Which element's 4+ ion would have the following electron configuration?
[Kr]4d^10
If I were an element, I would be the element of surprise! Just kidding. But let's find the element with that electron configuration: [Kr]4d^10.
Based on that electron configuration, we can deduce that the element belongs to the 4th period (row) of the periodic table and has 10 electrons in the 4d subshell.
The element that fits these criteria is the element with atomic number 48: Cadmium (Cd).
So, Cadmium (Cd) would have an ion with a 4+ charge and the electron configuration [Kr]4d^10.
That's right, folks! Cadmium stole some electrons and now it's rolling with a 4+ charge. Watch out for this electrifying element!
The electron configuration [Kr]4d^10 corresponds to the element Palladium (Pd) in its neutral state.
To determine the element's 4+ ion, we need to remove 4 electrons from the neutral atom.
Removing the 4s^2 electrons and one 4d^10 electron, we get the electron configuration for the Pd 4+ ion: [Kr]4d^8.
Therefore, the element's 4+ ion with the given electron configuration is Palladium (Pd) 4+.
To find the element's 4+ ion with the given electron configuration, we need to determine the element that has the electron configuration [Kr]4d^10.
The electron configuration [Kr]4d^10 belongs to the element with atomic number 38: Strontium (Sr). However, we need to find the 4+ ion of Strontium.
To do that, we can determine the number of electrons that have been removed to form the 4+ ion. Since the 4d subshell is full (contains 10 electrons), we need to remove all 10 electrons from the 4d subshell, as well as 2 electrons from the outermost 5s subshell. Therefore, for the 4+ ion, we would remove a total of 12 electrons from Strontium (Sr).
Hence, the element whose 4+ ion has the electron configuration [Kr]4d^10 is Strontium (Sr).
So the neutral atom will be
[Kr]4d^10 5s^2 5p^2 and that adds to 36 + 10 + 2 + 2 = ? as the atomic number and look that up on the periodic chart.