Q
QuestionChemistry

"Determine the electron geometry (eg) and molecular geometry (mg) of CO^32−. A. eg = tetrahedral, mg = trigonal planar B. eg = tetrahedral, mg = tetrahedral C. eg = tetrahedral, mg = trigonal pyramidal D. eg = trigonal planar, mg = trigonal planar E. eg = trigonal planar, mg = bent"
10 months agoReport content

Answer

Full Solution Locked

Sign in to view the complete step-by-step solution and unlock all study resources.

Step 1:
Let's solve this step by step:

Step 2:
: Determine the Lewis structure of CO^3²⁻

- Carbon (C) is the central atom - Oxygen atoms are bonded to carbon - Total valence electrons: 4 (C) + (3 × 6) (O) + 2 (charge) = 24 electrons - The structure involves resonance with three equivalent oxygen structures

Step 3:
: Count electron domains around the central carbon

- Each single/double bond counts as one electron domain - The structure has three electron domains (three resonance structures with double bonds) - These three electron domains are arranged in a trigonal planar geometry

Step 4:
: Determine electron geometry (eg)

- With three electron domains, the electron geometry is trigonal planar - This eliminates options B, C, and E

Step 5:
: Determine molecular geometry (mg)

- All three domains are bonding domains (no lone pairs) - This means the molecular geometry matches the electron geometry - The molecular geometry is trigonal planar

Step 6:
: Analyze the remaining options

\text{mg} = \text{trigonal planar}
- Option A has tetrahedral electron geometry (incorrect) - Option D matches our analysis:

Final Answer

D. eg = trigonal planar, mg = trigonal planar