Amongst SF , XeF , CF and H O, the number of species with two lone pairs of electrons is _____. 4 4 — Chemical Bonding Chemistry Question

💡 Solution & Explanation

**Step 1: Determine the Lewis structure and lone pairs for SF₄** SF₄: S has 6 valence electrons, 4 F atoms bonded. - Bonding pairs: 4 - Remaining electrons on S: 6 - 4 = 2 electrons = 1 lone pair - **Lone pairs on SF₄: 1** **Step 2: Determine lone pairs for XeF₄** XeF₄: Xe has 8 valence electrons, 4 F atoms bonded. - Bonding pairs: 4 - Remaining electrons on Xe: 8 - 4 = 4 electrons = 2 lone pairs - **Lone pairs on XeF₄: 2** ✓ **Step 3: Determine lone pairs for CF₄** CF₄: C has 4 valence electrons, 4 F atoms bonded. - Bonding pairs: 4 - Remaining electrons on C: 4 - 4 = 0 electrons - **Lone pairs on CF₄: 0** **Step 4: Determine lone pairs for H₂O** H₂O: O has 6 valence electrons, 2 H atoms bonded. - Bonding pairs: 2 - Remaining electrons on O: 6 - 2 = 4 electrons = 2 lone pairs - **Lone pairs on H₂O: 2** ✓ **Step 5: Count species with two lone pairs** Species with exactly 2 lone pairs: XeF₄ and H₂O Therefore, the answer is 2.00.