NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Chemical Kinetics

The reaction 2N₂O₅<--> 4NO₂ + O₂ follows Ist order kinetics. Hence the molecularity of reaction is

1. unimoleqular.
2. pseudo unimoleqular
3. bimoleqular.
4. none of these.

Option 1

Option 2

Option 3 Solution:

Solution

Two molecules of N2?O5? collide to give products. Hence, The molecularity of the reaction is Two.

Option 4

2 D and F Block Elements

Number of identical Cr-O bonds in dichromate ion Cr₂O₇²⁻ is :

1) 4
2) 6
3) 7
4) 8

Option 1

Option 2 Solution:

Solution

There are '6' identical Cr - O bonds in the dichromate ion Cr₂O₇²⁻

In a dichromate ion Cr₂O₇²⁻, there are 6 terminal Cr – O bonds. All these bonds are equivalent due to resonance.

The Dichromate ions Cr₂O₇²⁻. are oxoanions of chromium in the oxidation state +6.

Option 3

Option 4

3 D and F Block Elements

The solubility of silver bromide in hypo solution is due to the formation of

1) [Ag(S₂O₃)] ^-

2) [Ag(S₂O₃)₂] ^-3

3) Ag₂S₂O

4) Ag₂SO₃

Option 1

Option 2 Solution:

Solution

The solubility of silver bromide in hyposolution due to the formation of [Ag(S₂O₃)₂]^3-

Option 3

Option 4

4 Haloalkanes and Haloarenes

Liebermann's test is answered by :

1) methylamine
2) aniline
3) phenol
4) ethyl benzoate

Option 1

Option 2

Option 3 Solution:

When phenol is reacted with NaNO₂ and concentrated H₂SO₄,
it provides a deep green or blue colour which changes to red on dilution with water.
Generated substance in presence of NaOH / KOH restores original green or blue colour.
This reaction is termed as Liebermann's nitroso reaction

Option 4

5 Coordination Compounds

Atomic number of Cr and Fe are respectively 25 and 26. Which of the following is paramagnetic with the spin of the electron

1) [Cr(CO)_6]
2) [Fe(CO)₅]
3) [Fe(CN)₆]^4-
4) [Cr(NH₃)₆]³⁺

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Cr in [Cr(NH₃)6]³⁺ have three unpaired electrons

6 Haloalkanes and Haloarenes

In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e′) is :

1) 85 %

2) 15 %

3) 70 %

4) 60 %

Option 1

Option 2

Option 3 Solution:

SOLUTION

In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e) is 85−15=70 %

Note: Enantiomeric excess is the difference between the percent of two enantiomers.

Option 4

7 Purification and Characterisation of Organic Compounds

The techniques of gas liquid chromatography is suitable for compounds which are?

1) Soluble in water

2) Highly volatile

3) Vaporise without decompostion

4) Liquids

Option 1

Option 2

Option 3 Solution:

SOLUTION

Among the various types of gas chromatography, gas-liquid chromatography is the method most commonly used to separate organic compounds.

It follows the principle of partition. It can be used for compounds which vaporize without decomposition.

Option 4

8 Ionic Equilibrium

A solution contains a mixture of Ag⁺ (0.1M) and Hg₂²⁺ (0.1M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?
K_sp (AgI) = 8.5 $\times$ 10^–17

K_sp (Hg₂I₂) = 2.5 $\times$ 10^–26

1) 90.83%
2) 97.83%
3) 89.93%
4) 99.83%

Option 1

Option 2

Option 3

Option 4 Solution:

Solution:

Let us first calculate [I^–] to precipitate AgI and Hg₂I₂

K_sp[AgI] = [Ag⁺] [I^–]

8.5 $\times$ 10^–17 = (0.1) [I^–]

[I^–] to precipitate as AgI = (8.5 $\times$ 10^-17)

K_sp(Hg₂²⁺) = [Hg₂I₂][I^–] = 8.5 $\times$ 10^–16 M

2.5 $\times$ 10^–26 = 0.1 [I^–]²

[I^–] to precipitate Hg₂I₂ = 5.0 $\times$ 10^–13 M

[I^–] to precipitate AgI is smaller. Therefore, Ag I will start precipitating first. On further addition of I^– more AgI will precipitate and when [I^–] ³ 5.0 ´ 10^–13 J, Mg₂I₂ will start precipitating. The maximum concentration of Ag⁺ at this stage will thus be calculated as:

K_sp(AgI) = [Ag⁺] [I^–]

8.5 $\times$ 10^–17 = [Ag⁺] (5.0 ´ 10^–13)

or, [Ag⁺] = 1.7 $\times$ 10^–4 M

Percentage of Ag⁺ remained precipitated = [(1.7 $\times$ 10^–4 M)/0/1] $\times$ 100 = 0.17%

Thus percentage of Ag⁺ precipitated = 99.83%

9 Solid State

A solid X⁺Y^- has a bcc structure. If the distance of closest approach between the ions is 173 pm, the edge length of the cell is

1) √2 pm
2) (√3/2) pm
3) 200 pm
4) 400 pm

Option 1

Option 2

Option 3 Solution:

SOLUTION

In bcc
distance closest approach = a(√3/2)
173 = a(√3/2) a ≈200 pm

Option 4

10 Stoichiometry

Oxalic acid (H₂C₂O₄) is present in many plants and vegetables. If 24.0 mL of 0.01M KMnO4 solution is needed to titrate 1.0 g of H₂C₂O₄ to the equivalence-point, the mass percentage (as nearest integer) of H₂C₂O₄ in the sample is :
1) 3
2) 2
3) 5
4) 6

Option 1

Option 2

Option 3 Solution:

Solution

$$2$$ moles of $$KMnO_4$$ $$\equiv$$ $$5$$ moles of $$H_2C_2O_4$$.

Molar mass of oxalic acid = $$90$$ g/mol.

Number of moles of $$KMnO_4= \dfrac {24.0} {1000} \times 0.01 = 0.00024$$ mol.

Number of moles of $$H_2C_2O_4 = \dfrac {5}{2} \times 0.00024= 0.0006$$ mol.

Mass of $$H_2C_2O_4 = 0.0006 \times 90=0.054$$ g.

Mass percentage of $$H_2C_2O_4 = \dfrac {0.054}{1}\times 100 = 5.4\%$$.

The nearest integer is 5.

Option 4

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