Are the following ions diamagnetic or paramagnetic? C r + 3 , C a + 2 , N a + , C r , F e +
Mar 29, 2016
Notice how Cr is not an ion. :)
Anyways, we can start from the electron configuration of the neutral atoms. I'm assuming you meant Cr3+ , Ca2+ , and Fe3+ ...
Cr :[Ar]3d54s1 Ca :[Ar]4s2 Na :[Ne]3s1 Fe :[Ar]3d64s2
The rightmost orbitals listed here are highest in energy, so we ionize these atoms by booting off the highest-energy electrons. Thus:
Cr→Cr3++3e− [Ar]3d54s1→[Ar]3d3
Since there are 5 3d orbitals, they are not all at least singly filled yet, and thus, all three electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.
The original atom is also paramagnetic. The 4s subshell contains 1 electron (in one 4s orbital) and the 3d subshell contains 5 electrons, one in each 3d orbital. No valence electrons are paired here.
That's in agreement with our expectations from Hund's Rule (generally, for the lowest-energy configuration, maximize parallel spins where possible by singly-filling all orbitals of very similar energies first and then doubling up afterwards).
Ca→Ca2++2e− [Ar]4s2→[Ar]⇒1s22s22p63s23p6
This is a noble gas configuration, so no electrons are unpaired Thus, this is diamagnetic.
Na→Na++e− [Ne]3s1→[Ne]⇒1s22s22p6
This is a noble gas configuration, so no electrons are unpaired. Thus, this is diamagnetic.
Fe→Fe3++3e− [Ar]3d64s2→[Ar]3d5
Since there are 5 3d orbitals, in accordance with Hund's Rule, all five electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.
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