Question

7
n
nitrogen
14.007

8
o
oxygen
15.999

10
ne
neon
20.018

18
ar
argon
1.008

Explanation:

Step1: Analyze Nitrogen (N)

Nitrogen has an atomic number of 7, so it has 7 protons and 7 electrons. The electron configuration for N is \(1s^2 2s^2 2p^3\). The first shell (n=1) holds 2 electrons, the second shell (n=2) has \(2 + 3=5\) electrons. So the electron dot structure (Lewis structure) for N has 5 valence electrons (from the 2p and 2s, but valence is outermost shell, so 5).

Step2: Analyze Oxygen (O)

Oxygen has atomic number 8, so 8 electrons. Electron configuration \(1s^2 2s^2 2p^4\). First shell: 2, second shell: \(2 + 4 = 6\) valence electrons. So Lewis structure has 6 valence electrons.

Step3: Analyze Neon (Ne)

Neon has atomic number 10, electron configuration \(1s^2 2s^2 2p^6\). First shell: 2, second shell: \(2+6 = 8\) valence electrons (noble gas, full outer shell). Lewis structure has 8 valence electrons (octet).

Step4: Analyze Argon (Ar)

Argon has atomic number 18, electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^6\). The outermost shell (n=3) has \(2 + 6=8\) valence electrons (noble gas, full outer shell). Lewis structure has 8 valence electrons.

For the electron dot (Lewis) structures:

• Nitrogen: \(\cdot \underset{\cdot \cdot}{N} \cdot \) (5 valence electrons, represented as dots around N)
• Oxygen: \(\cdot \underset{\cdot \cdot}{O} \cdot \cdot\) (6 valence electrons)
• Neon: \(\underset{\cdot \cdot}{\overset{\cdot \cdot}{Ne}}\) (8 valence electrons, full octet)
• Argon: \(\underset{\cdot \cdot}{\overset{\cdot \cdot}{Ar}}\) (8 valence electrons, full octet)

Assuming the task is to write the Lewis structures (electron dot) for each:

Answer:

• Nitrogen (N): \(\boldsymbol{\cdot \underset{\cdot \cdot}{N} \cdot}\)
• Oxygen (O): \(\boldsymbol{\cdot \underset{\cdot \cdot}{O} \cdot \cdot}\)
• Neon (Ne): \(\boldsymbol{\underset{\cdot \cdot}{\overset{\cdot \cdot}{Ne}}}\)
• Argon (Ar): \(\boldsymbol{\underset{\cdot \cdot}{\overset{\cdot \cdot}{Ar}}}\)