Modern Periodic Law And The Present Form Of The Periodic Table

Step 1: Identifying the reaction mechanism
Toluene ( ) undergoes chlorination under boiling conditions, leading to the formation of Benzotrichloride ( ).
Step 2: Conversion to Benzoic Acid
Benzotrichloride ( ) undergoes acidic hydrolysis ( ), resulting in Benzoic Acid ( ).
Conclusion: Thus, is Toluene and is Benzoic Acid.

At the equivalence point, formic acid (HCOOH) is neutralized by NaOH, forming its conjugate base, formate . The pH at the equivalence point depends on the hydrolysis of the formate ion, which acts as a weak base.
Step 1: The neutralization reaction is:
Step 2: The concentration of formate ion at the equivalence point is the same as the initial concentration of formic acid (since the reaction is 1:1), i.e., 2M.
Step 3: The hydrolysis of formate ion in water:
Step 4: The of formate can be calculated from the relation:
Step 5: For a 2M solution of formate ion, the concentration of ions produced can be found using the equilibrium expression for .
Step 6: Solving the equilibrium gives the pOH, and the pH is then calculated as:

Hydrogen peroxide (H O ) has an interesting molecular structure where the two oxygen atoms are connected by a single bond, with each oxygen atom bonded to one hydrogen atom. The molecule adopts a non-planar, non-linear structure due to the repulsion between lone pairs of electrons on the oxygen atoms. 1. Bond angle: The bond angle in H O is approximately 111° (not 180° as in a linear molecule), which makes it non-linear. 2. Planarity: The oxygen-oxygen bond in H O is not in a plane due to the electron pair repulsion, which leads to a non-planar structure.
Thus, the correct answer is , non-planar, non-linear.

Viscosity is a measure of a fluid's resistance to flow. The effect of temperature on viscosity is different for liquids and gases: 1. Liquids: As the temperature increases, the kinetic energy of the molecules in the liquid increases. This leads to a decrease in the intermolecular forces and thus a reduction in viscosity. Therefore, the viscosity of liquids decreases with an increase in temperature. 2. Gases: In gases, increasing the temperature causes the molecules to move more rapidly, which increases the frequency of collisions between them. This leads to a reduction in the effect of intermolecular forces, and the viscosity of gases also decreases with an increase in temperature.
Thus, the correct answer is , as the viscosity decreases in both liquids and gases with an increase in temperature.

The compressibility factor is a measure of how much a real gas deviates from ideal behavior. For ideal gases, at all pressures. Real gases, however, exhibit deviations from ideality, and these deviations depend on the specific properties of the gas, including the parameters and , which are related to intermolecular forces and molecular size, respectively.
- Option A: For gas A, , which means no intermolecular attraction forces. The compressibility factor increases linearly with pressure, which is a characteristic behavior for gases without significant intermolecular interactions.
- Option B: For gas B, , which means no molecular size effects (ideal gas behavior). However, the dependence of on pressure is not linear at all pressures for real gases. This makes Option B incorrect, as it contradicts the graph, which shows a non-linear relationship at intermediate pressures.
- Option C: For gas C, which behaves as a typical real gas, both and are non-zero. By analyzing the minima and the intersection points with , both parameters and can indeed be determined.
- Option D: At high pressure, the compressibility factor for all real gases becomes positive, indicating that the gas particles are being compressed and interactions between molecules become more significant. Thus, the incorrect statement is Option B.

Let us examine the hybridization of the central atoms in each molecule: - : The central atom is carbon, which has 4 bond pairs and no lone pairs. This gives it an hybridization. - : The central atom is nitrogen, which has 3 bond pairs and 1 lone pair. This gives it an hybridization. - : The central atom is nitrogen, which has 4 bond pairs and no lone pairs due to the positive charge. This also gives it an hybridization. - : The central atom is oxygen, which has 2 bond pairs and 2 lone pairs. This gives it an hybridization. Thus, all these molecules have hybridization on their central atoms.
Therefore, the correct option is B.
- Option A: This is incorrect because the number of lone pairs is not the same for all molecules.
- Option C: This is incorrect because the bond angles are not the same for all molecules. For example, has 109.5° bond angles, while has 104.5° bond angles.
- Option D: This is incorrect because the number of bond pairs is different for each molecule.
Thus, the correct answer is Option B.

The complex ion involves chromium in the +3 oxidation state, i.e., . The electron configuration of is: In this case, chromium has three electrons in its -orbitals available for bonding. To form a coordination complex with six fluoride ions (F ), chromium must undergo hybridization to accommodate six bonding pairs. For six ligands (such as fluoride ions) to bond to chromium, six hybrid orbitals are required. This involves the promotion of electrons from the -orbitals to the higher energy and -orbitals, resulting in the hybridization. In this hybridization, two -orbitals, one -orbital, and three -orbitals combine to form six equivalent hybrid orbitals, which then overlap with the fluoride ions.
Thus, the hybridization of is .

The stability of an ionic bond depends on the difference in electronegativity between the two elements involved. A greater difference leads to a stronger ionic bond. Additionally, smaller ions tend to form more stable bonds due to the higher lattice energy.
- Na and Cl: Sodium (Na) has a lower ionization energy and chlorine (Cl) has a high electron affinity, but the ionic bond formed is relatively weaker than others because Na is a larger ion.
- Mg and F: Magnesium (Mg) has a higher ionization energy and fluorine (F) has a very high electron affinity. The small size of F and the high charge density of Mg result in a highly stable ionic bond.
- Li and F: Lithium (Li) and fluorine (F) also form a stable bond due to the small size of both ions, but magnesium and fluorine have a higher lattice energy due to magnesium's higher charge.
- Na and F: Sodium (Na) and fluorine (F) form an ionic bond, but it is less stable than the Mg-F bond because Na has a lower charge density than Mg. Thus, the most stable ionic bond is formed between Mg and F due to the higher lattice energy and stronger electrostatic attraction between the ions.