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Draw Lewis structures for the following compounds and

Chapter 1, Problem 6P

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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]^+

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 =

 

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