With regard to electron configuration, what do all the elements in Group IIA of the periodic table have in common?

Short Answer

Expert verified
Answer: All elements in Group IIA of the periodic table have 2 valence electrons in their outermost s-orbital.

Step by step solution

01

Identify Group IIA elements

Group IIA elements (alkaline earth metals) include Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), and Radium (Ra).
02

Determine electron configurations

Write down the electron configurations for each element: - Beryllium (Be): 1s^2 2s^2 - Magnesium (Mg): 1s^2 2s^2 2p^6 3s^2 - Calcium (Ca): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 - Strontium (Sr): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 - Barium (Ba): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 - Radium (Ra): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 6p^6 7s^2
03

Identify the common feature

Observe the electron configurations and look for similarities. In each electron configuration, the outermost (highest energy level) orbital always has two electrons. These electrons are called valence electrons and are represented by the "s^2" in the electron configuration. All Group IIA elements have 2 valence electrons in their outermost s-orbital. Therefore, the answer to the question is: All elements in Group IIA of the periodic table have 2 valence electrons in their outermost s-orbital.

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Most popular questions from this chapter

Calculate the force of attraction between a \(\mathrm{Ca}^{2+}\) and an \(\mathrm{O}^{2-}\) ion whose centers are separated by a distance of \(1.25 \mathrm{~nm}\).

For the \(K\) shell, the four quantum numbers for each of the two electrons in the \(1 s\) state, in the order of \(n l m_{l} m_{s}\), are \(100 \frac{1}{2}\) and \(100\left(-\frac{1}{2}\right)\). Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p\), and \(d\) subshells.

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The net potential energy between two adjacent ions, \(E_{N}\), may be represented by the sum of Equations \(2.9\) and \(2.11\); that is, $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B\), and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\), and then set the resulting expression equal to zero, because the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0}\). 2\. Solve for \(r\) in terms of \(A, B\), and \(n\), which yields \(r_{0}\), the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substituting \(r_{0}\) into Equation 2.17.

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