right]$

$\frac{{\mathrm{v}}_{2}}{{\mathrm{v}}_{1}}=4$

$\therefore$ v2=4v1=4x

8.

A solution contains 10 mL of 0.1 N NaOH and 10mL of 0.05 NH2SO4.pH of this solution is:

• zero

• less than 7

• greater than 7

• 7

C.

greater than 7

Number of milliequivalents of NaOH = 0.1 x 10 = 1

Number of milliequivalents of H2SO4
= 10 x 0.05 = 0.5

Milliequivalent of H2SO4 will neutralise NaOH so that (1-0.5) = 0.5 milliequivalent of NaOH is left unneutralised.

Therefore, pH of the resulting solution will be greater than seven i.e., pH > 7,

9.

The correct sets of quantum numbers which represents the highest energy level in an atomis:

D.

The order of energy of orbitals can be calculated from (n + 1) rule. The lower the value of (n +l) For an orbital, lower is its energy. If two orbitals have same (n+J) value, the orbital with lower value of n has lower energy.

(a) n+l = 4  (b) n+l = 3

(c) n+l = 4   (d) n+l = 4

$\therefore$ The correct set which represents the highest energy level is n = 3 , l = 2 , m = -2 and s = + $\frac{1}{2}$

10.

The hybridisation and shape of IF5 are respectively:

• sp3d , trigonal bipyramide

• sp3d , see saw

• sp3d , square pyramidal

• sp3d , pentagonal pyramidal

C.

sp3d , square pyramidal

IF5 has 5 bond pair and 1 lone pair

The geometry is square pyramidal with sp3d hybridisation