Why is Ni (CO) 4 tetrahedral and diamagnetic?
<p>Now the electronic <a href="https://chemisfast.blogspot.com/2020/09/balancing-chemical-equations.html" rel="noopener ugc nofollow" target="_blank">configuration</a> of nickel atom in ground state is 3d8 4s2. But in excited state, the 4s electron of central meal <strong>Ni-atom</strong> is shitted to 3d-sub shell.</p>
<p>Under this condition, the 3d sub shell occupied 10 electrons. That is, 3d-subshell has <strong>no unpaired</strong> electrons.</p>
<p>Now one vacant 4s orbital and three 4p orbital of outer shell are mixed to each other and form energetically <a href="https://chemisfast.blogspot.com/2020/09/balancing-chemical-equations.html" rel="noopener ugc nofollow" target="_blank">equivalent four hybridized</a> orbitals, involving <strong>sp3 </strong>hybridization.</p>
<p>These 4 vacant hybridized orbitals accept <a href="https://chemisfast.blogspot.com/2020/09/oxidation-and-reduction.html" rel="noopener ugc nofollow" target="_blank">four-electron pair</a> from four CO ligand forms L­­–M co-ordinate bond, resulting in the formation of tetrahedral <strong>Ni (CO) 4</strong> molecule.</p>
<p><a href="https://kgghosh1990.medium.com/why-is-ni-co-4-tetrahedral-and-diamagnetic-32fb112f75c4"><strong>Read More</strong></a></p>