# Question #2c7a3

Mar 24, 2016

Caesium fluoride $\text{CsF}$.

#### Explanation:

The electronic structure of caesium is $\left[X e\right] 6 {s}^{1}$.

The halogen in period 2 is fluorine.

The reaction is:

$2 C {s}_{\left(s\right)} + {F}_{2 \left(g\right)} \rightarrow 2 C s {F}_{\left(s\right)}$

Caesium fluoride is the binary compound formed.

Mar 24, 2016

Cesium fluoride, $\text{CsF}$.

#### Explanation:

This question can be answered by carefully inspecting the periodic table, so make sure that you have one near by.

So, the first important thing to notice here is that this reaction will produce an ionic compound. This is the case because the element located in group 1 is said to donate an electron to an element located in group 17, i.e. to a halogen.

Notice that the element located in group 1 donates a $\textcolor{red}{6} {s}^{1}$ electron. This should tell you that you're dealing with the group 1 element located in period $\textcolor{red}{6}$.

A quick look in the periodic table will reveal that this element is cesium, $\text{Cs}$. The halogen located in period 2 is fluorine, $\text{F}$, which means that the ionic compound will contain cesium and fluorine.

Since cesium donates one electron, it forms $1 +$ cations, ${\text{Cs}}^{+}$. Fluorine accepts one electron, so it form $1 -$ anions, ${\text{F}}^{-}$.

This means that the chemical formula for this ionic compound is $\text{CsF}$.

To name the compound, start with the name of the cation, cesium. The anion will take the suffix -ide, which means that the name of the compound will be cesium fluoride.