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# NEET Chemistry Solved Question Paper 2010

#### Multiple Choice Questions

1.

The structure of IF5 can be best described as :

• None of these

D.

None of these

Number of hybrid orbitals = number of bp + number of lp

= 5 + 1 = 6

Thus , hybridisation is sp3d2 but geometry , due to the presence of one lone pair , is square pyramidal , i.e. ,

2.

pH of a solution can be expressed as :

• -loge [H+]

• -log10 [H+]

• loge [H+]

• log10[H+]

B.

-log10 [H+]

pH is defined as the negative logarithm of the hydrogen ion concentration in solution .
pH = -log10 [H+]

3.

A solution contains 25% H2O , 25% C2H5OH and 50% CH3COOH by mass .The mole fraction of H2O would be :

• 0.25

• 2.5

• 0.502

• 5.03

C.

0.502

Mole fraction of H2O =

Let the total mass of solution = 100 g

Mass of H2O = 25 g

Mass of C2H5OH = 25g

Mass of CH3COOH = 50 g

Moles of H2O = $\frac{25}{18}$ = 1.388

($\because$ Molar mass of H2O = 18)

Moles of C2H5OH =$\frac{25}{46}$ = 0.543

($\because$ Molar mass of C2H5OH = 46)

Moles of CH3COOH = $\frac{50}{60}$ = 0.833

($\because$ Molar mass of CH3COOH = 60)

Total no. of moles = 1.388 + 0.543 + 0.833  = 2.764

$\therefore$ Mole fraction of H2O = $\frac{1.388}{2.764}$ = 0.502

4.

The percentage of p-character in the orbitals forming P-P bond in P4 is :

• 25

• 33

• 50

• 75

D.

75

Each P atom in P4 is sp3 hybridised , hence the percentage ofs and p character is 25% and 75% .

5.

Which of the following has largest atomic radii ?

• Al

• Al+

• Al2+

• Al3+

A.

Al

Atomic radii varies inversely with nuclear charge , ie ,

Atomic radii $\propto$

6.

If the solubility of PbCl2 at 25°C is 6.3 x 10-2 mol/L , its solubility product is :

• 1 x 10-6

• 1 x 10-3

• 1.1 x 10-6

• 1.1 x 10-5

B.

1 x 10-3

PbCl2 completely ionised in the solution as

PbCl2 $\to$ Pb2+ + 2Cl-

i.e , 1 mole of PbCl2 in the solution gives 1 mole of Pb2+ ion and 2 moles of Cl- ions .

Now , as the solubility of PbCl2

= 6.3 x 10-2 mol/L

$\therefore$ [Pb2+] = 6.3 x 10-2 mol/L

and [Cl-] = 2 X 6.3 X 10-2 mol/L

= 12.6 x 10-2 mol/L

$\therefore$Ksp for PbCl2 = [Pb2+][Cl-]2

= (6.3 x 10-2) x (12.6 x 10-2)2

= 1 x 10-3

# 7.Ionisation potential is lowest for :alkali metals inert gases halogens alkaline earth metals

A.

alkali metals

The atomic radii of alkali metals are the largest in their respective periods due to which the ionisation enthalpy of the alkali metals are the lowest as compared to the elements in other groups .

8.

From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion as :

• $\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\pi }$2p4$\mathrm{\sigma }$2p1

• $\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\sigma }$2p2$\mathrm{\pi }$2p3

• $\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\sigma }$2p3 , $\mathrm{\pi }$2p2

• $\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\sigma }$2p2$\mathrm{\pi }$2p4

A.

$\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\pi }$2p4$\mathrm{\sigma }$2p1

N2+ = (7 + 7 - 1 = 13)

$\mathrm{\sigma }$1s2$\stackrel{*}{\mathrm{\sigma }}$1s2$\mathrm{\sigma }$2s2$\stackrel{*}{\mathrm{\sigma }}$2s2$\mathrm{\pi }$2px2 $\approx$ $\mathrm{\pi }$2py2 , $\mathrm{\sigma }$2pz1

9.

The valence shell electronic configuration of Cr2+ ion is :

• 4s03d4

• 3p64s2

• 4s33d2

• 4s23d0

A.

4s03d4

The valence shell electronic configuration of Cr+ ion is :

= 1s2 , 2s2 2p6 , 3s2 3p6 3d4 , 4s0

10.

A double bond connecting two atoms .There is a sharing of :

• 2 electrons

• 1 electron

• 4 electrons

• All electrons

C.

4 electrons

A double bond connecting two atoms , thus there is a sharing of 4 electrons .We can explain it by taking example of C-C double bond .