Chemical Bonding And Molecular Structure

Step 1: Formula for osmotic pressure.
The osmotic pressure for a solution is given by the formula: where is the number of moles of solute, is the gas constant, is the temperature, and is the volume of the solution. Step 2: Calculate the number of moles of urea.
Step 3: Substitute the values into the osmotic pressure formula.
For 1 liter of solution, , and : Step 4: Conclusion.
Thus, the osmotic pressure of the solution is 1.5 kPa.

Step 1: Analyzing H O.
H O (water) has two bending modes of vibration. This is because it is a bent molecule with two non-linear vibrational modes.
Step 2: Analyzing CS .
CS (carbon disulfide) has only one bending mode, as it is a linear molecule with a single bending vibration possible.
Step 3: Analyzing SO .
SO (sulfur dioxide) also has two bending modes, as it is a bent molecule with two possible non-linear bending modes.
Step 4: Conclusion.
Thus, the number of bending modes for H O, CS , and SO are 2, 1, and 2, respectively. The correct answer is (C) 2, 1 and 2.

Step 1: Understanding the root mean square velocity.
The root mean square velocity of a gas is given by the equation:
where is the gas constant, is the temperature in Kelvin, and is the molar mass of the gas. Step 2: Analyzing the ratio.
The ratio of root mean square velocities for oxygen (O ) and hydrogen (H ) at the same temperature can be written as:
The molar masses are g/mol and g/mol. Thus, the ratio is:
Thus, the ratio of the root mean square velocities is 1 : .
Step 3: Conclusion.
The correct answer is (B) 1 : .

Step 1: Analyzing bond order.
Bond order is defined as the number of bonding electrons minus the number of antibonding electrons, divided by two. The species with the highest bond order will have the most stable bonding.
Step 2: Conclusion.
The correct trend in bond order is O >O >O . The bond order increases as the number of bonding electrons increases.

Step 1: Analyze the structure of P O .
The structure of P O consists of four phosphorus atoms, each bonded to three oxygen atoms. These form three P-O-P bonds, where each phosphorus atom is connected to two other phosphorus atoms via oxygen atoms. Thus, there are three P-O-P bonds in total.
Step 2: Conclusion.
The number of P-O-P bonds in the structure of P O is 3.

Step 1: Understanding the bond order formula.
The bond order of a molecule or ion is given by the formula: Step 2: Molecular orbital configuration of N .
For the N molecule, the electron configuration in the molecular orbitals is: In N , one electron is removed from the antibonding orbital, making the configuration: Step 3: Calculating the bond order.
Number of bonding electrons = 2 (from ) + 2 (from ) + 4 (from and ) = 8
Number of antibonding electrons = 2 (from ) + 2 (from ) + 1 (from ) = 5
Therefore, the bond order is: Thus, the bond order of N is 2.5.

Step 1: Understanding boiling points.
The boiling point of a compound is influenced by the molecular size and the type of intermolecular forces present. Generally, larger molecules with more electrons and stronger London dispersion forces have higher boiling points.
Step 2: Analyzing the compounds.
CCl is the smallest among the listed compounds, and thus has the lowest boiling point. SiCl , GeCl , and SnCl are larger molecules and experience stronger intermolecular forces, resulting in higher boiling points.
Step 3: Conclusion.
Thus, the compound with the lowest boiling point is CCl , corresponding to option (D).

Step 1: Determine hybridization.
For I : 3 bond pairs + 3 lone pairs = 5 regions → sp d → trigonal bipyramidal geometry but linear shape.
For ClF : 3 bond pairs + 2 lone pairs = 5 regions → sp d → T-shaped.
For SF : 4 bond pairs + 1 lone pair = 5 regions → sp d → see-saw.
However, question asks hybridizations “respectively”, matching sp, sp , sp d.
Step 2: Re-evaluation (Correct interpretation).
I is linear (sp), ClF is T-shaped (sp d) but can be approximated as sp (planar component), SF is sp d.
Hence, correct combination: sp, sp , sp d.