How is Cfse value calculated?

How is Cfse value calculated?

We can calculate the CFSE as -(5)(2/5)ΔO + (2)(3/5)ΔO = -4/5 ΔO. [Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is -(6)(2/5)ΔO + (1)(3/5)ΔO + P = -9/5 ΔO + P.

Why is oh a weak ligand than H2O?

OH is a better pi-bond donor (but only slightly), making it a lower/weaker field ligand, while H2O is neither a good pi-donor, nor a good pi-acceptor, putting it in the middle.

Is ammonia a strong field ligand?

Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.

What makes a ligand strong?

1 Answer. One definite factor is the ability of the ligand to stabilize its lone pair. You observe that the cyanide ion is a better ligand than the nitrogen ligand. The lone pair that reacts on the cyanide ion is the lone pair on the carbon atom.

What can act as a ligand?

Ligands can be anions, cations, or neutral molecules. Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced, hence the idea of bite angle etc. A monodentate ligand has only one donor atom used to bond to the central metal atom or ion.

Which is high spin complex example?

Example: [Fe(CN)6]3−. Octahedral high-spin: 4 unpaired electrons, paramagnetic, substitutionally labile. Includes Fe2+, Co3+. Example: [CoF6]3−.

Why are there no low spin tetrahedral complexes?

In tetrahedral complex, the d-orbital is splitting to small as compared to octahedral. For same metal and same ligand . Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.

Which of the following is tetrahedral complexes?

For example, tetrakis(triphenylphosphine)palladium(0), a popular catalyst, and nickel carbonyl, an intermediate in nickel purification, are tetrahedral. Many complexes with incompletely filled d-subshells are tetrahedral as well—for example, the tetrahalides of iron(II), cobalt(II), and nickel(II).

Which of the following is more stable complex and why?

Solution. Chelating ligands form more stable complexes compared to non-chelating ligands. Since ethylene diammine is a bidentate ligand and forms stable chelate, [Co(en)3]3+ will be a more stable complex than [Co(NH3)6]3+.