Questions: (1) What is the complete ground state electron configuration for the neon atom? (2) What is the complete ground state electron configuration for the vanadium atom?

(1) What is the complete ground state electron configuration for the neon atom?

(2) What is the complete ground state electron configuration for the vanadium atom?

Transcript text: (1) What is the complete ground state electron configuration for the neon atom? $\square$ (2) What is the complete ground state electron configuration for the vanadium atom? $\square$

Solution

Solution Steps

Step 1: Understanding Electron Configuration

Electron configuration describes the distribution of electrons in an atom's orbitals. The notation follows the format $ n\ell^{x} $, where $ n $ is the principal quantum number, $ \ell $ is the type of orbital (s, p, d, f), and $ x $ is the number of electrons in those orbitals.

Step 2: Electron Configuration for Neon (Ne)

Neon has an atomic number of 10, meaning it has 10 electrons. The electron configuration fills the orbitals in the order of increasing energy levels:

1s: 2 electrons
2s: 2 electrons
2p: 6 electrons

Thus, the complete ground state electron configuration for neon is: \[ 1s^2 2s^2 2p^6 \]

Step 3: Electron Configuration for Vanadium (V)

Vanadium has an atomic number of 23, meaning it has 23 electrons. The electron configuration fills the orbitals in the order of increasing energy levels:

1s: 2 electrons
2s: 2 electrons
2p: 6 electrons
3s: 2 electrons
3p: 6 electrons
4s: 2 electrons
3d: 5 electrons

Thus, the complete ground state electron configuration for vanadium is: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3 \]

Final Answer

The complete ground state electron configuration for the neon atom is: \[ \boxed{1s^2 2s^2 2p^6} \]
The complete ground state electron configuration for the vanadium atom is: \[ \boxed{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3} \]

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