Question

write the symbol of each neutral element based on its noble gas shorthand configuration: he2s^1 kr5s^24d^2 xe6s^24f^145d^10

Explanation:

Step1: Determine electrons from noble - gas core

The noble - gas core represents the electron configuration of the previous noble gas. For $[\text{He}]2s^{1}$, He has 2 electrons. The $2s^{1}$ adds 1 more electron, so the total number of electrons is $2 + 1=3$. The element with 3 electrons is Li.

Step2: For $[\text{Kr}]5s^{2}4d^{2}$

Kr has 36 electrons. The $5s^{2}$ adds 2 electrons and the $4d^{2}$ adds 2 more electrons. So the total number of electrons is $36+2 + 2=40$. The element with 40 electrons is Zr.

Step3: For $[\text{Xe}]6s^{2}4f^{14}5d^{10}$

Xe has 54 electrons. The $6s^{2}$ adds 2 electrons, the $4f^{14}$ adds 14 electrons and the $5d^{10}$ adds 10 electrons. So the total number of electrons is $54+2+14 + 10=80$. The element with 80 electrons is Hg.

Answer:

1. Li
2. Zr
3. Hg