Problem 3P Write Lewis structures for the following molecular formulas. (a) N 2 (b) HCN (c) HONO (d) CO 2 (e) CH 3 CHNH (f) HCO 2 H (g) C 2 H 3 Cl (h) HNNH (i) C 3 H 6(one double bond) (j) C3H4 (two double bonds) (k) C3H4 (one triple bond)
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Textbook Solutions for Organic Chemistry
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) \({\left[\mathrm{H}_{2} \mathrm{C}=\mathrm{OH}\right]^{+}}\)
Solution
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 =
full solution