NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Chemical Equilibrium

Which description applies to the reaction above?

AlCl₃ + Cl^------> AlCl₄^-

      1) amphotorism
        2) anion hydrolysis
        3) Arrhenius neutralization
        4) Lewis acid-base reaction

Option 1

Option 2

Option 3

Option 4 Solution:

2 Chemistry in Everyday Life

When sodium stearate CH3(CH2)16COONa, is added to hard water the formula of one possible precipitate is:

1) (CH₃(CH₂)₁₆COO)₂Ca

2) (CH₃(CH₂)₁₆COO)₂Ba

3) (CH₃(CH₂)₁₆COO)₂Cd

4) CH₃(CHv)₁₆COOK

Option 1 Solution:

Solution

When sodium stearate CH₃(CH₂)₁₆COONa, is added to hard water, calcium of hard water reacts with the soap and replaces its sodium and forms calcium precipitate, (CH₃(CH₂)₁₆COO)₂Ca

Option 2

Option 3

Option 4

3 Ionic Equilibrium

At 298K a 0.1 M CH3COOHsolution is 1.34% ionized. The ionization constant K_a for acetic acid will be

1) 0.182×10⁻⁵
2) 18.2×10⁻⁵
3) 1.82×10⁻⁵
4) none

Option 1

Option 2

Option 3 Solution:

Solution:

K = cα²= 0.1 × (1.34 / 100)² = 1.82×10⁻⁵

Option 4

4 Chemical Equilibrium

Buffer solution can be obtained by the mixing aqueous solutions of:

1) CH₃COONa and CH₃COOH
2) CH₃COOH and excess NaOH
3) CH3₃COONa and excess HCl
4) NaCl and HCl.

Option 1 Solution:

Solution

A mixture of acetic acid (a weak acid) and sodium acetate ( its salt with strong base sodium hydroxide) acts as acidic buffer.

Option 2

Option 3

Option 4

5 Chemical Kinetics

On increasing temperature from 200K to 220K, rate constant of reaction A increases by 3 times and rate constant of reaction B increases by 9 times then correct relationship between activation energy of A and B is:

1) E_A=3_EB
2) 3E_A=E_B
3) E_B=2E_A
4) E_A=2E_B

Option 1

Option 2

Option 3 Solution:

Solution

We have Arrhenius equation for calculation of energy of activation of reaction with rate constant $$K$$ and temperature $$T$$ is

$$K=A$$ $$e^{-Ea/RT}$$ $$- (i)$$

where, $$Ea$$= Arrhenius activation energy

$$A$$= pre exponential factor

$$R$$= Ideal gas constant

According to question,

For reaction $$A$$ :-

$$T_1=200K, T_2=220K$$

$$3(r_1)_A=(r_2)_A$$

where, $$(r_1)_A$$= rate of reaction $$A$$ at $$T_1 K$$

$$(r_2)_A=$$ rate of reaction $$A$$ at $$T_2K$$

$$\because$$ Rate of reaction $$\propto$$ Rate constant

$$\Rightarrow$$ $$3(K_1)_A=(K_2)_A$$

where, $$K_1$$ and $$K_2$$ are rate constants at $$T_1$$ and $$T_2K$$

Now, $$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$ $$- (ii)$$

Using $$(i)$$ in $$(ii)$$ and putting the values

$$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$

$$\Rightarrow \cfrac {1}{3}= \cfrac {A\quad e^{-Ea/RT_1}}{A\quad e^{-Ea/RT_2}}$$

$$\Rightarrow \cfrac {1}{3}=$$$$ e^{-Ea/R \left(\cfrac {1}{T_2}-\cfrac {1} {T_1}\right)}$$

$$= e^{-Ea/R \left(\cfrac {1}{220}- \cfrac {1}{200}\right)}$$

$$\Rightarrow\cfrac {1}{3}= e^{+\cfrac {E_a}{R} \times \cfrac {20}{220\times 200}}$$

$$\Rightarrow \log \left( \cfrac {1}{3}\right)=\cfrac {E_a}{R}\times \cfrac {20}{220\times 200}$$

$$E_a= 2200$$ $$R$$ $$\log (1/3)$$ $$- (A)$$

For reaction $$B$$:-

$$T_1=200K$$, $$T_2=220K$$

$$\cfrac {(K_1)_B} {(K_2)_B}=\cfrac {1}{9}$$

By following the similar procedure as above,

$$\log \left(\cfrac {1}{9}\right)=\cfrac {E_b}{R}\times \cfrac {1}{2200}$$

where, $$E_b=$$ activation energy of reaction $$B$$.

$$\Rightarrow E_b=2200$$ $$R$$ $$\log\left(\cfrac {1}{9}\right)$$ $$- (B)$$

Now, from $$(A)$$ and $$(B)$$:-

$$\cfrac {E_a}{E_b}=\cfrac {log (1/3)}{log (1/9)}=\cfrac {1}{2}$$

$$\Rightarrow 2E_a=E_b$$

$$\Rightarrow 2E_A=E_B$$

Option 4

6 Basic Principles of Organic Chemistry

The most stable free radical among the following is

1) C₆H₅CH₂
2) C₆H₅CHCH₃
3) CH₃CH₂
4) CH₃CHCH₃

Option 1

Option 2 Solution:

SOLUTION

The most stable free radical is C₆H₅CHCH₃ because of resonance.

Option 3

Option 4

7 Redox Reactions

In which of the following pair oxidation number of Fe is same:

1) K₃Fe(CN)₆,Fe₂O₃
2) Fe(CN)₅,Fe₂O₃
3) Fe₂O₃,FeO
4) Fe₂(SO₄)₃,K₄Fe(C)₆

Option 1 Solution:

SOLUTION

Let the oxidation state of Fe in K₃Fe(CN)₆and Fe₂O₃ be x and y respectively.

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Oxidation state of Fe in K₃Fe(CN)₆-

?CN is a negative ligand.

∴Oxidation number of CN is -1.

Oxidation number of K is +1.

∴3(+1)+x+6(−1)=0

⇒x−3=0

⇒x=3

Oxidation state of Fe in Fe₂O₃^-

Oxidation no. of O is -2.

2(y)+3(−2)=0

⇒2y=6

⇒y=3

Option 2

Option 3

Option 4

8 Thermodynamics

The change in entropy of an ideal gas during a reversible isothermal expansion is :

1) Negative
2) Positive
3) Zero
4) infinity

Option 1

Option 2 Solution:

In the reversible case, the change in entropy of the surroundings is equal and opposite to the change in the system, so the change in entropy of the universe is zero. In the free expansion, Q = 0, so the entropy of the surroundings does not change and the change in entropy of the universe is equal to ΔS for the system

Option 3

Option 4

9 Redox Reactions

Among the following, identify the species with an atom in +6 oxidation state.

1) [MnO₄]⁻
2) [Cr(CN)₆]³⁻
3) Cr₂O₃
4) CrO₂Cl₂

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

CrO₂Cl₂

Cr+2(O)+2(Cl)=0

Cr+2(−2)+2(1)=0

Cr=+6

10 Hydrocarbons

Which of the following compounds will not undergo Friedel -Craft's reaction easily?

1) Cumene
2) Xylene
3) Nitrobenzene
4) Toluene

Option 1

Option 2

Option 3 Solution:

SOLUTION

Nitro group being electron withdrawing, deactivates the benzene nucleus to such an extent so that it becomes incapable to give Friedel -Crafts reaction.
Nitrobenzene because of its unreactivity towards Friedel-Craft's reaction is used as a solvent for this reaction.

Option 4

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