Bond Parameters

To determine which pair of molecules, contain an odd electron molecule and an expanded octet molecule, let's analyze each option:

1. BCl₃ and SF₆:
   • BCl₃ (Boron Trichloride) is a compound in which boron has an incomplete octet. Boron has three valence electrons, bonding with three chlorine atoms without expanding its octet.
   • SF₆ (Sulfur Hexafluoride) has sulfur with six fluorine atoms bonded, leading to an expanded octet with 12 valence electrons around sulfur.
   • This option does contain an expanded octet molecule (SF₆), but BCl₃ does not have an odd electron.
2. NO and H₂SO₄:
   • NO (Nitric Oxide) is a molecule with an odd number of electrons (15 in total), classifying it as an odd electron molecule.
   • H₂SO₄ (Sulfuric Acid) has sulfur with an expanded octet due to the presence of more than 8 electrons in its valence shell.
   • This option has both an odd electron molecule (NO) and an expanded octet molecule (H₂SO₄), making it the correct choice.
3. SF₆ and H₂SO₄:
   • Both SF₆ and H₂SO₄ have expanded octets, but neither has an odd electron molecule.
4. BCl₃ and NO:
   • BCl₃ does not have an odd electron or an expanded octet.
   • NO has an odd electron; however, the combination does not have a molecule with an expanded octet.

Therefore, the pair of molecules that includes an odd electron molecule and an expanded octet molecule is: NO and H₂SO₄.

Step 1: Analyze the O–O bond length
In , the oxygen atoms are bonded to hydrogen atoms. In , the oxygen atoms are bonded to fluorine atoms. Fluorine is more electronegative than hydrogen. The higher electronegativity of fluorine in leads to a greater pull of electron density towards the fluorine atoms, which weakens the O–O bond and increases its bond length. Therefore, the O–O bond length in is longer than in . So, is longer.
Step 2: Analyze the O–H and O–F bond lengths
The O–H bond is formed between oxygen and hydrogen, while the O–F bond is formed between oxygen and fluorine. Fluorine has a smaller atomic radius than hydrogen. Also, oxygen and fluorine have much closer electronegativities, leading to a shorter O–F bond compared to the O–H bond where there’s a larger electronegativity difference. Therefore, the O–H bond length in is shorter than the O–F bond length in . So, is shorter.
Conclusion
The O–O bond length in is longer than in , and the O–H bond length is shorter than the O–F bond length. This corresponds to option (4).

Step 1: Consider the Trend Down the Group
Bond enthalpy generally decreases down a group in the periodic table. This is because as the atomic size increases, the bond length increases, and longer bonds are weaker.
Step 2: Analyze the Given Elements
The elements in question are C, Si, Ge, and Sn. They all belong to Group 14. Their atomic size increases down the group in the order:

Step 3: Determine the Bond Enthalpy Order
Since bond enthalpy decreases with increasing atomic size, the correct order of bond enthalpy is:

Concept: Bond length depends on bond order.

• Higher bond order shorter bond length.
• Lower bond order longer bond length.

In methyl acetate:

• : double bond
• : bond in ester group (partial double bond due to resonance)
• : single bond to methyl group

Step 1: {Analyze bond character} Bond length order: Thus Expressing in terms of : Which corresponds to option (B). However based on labeling in the given diagram the equivalent relation is written as