Question

what are the formal charges of both chlorines and iodine in icl₂⁻? select one: a. −1, 0 b. −1, −1 c. 0, 0 d. 0, −1

Explanation:

Step1: Recall Formal Charge Formula

The formula for formal charge (\(FC\)) is \(FC = V - N - \frac{B}{2}\), where \(V\) is the number of valence electrons in the free atom, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.

Step2: Analyze \(ICl_2^-\) Structure

• For Chlorine (\(Cl\)):
• Valence electrons of \(Cl\) (\(V\)): Chlorine is in group 17, so \(V = 7\).
• In \(ICl_2^-\), each \(Cl\) atom forms a single bond with \(I\). So, the number of bonding electrons (\(B\)) for each \(Cl\) is 2 (since a single bond has 2 electrons), and the number of non - bonding electrons (\(N\)): Each \(Cl\) has 6 non - bonding electrons (3 lone pairs).
• Calculate formal charge for \(Cl\): \(FC=7 - 6-\frac{2}{2}=7 - 6 - 1 = 0\)? Wait, no, wait. Wait, the Lewis structure of \(ICl_2^-\): Iodine is the central atom. Iodine has 7 valence electrons, each \(Cl\) has 7, and there is a - 1 charge, so total valence electrons \(=7 + 2\times7+1 = 22\).
• The Lewis structure of \(ICl_2^-\) has \(I\) in the center, with two single bonds to \(Cl\) atoms, and 3 lone pairs on \(I\) (since \(7 + 2+10 = 19\)? Wait, no, let's recalculate total valence electrons: \(I\) (group 17) has 7, \(Cl\) (group 17) has 7 each, and the negative charge adds 1. So total \(=7 + 2\times7+1=22\).
• For each \(Cl\): It forms a single bond with \(I\), so bonding electrons \(B = 2\), non - bonding electrons \(N=6\) (3 lone pairs). \(FC = 7-6 - \frac{2}{2}=0\)? Wait, no, that's wrong. Wait, maybe I made a mistake in the Lewis structure. Wait, \(ICl_2^-\) has a linear structure with \(I\) as the central atom, which has 5 lone pairs? No, wait, let's use the formula correctly.
• Wait, another way: The formal charge of an atom in a molecule/ion can also be thought of in terms of the octet rule and the charge of the ion.
• For \(Cl\) in \(ICl_2^-\): Each \(Cl\) is bonded to \(I\) with a single bond. The free \(Cl\) atom has 7 valence electrons. In the ion, each \(Cl\) has 6 non - bonding electrons (3 lone pairs) and 2 bonding electrons (from the single bond). So \(FC=7-(6 + \frac{2}{2})=7-(6 + 1)=0\)? No, that can't be. Wait, maybe the central \(I\):
• For \(I\) (central atom): Valence electrons \(V = 7\), non - bonding electrons \(N = 10\) (5 lone pairs), bonding electrons \(B = 4\) (2 single bonds, each with 2 electrons). So \(FC=7-10-\frac{4}{2}=7 - 10 - 2=- 5\)? No, that's wrong. Wait, I think I messed up the Lewis structure.
• Wait, the correct Lewis structure of \(ICl_2^-\): Iodine is in the 5th period, so it can have an expanded octet. The total number of valence electrons: \(I\) (7) + 2\(Cl\) (7 each) + 1 (negative charge) = 7 + 14+1 = 22.
• The central \(I\) forms two single bonds with \(Cl\) atoms. So the number of bonding electrons is \(2\times2 = 4\), and the number of non - bonding electrons on \(I\) is \(22-(4)-2\times6=22 - 4 - 12 = 6\)? Wait, no, each \(Cl\) has 6 non - bonding electrons (3 lone pairs), so 2\(Cl\) atoms have \(2\times6 = 12\) non - bonding electrons. The bonding electrons are 4 (2 single bonds). So non - bonding electrons on \(I\) are \(22-12 - 4=6\) (3 lone pairs). Then for \(I\): \(FC = 7-6-\frac{4}{2}=7 - 6 - 2=-1\).
• For each \(Cl\): \(V = 7\), \(N = 6\) (3 lone pairs), \(B = 2\) (single bond). So \(FC=7 - 6-\frac{2}{2}=0\). So the formal charge of each \(Cl\) is 0, and the formal charge of \(I\) is - 1. So the formal charges of \(Cl\) (both) and \(I\) are 0 and - 1.

Answer:

D. 0, -1