Which two elements is the second transition series (y through Cd) have four unpaired electrons in their 3+ ions?
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To find which elements in the second transition series have four unpaired electrons in their 3+ ions (y through Cd), we need to examine the electron configurations of these elements.
The second transition series consists of the following elements:
- Yttrium (Y)
- Zirconium (Zr)
- Niobium (Nb)
- Molybdenum (Mo)
- Technetium (Tc)
- Ruthenium (Ru)
- Rhodium (Rh)
- Palladium (Pd)
- Silver (Ag)
- Cadmium (Cd)
We need to determine the electron configurations of the 3+ ions of these elements. To do this, we need to consider the element's atomic number and the number of electrons it loses to form a 3+ ion.
To find the number of electrons lost by an element, subtract 3 from its atomic number. Then, distribute these lost electrons among the relevant subshells in the electron configuration.
For example, let's consider Molybdenum (Mo). Its atomic number is 42, so to form a 3+ ion, it loses three electrons. Molybdenum's full electron configuration is [Kr] 5s^24d^5. If it loses three electrons, it becomes [Kr] 5s^24d^2. In this configuration, there are two unpaired electrons.
By following this process, you can determine the electron configurations and the number of unpaired electrons for the 3+ ions of each element in the second transition series (y through Cd).