Question

draw the lewis structure created by the curved arrow. write in formal charges wherever they are not equal to zero.

Explanation:

Step1: Analyze the Curved Arrow Movement

The curved arrow shows the movement of an electron pair from the Cl atom (which has lone pairs) to form a bond with the positively charged C atom. Initially, the C - Cl bond is a double bond, and the C has a + charge. After the arrow (donation of a lone pair from Cl to C), we form a new bond.

Step2: Determine the New Bonding and Formal Charges

- The carbon atom: Initially, it had a + charge. After accepting the electron pair from Cl, the C - Cl bond becomes a triple bond? Wait, no, let's re - evaluate. The original structure: the C is bonded to another C (with H's) and has a double bond to Cl, and a + charge. The curved arrow is from the lone pair on Cl to the C. So, the Cl donates a lone pair to form a bond with C. So, the C - Cl bond becomes a triple bond? Wait, no, the C was positively charged, meaning it had 6 valence electrons (since C has 4 valence electrons normally; if it has a + charge, it has lost one, so 3 bonds? Wait, no, the original structure: the C is bonded to a C (with H3) and has a double bond to Cl, so that's 3 bonds (1 single to C, 2 double to Cl), so valence electrons around C: 3*2 = 6 (since each bond is 2 electrons), and C has 4 valence electrons, so formal charge = 4 - 6/2 - 0 (lone pairs) = 4 - 3 = +1.

After the curved arrow (Cl donates a lone pair to C), the C - Cl bond becomes a triple bond. Now, the C has 4 bonds (1 to C, 3 to Cl), so valence electrons around C: 4*2 = 8. Formal charge of C: 4 - 8/2 - 0 = 4 - 4 = 0.

For the Cl atom: Initially, Cl had a double bond to C, so number of bonds: 2, lone pairs: let's assume Cl has 3 lone pairs (since Cl has 7 valence electrons). After donating a lone pair to form a triple bond (3 bonds), the number of lone pairs on Cl: 7 - 3*2/2 - 3 (bonds: each bond is 2 electrons, 3 bonds is 6 electrons) Wait, no, formal charge of Cl: valence electrons of Cl is 7. Number of bonds: 3 (triple bond), number of lone pairs: let's calculate. After forming the triple bond, the number of electrons around Cl: 3*2 (from bonds) + 2*2 (lone pairs? Wait, the "Add 2 lone pairs" option is there. Wait, maybe the Cl initially had 2 lone pairs? No, Cl has 7 valence electrons. Let's do formal charge for Cl:

Formal charge = valence electrons - (number of bonds/2) - number of lone pair electrons.

Initially, Cl has a double bond to C (2 bonds), so bonds: 2, lone pairs: let's say x. 7 = 2*2/2 + x*2? No, each bond is 2 electrons, so 2 bonds contribute 4 electrons, lone pairs: each lone pair is 2 electrons. So 4 + 2x = 7? No, that's not possible. Wait, maybe the initial structure: the C is positively charged, so the C - Cl is a double bond, and Cl has 3 lone pairs (6 electrons) plus 2 electrons from the double bond (total 8 electrons around Cl, which is stable). So Cl has formal charge: 7 - 2 (bonds: double bond, so 2 bonds, 2*2/2 = 2) - 6 (lone pairs) = 7 - 2 - 6 = -1? Wait, no, that can't be. I think I messed up.

Wait, let's start over. The curved arrow is from the lone pair on Cl to the C (the positively charged C). So the C is electron - deficient (has a + charge). The Cl has lone pairs. So when Cl donates a lone pair to C, a new bond is formed. So the C - Cl bond becomes a triple bond. Now:

- Carbon atom: Now has 4 bonds (1 to the other C, 3 to Cl). So formal charge: 4 (valence electrons) - (8/2) (bonds: 4 bonds, 8 electrons) - 0 (lone pairs) = 4 - 4 = 0.

- Chlorine atom: Now has 3 bonds (triple bond) to C. The number of lone pairs on Cl: Cl has 7 valence electrons. The number of electrons in bonds: 3*2 = 6. So lone pair electrons: 7 - 6 = 1 lone pair? No, that can't be. Wait, the "Add 2 lone pairs" option is there. Oh, right, the Cl should have 3 lone pairs (6 electrons) and 1 bond? No, no. Wait, maybe the initial bond was a single bond? I think I misinterpreted the original structure.

Original structure: The C is bonded to a C (with H3) and has a double bond to Cl, and a + charge. Wait, no, the C with the + charge: let's count the valence electrons. The C is bonded to a C (single bond), double bonded to Cl, so that's 3 bonds (1 single, 2 double). So valence electrons around C: 3*2 = 6. Formal charge of C: 4 - 6/2 - 0 = +1 (since C has 4 valence electrons, 6 electrons from bonds, 0 lone pairs).

The Cl: double bonded to C, so 2 bonds (4 electrons), and let's say 3 lone pairs (6 electrons), so total electrons around Cl: 4 + 6 = 10? No, Cl can't have 10 electrons. Wait, no, Cl is in the third period, it can have an expanded octet, but usually, in Lewis structures, we try to follow octet for second - period elements, but Cl can have more. Wait, no, the C is a second - period element, so it follows octet.

After the curved arrow: The Cl donates a lone pair to C, so the C - Cl bond becomes a triple bond. Now, the C has 4 bonds (1 to C, 3 to Cl), so octet is satisfied (8 electrons). The Cl now has 3 bonds (triple bond) and 2 lone pairs (4 electrons). Let's calculate formal charges:

- For C: Valence electrons = 4. Number of bonds = 4 (each bond is 2 electrons, so 8 electrons from bonds), lone pairs = 0. Formal charge = 4 - (8/2) - 0 = 4 - 4 = 0.

- For Cl: Valence electrons = 7. Number of bonds = 3 (3*2 = 6 electrons from bonds), lone pairs = 2 (4 electrons). Formal charge = 7 - (6/2) - 4 = 7 - 3 - 4 = 0? Wait, no, 6/2 is 3 (number of bond pairs), 4 is lone pair electrons. 7 - 3 - 4 = 0. But initially, Cl had a double bond (2 bonds) and 3 lone pairs (6 electrons). Formal charge of Cl initially: 7 - (4/2) - 6 = 7 - 2 - 6 = -1. And C had +1. So the formal charges cancel out.

Now, to draw the Lewis structure: The central part is the two C atoms, one with H3 (CH3) and the other now bonded to Cl with a triple bond. The Cl has 2 lone pairs (as per the "Add 2 lone pairs" option? Wait, no, if Cl has a triple bond, it should have 2 lone pairs (4 electrons) to satisfy the octet? Wait, Cl has 7 valence electrons. Triple bond: 3 bonds (6 electrons), lone pairs: 1 lone pair (2 electrons) would give 6 + 2 = 8 electrons. Wait, I'm confused. Let's use the formal charge formula correctly.

Formal charge (FC) = Valence electrons (VE) - Non - bonding electrons (NBE) - Bonding electrons (BE)/2.

For the C that was positively charged:

Initial VE of C: 4, NBE: 0, BE: 6 (double bond to Cl (2 bonds, 4 electrons) and single bond to C (2 electrons)) → FC = 4 - 0 - 6/2 = 4 - 3 = +1.

After the arrow: C forms a triple bond with Cl (3 bonds, 6 electrons) and single bond with C (2 electrons), so BE = 8, NBE = 0 → FC = 4 - 0 - 8/2 = 4 - 4 = 0.

For Cl:

Initial VE of Cl: 7, NBE: 6 (3 lone pairs), BE: 4 (double bond) → FC = 7 - 6 - 4/2 = 7 - 6 - 2 = -1.

After the arrow: Cl forms a triple bond with C (BE = 6), NBE: 2 (1 lone pair? No, 7 - 6 = 1, but 1 lone pair is 2 electrons) → Wait, 7 - 6 (BE) = 1, so NBE = 2? No, 1 lone pair is 2 electrons. So FC = 7 - 2 - 6/2 = 7 - 2 - 3 = 2? No, that's wrong.

Wait, maybe the curved arrow is from the double bond? No, the curved arrow is from the lone pair on Cl. Let's look at the "Add 2 lone pairs" option. So we need to add 2 lone pairs to Cl. So Cl should have 2 lone pairs (4 electrons) and a triple bond? No, triple bond is 3 bonds (6 electrons), plus 4 lone pair electrons: 6 + 4 = 10 electrons around Cl, which is okay for Cl (third period, can have expanded octet).

So the Lewis structure: The C (CH3 - C) now has a triple bond to Cl. The Cl has 2 lone pairs (4 electrons). The formal charge of C is 0, and the formal charge of Cl: VE = 7, NBE = 4, BE = 6 → FC = 7 - 4 - 6/2 = 7 - 4 - 3 = 0. The other C (with H3) has formal charge 0, H's have 0.

Step3: Draw the Lewis Structure

- Draw the two C atoms: one with three H atoms (CH3) bonded to it, and the other C bonded to the first C and triple - bonded to Cl.
- Add 2 lone pairs to Cl (as per the "Add 2 lone pairs" option).
- Formal charges: C (the one bonded to Cl) has 0, Cl has 0, the other C has 0, H's have 0.

Answer:

The Lewis structure will have a carbon - carbon single bond, with one carbon bonded to three hydrogen atoms (CH₃) and the other carbon triple - bonded to chlorine. The chlorine atom has two lone pairs of electrons. The formal charge on the carbon (previously positive) is 0, and the formal charge on chlorine is 0. The structure can be represented as: $\ce{CH3 - C\equiv Cl}$, with Cl having two lone pairs (represented as two pairs of dots around Cl).