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Chapter 14: Q14.70 P (page 622)

Thallium forms the compound TlI3. What is the apparent oxidation state of Tl in this compound? Given that the anion is I3-, what is the actual oxidation state of Tl? Draw the shape of the anion, giving its VSEPR class and bond angles. Propose a reason why the compound does not exist as(Tl3+)(I-)3.

Short Answer

Expert verified

The apparent oxidation state of Tl in TlI3is +3.

The actual oxidation state of Tl in Tll3is +1.

The shape of the anion is given below:

The strength of the bond betweenTl3+andI-3is very low.

Step by step solution

01

Apparent oxidation state of Tl in TlI3

The apparent oxidation state of Tl in TlI3is +3 since iodine has -1 oxidation state.

Let the oxidation state of thallium be x. The calculation is given below:

x+-1脳3=0x-3=0x=+3

Thus, the apparent oxidation state of Tl in TlI3is +3.

02

Step 2: Actual oxidation state of Tl in TlI3

It is given that the anion is I3-. Let x be the oxidation state of thallium. So the oxidation state of Tl in TlI3can be calculated as follows:

x-1=0x=+1

So the oxidation state of thallium is +1.

03

Shape of anion

The shape of anion I3-is given below:

There is an excess of one electron at the second iodine, which is the reason for the negative charge. It belongs to AX2E3class and has a bond angle of 180掳.

04

Non-existence of (Tl3+)(I-)3

The bond between thallium and iodine has very low strength. Thus Tl3+I-3does not exist.

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