Chemistry Test 78

Result

Chemistry Test 78

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

A
K₂O < Na₂O < Al₂O₃ < MgO

B
Al₂O₃ < MgO < Na₂O < K₂O

C
MgO < K₂O < Al₂O₃ < Na₂O

D
MgO < K₂O < Na₂O < Al₂O

Solution

Basic nature of oxides increases with increase in the size of cation.

The increasing order of cations is: Al³⁺ < Mg²⁺ < Na⁺ < K⁺

Therefore the increasing correct order of basic strength is: Al₂O₃ < MgO < Na₂O < K₂O

Hence, the correct option is (B).
Question 2
4 / -1

The ratio of the energy difference between first and second, and that between second and third Bohr's orbits is

A
1/2

B
1/3

C
4/9

D
27/5

Solution
Question 3
4 / -1

Aryl halides are less reactive towards nucleophilic substitution than alkyl halides. Which of the following factors is not responsible for this fact?

A
Formation of more stable carbonium ion

B
Resonance stabilisation

C
Inductive effect

D
sp²-hybridised carbon attached to the halogen

Solution

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to resonance stabilization. In alkyl halides, the carbon atom is sp³-hybridized whereas, in aryl halides, it is sp²-hybridized and hence, more electronegative.

Hence, the correct option is (B).
Question 4
4 / -1

The correct order of increasing nucleophilicity is

A
Cl^– < Br^– < I^–

B
Br^– < Cl^– < I^–

C
I^– < Br^– < Cl^–

D
I^– < Cl^– < Br^–

Solution

As we come down the group atomic size increases, thus bond length decreases, bond strength increases and acidic strength increases and basic strength increases and nucleophiles having different donor atoms in the same group then nucleophilicity is anti-parallel to basic strength. Then order will be- Cl^– < Br^– < I^–

Hence, the correct option is (A).
Question 5
4 / -1

Consider the reaction, CH₃CH₂CH₂Br + NaCN→ CH₃CH₂CH₂CN +NaBr. This reaction will be the fastest in:

A
Ethanol

B
Methanol

C
N, N-dimethylformamide

D
Water

Solution

The solvent affects the rate of reaction because solvents may or may not surround a nucleophile, thus hindering or not hindering its approach to the carbon atom. Polar aprotic solvents, like tetrahydrofuran, are better solvents for this reaction than polar protic solvents because polar protic solvents will hydrogen bond to the nucleophile, hindering it from attacking the carbon with the leaving group. A polar aprotic solvent with a low dielectric constant or a hindered dipole end will favor SN² manner of nucleophilic substitution reaction. Examples: DMSO, DMF, acetone etc. In polar aprotic solvent, nucleophilicity parallels basicity.

Hence, the correct option is (C).
Question 6
4 / -1

A certain quantity of electricity is passed through an aqueous solution of AgNO₃ and CuSO₄ solution in series. If the mass of Ag deposited is 1.08 g, then the mass of copper deposited will be

A
0.635 g

B
6.35 g

C
0.3117 g

D
1.270 g

Solution
Question 7
4 / -1

Which of the following can possibly be used as an analgesic, without causing addiction and modification?

A
Morphine

B
N-acetyl-para-aminophenol

C
Diazepam

D
Tetrahydrocannabinol

Solution

N-acetyl-para-aminophenol(or paracetamol) is an antipyretic which can also be used as Analgesic to relieve pain without addiction and mood modification.

Hence, the correct option is (B).
Question 8
4 / -1

In which of the following changes is an electron added to antibonding (π^∗)orbital?

A

B

C

D
All of the above

Solution
Question 9
4 / -1

Calculate the mass of copper that will be deposited at the cathode in the electrolysis of a 0.2M solution of copper sulphate when the quantity of electricity equal to that required to liberate 2.24L of hydrogen at NTP from a 0.1M aqueous sulphuric acid is passed. (Atomic mass of Cu=63.5u)

A
63.5 g

B
6. 35 g

C
635 g

D
0. 635 g

Solution
Question 10
4 / -1

The bond dissociation energy of B−F in BF₃ is 646 kJ per mol whereas that of C−F in CF₄ is 515 kJ mol−1. The correct reason for higher bond dissociation energy of B−F bonds than that of C−F bond is

A
Lower degree of pπ−pπ interaction between B and F in BF₃ than that between C and F in CF_i

B
Smaller size of B atom than that of C atom

C
Stronger bond between B and F in BF₃ than that between C and F in CF₄

D
Significant pπ −pπ interactions between BF₃ whereas there is no possibility of such interactions between C and F in CF₄

Solution

BF₃ is a Lewis acid due to an incomplete octet of Boron. Among BF ₃ BCl₃,BBr₃ and so on, BF₃ is the weakest Lewis acid due to back bonding of lone pair electron of F to p-orbital of Boron.

B has vacant available p-orbital in BF₃ and hence it involves pπ - pπ back bonding which is not possible in CF₄ as C does not have any vacant orbital.

Hence, the correct option is (D).