Hybridisation

To determine the molecules/ions with hybridization, we examine the structure and electron pairs around the central atom of each species.

1. : Phosphorus has 5 bond pairs, no lone pairs. Hybridization is .
2. : Bromine has 5 bond pairs, 1 lone pair. Hybridization is (Octahedral).
3. : Phosphorus has 3 bond pairs, 1 lone pair. Hybridization is .
4. : Sulfur has 6 bond pairs, no lone pairs. Hybridization is (Octahedral).
5. : Iodine has 4 bond pairs, 2 lone pairs. Hybridization is (Square Planar).
6. : Chlorine has 3 bond pairs, 2 lone pairs. Hybridization is .
7. : Iodine has 5 bond pairs, 1 lone pair. Hybridization is (Octahedral).

After analyzing the molecules/ions, those with hybridization are , , , and , totaling 4.

The correct answer is (B) : Both Statement I and Statement II are false.
is square planar and diamagnetic with μ = 0 , its hybridisation is dsp²
is tetrahedral but diamagnetic.

To determine the truthfulness of the given statements about hybridization, let's analyze each statement separately.

1. Analysis of Statement I: "PF₅ and BrF₅ both exhibit sp³d hybridization."
   • : Phosphorus pentafluoride is a molecule where phosphorus is the central atom bonded to five fluorine atoms. The hybridization of phosphorus in PF₅ is sp³d, forming a trigonal bipyramidal geometry. This is correct.
   • : Bromine pentafluoride is a molecule where bromine is the central atom bonded to five fluorine atoms and one lone pair. The hybridization here is sp³d², resulting in a square pyramidal geometry. Statement I claims sp³d hybridization, which is incorrect for BrF₅.
2. Analysis of Statement II: "Both SF₆ and [Co(NH₃)₆]³⁺ exhibit sp³d² hybridization."
   • : Sulfur hexafluoride is where sulfur is bonded to six fluorine atoms. The hybridization of sulfur in SF₆ is indeed sp³d², forming an octahedral shape. This is correct.
   • : In this complex, cobalt is surrounded by six ammonia ligands. The metal-ligand bond involves d²sp³ hybridization when considering inner orbital complex formation (low-spin complex due to ligand field theory), not sp³d². Therefore, the statement is incorrect concerning this complex.

After analyzing both statements, we find that:

• Statement I is false since BrF₅ does not exhibit sp³d hybridization.
• Statement II is also false as [Co(NH₃)₆]³⁺ does not exhibit sp³d² but instead forms d²sp³ hybridization for this specific complex.

Thus, the correct answer is: Both Statement I and Statement II are false.

To solve the problem of determining which complex exhibits hybridization, we need to understand the concept of hybridization and apply it to each option.

Concept of Hybridization: Hybridization is the process of combining atomic orbitals to form new hybrid orbitals. The type of hybridization helps determine the geometry of the molecule.

1. [Co(NH₃)₆]³⁺

• Cobalt (Co) in this complex is in the +3 oxidation state.
• The electronic configuration of Co is [Ar] 3d⁷ 4s². For Co³⁺, it becomes [Ar] 3d⁶.
• The Co³⁺ ion will undergo hybridization, using its 3d, 4s, and 4p orbitals to form the octahedral complex.

2. SF₆

• In SF₆, the sulfur atom is surrounded by six fluorine atoms.
• Sulfur forms six bonds using hybridization, resulting in an octahedral geometry.

3. BrF₅

• In BrF₅, the central bromine atom is surrounded by five fluorine atoms and one lone pair.
• This molecule exhibits hybridization, leading to a square pyramidal shape.

4. [PtCl₄]^2-

• In this case, platinum (Pt) forms four coordinate bonds with chlorine atoms.
• The complex adopts a hybridization, resulting in a square planar geometry, typical for d⁸ electron count of Pt²⁺.

Hence, the correct answer is [Co(NH₃)₆]³⁺ as it forms a d²sp³ hybridization.

We are given that the concentration of the nitric acid solution is 75% by mass. This means that for every 100 g of solution, there are 75 g of nitric acid. We are asked to find the volume of the solution that contains 30 g of nitric acid. First, calculate the total mass of the solution required to get 30 g of nitric acid: Now, using the density of the solution, which is 1.25 g/mL, calculate the volume of the solution: Thus, the volume required is 40 mL.

The solubility product (Ksp) determines how soluble a compound is in water. The larger the Ksp, the more soluble the compound. In this case, we need to compare the solubility products for the compounds.
- Ca(OH)2 has a relatively high Ksp as it dissociates readily.
- AgBr has a lower Ksp compared to Ca(OH)2.
- HgS has an even lower Ksp, meaning it is less soluble than AgBr.
- PbS has the lowest Ksp among these compounds, making it the least soluble.
Thus, the increasing order of solubility products is: .

- Sucrose consists of glucose and fructose linked by an glycosidic bond, so it matches with (I).
- Maltose has two glucose units connected by an bond, and it also has the possibility of forming a linkage with , which matches with (II).
- Lactose is a disaccharide formed from one molecule of glucose and one of galactose, with an glycosidic bond, so it matches with (III).
- Amylopectin is a highly branched polymer of glucose with linkages in its chains, so it matches with (IV).
Thus, the correct matching is (A)-(I), (B)-(II), (C)-(III), (D)-(IV).