#1) The Lewis structure of PF3 shows that central phosphorus atom has ___ nonbonding and ___ bonding electron pairs.

#2) How many equivalent resonance forms can be drawn for CO3^-2? (carbon is the central atom)

#3) The bond angle in NF3 is slightly less than ___?

#4) Sample of gas (1.3mol) occupies ____L at 22 Celcius and 2.5 atm.

For PF3, I think P has 1 non-bonding pair and 3 bonding pairs. Here is a site that gives the Lewis structure for PF3. PF3 is closer to the bottom of the page.

(Broken Link Removed)

I am trying to study for my final exam. I am just trying to do my review packet and i just want to see if i am getting them correct.

but what i got was

1) 3,3
2) 2
3) 120
4) idk how to do that one.

#2. For CO3^-2, I would think there are three equivalent structures. You MAY have meant two + the original (which makes 3) but equivalent means all of them. All of them are resonance forms of each other.

We can't draw these things on the board but here is a site that does a nice job.
http://www.chem.ucla.edu/~harding/tutorials/resonance/draw_res_str.html

NF3, Lewis structure wise, looks much like PF3. The electronic geometry, therefore, is tetrahedral (1 lone pair, and 3 pairs of bonded electrons). [Four regions of high electron density gives a tetrahedral structure.] The lone pair occupies more space; therefore, the tetrahedral bond angle of 109o is reduced somewhat. So I think the answer should be "slightly less than 109o. Here is a site that talks about that.

http://answers.yahoo.com/question/index?qid=20061129055540AAOTWwD

#4. First, note the correct spelling of celsius.

Use PV = nRT.
P is in atmospheres (2.5), V is the only unknown (in liters), n is the number of moles (1.3), R is the universal gas constant of 0.08205 L*atm/mol*K, and T is the temperature in Kelvin (273+22 = 295 K

#1) To determine the number of bonding and nonbonding electron pairs on the central phosphorus atom in PF3, we need to draw the Lewis structure of PF3.

To do this, start by determining the total number of valence electrons for each atom. Phosphorus is in group 5 of the periodic table, so it has 5 valence electrons. Fluorine is in group 7, so each fluorine atom has 7 valence electrons.

The formula PF3 indicates there is one phosphorus atom and three fluorine atoms, so the total number of valence electrons is 5 (Phosphorus) + 3x7 (Fluorine) = 26.

Next, place the atoms in the structure, with the phosphorus atom in the center and the fluorine atoms spread around it.

Since each fluorine atom forms a single bond with the phosphorus atom, we use 1 bonding electron pair for each bond.

To distribute the remaining valence electrons, place them in lone pairs around the fluorine atoms, with each lone pair consisting of 2 nonbonding electron pairs.

After arranging all the atoms and distributing the valence electrons, count the number of nonbonding and bonding electron pairs on the central phosphorus atom. In the case of PF3, the central phosphorus atom has 1 nonbonding electron pair and 3 bonding electron pairs.

Therefore, the Lewis structure of PF3 shows that the central phosphorus atom has 1 nonbonding and 3 bonding electron pairs.

#2) To determine the number of equivalent resonance forms that can be drawn for CO3^-2, we need to understand the concept of resonance.

Resonance occurs when multiple Lewis structures can be drawn for a molecule or ion with the same arrangement of atoms but different electron distributions.

For CO3^-2, start by identifying the total number of valence electrons:

Carbon is in group 4, so it has 4 valence electrons.
Each oxygen is in group 6, so each oxygen atom has 6 valence electrons.
The -2 charge indicates that there are 2 more electrons than the sum of the valence electrons of carbon and oxygen.

So, the total number of valence electrons is 4 (Carbon) + 3x6 (Oxygen) + 2 (extra electrons) = 24.

Next, arrange the atoms with carbon as the central atom and oxygen atoms around it. Carbon forms double bonds with two of the oxygen atoms and a single bond with the third oxygen atom.

Now, we can draw resonance structures by redistributing the electrons.

Start by moving a lone pair from one of the oxygen atoms to form a double bond with carbon, while simultaneously moving one of the carbon-oxygen double bonds to a lone pair on the adjacent oxygen. This process will give us one resonance structure.

Repeat this process by moving lone pairs and double bonds around the atoms to create different arrangements while maintaining the same placement of atoms.

Count the number of equivalent resonance structures that can be drawn. In the case of CO3^-2, there are three equivalent resonance structures that can be drawn.

Therefore, the number of equivalent resonance forms that can be drawn for CO3^-2 is three.

#3) To determine the bond angle in NF3, we need to look at the molecular geometry of the molecule.

First, draw the Lewis structure of NF3 by determining the total number of valence electrons for each atom.

Nitrogen is in group 5, so it has 5 valence electrons.
Fluorine is in group 7, so each fluorine atom has 7 valence electrons.

The formula NF3 indicates there is one nitrogen atom and three fluorine atoms, so the total number of valence electrons is 5 (Nitrogen) + 3x7 (Fluorine) = 26.

Next, place the atoms in the structure, with the nitrogen atom in the center and the fluorine atoms spread around it.

Since each fluorine atom forms a single bond with the nitrogen atom, we use 1 bonding electron pair for each bond.

Distribute the remaining valence electrons in lone pairs around the fluorine atoms and the nitrogen atom.

After arranging all the atoms and distributing the valence electrons, we can determine the molecular geometry of NF3.

NF3 has a trigonal pyramidal shape, with the nitrogen atom at the center and three fluorine atoms arranged around it. This shape occurs when there is one lone pair and three bonding pairs around the central atom.

The bond angle in a trigonal pyramidal molecule like NF3 is less than the ideal tetrahedral angle of 109.5 degrees. In the case of NF3, the bond angle is slightly less than 109.5 degrees.

Therefore, the bond angle in NF3 is slightly less than 109.5 degrees.

#4) To determine the volume occupied by a sample of gas at a given temperature and pressure, we can use the ideal gas law:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

In the given problem, we are given the number of moles of gas (1.3 mol), the temperature (22 Celsius, which we need to convert to Kelvin), and the pressure (2.5 atm).

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. In this case, 22 Celsius + 273.15 = 295.15 Kelvin.

Now, we can plug in the values into the ideal gas law equation:

(2.5 atm)(V) = (1.3 mol)(0.0821 L·atm/mol·K)(295.15 K)

Solving for V:

V = (1.3 mol)(0.0821 L·atm/mol·K)(295.15 K) / (2.5 atm)

V ≈ 38.164 liters

Therefore, the sample of gas occupies approximately 38.164 liters at 22 Celsius and 2.5 atm.

I would be interested in knowing what you think about these questions and why? We are not in the habit of doing all of the homework. Show what you have done on a problem and we can help you through it.