## 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 2011

#### Multiple Choice Questions

1.

Which one of the following has the lowest ionisation energy?

• 1s2 , 2s2 , 2p6

• 1s2 , 2s2 , 2p6 , 3s1

• 1s2 , 2s2 , 2p5

• 1s2 , 2s2 , 2p3

B.

1s2 , 2s2 , 2p6 , 3s1

In the Periodic Table, the minimum ionisation energy is found at alkali metals. Hence sodium (electronic configuration = 1s2 2s2 2p6 3s1) has the lowest ionisation energy.

Note: Ionisation energy is the amount of energy required to remove an electron from the outermost shell an isolated atom.

2.

The amount of the heat released when 20 mL 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralisation is:

• - 100 x kJ/mol

• - 50 x kJ/mol

• + 100 x kJ/mol

• + 50 x kJ/mol

A.

- 100 x kJ/mol

Millimoles (or milliequivalents) of NaOH = 20 x 0.5 = 10

Millimoles (or milliequivalents) of HCl = 100 x 0.1 = 10

$\therefore$ $\underset{\underset{\mathrm{milliequivalents}}{10}}{\mathrm{NaOH}}$ + $\underset{\underset{\mathrm{milliequivalents}}{10}}{\mathrm{HCl}}$ $\to$ NaCl + H2O

Thus, heat released when 10 milliequivalents of HCl are. neutralised by 10 millimoles of NaOH = x kJ. But the heat of neutralisation is heat released when 1 equivalent of HCl is neutralised by 1 equivalent of NaOH.

$\therefore$ ${∆}_{\mathrm{neu}}$H = -  or - 100 x kJ/mol

3.

In the following electron-dot structure calculate the formal charge from left to right nitrogen atom; $\stackrel{\mathbf{.}\mathbf{.}}{\underset{\mathbf{.}\mathbf{.}}{\mathrm{N}}}=\mathrm{N}=\stackrel{\mathbf{.}\mathbf{.}}{\underset{\mathbf{.}\mathbf{.}}{\mathrm{N}}}$ :

• -1 , -1 , +1

• -1 , +1 , -1

• +1 , -1 , -1

• +1 , -1 , +1

B.

-1 , +1 , -1

Formal charge = total number of valence electrons - total number of nonbonding electrons - 1/2 x total number of bonding electrons

As for structure, $\stackrel{..}{\underset{..}{\mathrm{N}}}-\mathrm{N}-\underset{..}{\stackrel{..}{\mathrm{N}}}$ ;

For Ist and 3rd nitrogen atom

Formal charge = 5 - 4 - 1/2 x 4 = -1

For middle nitrogen atom

Formal charge = 5 - 0 - 1/2 x 8 = +1

4.

The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state)

• Li2+

• He+

• H

• H+

A.

Li2+

$\frac{1}{\mathrm{\lambda }}$ = R$\left(\frac{1}{{{\mathrm{n}}_{1}}^{2}}-\frac{1}{{{\mathrm{n}}_{2}}^{2}}\right)$ Z2

= R$\left(\frac{1}{\left(1{\right)}^{2}}-\frac{1}{\left(2{\right)}^{2}}\right)$ Z2

$\frac{1}{\mathrm{\lambda }}$ = $\frac{3}{4}$RZ2

$\therefore$For shortest $\mathrm{\lambda }$, Z must be maximum, which is for Li2+

# 5.If the equilibrium constants of the following equilibria,SO2 + 1/2O2 $\to$ SO3 and 2SO3 $\to$ 2SO2 + O2are given by K1 and K2 respectively, which of the following relations is correct?K2 = (1/K1)2 K1 = (1/K3)3 K2 = (1/K1) K2 = (K1)2

A.

K2 = (1/K1)2

SO2 + 1/2 O2 $\to$SO3

K1$\frac{\left[{\mathrm{SO}}_{3}\right]}{\left[{\mathrm{SO}}_{2}\right]\left[{\mathrm{O}}_{2}{\right]}^{1/2}}$

2SO3 $\to$2SO2 + O2

K2$\frac{\left[{\mathrm{SO}}_{2}{\right]}^{2}\left[{\mathrm{O}}_{2}\right]}{\left[{\mathrm{SO}}_{3}{\right]}^{2}}$

Obviously, K2 = (1/K1)2

6.

The representation of the ground state electronic configuration of He by box-diagram as

 $↑↑$

is wrong because it violates:

• Heisenberg's uncertainty principle

• Bohr's quantization theory of angular momenta

• Pauli exclusion principle

• Hund's rule

C.

Pauli exclusion principle

Pauli's exclusion principle states, "An orbital can have a maximum of two electrons and these must have opposite spins."

7.

2 g of metal carbonate is neutralised completely by 100 mL of 0.1 N HCI. The equivalent weight of metal carbonate is:

• 50

• 100

• 150

• 200

D.

200

As 2 g of metal carbonate is neutralised by 100 mL of 0.1 N HCl.

$\therefore$Number of gram-equivalents of HCl =

=0.01

$\therefore$ Number of gram-equivalents of metal carbonate= 0.01

Mass of 0.01 gram equivalent metal carbonate = 2 g

Mass of 1 gram equivalent metal carbonate = $\frac{2}{0.01}$

= 200 g

$\therefore$Equivalent mass of metal carbonate = 200

8.

The pH of an aqueous solution of CH3COONa of concentration C(M) is given by:

• 7 - 1/2 pKa - 1/2 log C

• 1/2 pKw + 1/2 pKb + 1/2 log C

• 1/2 pKw - 1/2 pKb - 1/2 log C

• 1/2 pKw - pKa + 1/2 log C

D.

1/2 pKw - pKa + 1/2 log C

CH3COONa + H2$⇌$  +

As sodium acetate is a salt of weak acid and strong base, in the hydrolysis of sodium acetate the pH is given by;

pH = 1/2 pKw + 1/2 pKa + 1/2 log C

9.

An electric current is passed through an aqueous solution of a mixture of alanine (isoelectric point 6.0), glutamic acid (3.2) and arginine (10.7) buffered at pH 6. What is the fate of the three acids?

• Glutamic acid migrates to anode at pH 6. Arginine is present as a cation and migrates to the cathode. Alanine in a dipolar ion remains uniformly distributed in solution

• Glutamic acid migrates to cathode and others remain uniformly distributed in solution

• All three remain uniformly distributed in solution

• All three move to cathode

A.

Glutamic acid migrates to anode at pH 6. Arginine is present as a cation and migrates to the cathode. Alanine in a dipolar ion remains uniformly distributed in solution

Amino acids exist as Zwitter ions which contain both positive and negative charge but overall electrically neutral. At their the isoelectric point, amino acids do not migrate to any electrode and remain uniformly distributed in the solution. Thus, alanine (isoelectric point 6.0) does not move to any electrode.

At a pH lower than their isoelectric point, they exist as cation and move towards the cathode. Thus, arginine (isoelectric point 10.7 exists as the cation.

At a pH higher than their isoelectric point, they exist as anion and move towards the anode. Thus, glutamic acid (isoelectric point 3.2) at pH 6.0 moves towards the anode.

10.

The energy of an electron in first Bohr orbit of H-atom is-13.6 eV. The possible energy value of electron in the excited state of Li2+ is:

• - 122.4 eV

• 30.6 eV

• - 30.6 eV

• 13.6 eV

C.

- 30.6 eV

En$\frac{{\mathrm{E}}_{1}}{{\mathrm{n}}^{2}}$ x Z2

For Li2+, the excited state, n = 2 and Z = 3

$\therefore$ En x (3)2

x 9

= - 30.6 eV