unpaired electrons of NiSO4.6H2O
What about them?
To determine the number of unpaired electrons in NiSO4.6H2O, we need to first find the electronic configuration of the nickel ion (Ni2+) in the compound.
The atomic number of nickel (Ni) is 28, which means it has 28 electrons. However, in NiSO4.6H2O, the nickel is in the +2 oxidation state (hence the Ni2+ ion), which means it has lost 2 electrons.
To find the electronic configuration of Ni2+, we need to remove 2 electrons from the neutral state electronic configuration of nickel (Ni):
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Removing two 4s electrons gives us:
1s2 2s2 2p6 3s2 3p6 3d8
Now, we need to determine the number of unpaired electrons. In the 3d orbital, there are five possible orbitals (3dx2-y2, 3dz2, 3dxz, 3dyz, and 3dxy), each capable of holding two electrons. Since there are 8 electrons in the 3d orbital, this means the electron configuration is d8, which implies that there are two unpaired electrons.
Therefore, NiSO4.6H2O has two unpaired electrons in the nickel ion (Ni2+).