QUESTION IMAGE
Question
practice: electron configurations
part 1: write out the electron configurations for the following atoms.
- be:
- f:
- p:
- cu:
- hg:
part 2: write out the electron configurations in noble gas notation for the following atoms.
- b:
- ca:
- cl:
- mn:
- pb:
13: in concept 1 we learned how to draw lewis structures. now go back and draw the lewis structure for each element above except the transition metals.
Step1: Find Be's atomic number (4)
Fill orbitals by Aufbau principle: $1s^2 2s^2$
Step2: Find F's atomic number (9)
Fill orbitals by Aufbau principle: $1s^2 2s^2 2p^5$
Step3: Find P's atomic number (15)
Fill orbitals by Aufbau principle: $1s^2 2s^2 2p^6 3s^2 3p^3$
Step4: Find Cu's atomic number (29)
Use stable full d-orbital exception: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$
Step5: Find Hg's atomic number (80)
Use stable full f/d-orbitals: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^{10} 6s^2$
Step6: Find B's noble gas core (He, Z=2)
Add valence electrons: $[He] 2s^2 2p^1$
Step7: Find Ca's noble gas core (Ar, Z=18)
Add valence electrons: $[Ar] 4s^2$
Step8: Find Cl's noble gas core (Ne, Z=10)
Add valence electrons: $[Ne] 3s^2 3p^5$
Step9: Find Mn's noble gas core (Ar, Z=18)
Add valence electrons: $[Ar] 3d^5 4s^2$
Step10: Find Pb's noble gas core (Xe, Z=54)
Add valence electrons: $[Xe] 4f^{14} 5d^{10} 6s^2 6p^2$
Step11: Identify non-transition elements
List: Be, F, P, B, Ca, Cl, Pb
Step12: Draw Lewis structures for each
Show valence electrons as dots around element symbol.
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Part 1:
- Be: $1s^2 2s^2$
- F: $1s^2 2s^2 2p^5$
- P: $1s^2 2s^2 2p^6 3s^2 3p^3$
- Cu: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$
- Hg: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^{10} 6s^2$
Part 2:
- B: $[He] 2s^2 2p^1$
- Ca: $[Ar] 4s^2$
- Cl: $[Ne] 3s^2 3p^5$
- Mn: $[Ar] 3d^5 4s^2$
- Pb: $[Xe] 4f^{14} 5d^{10} 6s^2 6p^2$
Part 3 (Lewis Structures):
- Be: $\cdot Be \cdot$ (2 valence electrons)
- F:
(7 valence electrons)
- P:
(5 valence electrons)
- B: $\cdot \dot{B} \cdot$ (3 valence electrons)
- Ca: $\cdot Ca \cdot$ (2 valence electrons)
- Cl:
(7 valence electrons)
- Pb:
(4 valence electrons)