See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Heat of Dissociation Analysis **Key Concept:** Heat of dissociation depends on bond strength in the adduct $A \cdot B$ complex. Stronger the B-N coordinate bond, higher the dissociation energy. **Step 1: Identify the adducts** All options form Lewis acid-base adducts where $(CH_3)_3N$ (Lewis base) coordinates to boron compounds (Lewis acids): - (A): $(CH_3)_3N \cdot BF_3$ - (B): $(CH_3)_3N \cdot B(CH_3)F_2$ - (C): $(CH_3)_3N \cdot B(CH_3)_2F$ - (D): $(CH_3)_3N \cdot B(CH_3)_3$ **Step 2: Assess boron's electron deficiency** The strength of N→B coordinate bond increases with boron's electron deficiency (acidity): - $BF_3$ is most electron-deficient (F is highly electronegative, withdraws electron density) - $B(CH_3)_2F$ is intermediate (mixed F and CH₃) - $B(CH_3)_3$ is least electron-deficient (CH₃ is electron-donating) **Step 3: Correlate with dissociation enthalpy** - **Strongest N-B bond** → **Highest dissociation energy** - **Weakest N-B bond** → **Minimum dissociation energy** **Step 4: Determine the answer** Since $B(CH_3)_3$ has the weakest electron-withdrawing power, the N-B coordinate bond in $(CH_3)_3N \cdot BF_3$ is **weakest**, requiring the **least energy** to break. **Answer: (A)** has the minimum heat of dissociation because $BF_3$ forms the weakest coordinate bond due to competition from three highly electronegative F atoms distributing boron's electron deficiency.