Chemistry Test 290

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Chemistry Test 290

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Weekly Quiz Competition

Question 1
4 / -1

In diborane ( B₂H₆ ), which of the following statements is incorrect?

A
The four terminal B-H bonds are 2-centre-2-electron bonds.

B
There are two 3-centre-2-electron bonds.

C
Both boron atoms are sp² hybridised.

D
The four terminal hydrogen atoms and the two boron atoms lie in one plane.

Solution

This statement is incorrect. In diborane, both boron atoms are actually sp³ hybridised to accommodate the four bonds (two terminal and two bridged).
Question 2
4 / -1

Which of the following reaction is not possible?

A
(CH₃)₂O + BF₃⟶ [(CH₃)₂O → BF₃]

B
NH₃+ BF₃⟶ [NH₃→ BF₃]

C
(CH₃)₂O + BF₄⁻¹⟶ [(CH₃)₂O → BF₄⁻¹]

D
N(CH₃)₃+ B(CH₃)₃⟶ [N(CH₃)₃→ B(CH₃)₃]

Solution

Step 1: Identify the type of reaction.
All four choices describe Lewis acid–base reactions in which a Lewis base (electron-pair donor) donates its lone pair to a Lewis acid (electron-pair acceptor) to form a coordinate (dative) bond. The arrow notation should therefore be a dative arrow (→) from the donor to the acceptor.
Question 3
4 / -1

A
can react with NH₃

B
dipole moment is zero

C
vacant orbital is not involved in bonding

D
electron deficient molecule

Solution

Given: The reaction between boron trifluoride (BF₃) and lithium aluminum hydride (LiAlH₄) in the presence of ether produces a compound X, along with lithium fluoride (LiF) and aluminum fluoride (AlF₃).

Solution:
In this context, the compound X is typically a borane derivative, such as diborane (B₂H₆).

The bonding in diborane is unique. It consists of four terminal B − H bonds, which are two-center two-electron bonds, and two bridging B − H − B bonds, which are three-center two-electron bonds (also known as banana bonds).

In the formation of the bridging bonds, the vacant orbital of one boron atom overlaps with the half-filled hybrid orbital of the other boron atom and the half-filled 1 s orbital of the bridging hydrogen atom. This process involves the vacant orbital in bonding. Therefore, "vacant orbital is not involved in bonding" statement is incorrect.
Question 4
4 / -1

Which period in the Periodic Table contains the 4f inner transition series?

A
5^th period

B
6^th period

C
7^th period

D
4^th period

Solution

The 4f inner transition series (lanthanoid series) starts at cerium (Z = 58) and ends at lutetium (Z = 71), which belongs to the 6th period. The 6th period includes the filling of 4f orbitals after the 6s orbital.
Question 5
4 / -1

Two solutions A and B are separated by semi- permeable membrane. If liquid flows form A to B then

A
A is less concentrated than B

B
A is more concentrated than B

C
Both have same concentration

D
None of these

Solution

Step 1: Liquid (solvent) flows from solution A to solution B through the semi-permeable membrane.

Step 2: Osmosis dictates that solvent always moves from the less concentrated (lower solute concentration) side to the more concentrated (higher solute concentration) side.

Step 3: Because the solvent is leaving A and entering B, A must initially have the higher solute concentration.

Conclusion: A is more concentrated than B.
Question 6
4 / -1

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20^∘C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be (write the answer upto two decimal places):

A
0.549

B
0.200

C
0.786

D
0.478

Solution

Key Concepts and Formulas

1. Raoult's Law: The partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
Question 7
4 / -1

Volume of water needed to mix with 10 ml 10 N HNO₃ to get 0.1 N HNO₃ :

A
1000 ml

B
990 ml

C
1010 ml

D
10 ml

Solution
Question 8
4 / -1

Which of the following amines has the lowest pKb value?

A
Methanamine

B
N-Methylmethanamine

C
N,N-Dimethylmethanamine

D
Benzenamine

Solution
Question 9
4 / -1

Which of the following is by product of saponification :

A
Glycol

B
Glycerol

C
Bithionol

D
Phenol

Solution

Step 1: Identify the products of saponification.

The products of saponification are soap and glycerol.

Step 2: Define saponification.

Saponification is the process of turning fats or oils (triglycerides) into a salt of a fatty acid (soap) and glycerol by reacting them with a strong base, such as sodium hydroxide (NaOH) or potassium hydroxide (KOH).
Question 10
4 / -1

Which block does hydrogen strictly belong to based on its electronic configuration?

A
p−block

B
d−block

C
s−block

D
f−block

Solution

Given: Hydrogen has a single electron in the 1s orbital (1s¹), which places it in the s− block. However, it is often positioned separately due to its unique properties.