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This Lewis structure for \(\mathrm{SF}_{5}^{+}\) is drawn incorrectly. What error was made when determining the number of valence electrons?

Short Answer

Expert verified
The total valence electrons should be 40.

Step by step solution

01

Determine the Elemental Valence Electrons

First, identify the number of valence electrons for each element in the species based on the periodic table. Sulfur typically has 6 valence electrons, and each fluorine atom has 7 valence electrons.
02

Sum the Valence Electrons

As \(\mathrm{SF}_5^{+}\) has one sulfur atom and five fluorine atoms, calculate their total valence electrons. Add sulfur's 6 electrons to fluorine's total, which is \5 \times 7 = 35\, resulting in \6 + 35 = 41\ valence electrons.
03

Account for the Positive Charge

For a positively charged ion (\(\mathrm{+1})\), subtract one electron from the sum calculated in Step 2. This gives \41 - 1 = 40\ valence electrons.
04

Identify the Error in the Lewis Structure

The mistake in drawing the Lewis structure was using an incorrect total of valence electrons. The correct number of electrons should be 40 after accounting for the positive charge, not 41 or any other number.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the electrons in the outermost shell of an atom. These electrons are crucial because they determine how atoms interact, bond, and form molecules. For a Lewis structure, knowing the number of valence electrons is essential as it helps create an accurate depiction of the molecule’s electronic structure. Each element in the periodic table has a specific number of valence electrons. For example:
  • Sulfur, found in group 16, typically has 6 valence electrons.
  • Fluorine, found in group 17, usually has 7 valence electrons.
When constructing a Lewis structure, it’s important to accurately sum these valence electrons. Any error in this sum can lead to an incorrect representation of the molecule, as seen in the case of \(\mathrm{SF}_{5}^{+}\), where the mistake was a simple miscalculation of valence electrons due to not considering the charges properly.
Molecular Ions
Molecular ions are molecules that have gained or lost electrons, resulting in an overall charge. In the case of \(\mathrm{SF}_{5}^{+}\), the molecule is a cation, meaning it has lost an electron and carries a positive charge. Calculating the correct number of valence electrons for a molecular ion involves:
  • Adding electrons if the ion is negatively charged (anion).
  • Subtracting electrons if the ion is positively charged (cation).
For \(\mathrm{SF}_{5}^{+}\), after summing the valence electrons of sulfur and fluorine, you must subtract one electron to account for the positive charge. This adjustment is straightforward yet critical, as overlooking it can result in an incorrect Lewis structure.
Chemical Bonding
Chemical bonding is the force that holds atoms together in a molecule. For \(\mathrm{SF}_{5}^{+}\), these bonds involve sharing valence electrons between sulfur and fluorine atoms, forming covalent bonds. In covalent bonding:
  • Atoms share pairs of electrons to reach a stable electronic configuration.
  • Sulfur, with its six valence electrons, needs two more to complete its octet, while each fluorine atom with seven valence electrons needs one more.
  • When sulfur bonds with five fluorine atoms, the molecule must ensure all atoms achieve stable configurations, following the octet rule for fluorine and an expanded octet possibility for sulfur due to its ability to utilize d orbitals.
Drawing a Lewis structure involves pairing these valence electrons correctly, factoring in shared pairs, to visualize the molecule’s configuration accurately.
Sulfur Fluoride Compounds
Sulfur fluoride compounds are a class of chemicals that include various sulfur and fluorine bonded units. These compounds can exhibit different properties and structures depending on their electron count and the specific atoms involved. Some key points include:
  • Sulfur can form multiple bonds with fluorine, creating compounds such as \(\mathrm{SF}_{2}\), \(\mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\), besides \(\mathrm{SF}_{5}^{+}\).
  • In compounds like \(\mathrm{SF}_{5}^{+}\), sulfur uses its ability to expand its valence shell and form more than four bonds, accommodating five fluorines, which is unique compared to typical main-group elements.
  • The chemical behavior of these compounds often involves interactions due to the highly electronegative fluorine atoms, affecting properties like reactivity and molecular geometry.
Understanding how sulfur bonds with fluorine provides insight into the diversity and complexity of its compounds, including factors like bond angles and lengths in the structure.

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Most popular questions from this chapter

Elemental phosphorus has the formula \(\mathrm{P}_{4}\). Propose a Lewis structure for this molecule. [Hints: (1) Each phosphorus atom is bonded to three other phosphorus atoms. (2) Visualize the structure three-dimensionally, not flat on a page.]

Write Lewis structures for these molecules. (a) Formic acid, \(\mathrm{HCOOH}\), in which atomic arrangement is (b) Acetonitrile, \(\mathrm{CH}_{3} \mathrm{CN}\) (c) Vinyl chloride, \(\mathrm{CH}_{2} \mathrm{CHCl}\), the molecule from which PVC plastics are made

Three known isomers exist of \(\mathrm{N}_{2} \mathrm{CO},\) with the atoms in these sequences: \(\mathrm{NOCN} ; \mathrm{ONNC} ;\) and ONCN. Write resonance structures for each isomer and use formal charge to predict which isomer is the most stable.

Write Lewis structures for these molecules or ions. (a) \(\mathrm{CH}_{3} \mathrm{Cl}\) (b) \(\mathrm{SiO}_{4}^{4-}\) (c) \(\mathrm{ClF}_{4}^{+}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{6}\)

Which of these molecules can have cis and trans iso- mers? For those that do, write the structural formulas of the two isomers and label each cis or trans. For those that cannot have these isomers, explain why. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{BrC}=\mathrm{CBrCH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{IC}=\mathrm{CICH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{ClC}=\mathrm{CHCH}_{3}\) (e) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}\)

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