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# NEET Chemistry Solved Question Paper 2007

#### Multiple Choice Questions

1.

The number of moles of KMnO4reduced by one mole of KI in alkaline medium is

• one-fifth

• five

• one

• two

D.

two

In alkaline solution, KMnO4 is reduced to MnO2 (colourless).

$2{\mathrm{KMnO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}\to 2{\mathrm{MnO}}_{2}+2\mathrm{KOH}\phantom{\rule{0ex}{0ex}}\overline{)+3\left[\mathrm{O}\right]\mathrm{Ki}+3\left[\mathrm{O}\right]\to {\mathrm{KIO}}_{3}}\phantom{\rule{0ex}{0ex}}2{\mathrm{KMnO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}+\mathrm{KI}\to 2{\mathrm{MnO}}_{2}+2\mathrm{KOH}+{\mathrm{KIO}}_{3}$

Hence, two moles of KMnO4 are reduced by one mole of KI.

2.

Equilibrium constants K1 and K2 for the following equilibria
NO (g) +$\frac{1}{2}{\mathrm{O}}_{2}$$\stackrel{{\mathrm{K}}_{1}}{⇌}$NO2(g) and

2NO2(g)$\stackrel{{\mathrm{K}}_{2}}{⇌}$2NO (g) + O2(g)

are related as

D.

(i) NO (g) +$\frac{1}{2}{\mathrm{O}}_{2}$(g)$\stackrel{{\mathrm{K}}_{1}}{⇌}$NO2(g)

So, K1 =$\frac{\left[{\mathrm{NO}}_{2}\right]}{\left[\mathrm{NO}\right]\left[{\mathrm{O}}_{2}{\right]}^{1}{2}}}$ ...(1)

(ii) 2NO2 (g)$\stackrel{{\mathrm{K}}_{2}}{⇌}$2NO + O2 (g)

So, K2 =$\frac{\left[\mathrm{NO}{\right]}^{2}\left[{\mathrm{O}}_{2}\right]}{\left[{\mathrm{NO}}_{2}\right]}$ ...(2)

From equation (1)

${\mathrm{K}}_{1}^{2}=\frac{\left[{\mathrm{NO}}_{2}{\right]}^{2}}{\left[{\mathrm{NO}}_{2}\right]}$ or

$\frac{1}{{\mathrm{K}}_{1}^{2}}=\frac{\left[\mathrm{NO}{\right]}^{2}\left[{\mathrm{O}}_{2}\right]}{\left[{\mathrm{NO}}_{2}{\right]}^{2}}$ ...(3)

from equation (2) and (3)

K2 =$\frac{1}{{\mathrm{K}}_{1}^{2}}$

3.

Which of the following is the electron deficient molecule?

• B2H6

• C2H6

• PH3

• SiH4

A.

B2H6

B2H6 is electron deficient molecule because boron atom has three half filled orbitals in excited state. The structure of B2H6 is represented as follows :

In it two electrons of a B - H bond are involved in formation of three centre bond, these bonds are represented as dotted lines.

4.

The correct sequence of increasing covalent character is represented by

• LiCl < NaCl < BeCl2

• BeCl2 < NaCl < LiCl

• NaCl < LiCl < BeCl2

• BeCl2 < LiCl < NaCl

C.

NaCl < LiCl < BeCl2

According to Fajan's rule,lower the size of cation higher will be its polarising power and higher will be covalent character.

$\therefore$Polarising power$\propto$$\frac{1}{\mathrm{size}\mathrm{of}\mathrm{action}}$

Covalent character$\propto$Polarising power

So, the correct order is NaCl < LiCl < BeCl2

# 5.Which of the following molecules has trigonal planar geometry?IF3 PCl3 NH3 BF3

D.

BF3

Among all the given options, IF3 has trigonal planar geometry.

PCl3 has pyramidal geometry.

NH3 has pyramidal geometry.

BF3 has trigonal planar geometry

6.

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?

• Exothermic and decreasing disorder

• Endothermic and increasing disorder

• Exothermic and increasing disorder

• Endothermic and decreasing disorder

C.

Exothermic and increasing disorder

If a reaction is exothermic,$∆$H will be -ve and in increasing order, $∆$S will be +ve thus, at this condition, $∆$G is negative according to following equation:
$∆$G = $∆$H - T$∆$S

$∆$G = -ve.

Hence, for spontaneous reaction $∆$G must be -ve

7.

The absolute enthalpy of neutralisation of the reaction

MgO(s) + 2HCl(aq)$\to$ MgCl2(aq) + H2O(l)

will be

• less than - 57.33 kJ mol-1

• - 57.33 kJ mol-1

• greater than-- 57.33 kJ mol-1

• 57.33 kJ mol-1

A.

less than - 57.33 kJ mol-1

Heat of neutralisation of strong acid and strong base is -57.33 kJ. MgO is weak base while HCl is strong acid, so the heat of neutralisation of MgO and HCl is lower to -57.33 kJ because MgO requires some heat in ionisation, then net released amount of heat is decreased.

8.

A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three $\mathrm{\alpha }$-particles in succession. The group of the periodic table to which the resulting daughter element would belong to

• Group 14

• Group 16

• Group 4

• Group 6

A.

Group 14

After decay of radioactive element, the last element Pb is obtained which belongs to group 14 of periodic table.

9.

A reaction occurs spontaneously if

• T$∆$S < $∆$H and both $∆$H and $∆$S are +ve

• T$∆$S > $∆$H and both $∆$H and $∆$S are +ve

• T$∆$S = $∆$H and both $∆$H and $∆$S are +ve

• T$∆$S > $∆$H and $∆$H is +ve and $∆$S is - ve

B.

T$∆$S > $∆$H and both $∆$H and $∆$S are +ve

The spontaneity of reaction is based upon the negative value of $∆$G, $∆$G is based upon T, $∆$S and $∆$H according to the following equation (Gibbs-Helmholtz equation):

$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

If the magnitude of $∆$H - T$∆$S is negative, then the reaction is spontaneous.

When T$∆$S > $∆$H and $∆$H and $∆$S are +ve, then $∆$G is negative.

10.

The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is(Atomic mass Al= 27)

• 180 kg

• 270 kg

• 540 kg

• 90 kg

D.

90 kg

In Hall and Heroult process

2Al2O3$\to$4Al + 3O2

4C + 3O2$\to$2CO2 + 2CO$↑$

2Al2O3 + 4C$\to$4Al + 2CO2 + 2CO

Only for removal of CO2, following equation is possible

2Al2O3 + 3C$\to$4Al + 3CO2

Molecular weight of Carbon = 12

Molecular weight of Aluminium = 27

$\therefore$For Carbon/ C = 3 x 12 = 36

For Aluminium/ Al = 4 x 27 = 108

$\because$For 108 gm of Al, 36 gm of C is required in above reaction

$\therefore$For 270 gm of Al require amount of C =$\frac{36}{108}×270=90\mathrm{gm}$