Question

question 14 (2 points) write a shorthand electron configuration for the lodine ion (i-).

Explanation:

Step1: Identify atomic number of iodine

Iodine (I) has atomic number 53. The iodide ion ($I^-$) has gained 1 electron, so it has 54 electrons.

Step2: Use noble - gas shorthand

The nearest noble gas before iodine is krypton (Kr) which has 36 electrons.

Step3: Write the remaining electron configuration

We need to account for the remaining $54 - 36=18$ electrons. The electron - configuration after Kr is $[Kr]5s^{2}4d^{10}5p^{6}$.

Answer:

$[Kr]5s^{2}4d^{10}5p^{6}$