Answer:
D. 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10,5s2,5p6
Explanation:
Hello!
In this case, since the standard iodine atom has 53 electrons, when it forms the iodide ion it is known it gains one spare electron so now it has 54; it means we need to write the new electron configuration up to 54 as shown below:
[tex]I^-:1s^2,2s^2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10},5p^6.[/tex]
Thus, the answer should be:
D. 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10,5s2,5p6
Even when the order is not the adequate one.
Regards!
