Question

draw the lewis structure for the becl4^2 - ion. predict its geometry and describe the hybridization state of the be atom. part 1 of 3 draw the lewis structure for the becl4^2 - ion. be sure to add in all non - zero formal charges.

Explanation:

Step1: Determine valence - electrons

Beryllium (Be) has 2 valence electrons, each chlorine (Cl) has 7 valence electrons, and there are 2 extra electrons due to the 2 - charge. So the total number of valence electrons is \(2 + 4\times7+ 2=32\).

Step2: Place the central atom

Be is the central atom. Connect the four Cl atoms to Be with single bonds. Each single bond uses 2 electrons, so \(4\times2 = 8\) electrons are used for the bonds.

Step3: Distribute remaining electrons

We have \(32 - 8=24\) electrons left. Place these electrons as lone - pairs on the Cl atoms. Each Cl atom gets 3 lone - pairs (\(4\times6 = 24\)).

Step4: Calculate formal charges

The formal charge on Be is \(FC_{Be}=2 - 4= - 2\). The formal charge on each Cl is \(FC_{Cl}=7-(6 + 1)=0\).
The Lewis structure has Be in the center with four single bonds to Cl atoms, and each Cl atom has 3 lone - pairs. The Be atom has a formal charge of - 2.

For the geometry and hybridization:

Step5: Determine the electron - pair geometry

The central Be atom has 4 bonding pairs and 0 lone pairs. According to VSEPR theory, the electron - pair geometry is tetrahedral.

Step6: Determine the molecular geometry

Since there are no lone pairs on the central atom, the molecular geometry is also tetrahedral.

Step7: Determine hybridization

The central Be atom forms 4 sigma bonds. To form 4 sigma bonds, Be must be \(sp^{3}\) hybridized.

Answer:

Lewis structure: Be in the center with four single bonds to Cl atoms, each Cl has 3 lone - pairs and Be has a formal charge of - 2. Geometry: Tetrahedral. Hybridization of Be: \(sp^{3}\)