cbse 11th chemistry | Online Objective Test

Chemical Bonding and Molecular Structure Prepare / Learn

Q1. When

O_{2}

is converted into

{O_{2}}^{+}

A. paramagnetic character decreases and the bond order increases

B. both paramagnetic character and bond order increase

C. paramagnetic character increases

D. bond order decreases

Answer : Option A
Explaination / Solution:

Bond order = $\frac{N_{b} - N_{a}}{2}$

For O2 = $\frac{10 - 6}{2}$ = 2

For O2+ = $\frac{10 - 5}{2}$ = 2.5

The s-Block Elements Prepare / Learn

Q2. Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down the group?

A. lattice enthalpy decreases down the group

B. size of anions being much shorterer compared to cations

C. hydration enthalpy decreases down the group

D. coordination numbesr more than four

Answer : Option C
Explaination / Solution:

Solubility of sulphates and carbonates decreases down the group because of decrease in hydration energy. which is insufficient to over come the lattice energy of ions.

Organic Chemistry Some Basic Principles and Techniques Prepare / Learn

Q3. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.

H_{3} C - - - HC = C H_{2} + H^{+} \to ?

1^{\circ}

Carbanion

1^{\circ}

Carbocation

2^{\circ}

Carbanion

1^{\circ}

Carbocation

Answer : Option B
Explaination / Solution:

A secondary carbocation is formed.

States of Matter Prepare / Learn

Q4. 2.9 g of a gas at 95

^{\circ} C

occupied the same volume as 0.184 g of dihydrogen at 17

^{\circ} C

, at the same pressure. What is the molar mass of the gas?

A. 40 g

mo l^{- 1}

B. 50 g

mo l^{- 1}

C. 43 g

mo l^{- 1}

D. 46 g

mo l^{- 1}

Answer : Option A
Explaination / Solution:

PV=(

m_{1}

/M)R

T_{1}

where m 1 =2.9g and

T_{1}

=368K PV= (

m_{2}

/2)

R T_{2}

where

m_{2}

=0.184 and

T_{2}

= 290K

Redox Reactions Prepare / Learn

Q5. The oxidizing power of halogens increase in the order of

A. F2<I2<Br2<Cl2

B. Cl2<F2< I2< Br2

C. II2<Br2<Cl2<F2

D. Br2<Cl2<F2< I2

Answer : Option C
Explaination / Solution:

Halogens have high electronegativity and electron affinity. They have greater tendency to accept electrons or easily reduced, therefore they are strong oxidizing agent. As the reduction potential decrease down the group,the oxidizing power decrease down the group the order of the oxidizing power will be as under F2 > Cl2> Br2> I2 ¥ The oxidizing power depends on, Heat of dissociation of halogen molecule. Electron affinity of atom. Hydration energy of the ion. Heat of vaporization If a halogen has low energy of dissociation, a high electron affinity, and higher hydration of its ion, it will have high oxidizing power. F has although low electron affinity than Cl but low dissociation energy and have high hydration energy of its ion, therefore Fluorine is strongest oxidizing agent.

Some Basic Concepts of Chemistry Prepare / Learn

Q6.

How many molecules of H2O are there in 18g of water?

(Hint: Avagadro’s Number = 6.02 x 1023 atoms/mol)

A. 6.022 x 1024

B. 8.02 x 1023

C. 7.02 x 1023

D. 6.022 x 1023

Answer : Option D
Explaination / Solution:

Since , the number of molecules contained in one mole of a substance

= Avogadro's number (N)

= 6.022 x 1023 molecules ,

and , gram molar mass of water

= 18 g

$∴$ 18g H2O = 1mol water

$∴$ one mol of water would contain

= 6.022 x 1023 molecules of water

Equilibrium Prepare / Learn

Q7.

B F_{3}

does not have proton but still acts as an acid and reacts with NH3. Why is it so? What type of bond is formed between the two?

B F_{3}

is a Lewis acid and coordinate bond is formed

B F_{3}

is a Lewis acid and coordinate bond is formed

B F_{3}

is a Brönsted acid and coordinate bond is formed

B F_{3}

is a Brönsted acid and coordinate bond is formed

Answer : Option B
Explaination / Solution:

B F_{3}

is a Lewis acid as B has its vacant p orbital and coordinate bond is formed as

N H_{3}

donates its lp of electrons to the vacant orbital of B.

Hydrocarbons Prepare / Learn

Q8. Which one of these is NOT TRUE for benzene? Explanation: Benzene forms mono substituted product.

A. Heat of hydrogenation of benzene is less than the theoretical value

B. The bond angle between carbon-carbon bonds is

12 0^{\circ}

There are three carbon-carbon single bonds and three carbon-carbon double bonds

It forms only one type of monosubstituted product

Answer : Option C
Explaination / Solution:
No Explaination.

Classification of Elements and Periodicity in Properties Prepare / Learn

Q9. The increasing order of bond dissociation energies of following species is -----?

A. Cl2< O2< N2< F2

B. N2< O2< F2< Cl2

C. F2< Cl2< O2< N22

D. F2< N2< Cl2< O2

Answer : Option C
Explaination / Solution:

As bond order increases bond dissociation energy also increases .

Thus $N_{2}$

(i) the triple bonded $N_{2}$ with bond order

= 3

has the highest bond dissociation energy .

(ii) The doubly bonded $O_{2}$ with bond order

= 2

has lesser bond dissociation energy than $N_{2}$

(iii) $F_{2}$ and $C l_{2}$ both are single bonded having the same bond order

= 1

but , the electron - electron repulsive force ( e - e repulsion ) is maximum in a $F_{2}$ molecule due to its small size. This accounts for the lower bond dissociation energy of $F_{2}$ as compared to $C l_{2}$

Environmental Chemistry Prepare / Learn

Q10. The balance of carbon dioxide and oxygen has been upset by :

A. Increasing human population

B. Burning fossil fuels and deforestation

C. Large scale deforestation

D. Excessive burning of fossil fuels

Answer : Option B
Explaination / Solution:

The balance of carbon dioxide and oxygen has been upset by Burning fossil fuels and deforestation which causes release of excessive CO2 and reduced O2

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