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QuestionChemistry

Draw the Lewis structure for [SeF^5]−.
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Answer

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Step 1:
I'll solve this Lewis structure problem step by step, following the specified LaTeX formatting guidelines:

Step 2:
: Determine the total number of valence electrons

Total valence electrons: $$6 + (7 \times 5) + 1 = 6 + 35 + 1 = 42$$ electrons
- Fluorine (F): 7 valence electrons × 5 fluorines

Step 3:
: Arrange central atom and surrounding atoms

- Selenium will be the central atom - Five fluorine atoms will be arranged around selenium

Step 4:
: Connect atoms with single bonds

Remaining electrons: $$42 - 10 = 32$$ electrons
- Draw single bonds between selenium and each fluorine

Step 5:
: Add lone pairs

- Place $$2$$ lone pairs on selenium
- Place 6 lone pairs on each fluorine atom (30 electrons)

Step 6:
: Check octet rule and formal charges

- Each fluorine has $$8$$ electrons
- Selenium has expanded octet (more than 8 electrons) - Formal charge calculation confirms stable structure

Final Answer

The Lewis structure for [SeF_{5}]^{-} will have: - Selenium at center - Five fluorine atoms bonded to selenium - Each fluorine with 6 lone pair electrons - 2 lone pairs on selenium - Single bonds between selenium and fluorines