Using curved-arrow notation, show the formation of reactive intermediates when convalent bonds in the following substance undergo heterolytic (polar) cleavage.
(i) `CH_(3)-SCH_(3)` , (ii) `CH_(3)-CN`
(iv) `CH_(3)-Cu` , (iv) `Ag-I`
(v) `H_(3)N^(+)-BF_(3)` , (vi) `Cu(-OH_(2))_(4)^(2+)`
Stratergy: Both bonding electrons remain with the more electronegative atom. If the bonding atom have same electronegetivities,, then the bonding electrons are taken away by the larger atom as it can sustain the negative charge more effectively.

Explanation

(i)
Both `C` and `S` have identical electronegativities `(2.5)` on the paulling's scale. Since `S` atom is bigger atom can disperse the negative charge more effectively.
(ii)
Due to the presence of `N` atom which is more electronegative than `C` atom, the `CN` group takes away the bonding `e^(-)` pair.
(iii)
`Cu` (metal) is less electronegative than `C` (a nonmetal) Thus, `CH_(3)` group takes away the shar `e^(-)` pair.
(iv)
`I` ( a nonmetal) is more electronegative than `Ag` ( a metal).
(v)
`N( 3.0)` is more electronegative than `B( 2.0)`. Thus, the shared `e^(-)` pair is taken away by the `N` atom.
(vi)
`Cu` is a metal while `O` is a nonmetal. Thus, bonding electron pairs moves with `O` atom.