### Universal Decision Tree for Chemical Bonding Problems
**Step 1: Identify the question type**
- Shape/geometry question → Use VSEPR (SN method)
- Bond order/magnetic → Use MOT
- Ionic character/polarity → Use Fajan's rules / dipole moment
- Lattice energy → Use Born-Haber cycle
- Sigma/pi bond count → Use bond type rules
**Step 2: For VSEPR questions — SN Method**
```
1. Count valence electrons of central atom
2. SN = σ bonds + lone pairs on central atom
3. SN → hybridization (2=sp, 3=$sp^{2}$, 4=$sp^{3}$, 5=$sp^{3}$d, 6=$sp^{3}$$d^{2}$)
4. SN → electron geometry
5. Subtract lone pairs → molecular shape
6. Apply lone pair compression to get bond angles
```
**Step 3: For MOT questions — Bond Order Method**
```
1. Count total electrons in the species
2. Identify Z (for homonuclear): Z≤7 → π2p before σ2p; Z>7 → σ2p before π2p
3. Fill MOs in correct order (Aufbau, Pauli, Hund's)
4. Count Nb (bonding) and Na (antibonding)
5. BO = (Nb − Na) / 2
6. If any unpaired electrons → paramagnetic; else diamagnetic
```
**Step 4: For dipole moment questions**
```
1. Draw the Lewis structure
2. Identify molecular shape (via VSEPR)
3. Check: is the molecule symmetric? (All bonds to identical atoms, no lone pairs disturbing symmetry)
4. If symmetric → μ = 0; if asymmetric → μ ≠ 0
5. For qualitative comparison: higher electronegativity difference + more asymmetric shape → higher μ
```
**Step 5: For Fajan's rules questions**
```
1. Identify the cation and anion
2. Compare: smaller cation or higher cation charge → more covalent
3. Compare: larger anion → more covalent
4. More covalent → lower melting point, more soluble in organic solvents, lower lattice energy
```
### Worked Example Using the Chain
**Question:** Is $PCl_{3}$ polar or non-polar?
Chain:
1. Shape question → VSEPR
2. P (Group 15): 5 valence $e^{-}$; 3 bonds to Cl; lone pairs = (5−3)/2 = 1
3. SN = 3+1 = 4 → $sp^{3}$ → tetrahedral electron geometry → trigonal pyramidal molecular shape
4. Trigonal pyramidal = asymmetric (lone pair at apex)
5. P–Cl bonds are polar ($\Delta E$N ≠ 0) AND asymmetric → μ ≠ 0
6. **$PCl_{3}$ is POLAR** (μ ≈ 0.97 D)$
Part of PC-03 — Chemical Bonding & Molecular Structure
Reasoning Chain: How to Solve Any NEET Chemical Bonding Question
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