Question

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) Water, \(\mathrm{H}_{2} \mathrm{O}\), is a bent molecule. Predict the shape of the molecular ion formed from the water molecule if you were able to remove four electrons to make \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\).

Short answer

The molecular geometry of BrF4- is square planar because it has 4 bond pairs and 2 lone pairs around the central Br atom, which arrange themselves in an octahedral electronic geometry. The molecular geometry of BF4- is tetrahedral due to the presence of 4 bond pairs and no lone pairs around the central B atom. For the (H2O)4+ ion, after losing 4 electrons, it has 2 bond pairs and no lone pairs, so its molecular geometry would be linear.

Step-by-step solution

Determine the electronic geometry for BrF4- and BF4- ions

First, we need to determine the electronic geometry of BrF4- and BF4- ions. Electronic geometry is defined by the arrangement of electron pairs (lone pairs and bonded pairs) around the central atom. For BrF4-, the central atom is Br with 7 valence electrons. Since it gains one electron, it has 8 valence electrons in total. The ion forms 4 bonds with F atoms which results in 4 single bonds. As there are 8-4=4 electrons left, there are 2 lone pairs on the Br atom. For BF4-, the central atom is B with 3 valence electrons. It forms 4 single bonds with F atoms resulting in a total of 4 bond pairs. There are no lone pairs.

Determine the molecular geometry of BrF4- and BF4- ions

Now, we will use the VSEPR theory to determine the molecular geometry of the ions. For BrF4-, there are 4 bond pairs and 2 lone pairs around the central Br atom. These six electron pairs arrange themselves in the most stable geometry: an octahedral arrangement. However, we only account for the positions of bond pairs to define the molecular geometry. Since the bond pairs form a square in the middle of the octahedron, the molecular geometry of BrF4- is square planar. For BF4-, there are 4 bond pairs and no lone pairs around the central B atom. These four electron pairs arrange themselves in a tetrahedral geometry, so the molecular geometry of BF4- is also tetrahedral.

Determine the electronic geometry for H2O

Now, we will determine the electronic geometry for H2O molecule. The central atom in H2O is O, which has 6 valence electrons. Oxygen forms 2 single bonds with hydrogen atoms, leaving 4 electrons as 2 lone pairs. There are 2 bond pairs and 2 lone pairs around the oxygen atom.

Determine the molecular geometry for (H2O)4+

We are asked to predict the shape of the H2O molecule when it loses 4 electrons. As H2O originally has 2 lone pairs and 2 bond pairs, losing 4 electrons would remove both lone pairs, leaving only 2 bond pairs. With 2 bond pairs and no lone pairs, the electron pairs arrange themselves in a linear geometry. Therefore, the shape of (H2O)4+ would be linear.