Solution Found!
Draw Lewis structures for the following compounds and
Chapter 1, Problem 6P(choose chapter or problem)
Draw Lewis structures for the following compounds and ions, showing appropriate formal
charges.
(a) \({\left[\mathrm{CH}_{3} \mathrm{OH}_{2}\right]^{+}}\)
(b) \(\mathrm{NH}_{4} \mathrm{Cl}\)
(c) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{NCl}\)
(d) \(\mathrm{NaOCH}_{3}\)
(e) \({ }^{+} \mathrm{CH}_{3}\)
(f) \({ }^{-} \mathrm{CH}_{3}\)
(g) \(\mathrm{NaBH}_{4}\)
(h) \(\mathrm{NaBH}_{3} \mathrm{CN}\)
(i) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}-\mathrm{BF}_{3}\)
(j) \({\left[\mathrm{HONH}_{3}\right]^{+}}\)
(k) \(\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}\)
(l) \({\left[\mathrm{H}_{2} \mathrm{C}=\mathrm{OH}\right]^{+}}\)
Equation Transcription:
Text Transcription:
[CH_{3}OH_{2}]+
NH_{4}Cl
(CH_{3})_{4}NCl
NaOCH_{3}
^{+}CH_{3}
^{-}CH_{3}
NaBH_{4}
NaBH_{3}CN
(CH_{3})_{2}O-BF_{3}
[HONH_{3}]^+
KOC(CH_{3})_{3}
[H_{2}C=OH]^+
Questions & Answers
QUESTION:
Draw Lewis structures for the following compounds and ions, showing appropriate formal
charges.
(a) \({\left[\mathrm{CH}_{3} \mathrm{OH}_{2}\right]^{+}}\)
(b) \(\mathrm{NH}_{4} \mathrm{Cl}\)
(c) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{NCl}\)
(d) \(\mathrm{NaOCH}_{3}\)
(e) \({ }^{+} \mathrm{CH}_{3}\)
(f) \({ }^{-} \mathrm{CH}_{3}\)
(g) \(\mathrm{NaBH}_{4}\)
(h) \(\mathrm{NaBH}_{3} \mathrm{CN}\)
(i) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}-\mathrm{BF}_{3}\)
(j) \({\left[\mathrm{HONH}_{3}\right]^{+}}\)
(k) \(\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}\)
(l) \({\left[\mathrm{H}_{2} \mathrm{C}=\mathrm{OH}\right]^{+}}\)
Equation Transcription:
Text Transcription:
[CH_{3}OH_{2}]+
NH_{4}Cl
(CH_{3})_{4}NCl
NaOCH_{3}
^{+}CH_{3}
^{-}CH_{3}
NaBH_{4}
NaBH_{3}CN
(CH_{3})_{2}O-BF_{3}
[HONH_{3}]^+
KOC(CH_{3})_{3}
[H_{2}C=OH]^+
ANSWER:
Step 1 of 12
a)
No of valence electrons = 4 from Carbon, 5 from hydrogen, 6 from Oxygen = 15.
Due to positive ion no of electrons become 15-1 = 14.
Formal charge on Carbon = = 0
Formal charge on Hydrogen = = 0
Formal charge on Oxygen = Oxygen has 6 valence electrons. 3 electrons are shared with Carbon, and 2 hydrogen atoms. Out of left 3 electrons one electron is relieved to give a positive charge.
Hence FC on Oxygen =