Write Lewis structure for the following
(a) Ethene `(C_(2)H_(4))` the most important reactant in polymer manufacture
(b) Nitrogen `(N_(2))` the most abundant atmosheric gas
(c ) Methanol `(CH_(4)O)` an important industrial alcohol that is beign used as a gasoline alternative in car engines .

Explanation

For `C_(2)H_(4)` after steps (a) to (d) we have

Change a lone pair to a bonding pair The `C` on the right has an octet but the `C` on the left has only `6e^(-)` so we convert the lone pair to another bonding pair between the two `C` atoms

(b) For `N_(2)` after steps (a) to (d) we have `:overset(..)N-overset(..)N:` , Neither `N` has an octet, so we change a lone pair to bonding pair `:overset(..)N=N:`. In this case moving one lone pair to make a double bond strill does not give each `N` an octet, so we move another lone pair to make a triple bond `:N-=N:`.
(c) Plance the atoms relative to each other The `H` atoms can have only one bond so `C` and `H` must be adjacent to each other In nearly all their compounds `C` has four bonds and `H` has two, so we arrange the `H` atoms
(ii) Find the sum of valence electrons
`[1 xx C (4e^(-))]+[1 xx O (6e^(-))] +[4 xx H(1e^(-))] = 14e^(-)`
(iii) Ad single bonds subtract `2e^(-)` for each bond
`H-underset(H)underset(|)overset(H)overset(|)C-O-H`
Five bonds use `10e^(-),so14e^(-)-10e^(-)` leaves `4e^(-)` reamining
(iv) Add the remaining electrons in pairs
`H-underset(H)underset(|)overset(H)overset(|)C-underset(..)overset(..)O-H`

Carbon already has an octet, so the four remaining valence `e^(-)` from two lone pairs on `O` give the Lewis structure for methanol .