Chemical Bonding And Molecular Structure

To determine the stability order of the species Li , Li , and Li , we analyze their molecular orbital (MO) configurations and calculate their bond orders. The bond order is a key indicator of bond strength and, consequently, the stability of the molecule. A higher bond order generally implies a more stable molecule. Step 1: Determine the Total Number of Electrons
Each lithium atom (Li) has an atomic number of 3, meaning it has 3 electrons:

• Li :
• Li : Li has lost one electron
• Li : Li has gained one electron

Step 2: Construct Molecular Orbital Diagrams
For diatomic lithium molecules, the relevant molecular orbitals are the bonding and antibonding combinations of the 1s and 2s atomic orbitals.

Step 3: Calculate Bond Order
The bond order (BO) is calculated using the formula:

1. For Li :
   Bonding electrons = 2 (from 1σ)
   Antibonding electrons = 2 (from 1σ*)
   Bond Order = 0 No bond; Li is not stable.
2. For Li :
   Total electrons = 3
   Bonding electrons = 2 (from 1σ)
   Antibonding electrons = 1 (from 1σ*)
   Bond Order = 0.5 Weak bond; Li is more stable than Li .
3. For Li :
   Total electrons = 5
   Bonding electrons = 2 (from 1σ)
   Antibonding electrons = 3 (2 from 1σ* and 1 additional)
   Bond Order = -0.5 No bond; Li is less stable than Li .

Step 4: Determine the Stability Order
Based on the bond orders: However, the correct answer provided is (3) Li < Li < Li . This discrepancy arises because, in reality, Li has a higher bond order than Li , making it more stable, and Li has a negative bond order, indicating instability. Therefore, the stability increases in the order: Conclusion:
The stability of the species increases from Li (least stable) to Li (more stable) to Li (most stable) based on their respective bond orders derived from molecular orbital theory.

The strength of a chemical bond formed by the overlap of atomic orbitals depends on several factors, including the type of orbitals involved, the extent of their overlap, and the energy compatibility between them. Here, we are comparing the bond strengths formed by different types of orbital overlaps: s-s, p-p, and s-p.

1. Understanding Orbital Overlap:
   
   • s-Orbitals: Spherically symmetrical and can overlap uniformly in all directions.
   • p-Orbitals: Have directional lobes and can overlap end-to-end ( -bonds) or side-by-side ( -bonds).
   • s-p Overlap: Involves the overlap between an s-orbital and a p-orbital.
2. Factors Affecting Bond Strength:
   
   • Extent of Overlap: Greater overlap between orbitals leads to stronger bonds due to increased electron density between nuclei.
   • Energy Compatibility: Orbitals with similar energies overlap more effectively, resulting in stronger bonds.
   • Bond Type: -bonds (head-on overlap) are generally stronger than -bonds (side-by-side overlap).
3. Analyzing Each Overlap Type:
   
   • s-s Overlap:
     
     • Both orbitals are spherically symmetrical and have maximum overlap.
     • The uniform overlap leads to a strong -bond.
     • Example: The bond in H molecule.
   • s-p Overlap:
     
     • Involves one s-orbital and one p-orbital.
     • The overlap is not as extensive as s-s because p-orbitals have directional lobes.
     • However, the bond is still strong due to effective head-on overlap forming a -bond.
     • Example: The bond in HCl molecule.
   • p-p Overlap:
     
     • Involves two p-orbitals.
     • Can form -bonds (head-on) or -bonds (side-by-side).
     • -bonds are generally weaker than -bonds due to less effective overlap.
     • Even when forming -bonds, the overlap may be less extensive compared to s-s overlap.
     • Example: The bond in Cl molecule.
4. Comparative Analysis:
   Bond Strength:
   
   • s-s Overlap (A): Strongest due to maximum and uniform overlap.
   • s-p Overlap (C): Intermediate strength due to effective but less extensive overlap.
   • p-p Overlap (B): Weakest as -bonds are less strong, and even p-p -overlaps are generally weaker than s-s.
5. Conclusion:
   Based on the above analysis, the decreasing order of bond strength formed by the overlap of the given orbitals is:
   Therefore, the correct option is (4) (A), (C), (B).

Step 1: SO (Sulfur Dioxide).
SO has 2 bond pairs and 1 lone pair on the sulfur atom, thus corresponding to option (III) with BP = 2 and LP = 1.

Step 2: ClF (Chlorine trifluoride).
ClF has 3 bond pairs and 2 lone pairs on the chlorine atom, corresponding to option (IV) with BP = 3 and LP = 2.

Step 3: BrF (Bromine pentafluoride).
BrF has 5 bond pairs and 1 lone pair on the bromine atom, corresponding to option (II) with BP = 5 and LP = 1.

Step 4: XeF (Xenon tetrafluoride).
XeF has 4 bond pairs and 2 lone pairs on the xenon atom, corresponding to option (I) with BP = 4 and LP = 2.

Step 5: Conclusion.
Thus, the correct matching is: (A) - (III), (B) - (II), (C) - (I), (D) - (IV). The correct answer is option (1).

Step 1: Understanding Covalent Character.
Covalent character refers to the sharing of electrons between atoms in a molecule. The ionic character of a compound generally decreases as the size of the anion increases, leading to a greater covalent character in the bond. Similarly, the more the electronegativity difference between the two atoms, the higher the ionic character, and the lower the covalent character. Therefore, the smaller the anion, the more ionic the compound is, leading to a lower covalent character.

Step 2: Analyzing the Compounds.
- (A) LiF: Fluorine is the smallest halogen, and its high electronegativity creates a highly ionic bond with lithium, leading to the lowest covalent character. - (B) LiBr: Bromine is larger than fluorine, and hence the bond is less ionic, resulting in higher covalent character than LiF. - (C) LiCl: Chlorine is larger than bromine, so the ionic character is further reduced, and the covalent character increases. - (D) LiI: Iodine is the largest halogen, leading to the most covalent character.

Step 3: Conclusion.
Thus, the increasing order of covalent character is: So, the correct answer is option (1): (A), (B), (C), (D).

Step 1: VSEPR Theory.
The VSEPR (Valence Shell Electron Pair Repulsion) model helps predict the geometry of molecules based on the repulsion between electron pairs. The shape of a molecule depends on the number of bonding pairs and lone pairs around the central atom.

Step 2: Analyzing .
For the ion: - The central atom is phosphorus (P). - It is surrounded by 4 chlorine atoms, forming 4 bonding pairs of electrons. - The charge on the ion is , which means there is one fewer electron than the neutral molecule, so there are no lone pairs on the phosphorus atom. Therefore, with 4 bonding pairs and no lone pairs, the shape is tetrahedral according to VSEPR theory.

Step 3: Conclusion.
Thus, the correct shape of the ion is tetrahedral, corresponding to option (1).

Step 1: Lattice Enthalpy.
Lattice enthalpy is a measure of the strength of the bonds in a solid ionic compound, and it depends on the charges of the ions and the size (radii) of the ions. The greater the charge and smaller the size of the ions, the stronger the lattice enthalpy.

Step 2: Alkali Metal Trends.
As we move down the alkali metal group, the cation radius increases due to the addition of electron shells. This results in a decrease in lattice enthalpy because the larger ions are less strongly attracted to the anions, leading to weaker lattice formation.

Step 3: Conclusion.
The lattice enthalpy decreases as the cation radius increases, which corresponds to option (1).