Need NEET 2010 Chemistry Solved Previous Year Question Paper

Find solved questions and answers for the paper 2010 with download option

Subject

Chemistry

Class

NEET Class 12

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 2010

Multiple Choice Questions

The valence shell electronic configuration of Cr²⁺ ion is :

4s⁰3d⁴
3p⁶4s²
4s³3d²
4s²3d⁰

4s⁰3d⁴

The valence shell electronic configuration of Cr⁺ ion is :

= 1s² , 2s² 2p⁶ , 3s² 3p⁶ 3d⁴ , 4s⁰

The structure of IF₅ can be best described as :

None of these

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. ,

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

0.502

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

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

( $∵$ Molar mass of H₂O = 18)

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

( $∵$ Molar mass of C₂H₅OH = 46)

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

( $∵$ Molar mass of CH₃COOH = 60)

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

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

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

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

$∴$ [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

$∴$ K_sp for PbCl₂ = [Pb²⁺][Cl^-]²

= (6.3 x 10^-2) x (12.6 x 10^-2)²

= 1 x 10^-3

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

2 electrons
1 electron
4 electrons
All electrons

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 .

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

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

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

$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $π$ 2p⁴ , $σ$ 2p¹
$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $σ$ 2p² , $π$ 2p³
$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $σ$ 2p³ , $π$ 2p²
$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $σ$ 2p² , $π$ 2p⁴

$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $π$ 2p⁴ , $σ$ 2p¹

N₂⁺ = (7 + 7 - 1 = 13)

$σ$ 1s² , $\overset{*}{σ}$ 1s² , $σ$ 2s² , $\overset{*}{σ}$ 2s² , $π$ 2p_x² $\approx$ $π$ 2p_y² , $σ$ 2p_z¹

Ionisation potential is lowest for :

alkali metals
inert gases
halogens
alkaline earth metals

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 .