Question

1. draw valid lewis structures, and determine the molecular geometries of the following molecules and ions using the vsepr theory. for cations subtract one electron from the total of valence electrons, for anions add one electron to the total of valence electrons. you will need an extra sheet of paper to complete this assignment (please staple it to this one).

sih₄; pcl₃; co₂; ccl₄; h₂o; nf₃; o₃; of₂; nh₄⁺; nh₂⁻; alh₄⁻; h₃o⁺; bf₄⁻; brf₂⁺; cl₄.
very important: in bf₄⁻ b obeys the octet rule;
in brf₂⁺ br is the central atom and it is allowed to have more than 1 bond.

Answer:

To solve the problem of drawing Lewis structures and determining molecular geometries, we'll take $\ce{SiH4}$ as an example (the process is similar for other molecules/ions):

Step 1: Calculate total valence electrons

• $\ce{Si}$ is in group 14, so it has 4 valence electrons.
• Each $\ce{H}$ is in group 1, so each has 1 valence electron. There are 4 $\ce{H}$ atoms.
• Total valence electrons: $4 + (4\times1) = 8$.

Step 2: Draw the Lewis structure

• The central atom is $\ce{Si}$ (less electronegative than $\ce{H}$).
• Connect $\ce{Si}$ to each of the 4 $\ce{H}$ atoms with single bonds. Each single bond uses 2 electrons, so 4 bonds use $4\times2 = 8$ electrons, which is exactly the total valence electrons.
• The Lewis structure is: $\ce{H - Si - H}$ (with two more $\ce{H}$ atoms bonded to $\ce{Si}$, forming a tetrahedral arrangement).

Step 3: Determine molecular geometry (using VSEPR theory)

• The central $\ce{Si}$ atom has 4 bonding pairs (no lone pairs).
• For a central atom with 4 bonding pairs and 0 lone pairs, the electron - pair geometry and molecular geometry are both tetrahedral (according to VSEPR theory, the formula for this case is $AX_4$, where $A$ is the central atom and $X$ are the surrounding atoms).

Let's take another example, $\ce{CO2}$:

Step 1: Calculate total valence electrons

• $\ce{C}$ is in group 14, so it has 4 valence electrons.
• Each $\ce{O}$ is in group 16, so each has 6 valence electrons. There are 2 $\ce{O}$ atoms.
• Total valence electrons: $4+(2\times6)=16$.

Step 2: Draw the Lewis structure

• The central atom is $\ce{C}$.
• We start by connecting $\ce{C}$ to each $\ce{O}$ with a single bond. But if we do that, we will have remaining electrons. We find that a double - bond between $\ce{C}$ and each $\ce{O}$ is needed. The Lewis structure is $\ce{O = C = O}$. Each double bond has 4 electrons, and the total number of electrons in the bonds is $2\times4 = 8$, and each $\ce{O}$ has 2 non - bonding electron pairs (4 electrons per $\ce{O}$), so $2\times4 = 8$. The total is $8 + 8=16$, which matches the total valence electrons.

Step 3: Determine molecular geometry (using VSEPR theory)

• The central $\ce{C}$ atom has 2 bonding pairs (each double bond is counted as one bonding domain) and 0 lone pairs.
• For a central atom with 2 bonding pairs and 0 lone pairs, the electron - pair geometry and molecular geometry are both linear (the VSEPR formula is $AX_2$).

For a cation like $\ce{NH4+}$:

Step 1: Calculate total valence electrons

• $\ce{N}$ is in group 15, so it has 5 valence electrons.
• Each $\ce{H}$ has 1 valence electron, and there are 4 $\ce{H}$ atoms.
• Since it is a cation with a + 1 charge, we subtract 1 electron from the total.
• Total valence electrons: $5+(4\times1)-1 = 8$.

Step 2: Draw the Lewis structure

• The central atom is $\ce{N}$.
• Connect $\ce{N}$ to each of the 4 $\ce{H}$ atoms with single bonds. Each single bond uses 2 electrons, so 4 bonds use $4\times2 = 8$ electrons, which is equal to the total valence electrons. The Lewis structure is a tetrahedral arrangement of $\ce{H}$ atoms around $\ce{N}$.

Step 3: Determine molecular geometry (using VSEPR theory)

• The central $\ce{N}$ atom has 4 bonding pairs and 0 lone pairs. So the molecular geometry is tetrahedral (VSEPR formula $AX_4$).

For an anion like $\ce{BF4-}$:

Step 1: Calculate total valence electrons

• $\ce{B}$ is in group 13, so it has 3 valence electrons.
• Each $\ce{F}$ is in group 17, so each has 7 valence electrons. There are 4 $\ce{F}$ atoms.
• Since it is an anion with a - 1 charge, we add 1 electron to the total.
• Total valence electrons: $3+(4\times7)+1=3 + 28+1 = 32$.

Step 2: Draw the Lewis structure

• The central atom is $\ce{B}$.
• Connect $\ce{B}$ to each of the 4 $\ce{F}$ atoms with single bonds. Each single bond uses 2 electrons, so 4 bonds use $4\times2 = 8$ electrons. Each $\ce{F}$ atom has 3 non - bonding electron pairs (6 electrons per $\ce{F}$), so $4\times6 = 24$ electrons. The total number of electrons in bonds and non - bonding pairs is $8 + 24=32$, which matches the total valence electrons.

Step 3: Determine molecular geometry (using VSEPR theory)

• The central $\ce{B}$ atom has 4 bonding pairs and 0 lone pairs. So the molecular geometry is tetrahedral (VSEPR formula $AX_4$).

For a molecule with a central atom having lone pairs, like $\ce{H2O}$:

Step 1: Calculate total valence electrons

• $\ce{O}$ is in group 16, so it has 6 valence electrons.
• Each $\ce{H}$ has 1 valence electron, and there are 2 $\ce{H}$ atoms.
• Total valence electrons: $6+(2\times1)=8$.

Step 2: Draw the Lewis structure

• The central atom is $\ce{O}$.
• Connect $\ce{O}$ to each $\ce{H}$ with a single bond. This uses $2\times2 = 4$ electrons. The remaining $8 - 4 = 4$ electrons form 2 non - bonding electron pairs on $\ce{O}$. The Lewis structure has $\ce{O}$ in the center with two $\ce{H}$ atoms bonded and two lone pairs.

Step 3: Determine molecular geometry (using VSEPR theory)

• The central $\ce{O}$ atom has 2 bonding pairs and 2 lone pairs. The electron - pair geometry is tetrahedral (because the total number of electron domains is $2 + 2=4$), but the molecular geometry (the arrangement of the atoms) is bent (or angular). The VSEPR formula is $AX_2E_2$, where $E$ represents a lone pair.

The process for other molecules and ions (such as $\ce{PCl3}$, $\ce{CCl4}$, $\ce{NF3}$, $\ce{O3}$, $\ce{OF2}$, $\ce{NH2-}$, $\ce{AlH4-}$, $\ce{H3O+}$, $\ce{BrF2+}$, $\ce{Cl4}$) follows similar steps:

1. Calculate the total number of valence electrons (adjusting for charge in the case of ions: subtract 1 for cations, add 1 for anions).
2. Draw the Lewis structure by placing the least electronegative atom (except $\ce{H}$) in the center, connecting it to the surrounding atoms with single bonds initially, and then distributing the remaining electrons as non - bonding pairs (and converting single bonds to multiple bonds if necessary to satisfy the octet rule, except for atoms that can have an expanded octet).
3. Determine the molecular geometry using VSEPR theory by identifying the number of bonding pairs and lone pairs around the central atom. The general rules for VSEPR theory are:

• The electron - pair geometry depends on the total number of electron domains (bonding + lone pairs) around the central atom.
• The molecular geometry depends on the number of bonding pairs and lone pairs (lone pairs occupy more space than bonding pairs and can distort the molecular geometry).

For example, for $\ce{PCl3}$:

• Total valence electrons: $5+(3\times7)=5 + 21 = 26$.
• Lewis structure: $\ce{P}$ in the center, bonded to 3 $\ce{Cl}$ atoms with single bonds (using $3\times2 = 6$ electrons), and $\ce{P}$ has 1 non - bonding electron pair (2 electrons). The remaining electrons: $26-(6 + 2)=18$ electrons, which are distributed as 3 non - bonding electron pairs on each $\ce{Cl}$ atom (6 electrons per $\ce{Cl}$).
• Molecular geometry: The central $\ce{P}$ has 3 bonding pairs and 1 lone pair. The electron - pair geometry is tetrahedral, and the molecular geometry is trigonal pyramidal (VSEPR formula $AX_3E$).