Position Of Hydrogen In The Periodic Table

We are asked to analyze the order of isotopes , , and with respect to four properties: boiling point, bond energy, bond length, and reactivity with chlorine.
I. Order of boiling point: The boiling point of hydrogen isotopes increases with increasing mass. Therefore, the order of boiling points is: This is because (tritium) has the heaviest atoms, followed by (deuterium), and (protium) has the lightest atoms.
II. Order of bond energy: The bond energy increases as the mass of the hydrogen isotope decreases. Since the bond energy is inversely related to the mass of the isotopes, the order of bond energy is: This follows because (protium) has the weakest bond due to its lighter mass, and (tritium) has the strongest bond.
III. Order of bond length: The bond length remains the same for all three isotopes because the bond length is determined by the type of atom, not the mass. Therefore, the order of bond lengths is:
IV. Order of reactivity with : Reactivity with chlorine decreases as the mass of the hydrogen isotope increases. Therefore, the order of reactivity is: This is because the lighter the isotope, the more reactive it is with chlorine.
Conclusion: All of the given orders are correct. Therefore, the correct answer is .

To answer this question, we need to consider the bond order and the electronic configurations of the species involved. Bond order is a measure of the strength of a bond, and it is determined by the number of bonding electrons minus the number of anti-bonding electrons.
1. Pair I: O , O : - O (oxygen molecule) has 16 electrons. The bond order is calculated as: - O (oxide ion) has 18 electrons. The bond order is: Therefore, O has a stronger bond than O , meaning the first species has the weaker bond.
2. Pair II: N , N : - N has 10 electrons. The bond order is: - N has 9 electrons. The bond order is: Therefore, N has a stronger bond than N , meaning the first species has the stronger bond.
3. Pair III: NO , NO : - NO has 10 electrons. The bond order is: - NO has 11 electrons. The bond order is: Therefore, NO has a stronger bond than NO , meaning the first species has the stronger bond.
Conclusion: The stronger bonds are found in the first species in pairs II and III. Thus, the correct answer is .

In the hydrogen atom, the energy levels of orbitals are determined solely by the principal quantum number . The hydrogen atom does not have electron-electron repulsion, so the energy of orbitals with the same principal quantum number (like 3s, 3p, and 3d) are degenerate, meaning they have the same energy.
- Option A: This is correct because in the hydrogen atom, orbitals with the same (such as 3s, 3p, and 3d) have the same energy, as the energy depends only on the principal quantum number, not the type of orbital (s, p, or d).
- Option B: This is incorrect. While in multi-electron atoms, the 3s and 3p orbitals are lower in energy than the 3d orbital, in the case of the hydrogen atom, all the 3 orbitals (3s, 3p, and 3d) have the same energy.
- Option C: This is incorrect for the hydrogen atom. All the orbitals with the same principal quantum number have the same energy.
- Option D: This is incorrect for the hydrogen atom. The energy of the 3s orbital is not necessarily lower than the 3p orbital; both have the same energy in a hydrogen atom. Thus, the correct statement is Option A.

When glucose reacts with phenylhydrazine, it forms an osazone. Glucose, being an aldose sugar, contains both an aldehyde group and a hydroxyl group. The formation of an osazone involves the reaction of glucose with phenylhydrazine at both the aldehyde group and one of the hydroxyl groups. Glucose undergoes a reaction with three molecules of phenylhydrazine to form the corresponding osazone, which is a crystalline compound. This is a characteristic feature of reducing sugars like glucose, which form osazones in the presence of excess phenylhydrazine.
Thus, the value of is .

Let’s match the species from Column I to their correct types in Column II:
1. DDT (A): DDT (Dichlorodiphenyltrichloroethane) is a well-known pesticide, so it matches with 4. Pesticides.
2. NaClO (B): Sodium chlorate (NaClO ) is commonly used as a herbicide, so it matches with 3. Herbicides.
3. Cl (C): Chlorine gas (Cl ) is a disinfectant, so it matches with 2. Disinfectant.
4. PAN (D): PAN (Peroxyacetyl nitrate) is involved in the formation of photochemical smog, so it matches with 1. Photochemical smog.
Thus, the correct matching is , A → 4, B → 3, C → 2, D → 1.