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Step 1:I'll solve this step by step, following the precise LaTeX formatting guidelines:
For NOBr:
Step 2:: Determine the central atom
- The central atom is N (nitrogen)
Step 3:: Count valence electrons
- Total: $$5 + 6 + 7 = 18$$ valence electrons
- N: 5 valence electrons - O: 6 valence electrons - Br: 7 valence electrons
Step 4:: Draw Lewis structure
- Arrange electrons to satisfy octet rule - Single bonds between N-O and N-Br - Remaining electrons as lone pairs
Step 5:: Electron-pair geometry
- Electron-pair geometry is $$\mathrm{tetrahedral}
- Count total electron domains around N - 3 bonding domains (N-O, N-Br) - 1 lone pair - Total: 4 electron domains
Step 6:: Molecular geometry
For $$\mathrm{NO}_{2}\mathrm{F}$$:
- 1 lone pair reduces symmetry
Step 7:: Determine the central atom
- The central atom is N (nitrogen)
Step 8:: Count valence electrons
- Total: $$5 + 12 + 7 = 24$$ valence electrons
- N: 5 valence electrons - O (×2): 6 × 2 = 12 valence electrons - F: 7 valence electrons
Step 9:: Draw Lewis structure
- Double bonds between N-O - Single bond between N-F - Remaining electrons as lone pairs
Step 10:: Electron-pair geometry
- Electron-pair geometry is $$\mathrm{tetrahedral}
- Count total electron domains around N - 3 bonding domains (N-O×2, N-F) - 1 lone pair - Total: 4 electron domains
Step 11:: Molecular geometry
- Molecular geometry is $$\mathrm{bent}
- 1 lone pair reduces symmetry
Final Answer
1. NOBr: - Electron-pair geometry: \mathrm{tetrahedral} - Lone pairs: 1 - Molecular geometry: \mathrm{bent} 2. \mathrm{NO}_{2}\mathrm{F}: - Electron-pair geometry: \mathrm{tetrahedral} - Lone pairs: 1 - Molecular geometry: \mathrm{bent}
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