Textbook Problem

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Depict the electron configuration for magnesium using an orbital box diagram and noble gas notation. Give a complete set of four quantum numbers for each of the electrons beyond those of the preceding noble gas.

Interpretation Introduction

Interpretation:

The electronic configuration and quantum number of Magnesium (Mg) has to be derived.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons (e^-) of an given molecule or respective atoms in atomic or molecular orbitals. We consider the electronic configuration of neon atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$, the electronic configurations describes each electrons as moving independently orbitals, an average filed created ay all other orbitals.

The aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Quantum numbers: These terms are explained for the distribution of electron density in an atom. They are derived from the mathematical solution of Schrodinger’s equation for the hydrogen atom.  The types of quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m_l) and the electron spin quantum number (m_s). Each atomic orbital in an atom is categorized by a unique set of the quantum numbers.

Explanation of Solution

Electronic configuration method of Magnesium (Mg=12) atom: According to the periodic table explain that (Mg) atomic number (Z=12) it is a second alkaline element of (s-block elements) after the noble gas neon (Ne).

Moreover the (Mg) is the third period position in periodic table, it has two electrons with corresponding orbital shells (n=2). Hence the two electrons are assigned to (3s) and remaining electrons is assigned to (2p) electron shells the all p-electrons are empty shells there is no electrons.

So the element is precedes by the noble gas argon, so the electron configuration is $[\text{Ne}]3s^2$. The electronic configuration notations methods are shown below.

      $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Magnesium}\text{(Mg)}\text{=12}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Mg)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}\\ \\ spdf\text{with noble gas notation}=[\text{Ne}]{\text{3s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Neon}=10]\\ \text{Orbital}\text{box}\text{notation }\text{\&}\end{array}$