Question

Which of the following electron configurations most likely represents an atom that will form an anion?

Which of the following electron configurations most likely represents an atom that will form an anion?

1-[Kr] 5s2 4d9
2-[Kr] 5s2
3-[Kr] 5s1
4-[Kr] 5s2 4d10 5p5

Answer 1

4

Explanation:

An anion will have a negative charge. Only 4 has a structure that will readily form a negative ion.

1 is silver, atomic number 47. Silver and does move one of the 5s electrons to the 4d sub orbital. This gives an electron configuration of `4d^10 5s^1` The single 5s electron being on the outside valance shell is easily lost resulting in a +1 charge. not an anion.

2 is Strontium 38, Strontium will easily lose the two 5s electrons, this will result in a +2 charge not an anion.

3 is Rb Rubidium 37 Rubidium will easily lose the single 5s electron, this will result in a +1 charge not an anion.

4 is Iodine 53. Iodine often gains one electron completing the fifth energy level resulting in the electron configuration of Xeon 54 a stable inert gas. This will result in a -1 ion which is an anion.