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# NEET Chemistry Solved Question Paper 2010

#### Multiple Choice Questions

1.

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 , thusthere is a sharing of 4 electrons .We can explainit by taking example of C-C double bond .

2.

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

3.

A solution contains 25% H2O , 25% C2H5OH and 50% CH3COOHby mass .The mole fractionof H2O would be :

• 0.25

• 2.5

• 0.502

• 5.03

C.

0.502

Mole fraction of H2O =$\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_{2}\mathrm{O}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{all}\mathrm{components}}$

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.

pH of a solution can be expressed as :

• -loge [H+]

• -log10 [H+]

• loge [H+]

• log10[H+]

B.

-log10 [H+]

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

5.

Which of the following has largest atomic radii ?

• Al

• Al+

• Al2+

• Al3+

A.

Al

Atomic radii varies inversely with nuclearcharge , ie ,

Atomic radii$\propto$$\frac{1}{\mathrm{nuclear}\mathrm{charge}}$

6.

If the solubility of PbCl2 at25°C is 6.3 x 10-2mol/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

PbCl2completely ionised in the solution as

PbCl2$\to$Pb2++ 2Cl-

i.e ,1 mole of PbCl2 in the solution gives 1 moleof 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.

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

# 8.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 sp3hybridised , hence thepercentage ofs and p character is25% and 75% .

9.

Ionisation potential is lowest for :

• alkali metals

• inert gases

• halogens

• alkaline earth metals

A.

alkali metals

The atomic radii of alkali metals are the largestin their respective periods due to which theionisation enthalpy of the alkali metals are thelowest as compared to the elements in othergroups .

10.

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 sp3d2but geometry , dueto the presence of one lone pair , is squarepyramidal , i.e. ,