Irrerbthist6n

2022-01-08

What monatomic ions would you expect radium ($Z=88$) and selenium ($Z=34$) to form?

sukljama2

We expect radium to form a positively charged ion (cation , $R{a}^{2+}$) since it has two electrons it its outer shell of electronic sheath. By giving away the electrons it becomes more stable as it accomplishes the configuration of its nearby noble gas neighbor.
On the other hand, we expect selenium to form a negatively charged ion (anion, $S{e}^{2-}$) since its missing two electrons to fill its outer shell of electronic sheath.
We expect radium to form $R{a}^{2+}$ and selenium to form $S{e}^{2-}$

Medicim6

1) Radium will form $R{a}^{2+}$ ions, because it is metal with two valence electrons ($7{s}^{2}$) and it will lost two electrons to have electric configuration like noble gas radon ($Z=86$).
${}_{\left\{88\right\}}Ra1{s}^{22}{s}^{22}{p}^{63}{s}^{23}{p}^{63}{d}^{10}4{s}^{24}{p}^{64}{d}^{10}4{f}^{14}5{s}^{25}{p}^{65}{d}^{10}6{s}^{26}{p}^{67}{s}^{2}$.
2) Selenium wil form $S{e}^{2-}$ ions, because it is nonmetal and it has six valence electrons ($4{s}^{24}{p}^{4}$), it will gain two electrons to have electron configuration like noble gas krypton ($Z=36$).
Electronic configuration of selenium atom: ${}_{\left\{34\right\}}Se1{s}^{22}{s}^{22}{p}^{63}{s}^{23}{p}^{63}{d}^{10}4{s}^{24}{p}^{4}$.

fudzisako

Step 1. Radium ($\text{Ra}$):
Radium is an alkaline earth metal located in Group 2 of the periodic table. It has an atomic number of 88, indicating the presence of 88 protons and electrons. The electron configuration of radium is $\text{[Rn]}7{s}^{2}$, where $\text{[Rn]}$ represents the electron configuration of the previous noble gas, radon.
Since radium is a metal, it tends to lose electrons to achieve a stable electron configuration. In doing so, it forms a 2+ cation by losing both of its valence electrons from the $7s$ orbital. Therefore, the monatomic ion formed by radium is ${\text{Ra}}^{2+}$.
Step 2. Selenium ($\text{Se}$):
Selenium is a nonmetal located in Group 16 of the periodic table. It has an atomic number of 34, indicating the presence of 34 protons and electrons. The electron configuration of selenium is $\text{[Ar]}3{d}^{10}4{s}^{2}4{p}^{4}$, where $\text{[Ar]}$ represents the electron configuration of the previous noble gas, argon.
Selenium tends to gain two electrons to achieve a stable electron configuration by filling its $4p$ orbital. By gaining two electrons, selenium forms a 2- anion. Therefore, the monatomic ion formed by selenium is ${\text{Se}}^{2-}$.
In summary, the monatomic ion formed by radium is ${\text{Ra}}^{2+}$, and the monatomic ion formed by selenium is ${\text{Se}}^{2-}$.

xleb123

- Radium (Z=88) forms the ${\mathbf{R}\mathbf{a}}^{2+}$ cation.
- Selenium (Z=34) forms the ${\mathbf{S}\mathbf{e}}^{2-}$ anion.
Explanation:
To determine the monatomic ions formed by radium (Z=88) and selenium (Z=34), we need to consider their electron configurations.
Radium, with an atomic number of 88, has the electron configuration $1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}4{s}^{2}3{d}^{10}4{p}^{6}5{s}^{2}4{d}^{10}5{p}^{6}6{s}^{2}4{f}^{14}5{d}^{10}6{p}^{6}7{s}^{2}$.
To achieve a more stable electron configuration, radium is likely to lose its two outermost electrons from the $7s$ orbital, forming a $2+$ cation. Therefore, the monatomic ion formed by radium is ${\mathbf{R}\mathbf{a}}^{2+}$.
Selenium, with an atomic number of 34, has the electron configuration $1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}4{s}^{2}3{d}^{10}4{p}^{4}$.
Selenium can either gain two electrons to achieve a stable noble gas configuration (similar to argon) or lose six electrons to achieve a stable $3{s}^{2}3{p}^{6}$ configuration. However, it is more likely for selenium to gain two electrons rather than losing six due to the high energy required for the latter process.
Thus, selenium is likely to gain two electrons, forming a $2-$ anion. Therefore, the monatomic ion formed by selenium is ${\mathbf{S}\mathbf{e}}^{2-}$.

Andre BalkonE

To determine the monatomic ions formed by radium ($Z=88$) and selenium ($Z=34$), we need to consider their electron configurations.
Radium belongs to Group 2 (or alkaline earth metals) of the periodic table. The electron configuration of radium is $\left[Rn\right]7{s}^{2}$. By losing its two valence electrons, radium will achieve a stable noble gas configuration and form a $2+$ cation. The equation representing the ionization of radium can be written as:
$\mathrm{R}\mathrm{a}⟶{\mathrm{R}\mathrm{a}}^{2+}+2{e}^{-}$
Selenium, on the other hand, is located in Group 16 (or chalcogens) of the periodic table. The electron configuration of selenium is $\left[Ar\right]3{d}^{10}4{s}^{2}4{p}^{4}$. Selenium requires two additional electrons to achieve a stable noble gas configuration. Therefore, it tends to gain two electrons and form a $2-$ anion. The equation representing the gain of electrons by selenium can be written as:
$\mathrm{S}\mathrm{e}+2{e}^{-}⟶{\mathrm{S}\mathrm{e}}^{2-}$
Thus, the monatomic ion formed by radium is ${\mathrm{R}\mathrm{a}}^{2+}$ (a $2+$ cation), and the monatomic ion formed by selenium is ${\mathrm{S}\mathrm{e}}^{2-}$ (a $2-$ anion).

Do you have a similar question?