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6. Rubidium is in Group I of the Periodic Table and bromine is in Group VII. Rubidium reacts with bromine to form an ionic compound.

Which row shows the electron change taking place for rubidium and the correct formula of the rubidium ion?

\begin{tabular}{|l|l|l|}
\hline
& Electron change & Formula of ion formed \\
\hline
A & Electron gained & [tex]$Rb^{+}$[/tex] \\
\hline
B & Electron gained & [tex]$Rb^{-}$[/tex] \\
\hline
C & Electron lost & [tex]$Rb^{+}$[/tex] \\
\hline
D & Electron lost & [tex]$Rb^{-}$[/tex] \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
To understand the changes that occur when rubidium (Rb) reacts with bromine (Br), let's examine the general behavior of elements in these groups.

### Properties of Rubidium (Group I):
- Rubidium is an alkali metal found in Group I of the Periodic Table.
- Elements in Group I have one electron in their outermost shell.
- In forming ionic compounds, these elements tend to lose their one outer electron, resulting in a positively charged ion (a cation).

### Properties of Bromine (Group VII):
- Bromine is a halogen found in Group VII of the Periodic Table.
- Elements in Group VII have seven electrons in their outermost shell.
- These elements typically gain one electron to complete their outer shell, forming a negatively charged ion (an anion).

### Electron Change for Rubidium:
- Rubidium loses one electron to achieve a stable electronic configuration (similar to the noble gas preceding it).
- By losing one electron, rubidium forms a positively charged ion with a charge of +1, denoted as [tex]\( Rb^+ \)[/tex].

### Matching to the Correct Row:
Given the information about how rubidium behaves:
1. Electron Change: Rubidium loses an electron.
2. Formula of Ion Formed: [tex]\( Rb^+ \)[/tex].

We can now match this information to the provided table:

\begin{tabular}{|l|l|l|}
\hline & Electron change & Formula of ion formed \\
\hline A & Electron gained & [tex]\( Rb^+ \)[/tex] \\
\hline B & Electron gained & [tex]\( Rb^- \)[/tex] \\
\hline C & Electron lost & [tex]\( Rb^+ \)[/tex] \\
\hline D & Electron lost & [tex]\( Rb^- \)[/tex] \\
\hline
\end{tabular}

- Row A is incorrect because it states that rubidium gains an electron and forms [tex]\( Rb^+ \)[/tex], which contradicts the fact that gaining an electron would result in a negatively charged ion.
- Row B is incorrect because it suggests rubidium gains an electron and forms [tex]\( Rb^- \)[/tex], which is not possible for Group I elements.
- Row C is correct because it accurately reflects that rubidium loses an electron and forms [tex]\( Rb^+ \)[/tex].
- Row D is incorrect because although it correctly states that rubidium loses an electron, it incorrectly suggests that the resulting ion is [tex]\( Rb^- \)[/tex].

Thus, the correct row is:
- Row C: Electron lost & Formula of ion formed [tex]\( Rb^+ \)[/tex].
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