Question

determine the electron configuration of:

1. o²⁻ e⁻= 10 p⁺= 8 n⁰= 10-8=

electron configuration:
draw the bohr’s atomic model of the following atoms.

1. n

e⁻= __ p⁺= n⁰= __
electron configuration:
no. of shells = __ valence e⁻ = __
image of bohr model template for n

1. n³⁻

e⁻= __ p⁺= n⁰= __
electron configuration:
no. of shells = __ valence e⁻ = __
image of bohr model template for n³⁻

1. mg²⁺

Explanation:

1) For $\boldsymbol{O^{2-}}$:

Step1: Find number of electrons ($e^-$)

Oxygen has atomic number 8. $O^{2-}$ has gained 2 electrons, so $e^- = 8 + 2 = 10$.

Step2: Find number of protons ($p^+$)

Protons equal atomic number, so $p^+ = 8$.

Step3: Find number of neutrons ($n^0$)

Atomic mass of O is ~16. $n^0 = 16 - 8 = 8$.

Step4: Electron Configuration

Using Aufbau principle: $1s^2 2s^2 2p^6$.

Step1: Find $e^-$, $p^+$

Atomic number of N is 7, so $e^- = 7$, $p^+ = 7$.

Step2: Find $n^0$

Atomic mass of N is ~14. $n^0 = 14 - 7 = 7$.

Step3: Electron Configuration

$1s^2 2s^2 2p^3$.

Step4: No. of Shells and Valence $e^-$

Shells: 2 (n=1, n=2). Valence electrons: 5 (in 2p).

Step1: Find $e^-$

N has 7 electrons, $N^{3-}$ gains 3, so $e^- = 7 + 3 = 10$.

Step2: Find $p^+$

Protons = atomic number = 7.

Step3: Find $n^0$

Atomic mass ~14, $n^0 = 14 - 7 = 7$.

Step4: Electron Configuration

$1s^2 2s^2 2p^6$.

Step5: No. of Shells and Valence $e^-$

Shells: 2 (n=1, n=2). Valence electrons: 8 (full outer shell).

Answer:

$e^- = 10$, $p^+ = 8$, $n^0 = 8$, Electron Configuration: $1s^2 2s^2 2p^6$

1. For $\boldsymbol{N}$ (Nitrogen):