How many $π$ $pi$ bonds are in ${[Ag {(CN)}_{2}]}_{2}^{-}$ $["Ag"("CN")_2]^(-)$ ?

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How many $π$ $pi$ bonds are in ${[Ag {(CN)}_{2}]}_{2}^{-}$ $["Ag"("CN")_2]^(-)$ ?

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Answer 1 · 2016-04-17T21:08:41

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

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

The structure of ${[Ag {(CN)}_{2}]}_{2}^{-}$ $["Ag"("CN")_2]^(-)$ looks like this:

$: N \equiv (-) C : - (+) Ag - : (-) C \equiv N :$ $:"N"-=stackrel((-))("C"):-stackrel((+))("Ag")- :stackrel((-))("C")-="N":$

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

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How many $π$ $pi$ bonds are in ${[Ag {(CN)}_{2}]}_{2}^{-}$ $["Ag"("CN")_2]^(-)$ ?

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How many piπpi bonds are in ["Ag"("CN")_2]^(-)[Ag(CN)2]−["Ag"("CN")_2]^(-)?

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How many $π$ $pi$ bonds are in ${[Ag {(CN)}_{2}]}_{2}^{-}$ $["Ag"("CN")_2]^(-)$ ?