how many unpaired electrons are in this complex? Ni(en)2(CN)2 where en is the neutral ethylenediamine and CN is cyanide as the counter ion, I want to say there are no unpaired e- but I'd like a check
To determine the number of unpaired electrons in the complex [Ni(en)2(CN)2], we need to consider the electronic configuration of nickel (Ni).
The atomic number of nickel is 28, so its electronic configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d8.
Since we are focusing on the complex [Ni(en)2(CN)2], the nickel atom in this complex is in a +2 oxidation state.
To assign the electron configuration for the nickel ion (Ni2+), we remove two electrons from the outermost shell:
1s2 2s2 2p6 3s2 3p6 3d8 → 1s2 2s2 2p6 3s2 3p6 3d8 → 1s2 2s2 2p6 3s2 3p6 3d8.
Now, let's determine the number of unpaired electrons. We need to fill the t2g and eg orbitals based on Hund's rule.
In the d-orbital, nickel has five unpaired 3d electrons before bonding. In the complex, each ethylenediamine (en) ligand donates two electrons to the metal, and each cyanide (CN) ligand donates one electron.
So, the coordination number (CN) of the complex is 6, and it consists of two en ligands (each contributes 2 electrons) and two CN ligands (each contributes 1 electron). Therefore, the total number of electrons donated by the ligands is: (2 × 2) + (2 × 1) = 6.
Now, let's distribute the 6 ligand electrons to the available d-orbitals. The t2g orbitals can hold a maximum of 6 electrons, and the eg orbitals can hold a maximum of 4 electrons.
Since we have only 6 ligand electrons, they all fill the t2g orbitals, leaving the eg orbitals empty. Therefore, there are no unpaired electrons in the complex [Ni(en)2(CN)2].
In conclusion, you are correct in saying that there are no unpaired electrons in the complex.