This compound is characterized by ionic bonding between a group 1 metal cation and a tetrahedral anion. Write an appropriate Lewis structure for the anion and a specify a formal charge if there is one.
NaBF4
I looked up the answer and I have a few questions on why it is the answer. Why is the Na+ left on in the tetrahedral? I know a tetrahedral consists of 5 spots but I have 6 atoms I need to place. Also, how do I know that B is surrounded by the F rather than the Na?
1. The problem specifies that it is an ionic bond; therefore, the Na exists as a Na+ and the BF4^- as the anion. Is the Na^+ left on the tetrahedron in a drawing which you are looking at. I can't see the structure; perhaps the drawing you are looking at just as the Na^+ very close.The tetrahedral structure, then, consists of the B in the middle of the tetrahedron with a B at each of the four corners. (Does a tetrahedron have five "spots"? It has four sides; i.e., three sides that are triangles plus the bottom of the triangle (another triangle) so it has four corners or four "spots." I also don't see 6 atoms to place. You have A separate Na^+ and a separate BF4^- ion. The BF4^- then, as I said above, has the B in the middle of the tetrahedron with a F at each of the four corners.
2. Why is B surrounded by F than than the Na. Na is not a part of the structure. Note the problem states that NaBF4 is a compound characterised by an ionic bond between a metal I cation and a tetrahedral anion. Na is the metal I cation. BF4^- is the tetrahedral anion.
If this isn't clear to you, please try rephrasing your question and I'll take another crack at it.
To determine the appropriate Lewis structure for the anion in NaBF4, let's break it down step by step:
1. Start by identifying the central atom. In this case, Beryllium (B) is the central atom because it is surrounded by four Fluorine (F) atoms.
2. Next, count the total number of valence electrons. Sodium (Na) contributes 1 valence electron, Beryllium (B) contributes 3, and each Fluorine (F) contributes 7 (since there are four of them). Therefore, the total number of valence electrons is:
1 electron (Na) + 3 electrons (B) + 4 electrons (F) = 8 electrons
3. Distribute the remaining electrons around the central atom (B) and between the atoms (Na and F) according to octet rule. Since Fluorine readily forms single bonds, we can connect each F atom to B with a single bond. This will use up 4 electrons (1 electron from each of the 4 F atoms).
4. Now, place remaining electrons on the outer atoms (F). Each F atom should have 8 electrons, but in this case, each F atom has 6 electrons. Therefore, we need to add two more electrons to each F atom.
5. Since we added 4 electrons to the F atoms, 8 - 4 = 4 remaining electrons. Place those remaining electrons around the central atom (B) as lone pairs.
The final structure:
F
/
F -- B -- F
\
F
In this Lewis structure, the Beryllium (B) atom is in the center and is connected to four Fluorine (F) atoms in a tetrahedral arrangement. It is important to note that the position of the Sodium (Na) atom is not shown in the Lewis structure of the anion. This is because it doesn't directly participate in forming the anion and is instead associated with the overall compound.
Regarding the formal charges, the Lewis structure does not show any formal charges. However, we can calculate the formal charge for each atom in the structure by using the formula:
Formal Charge = Valence electrons - Non-bonding electrons - 1/2*(Bonding electrons)
Keep in mind that the formal charge represents an atom's charge in a molecule or ion relative to its original, isolated state.