In SO₂$\text{SO}_2$, NO₂^-$\text{NO}_2^-$ and N₃^-$\text{N}_3^-$ the hybridizations at the central atom are respectively:

Core Logic

The ionic character of a bond is directly proportional to the electronegativity difference (Δ EN$\Delta EN$) between the two bonded atoms. Larger Δ EN$\Delta EN \implies$ higher ionic character.

Step 1: Assess Electronegativity Differences

• N₂$N_2$: Both atoms are Nitrogen. Δ EN = 0$\Delta EN = 0$. Purely covalent. (Lowest ionic character)
• SO₂$SO_2$: Bond between S and O. Moderate Δ EN$\Delta EN$. Covalent with some polarity.
• ClF₃$ClF_3$: Bond between Cl and F. Δ EN$\Delta EN$ is higher than S-O as F is the most electronegative element.
• K₂O$K_2O$: Bond between K (alkali metal, very low EN) and O. Very high Δ EN$\Delta EN$. Ionic.
• LiF$LiF$: Bond between Li (alkali metal) and F (highest EN). Maximum Δ EN$\Delta EN$ possible among these options. Most ionic.

Step 2: Order Derivation

Increasing order of ionic character (or Δ EN$\Delta EN$): N₂ < SO₂ < ClF₃ < K₂O < LiF$N_2 < SO_2 < ClF_3 < K_2O < LiF$

Pattern Recognition

Homodiatomic (N₂$N_2$) is always 0% ionic. Alkali metal + Halogen (LiF$LiF$) represents the extreme of the ionic spectrum. Sorting non-metals by group distance yields the middle ranks.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Using VSEPR theory to find the hybridization and shape:

• PF₅$PF_5$: P has 5 valence electrons, forms 5 single bonds with F. Steric number = 5 (sp3d). 0 lone pairs. Shape = Trigonal bipyramidal.
• BrF₅$BrF_5$: Br has 7 valence electrons, forms 5 single bonds, 1 lone pair. Steric number = 6 (sp3d2). Shape = Square pyramidal.
• PCl₅$PCl_5$: P has 5 valence electrons, 5 bonds, 0 lone pairs. Steric number = 5 (sp3d). Shape = Trigonal bipyramidal.
• [PtCl₄]²⁻$[PtCl_4]^{2-}$: Pt²⁺$Pt^{2+}$ is a d⁸$d^8$ system. With Cl^-$Cl^-$ (but 4d/5d transition metals always form low spin square planar complexes), it's dsp²$dsp^2$ hybridized. Shape = Square planar.
• BF₃$BF_3$: B has 3 valence electrons, 3 bonds, 0 lone pairs. Steric number = 3 (sp2). Shape = Trigonal planar.
• Fe(CO)₅$Fe(CO)_5$: Fe (d6s2 -> d8 under strong field CO$CO$). Carbonyls strongly prefer 5-coordinate trigonal bipyramidal geometry for d⁸$d^8$ (dsp³$dsp^3$ hybridization). Shape = Trigonal bipyramidal.

Step 1: Count Trigonal Bipyramidal Molecules

Molecules with trigonal bipyramidal shape:

• PF₅$PF_5$
• PCl₅$PCl_5$
• Fe(CO)₅$Fe(CO)_5$

Total count = 3.

Pattern Recognition

Steric Number = 5 with 0 lone pairs ALWAYS yields Trigonal Bipyramidal geometry. Watch out for BrF₅$BrF_5$ which has 5 bonds but 1 lone pair (SN = 6, Square Pyramidal).

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Related Formula

Core Logic

When atomic orbitals combine, they form an equal number of molecular orbitals:

• Two 2s$2s$ atomic orbitals combine to form 1 bonding orbital (σ₂ₛ$\sigma_{2s}$) and 1 antibonding orbital (σ^*₂ₛ$\sigma^*_{2s}$).
• Six 2p$2p$ atomic orbitals combine to form 3 bonding orbitals (σ2pz, π2pₓ, π2py$\sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}$) and 3 antibonding orbitals (σ^2pz, π^2pₓ, π^*2py$\sigma^_{2p_z}, \pi^_{2p_x}, \pi^*_{2p_y}$).

Step 1: Total Summation

Summing the antibonding orbitals from both subshells:

Pattern Recognition

The linear combination of N$N$ atomic orbitals always yields exactly (N)/(2)$\frac{N}{2}$ antibonding molecular orbitals.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Related Formula

Core Logic

Analyze the molecular geometry and symmetry of each molecule:

• CH₄$\text{CH}_4$: Symmetrical tetrahedral geometry μ = 0$\implies \mu = 0$.
• BF₃$\text{BF}_3$: Symmetrical trigonal planar geometry μ = 0$\implies \mu = 0$.
• H₂O$\text{H}_2\text{O}$: Bent shape due to lone pairs μ ≠ 0$\implies \mu \neq 0$.
• HF$\text{HF}$: Linear asymmetric diatomic molecule μ ≠ 0$\implies \mu \neq 0$.
• NH₃$\text{NH}_3$: Trigonal pyramidal shape due to a lone pair μ ≠ 0$\implies \mu \neq 0$.
• CO₂$\text{CO}_2$: Symmetrical linear structure (O=C=O$\mathrm{O}=\mathrm{C}=\mathrm{O}$) where dipoles cancel out μ = 0$\implies \mu = 0$.
• SO₂$\text{SO}_2$: Bent angular geometry due to a lone pair μ ≠ 0$\implies \mu \neq 0$.

Step 1: Final Counting

The molecules with a net zero dipole moment are CH₄$\text{CH}_4$, BF₃$\text{BF}_3$, and CO₂$\text{CO}_2$. This gives a total count of 3.

Pattern Recognition

Molecules with a symmetrical arrangement of identical bonds and no lone pairs on the central atom (e.g., tetrahedral CH₄$\text{CH}_4$, trigonal planar BF₃$\text{BF}_3$, linear CO₂$\text{CO}_2$) always have a net dipole moment of zero.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

For CHCl₃$CHCl_3$, the individual dipole vectors do not cancel due to differing electronegativities of H and Cl, leading to a permanent non-zero dipole moment (μ ≠ 0$\mu \neq 0$).

Pattern Recognition

Symmetry yields vector cancellation arrow μ=0$\rightarrow \mu=0$. Asymmetry in CHCl₃$CHCl_3$ prevents cancellation arrow$\rightarrow$ polar.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure