Question

electron configuration practice
chemistry

1. write a ground - state electron configuration for these ions. remember that ions have a change in the total number of electrons (positive have lost electrons and negative have gained). example: n³⁻ is 1s² 2s² 2p⁶. it has three extra electrons
2. o²⁻:
3. fe²⁺:
4. b³⁺:
5. n³⁺:
6. k⁺:
7. co³⁺:

Explanation:

Step1: Determine neutral - atom electrons

For oxygen ($O$), the atomic number is 8. In $O^{2 - }$, it has gained 2 electrons, so total electrons = 8 + 2=10.
The electron - configuration is $1s^{2}2s^{2}2p^{6}$.

Step2: Determine electrons for $Fe^{2+}$

Iron ($Fe$) has an atomic number of 26. In $Fe^{2+}$, it has lost 2 electrons, so total electrons = 26 - 2 = 24.
The electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}$.

Step3: Determine electrons for $B^{3+}$

Boron ($B$) has an atomic number of 5. In $B^{3+}$, it has lost 3 electrons, so total electrons = 5 - 3 = 2.
The electron - configuration is $1s^{2}$.

Step4: Determine electrons for $N^{2 - }$

Nitrogen ($N$) has an atomic number of 7. In $N^{2 - }$, it has gained 2 electrons, so total electrons = 7+2 = 9.
The electron - configuration is $1s^{2}2s^{2}2p^{5}$.

Step5: Determine electrons for $K^{+}$

Potassium ($K$) has an atomic number of 19. In $K^{+}$, it has lost 1 electron, so total electrons = 19 - 1 = 18.
The electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$.

Step6: Determine electrons for $Ca^{2+}$

Calcium ($Ca$) has an atomic number of 20. In $Ca^{2+}$, it has lost 2 electrons, so total electrons = 20 - 2 = 18.
The electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$.

Answer:

1. $1s^{2}2s^{2}2p^{6}$
2. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}$
3. $1s^{2}$
4. $1s^{2}2s^{2}2p^{5}$
5. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$
6. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$