Question

Phosphoryl chloride, \(\mathrm{POCl}_{3}\), has the skeleton structure Write (a) a Lewis structure for \(\mathrm{POCl}_{3}\) following the octet rule. Calculate the formal charges in this structure. (b) a Lewis structure in which all the formal charges are zero. (The octet rule need not be followed.)

Short answer

Question: Draw two Lewis structures for phosphoryl chloride (POCl3). In one structure, follow the octet rule and calculate the formal charges, and in the other, draw a structure in which all formal charges are zero (does not need to follow the octet rule). Answer: Structure 1 (following the octet rule): - Lewis Structure: ``` O || P - Cl | Cl ``` - Formal Charges: - P: +1 - O: 0 - Cl(1): 0 - Cl(2): 0 - Cl(3): 0 Structure 2 (zero formal charges): - Lewis Structure: ``` O || P - Cl || Cl ``` - Formal Charges: - P: 0 - O: 0 - Cl(1): 0 - Cl(2): 0 - Cl(3): 0

Step-by-step solution

Count Total Valence Electrons

To draw the Lewis structure, we need to find the total number of valence electrons in the molecule. For POCl3, we have the following valence electrons: - Phosphorus (P) has 5 valence electrons - Oxygen (O) has 6 valence electrons - Each Chlorine (Cl) has 7 valence electrons, and there are three of them Total valence electrons: \(5 + 6 + (3*7) = 5 + 6 + 21 = 32\)

(a): Draw Lewis Structure following the Octet Rule

Using the electron-dot formula, we can represent the 32 valence electrons in the molecule. Place the P in the center and connect it with single bonds (2 electrons) to the O and three Cl atoms. We are left with (32-8)=24 electrons. Distribute the remaining electrons by giving each atom an octet (or for Hydrogen, just a doublet) without exceeding eight electrons for each atom. The Lewis structure is: ``` O || P - Cl | Cl ```

(a): Calculate Formal Charges

Formal charge can be calculated using the formula: Formal charge = (valence electrons of an atom) - (non-bonding electrons) - (1/2 * bonding electrons) Calculate formal charge for P: \((5 - 0 - (1/2 * 8)) = 5 - 4 = 1\) Calculate formal charge for O: \((6 - 4 - (1/2 * 4)) = 6 - 4 - 2 = 0\) Calculate formal charge for Cl(1), Cl(2), and Cl(3): \((7 - 6 - (1/2 * 2)) = 7 - 6 - 1 = 0\) So, the formal charges are: P: +1 O: 0 Cl(1): 0 Cl(2): 0 Cl(3): 0

(b): Draw Lewis Structure with Zero Formal Charges

Since we don't need to follow the octet rule in this case, we can try a Lewis structure such that all the formal charges are 0. We achieve that by adding an extra bond between P and Cl(2) (for example): ``` O || P - Cl || Cl ```

(b): Calculate Formal Charges

Calculate formal charges for the new structure: - P: \((5 - 0 - (1/2 * 10)) = 5 - 5 = 0\) - O: \((6 - 4 - (1/2 * 4)) = 6 - 4 - 2 = 0\) - Cl(1): \((7 - 6 - (1/2 * 2)) = 7 - 6 - 1 = 0\) - Cl(2): \((7 - 4 - (1/2 * 4)) = 7 - 4 - 2 = 1\) (this value will be the same for all Cl atoms) Thus, all formal charges in this structure are zero.