Question

name: ____________ teacher: __________ q1 chemistry: u2l16 excited electrons date: ____________ for each of the following atoms, write the abbreviated electron configuration (noble gas notation) 7. chlorine atom 8. chlorine ion with a charge of - 1 9. oxygen atom 10. oxygen ion with a charge of - 2 stretch it! write the full electron configuration for each below. 11. cobalt atom 12. cobalt ion with a charge of +2 for each of the following atoms, write the orbital notation. 13. oxygen atom 14. oxygen ion with a charge of - 2 15. carbon atom 16. carbon ion with a charge of +4 17. carbon ion with a charge of +2 18. carbon ion with a charge of - 4

Explanation:

Step1: Recall noble - gas notation rules

Noble - gas notation uses the electron configuration of the nearest noble gas with fewer electrons and then adds the remaining electrons.

Step2: Chlorine atom (atomic number = 17)

The nearest noble gas is neon (atomic number = 10). So, the abbreviated electron configuration is $[Ne]3s^{2}3p^{5}$.

Step3: Chlorine ion ($Cl^{-}$, 18 electrons)

It has one more electron than the chlorine atom. The abbreviated electron configuration is $[Ar]$, since argon has 18 electrons.

Step4: Oxygen atom (atomic number = 8)

The nearest noble gas is helium (atomic number = 2). The abbreviated electron configuration is $[He]2s^{2}2p^{4}$.

Step5: Oxygen ion ($O^{2 -}$, 10 electrons)

The abbreviated electron configuration is $[Ne]$, as neon has 10 electrons.

Step6: Cobalt atom (atomic number = 27)

The full electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}$.

Step7: Cobalt ion ($Co^{2+}$, 25 electrons)

Two electrons are removed from the 4s orbital first. The full electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$.

Step8: Oxygen atom orbital notation

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow$ $\uparrow$

Step9: Oxygen ion ($O^{2 -}$) orbital notation

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$

Step10: Carbon atom (atomic number = 6) orbital notation

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow$ $\uparrow$

Step11: Carbon ion ($C^{4+}$, 2 electrons) orbital notation

1s: $\uparrow\downarrow$

Step12: Carbon ion ($C^{2+}$, 4 electrons) orbital notation

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$

Step13: Carbon ion ($C^{4 -}$, 10 electrons) orbital notation

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$

Answer:

1. $[Ne]3s^{2}3p^{5}$
2. $[Ar]$
3. $[He]2s^{2}2p^{4}$
4. $[Ne]$
5. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}$
6. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$
7. 1s: $\uparrow\downarrow$; 2s: $\uparrow\downarrow$; 2p: $\uparrow\downarrow$ $\uparrow$ $\uparrow$
8. 1s: $\uparrow\downarrow$; 2s: $\uparrow\downarrow$; 2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
9. 1s: $\uparrow\downarrow$; 2s: $\uparrow\downarrow$; 2p: $\uparrow$ $\uparrow$
10. 1s: $\uparrow\downarrow$
11. 1s: $\uparrow\downarrow$; 2s: $\uparrow\downarrow$
12. 1s: $\uparrow\downarrow$; 2s: $\uparrow\downarrow$; 2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$