Question

1. complete the electron dot structures below to show how beryllium fluoride (bef2) is formed. use the diagram on page 203 as a model. be + f f → be f f

Explanation:

Step1: Determine valence electrons

Beryllium (Be) has 2 valence electrons, and fluorine (F) has 7 valence electrons.

Step2: Show electron - transfer

Be loses 2 electrons to form a $Be^{2 + }$ ion. Each F atom gains 1 electron to form $F^-$ ions.

Step3: Draw electron - dot structure

For the reactants, Be has 2 dots around it and each F has 7 dots around it. For the product, $Be^{2+}$ has no dots (as it has lost its valence electrons), and each $F^-$ has 8 dots around it (full - octet). The final structure shows $Be^{2+}$ ion surrounded by two $F^-$ ions.

Answer:

Draw 2 dots around Be for reactant Be, 7 dots around each F for reactant Fs. For the product, no dots around $Be^{2+}$, 8 dots around each $F^-$ to represent full - octet of fluoride ions.