Chemical Bonding and Molecular Structure — Practice Quiz

Axial bond pairs face more repulsion from equatorial bond pairs (at 90°) than equatorial bonds face from each other (at 120°)

Step 1: In PCl₅ (sp³d hybridisation), three Cl atoms occupy equatorial positions (120° apart) and two Cl atoms occupy axial positions (90° from equatorial plane). Step 2: Each axial bond pair is at 90° from all three equatorial bond pairs — this is a strong, close-range repulsion. Step 3: Each equatorial bond pair is at 120° from the other two equatorial pairs — a weaker, wider-angle repulsion. Step 4: Because axial bonds suffer greater repulsion, they are pushed outward slightly, making them longer (and weaker). Step 5: This makes PCl₅ more reactive because the axial Cl atoms are easier to di

ClF₃ → sp³

Step 1: ClF₃ has Cl as the central atom. Cl has 7 valence electrons and forms 3 bonds with F, leaving 2 lone pairs. Step 2: Total electron pairs = 3 (bond pairs) + 2 (lone pairs) = 5. Step 3: With 5 electron pairs, the hybridisation is sp³d (not sp³ which only accounts for 4 pairs). Step 4: sp³d gives a trigonal bipyramidal electron geometry; with 2 lone pairs in equatorial positions, the molecular shape is T-shaped. Step 5: sp³ would only be correct for 4 electron pairs. So ClF₃ → sp³d is correct, making ClF₃ → sp³ the wrong match.

KK(σ2s)²(σ*2s)²(σ2pz)²(π2px)²(π2py)²

Step 1: N has 7 electrons; N₂ has 14 electrons total. KK accounts for 4 electrons (σ1s² and σ*1s²). Step 2: For N₂ (lighter molecule), the correct MO order puts (π2p) BELOW (σ2pz): KK(σ2s)²(σ*2s)²(π2px)²(π2py)²(σ2pz)². Step 3: Both orderings give the same electron population here; Nb = 10, Na = 4. Step 4: Bond order = ½(10 – 4) = 3. The configuration KK(σ2s)²(σ*2s)²(σ2pz)²(π2px)²(π2py)² also gives the same count: Nb=10, Na=4. Step 5: All electrons are paired, confirming diamagnetic nature. Option A correctly lists all bonding MOs fully filled with no antibonding electrons beyond σ*2s.

CO₂ is a linear molecule; the two equal and opposite C=O bond dipoles cancel each other vectorially

Step 1: Each C=O bond is polar because oxygen is more electronegative than carbon, so each bond has a dipole moment pointing from C toward O. Step 2: In CO₂, the carbon is sp hybridised, making the molecule perfectly linear (bond angle = 180°). Step 3: The two C=O dipoles point in exactly opposite directions along the same axis. Step 4: Since dipole moment is a vector quantity, these two equal but opposite vectors cancel completely: μ(net) = μ₁ – μ₂ = 0. Step 5: This is why CO₂ is nonpolar overall despite having polar bonds — symmetry of the molecule is the key factor. Compare this with H₂O (b