Atomic Masses And Composition Of Nucleus

Step 1: The bond order of a species is calculated using the molecular orbital theory. The formula for bond order is:
Step 2: For , the electron configuration is , resulting in a bond order of 2.5.
Step 3: For , the bond order is 2.5, while for , the bond order is 2.0, and for , the bond order is 2.
Step 4: The species with the smallest bond order is with a bond order of 2.0.

Step 1: The basic strength of oxides generally increases as we move down a group in the periodic table. This is because the oxides become more ionic and thus more basic.
Step 2: The order of basicity is inversely related to the electronegativity of the metal. As we move from Be to Ba, the metal's electronegativity decreases, and the oxide becomes more basic.
Step 3: The correct order of increasing basic strength is:

Step 1: Understanding Clemmensen Reduction.
- Clemmensen reduction is a method used to reduce carbonyl compounds (aldehydes and ketones) to alkanes.
- The reaction is carried out in strongly acidic conditions.
Step 2: Reaction Mechanism.

- Zinc amalgam (Zn(Hg)) acts as the reducing agent.
- Hydrochloric acid provides the protons needed for reduction.
- The carbonyl oxygen is removed as water, and the carbonyl carbon is converted to an alkane.
Step 3: Identifying the correct answer.
- Option (A) is correct because Clemmensen reduction specifically requires zinc amalgam and HCl.
- Other options involve sodium amalgam or nitric acid, which are not used in this reaction.

Hydrolysis of diborane.
- Diborane reacts with water to form boric acid ( ) and hydrogen gas ( ).

Step 1: Understanding the bcc structure. In a body-centred cubic (bcc) unit cell, there are 2 atoms per unit cell. This is because there is 1 atom at the center of the cell and 1/8th of an atom at each of the 8 corners.
Step 2: Volume of the unit cell. The edge length of the unit cell is given as . The volume of the unit cell is:
Step 3: Density of the element. The density of the element is given as 5 g cm . Convert this to kg m :
Step 4: Number of unit cells in 300 g of the element. The molar mass of the element can be calculated using the density and the volume of the unit cell. The number of moles in 300 g of the element is: The total volume occupied by 1 mole of the element is: Using the relationship between molar mass and density:
Step 5: Conclusion. Given the setup, solving these steps yields that the number of atoms in 300g is . Thus, the correct answer is (D) .