Yes! It's actually a great pi acceptor.
ligands which can accept electrons from the metal d orbital into there anti bonding orbital such as CO, or C=C
The CO ligand can easily back-bond, accepting electron density from the metal centre through pi bonds. This is because of the empty anti-bonding orbitals.
Ionic Bond between K+ and CN- ions. The C and N in CN_ ion are bonded by covalent sigma and pi bonds.
Ionic between the Na+ and NO3- ions, and covalent N-O bonds within the nitrate ion.
NB: I have changed the question from electron donor to electron acceptor as BF3 has no electron donating properties and so it must be assumed the original question was erroneous.So NF3. Nitrogen has 8 electrons and a full shell. Each Fluorine has 8 electrons and a full shell. That's a stable, satisfied molecule. Accepting or donating electrons is thermodynamically unfavourable.BF3, however...is very electron poor. The fluorines actually donate electron density to the boron via weak pi bonding to help compensate the molecule's strong electron poverty. The boron is 2 electrons from completing it's octet (and thus becoming very stable) and so is a strong Lewis acid (lone pair acceptor). By accepting two electrons, BF3 can become electronically equivalent to NF3.
ligands which can accept electrons from the metal d orbital into there anti bonding orbital such as CO, or C=C
A pi-H ion doesn't exist. It is an error or confusion.
There are two pi bonds.
Ionic, between K+ and pi-bonded cyanide, CN-.
The CO ligand can easily back-bond, accepting electron density from the metal centre through pi bonds. This is because of the empty anti-bonding orbitals.
its not 4.
Stephen Robert Ely has written: 'Novel lanthanide compounds with [pi]-ligands' -- subject(s): Rare earth metal compounds 'Transmission of digital information'
When a positive ion (electrophile) attacks on a pi bond or partially negative carbon atom and replace H as a positive ion then it is electrophilic substitution reaction.
Yes, phosphorous (and sulfur) have access to a d orbital. It's a bit weird (as is most chemistry), in the ground state phosphorous does not have any d orbital electrons, however, d orbital hybridization is used to explain why phosphorous can form more than the "octet" number of bonds, such as PCl5. This d orbital is also used when describing phosphorous as a pi-acceptor ligand, and the reason it can be considered a pi-acceptor ligand is because it does have access to that d orbital, which can accept the metal's e- density. Hope that helped.
Ionic Bond between K+ and CN- ions. The C and N in CN_ ion are bonded by covalent sigma and pi bonds.
There are 5 sigma bonds along with 2 pi bonds in oxalate ion and 2 ionic bonds with potassium.
Bonding in π-complexes is strongest when both the filled π-bonding orbital of the π-bonded ligand donates TO the metal and the empty π* orbital on the ligand can accept electron density FROM the metal. A metal with a partially-filled set of d orbitals is able to participate in this synergistic mode of bonding; main group atoms virtually never have filled pπ orbitals available for donating electron density to π-complexed ligand, hence this kind of complex occurs only with transition metals.