Question

The experimental result says that two iodine atoms are in different environments and predict the all possible INCORRECT arrangement for I2 Cl4 Br2

• A. (ClBr)I(μ-Cl)2I(ClBr)
• B. (Cl2)I(μ-Br)2I(Cl2)
• C. (Cl2)I(μ-Cl)2I(Br2)
• D. (Cl2)I(μ-Cl)(μ-Br)I(ClBr)

Answer 1

Answer: (C), (D)

1, 2

Answer 2

The problem states that in the molecule $I_2Cl_4Br_2$, the two iodine atoms are in different environments. We need to identify all possible incorrect arrangements, which means arrangements where the two iodine atoms are in the same environment.

First, let's determine the oxidation state of iodine. Assuming Cl and Br are -1: $2(I) + 4(-1) + 2(-1) = 0$ $2I - 4 - 2 = 0$ $2I = 6$ $I = +3$.

Iodine in the +3 oxidation state typically forms compounds like $ICl_3$, which exists as a dimer $I_2Cl_6$. This dimer has a planar structure where two iodine atoms are bridged by two chlorine atoms, and each iodine atom is also bonded to two terminal chlorine atoms. The geometry around each iodine is square planar ($AX_4E_2$).

Following this established chemistry, we can propose a dimeric structure for $I_2Cl_4Br_2$ of the form $(Y_2)I(\mu-X)_2I(Y_2)$, where $X$ are bridging atoms and $Y$ are terminal atoms. We have 4 Cl atoms and 2 Br atoms in total.

We need to consider the possible combinations for the two bridging atoms ($\mu-X_1, \mu-X_2$) and the four terminal atoms ($Y_1, Y_2, Y_3, Y_4$).

Case 1: Two Cl atoms are bridging ($\mu-Cl, \mu-Cl$). The remaining atoms are $Cl_2Br_2$ for terminal positions. Each iodine atom must be bonded to two terminal atoms.

• Arrangement A: Each iodine is bonded to one terminal Cl and one terminal Br. Structure: $(ClBr)I(\mu-Cl)_2I(ClBr)$

  • Environment of $I_A$: (2 bridging Cl, 1 terminal Cl, 1 terminal Br)
  • Environment of $I_B$: (2 bridging Cl, 1 terminal Cl, 1 terminal Br)
  • Since the environments are identical, this is an INCORRECT arrangement.

• Arrangement B: One iodine is bonded to two terminal Cls, and the other iodine is bonded to two terminal Brs. Structure: $(Cl_2)I(\mu-Cl)_2I(Br_2)$

  • Environment of $I_A$: (2 bridging Cl, 2 terminal Cl)
  • Environment of $I_B$: (2 bridging Cl, 2 terminal Br)
  • Since the environments are different, this is a correct arrangement according to the experimental result.

Case 2: One Cl and one Br atom are bridging ($\mu-Cl, \mu-Br$). The remaining atoms are $Cl_3Br_1$ for terminal positions. Each iodine atom must be bonded to two terminal atoms.

• Arrangement C: One iodine is bonded to two terminal Cls, and the other iodine is bonded to one terminal Cl and one terminal Br. Structure: $(Cl_2)I(\mu-Cl)(\mu-Br)I(ClBr)$

  • Environment of $I_A$: (1 bridging Cl, 1 bridging Br, 2 terminal Cl)
  • Environment of $I_B$: (1 bridging Cl, 1 bridging Br, 1 terminal Cl, 1 terminal Br)
  • Since the environments are different, this is a correct arrangement.

Case 3: Two Br atoms are bridging ($\mu-Br, \mu-Br$). The remaining atoms are $Cl_4$ for terminal positions. Each iodine atom must be bonded to two terminal atoms.

• Arrangement D: Each iodine is bonded to two terminal Cls. Structure: $(Cl_2)I(\mu-Br)_2I(Cl_2)$

  • Environment of $I_A$: (2 bridging Br, 2 terminal Cl)
  • Environment of $I_B$: (2 bridging Br, 2 terminal Cl)
  • Since the environments are identical, this is an INCORRECT arrangement.

Therefore, the possible incorrect arrangements are those where the two iodine atoms are in the same environment. These are:

1. $(ClBr)I(\mu-Cl)_2I(ClBr)$
2. $(Cl_2)I(\mu-Br)_2I(Cl_2)$