QQuestionChemistry
QuestionChemistry
Write a Lewis structure for methyl bromide, $\mathrm{CH}_{3} \mathrm{Br}$, showing all valence electrons.
- You do not have to consider stereochemistry.
- Explicitly draw all H atoms.
- Include all valence lone pairs in your answer.
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Answer
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Step 1:: Determine the total number of valence electrons in the molecule.
Total valence electrons in $\mathrm{CH}_{3} \mathrm{Br} = 4 \text { (from C) }+3(1 \text { (from H) })+7 \text { (from Br) }=15$ valence electrons
Carbon is in the 4th group, hydrogen is in the 1st group, and bromine is in the 7th group. Thus, carbon has 4 valence electrons, hydrogen has 1 valence electron, and bromine has 7 valence electrons.
Step 2:: Draw the skeletal structure of the molecule.
Place the central atom, carbon, in the center and connect it to the other atoms with single bonds. In this case, there will be one bromine atom and three hydrogen atoms bonded to the carbon atom.
Step 3:: Distribute the remaining valence electrons as lone pairs on the atoms.
- Distribute the remaining electrons around the bromine atom first, as bromine is more electronegative than carbon. Place 3 lone pairs on the bromine atom. - Next, place the remaining electrons around the carbon atom. Distribute the electrons to complete the octet for each atom. In this case, there will be no remaining electrons, and carbon will have 8 electrons (4 from bonds and 4 as lone pairs).
Step 4:: Check the formal charges on the atoms.
- Formal charge on bromine $=7 \text { valence electrons }-1 \text { (from bond) }-6 \text { (lone pairs) }=0$
Since all atoms have a formal charge of 0, the Lewis structure is valid.
Step 5:: Draw the final Lewis structure for $\mathrm{CH}_{3} \mathrm{Br}$.
\chemfig{H-C(=[6]Br)(-[:30]H)(-[:-30]H)(-[:-90]H)} \qquad \text{(15 valence electrons)}
Final Answer
The Lewis structure for methyl bromide, $\mathrm{CH}_{3} \mathrm{Br}$, is shown above, with all valence electrons explicitly displayed.
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