Chemical Bonding

Bond order is exactly 1.5, confirmed because 128 pm lies between 121 pm and 148 pm

Step 1: Ozone has two canonical structures — one with O=O (double bond, 121 pm) on the left, and one with O=O on the right. Each structure has one double and one single bond. Step 2: The resonance hybrid is not one structure rapidly switching — it is a SINGLE stable structure that is an average of all canonical structures. Step 3: The average bond order = (1 + 2)/2 = 1.5. This means each O-O bond in O₃ has characteristics intermediate between single and double bond. Step 4: The experimental bond length 128 pm lies between 121 pm (double) and 148 pm (single), confirming bond order ≈ 1.5. Step 5

BF₃

Step 1: Dipole moment is a vector quantity — it has both magnitude and direction. The NET dipole moment of a molecule depends on the vectorial sum of all individual bond dipoles. Step 2: NH₃ has a pyramidal shape (3 bond pairs + 1 lone pair). The three N-H bond dipoles do NOT cancel, and the lone pair adds its orbital dipole — net dipole = 1.47 D (non-zero). Step 3: H₂O is bent/V-shaped (2 bond pairs + 2 lone pairs). The two O-H bond dipoles point in similar directions and don't cancel — net dipole = 1.85 D (non-zero). Step 4: HCl is a diatomic polar molecule with electronegativity difference

3 sigma bonds and 2 pi bonds

Step 1: In ethyne (H-C≡C-H), each carbon is sp hybridised. sp hybridisation means one 2s orbital mixes with one 2p orbital to give 2 sp hybrid orbitals, leaving 2 unhybridised p orbitals on each carbon. Step 2: Bond 1 — C-H sigma bond on the left carbon: sp hybrid orbital of C overlaps with 1s of H (σ bond). Step 3: Bond 2 — C-C sigma bond: sp hybrid orbital of C₁ overlaps with sp hybrid orbital of C₂ along the internuclear axis (σ bond). Step 4: Bond 3 — C-H sigma bond on the right carbon (σ bond). So total sigma bonds = 3. Step 5: The 2 unhybridised p orbitals on each carbon (say p_y and p_z

Bond order = 3; diamagnetic because all electrons in MOs are paired

Step 1: N₂ has 14 electrons total (7 from each N). Full MO configuration: σ1s², σ*1s², σ2s², σ*2s², π2p²(x), π2p²(y), σ2p². Step 2: Count bonding electrons (in σ1s, σ2s, π2p_x, π2p_y, σ2p) = 2+2+2+2+2 = 10. Count antibonding electrons (in σ*1s, σ*2s) = 2+2 = 4. Step 3: Bond order = ½(bonding - antibonding) = ½(10 - 4) = ½(6) = 3. This confirms the triple bond N≡N. Step 4: For magnetic behaviour: look at ALL molecular orbitals. Every orbital has 2 electrons (completely filled, paired). There are NO singly occupied (half-filled) MOs. Step 5: Since all electrons are paired, there are no unpaired