Chemical Bonding And Molecular Structure

Dipole moment,

where, magnitude of electric charge
distance between particles (or bond length)

or

Sulphuric anhydride is and its structure is as follows:

bonds are present.

Step 1: Understand the molecular comparison.
Comparing the stability of diatomic molecules or ions, H is more stable than He due to the higher bond order in H .
Step 2: Conclusion.
Thus, the correct answer is (D) H >He .
Final Answer:

The structure with the double bonds between the sulfur and oxygen atoms (i.e., ) is the most stable due to the resonance structures and the least repulsion between atoms.

Step 2: Conclusion.
The most preferred structure is option (d), where sulfur is double-bonded to each oxygen.

The hybridisation of the central atoms in and is different. In , the nitrogen atom is sp² hybridised, while in , the oxygen atom is sp³ hybridised.

Step 2: Conclusion.
The correct statement is that they are dissimilar in hybridisation for the central atom with different structures, corresponding to option (a).

The oxygen molecule is paramagnetic, but the ion is diamagnetic. Therefore, option (c) presents the correct description.

Step 2: Conclusion.
The correct description is - diamagnetic, - paramagnetic, corresponding to option (c).

The bond angles in these molecules are influenced by the electron pair geometry around the central atom. has a 107° bond angle, has a slightly larger bond angle, and has a 120° bond angle due to its tetrahedral geometry.

Hydrogen bonds are strongest when the hydrogen is bonded to a highly electronegative atom like fluorine. The HF . . . HF interaction is the strongest in the vapour phase.

In dicyanoethene, the sigma and pi bonds can be counted based on the bonding between carbon atoms and the number of double bonds.

The species and are non-planar due to their molecular geometries. and other species are planar.

The compound does not exist as is a highly reactive diatomic molecule that does not readily combine with .

1,3,4 are coloured ions hence the answer is b.

Step 1: Phosphorus can expand its valency by utilizing vacant -orbitals in its outer shell, enabling it to form , while nitrogen cannot due to the absence of such -orbitals.

Final Answer:

Step 1: Xenon in , , and exhibits different numbers of lone pairs depending on the oxidation state of Xenon and its coordination.
Step 2: In , has 2 lone pairs, in it has 3 lone pairs, and in , it has 4 lone pairs.

Final Answer:

Step 1: Understanding Polarizability
Polarizability refers to the ability of an ion or atom to have its electron cloud distorted by an external electric field. Larger ions with more electrons are generally more polarizable.
Step 2: Analyzing Halide Ions
The halide ions are , , , and . Their polarizability increases with increasing size and number of electrons.
Step 3: Order of Polarizability
The order of polarizability for halide ions is:
This is because is the largest and has the most electrons, making it the most polarizable, while is the smallest and least polarizable.
Step 4: Matching with the Options
The correct order corresponds to option (D). Final Answer: The polarizability of halide ions increases in the order .

Step 1: Understanding Acidic, Basic, and Amphoteric Oxides
Acidic Oxides: Oxides of non
metals that react with water to form acids.
Basic Oxides: Oxides of metals that react with water to form bases.
Amphoteric Oxides: Oxides that can react with both acids and bases.
Step 2: Analyzing the Options
Option (A): (basic), (acidic), (amphoteric). This option is incorrect because the order is not as required.
Option (B): (acidic), (basic), (acidic). This option is incorrect because it does not include an amphoteric oxide.
Option (C): (acidic), (basic), (amphoteric). This option is correct.
Option (D): (basic), (acidic), (amphoteric). This option is incorrect because the order is not as required.
Step 3: Matching with the Options
The correct order of acidic, basic, and amphoteric oxides is given in option (C). Final Answer: The correct option is (C) .