Question

What is the electron configuration of the iodide ion?

A. 1s22s22p63s23p63d104s24p64d105s2

B. 1s22s22p63s23p63d104s24p6

C. 1s22s22p63s23p63d104s24p64d10

D. 1s22s22p63s23p63d104s24p64d105s25p6


Please try and put the answer first and then give an explanation.

Answer 1

1s2,2s2,2p6,3s2,3p6,4s2, 3d10,4p6,5s2,4d10,5p6 is techically more correct, but in this case the closest answer given by the problem is D.

Step-by-step explanation:

Note that in thereory, 4s2 is before 3d10 and 5s2 is before 4d10. However, this is not one of the choices, so the answer is D.

Answer 2

D. 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10,5s2,5p6

Step-by-step 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:

`I^-:1s^2,2s^2p^6,3s^2,3p^6,4s^2,3d^(10),4p^6,5s^2,4d^(10),5p^6.`

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!