Question

general steps

1. count all bonds to get the number of sigma bonds
2. add 1 pi bond for each double bond
3. add 2 pi bonds for each triple bond

sulfuryl chloride, so₂cl₂ (~5:03)
image of sulfuryl chloride molecule

Explanation:

Step1: Determine the structure of \( SO_2Cl_2 \)

The Lewis structure of \( SO_2Cl_2 \) has the sulfur atom bonded to two chlorine atoms (single bonds) and two oxygen atoms. The sulfur - oxygen bonds: one double bond and one double bond? Wait, actually, the correct structure: Sulfur is the central atom. It forms single bonds with 2 Cl atoms (\( S - Cl \)) and double bonds with 2 O atoms (\( S = O \)).

First, count the sigma bonds: All single bonds and one of the bonds in a double bond is a sigma bond. So, the \( S - Cl \) bonds: 2 sigma bonds. The \( S = O \) bonds: each double bond has 1 sigma bond, so 2 sigma bonds from the double bonds. Also, wait, no: Let's re - examine. The formula for \( SO_2Cl_2 \): S is bonded to 2 Cl (single bonds, so 2 sigma bonds) and 2 O. The S - O bonds: in \( SO_2Cl_2 \), the sulfur - oxygen bonds are double bonds? Wait, the correct Lewis structure: S has a steric number of 4 (2 single bonds to Cl, 2 double bonds to O? No, steric number calculation: number of bonding groups + number of lone pairs. For S in \( SO_2Cl_2 \), valence electrons of S: 6, Cl: 7 each (2 Cl: 14), O: 6 each (2 O: 12). Total valence electrons: \( 6 + 14+12 = 32 \).

Let's distribute the electrons: S is central. Connect S to 2 Cl and 2 O. Each S - Cl bond: 2 electrons (1 bond), so 2 bonds: 4 electrons. Each S - O bond: let's start with single bonds. S - O single bonds: 2 bonds, 4 electrons. Now, S has used \( 2 + 2=4 \) bonds (8 electrons), and has \( 6 - 4 = 2 \) electrons left (1 lone pair? No, wait, valence electrons of S: 6. After forming 4 single bonds (to 2 Cl and 2 O), it has \( 6-4 = 2 \) electrons (1 lone pair). But O has 6 valence electrons. If S - O is a single bond, O has \( 6 - 1=5 \) electrons (needs 1 more to complete octet). So we can form double bonds between S and O. So, convert two of the single S - O bonds to double bonds. So, each S - O double bond: now, for each S - O double bond, we have 1 sigma and 1 pi bond.

Now, count sigma bonds:
- \( S - Cl \) bonds: 2 sigma bonds (single bonds)
- \( S - O \) bonds: each double bond has 1 sigma bond, so 2 sigma bonds (from the two \( S = O \) double bonds)
Wait, no, actually, when we have a double bond (\( S = O \)), one bond is sigma and one is pi. So total sigma bonds: \( S - Cl \) (2) + \( S - O \) (2, one from each double bond) + wait, no, the S is bonded to 2 Cl (single bonds, 2 sigma) and 2 O (double bonds, each double bond has 1 sigma). So total sigma bonds: \( 2 + 2=4 \)? Wait, no, maybe I made a mistake. Let's use the general steps:

Step 1: Count all bonds to get the number of sigma bonds. In a single bond, there is 1 sigma bond. In a double bond, there is 1 sigma and 1 pi bond. In a triple bond, 1 sigma and 2 pi bonds.

For \( SO_2Cl_2 \):
- Bonds: 2 \( S - Cl \) (single bonds, so 2 sigma bonds)
- 2 \( S = O \) (double bonds, each has 1 sigma bond, so 2 sigma bonds)
Total sigma bonds: \( 2 + 2=4 \)? Wait, no, wait the correct structure: Let's check the molecular geometry. The \( SO_2Cl_2 \) has a tetrahedral geometry around S. The bonds: S - Cl (2), S - O (2). The S - O bonds: are they single or double? The formal charge on S: valence electrons of S is 6. In \( SO_2Cl_2 \), if S is bonded to 2 Cl (each single bond: 2 electrons from S) and 2 O (each double bond: 4 electrons from S? No, formal charge formula: \( FC=V - N - \frac{B}{2} \), where V is valence electrons, N is non - bonding electrons, B is bonding electrons.

For S: V = 6, N = 0 (in the correct structure, S has no lone pairs), B: if S is bonded to 2 Cl (single bonds: 2 bonds, 4 electrons) and 2 O (double bonds: 2 bonds, 4 electrons), total B = 8. Then \( FC = 6-0-\frac{8}{2}=6 - 4=2 \). That's not good. Wait, maybe the S - O bonds are single bonds? Let's recalculate. If S is bonded to 2 Cl (single bonds, B = 4) and 2 O (single bonds, B = 4), total B = 8. N (lone pairs on S): \( 6 - 8/2=6 - 4 = 2 \) (1 lone pair). Then for O: V = 6, N: if S - O is single bond, O has \( 6 - 1=5 \) non - bonding electrons (needs 1 more). So we can form a double bond between S and O. Let's take one O: form a double bond (so B for that O: 4, N: \( 6 - 4/2=6 - 2 = 4 \)? No, formal charge for O in double bond: \( FC = 6-4-\frac{4}{2}=6 - 4 - 2 = 0 \). For O in single bond: \( FC=6 - 6-\frac{2}{2}=6 - 6 - 1=-1 \). So to minimize formal charges, we have two double bonds between S and O. So S has B = 2 (Cl) + 4 (O, two double bonds) = 6? Wait, no, each double bond is 2 bonds (sigma and pi), so S is bonded to 2 Cl (single bonds: 2 bonds) and 2 O (double bonds: 2 bonds each? No, a double bond is one bond with two electron pairs. I think I'm overcomplicating. Let's use the general method for counting sigma and pi bonds in \( SO_2Cl_2 \):

The correct way:
- Sigma bonds: All single bonds and one bond in each double/triple bond.
- Pi bonds: One bond in each double bond, two in each triple bond.

In \( SO_2Cl_2 \), the structure is: S is central, bonded to 2 Cl (single bonds, so 2 sigma bonds) and 2 O (double bonds, so 2 sigma bonds from the double bonds (one per double bond) and 2 pi bonds (one per double bond)).

So step 1: Count sigma bonds. The number of sigma bonds: \( 2(S - Cl) + 2(S - O \text{ sigma in double bond})=4 \)? Wait, no, wait the actual number: Let's look up the structure of \( SO_2Cl_2 \). The sulfur atom is bonded to two chlorine atoms (single bonds, sigma) and two oxygen atoms (double bonds). Each double bond has one sigma and one pi bond. So:

- Number of sigma bonds: \( 2(S - Cl) + 2(S - O \text{ sigma})= 2 + 2=4 \)
- Number of pi bonds: Each double bond has one pi bond, so 2 pi bonds (from the two \( S = O \) double bonds)

Wait, let's verify with the general steps:

Step 1: Count all bonds to get the number of sigma bonds. The bonds are: 2 \( S - Cl \) (single, sigma), 2 \( S = O \) (each has 1 sigma). So total sigma bonds: \( 2 + 2=4 \)

Step 2: Add 1 pi bond for each double bond. There are 2 double bonds (\( S = O \)), so pi bonds: \( 2\times1 = 2 \)

Step 3: Add 2 pi bonds for each triple bond. There are no triple bonds, so 0.

So sigma bonds: 4, pi bonds: 2.

Wait, but let's re - check. The formula \( SO_2Cl_2 \): S is bonded to 2 Cl (single bonds, 2 sigma), 2 O (double bonds, each with 1 sigma and 1 pi). So sigma bonds: \( 2 + 2=4 \), pi bonds: \( 2\times1 = 2 \)

Step 1: Count sigma bonds

In \( SO_2Cl_2 \), the bonds are:
- \( S - Cl \) bonds: 2 (single bonds, so 2 sigma bonds)
- \( S = O \) bonds: 2 (each double bond has 1 sigma bond, so 2 sigma bonds)
Total sigma bonds \(=2 + 2=4\)

Step 2: Count pi bonds

Each \( S = O \) double bond has 1 pi bond. There are 2 \( S = O \) double bonds. So pi bonds \(=2\times1 = 2\) (since we add 1 pi bond for each double bond)

Answer:

Sigma bonds: \(\boldsymbol{4}\), Pi bonds: \(\boldsymbol{2}\)