in the space below, write the abbreviated electron configurations of the following elements:
6) cobalt
7) silver
8) tellurium
9) radium
10) lawrencium

Question

Accepted Answer

# Explanation:
## Step1: Determine the atomic number
- Cobalt ($Co$) has an atomic number of $27$. The noble gas before it is argon ($Ar$, atomic number $18$).
- Silver ($Ag$) has an atomic number of $47$. The noble gas before it is krypton ($Kr$, atomic number $36$).
- Tellurium ($Te$) has an atomic number of $52$. The noble gas before it is krypton ($Kr$, atomic number $36$).
- Radium ($Ra$) has an atomic number of $88$. The noble gas before it is radon ($Rn$, atomic number $86$).
- Lawrencium ($Lr$) has an atomic number of $103$. The noble gas before it is radon ($Rn$, atomic number $86$).

## Step2: Write the abbreviated electron configuration
- For cobalt: After $Ar$ ($[Ar]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$), we have $27 - 18=9$ electrons left. The configuration is $[Ar]3d^{7}4s^{2}$.
- For silver: After $Kr$ ($[Kr]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}$), we have $47 - 36 = 11$ electrons. Due to the stability of the $d^{10}$ configuration, it is $[Kr]4d^{10}5s^{1}$.
- For tellurium: After $Kr$, we have $52-36 = 16$ electrons. The configuration is $[Kr]4d^{10}5s^{2}5p^{4}$.
- For radium: After $Rn$ ($[Rn]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{2}6p^{6}$), we have $88 - 86=2$ electrons. The configuration is $[Rn]7s^{2}$.
- For lawrencium: After $Rn$, we have $103 - 86=17$ electrons. The configuration is $[Rn]5f^{14}6d^{1}7s^{2}$.

# Answer:
6) $[Ar]3d^{7}4s^{2}$
7) $[Kr]4d^{10}5s^{1}$
8) $[Kr]4d^{10}5s^{2}5p^{4}$
9) $[Rn]7s^{2}$
10) $[Rn]5f^{14}6d^{1}7s^{2}$

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Question

Explanation:

Step1: Determine the atomic number

Step2: Write the abbreviated electron configuration

Answer: