how Sm many upaired electrons are in (Z=62) and write the electron configuration of Sm3+ ion.
The element with atomic number Z=62 is samarium (Sm).
To determine the number of unpaired electrons, we need to find the electron configuration of neutral samarium (Sm):
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6
The valence shell for samarium is the 6s shell, and it contains 2 electrons. All the other shells are filled. So, there are two unpaired electrons in samarium.
To write the electron configuration of the Sm3+ ion, we need to remove 3 electrons from the neutral configuration to match the 3+ charge.
The electron configuration of Sm3+ ion is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f5
To determine the number of unpaired electrons in an element with atomic number Z, we need to examine its electron configuration. The element in question is Samarium (Sm), with atomic number Z = 62.
The electron configuration of Samarium (Sm) is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^6
To determine the number of unpaired electrons, we focus on the last orbital, which is the 4f orbital in the case of Samarium. In the neutral atom (Sm), there are 6 electrons in the 4f orbital. As per Hund's Rule, each orbital is occupied by a single electron before any pairing occurs.
Since all 6 electrons in the 4f orbital of Samarium are paired, there are no unpaired electrons in the neutral atom.
To write the electron configuration of the Sm3+ ion, we need to remove three electrons from the neutral atom. The electron configuration of Sm3+ will be:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10
Note: The 4f electrons are not removed because they occupy a higher energy level and are not involved in ion formation for Samarium.