which element has the ground-state electron configuration [Kr]5s^2 4d^10 5p^4
You know what I do? I add up the electrons (I count 52) and look on the periodic table for element # 52.
The element with the ground-state electron configuration [Kr] 5s^2 4d^10 5p^4 is iodine (I), which has an atomic number of 53.
To determine which element has the ground-state electron configuration [Kr]5s^2 4d^10 5p^4, we need to identify the element in the periodic table that matches this electron configuration.
The electron configuration [Kr]5s^2 4d^10 5p^4 corresponds to the valence shell configuration of an element. In this case, the valence shell is the outermost electron shell, which for this electron configuration is the 5s and 5p orbitals.
By examining the electron configuration, we can see that the element has:
- 2 electrons in the 5s orbital,
- 10 electrons in the 4d orbital, and
- 4 electrons in the 5p orbital.
To find the element, we need to count down the periodic table until we reach an element with these electron configurations.
Counting down from [Kr] (krypton), we find that the next element is selenium (Se), which has the electron configuration [Ar]4s^2 3d^10 4p^4. However, we can clearly see that it does not match the given electron configuration.
The element following selenium is bromine (Br), which has the electron configuration [Ar]4s^2 3d^10 4p^5. Again, it doesn't match completely.
Finally, the next element after bromine is iodine (I). Iodine's electron configuration is [Kr]5s^2 4d^10 5p^5, which is very close to the given electron configuration. We just need to remove one electron from iodine, leaving us with [Kr]5s^2 4d^10 5p^4. Hence, the element with the ground-state electron configuration [Kr]5s^2 4d^10 5p^4 is iodine (I).