Question

How many `pi` bonds are in `["Ag"("CN")_2]^(-)`?

Answer 1

A `pi` bond is always made by sidelong interaction between two orbitals. If it is between two `p` orbitals, then two out of two lobes overlap. If it is between two `d` orbitals, then two out of four lobes interact.

You can imagine it as the second and third bonds in a triple bond. `"CN"^(-)` has a `-1` charge, as is implied by the `+1` charge of silver and the overall `-1` charge of dicyanosilver(I).

The structure of `["Ag"("CN")_2]^(-)` looks like this:

`:"N"-=stackrel((-))("C"):-stackrel((+))("Ag")- :stackrel((-))("C")-="N":`

So this complex will have four `\mathbf(pi)` bonds.