Question

The \[{{\text{I}}_{\text{4}}}{{\text{O}}_{\text{9}}}\] is a/ an:
A.Covalent compound
B.Coordinate compound
C.Ionic compound
D.Double salt

Answer 1

Hint: \[{{\text{I}}_{\text{4}}}{{\text{O}}_{\text{9}}}\] is also known as tetraiodine nonoxide having iodine in two oxidation states \[\left( { + 3} \right)\] and \[\left( { + 5} \right)\] and it is a dark yellow solid. This solid oxide of iodine is not stable at high temperatures.

Complete answer:
In \[{{\text{I}}_{\text{4}}}{{\text{O}}_{\text{9}}}\] or tetraiodine nonoxide, there is one iodine atom at the centre which is linked to two $ {\text{OI}}{{\text{O}}_{\text{2}}} $ units on the two sides of the iodine atom. This has been prepared early by the reaction of ozone with iodine. When tetraiodine nonoxide is heated then it decomposes to form the tripositive iodide ion and iodate anion.
 \[{{\text{I}}_{\text{4}}}{{\text{O}}_{\text{9}}}\xrightarrow{\Delta }{\text{ }}{{\text{I}}_{\text{3}}}^{\text{ + }}{\text{ + 3I}}{{\text{O}}_{\text{3}}}^{\text{ - }}\]
As the decomposition of the compound leads to the formation of the ionic compounds so this compound is an ionic compound.

So, the correct answer is option C.
Note:
There are various compounds of iodine in which iodine exists in various oxidation states that can range from \[\left( { + 1{\text{ to }} + 7} \right)\] . The appearance of Diiodine monoxide is unknown but it is not isolable. There is iodine monoxide which is a purple in colour but it cannot be isolated. Iodine dioxide is a in which the oxidation state of iodine is \[\left( { + 3} \right)\] and \[\left( { + 5} \right)\] and it decomposes at 100 $ ^{\text{0}}{\text{C}} $ . It decomposes to form hydroiodic acid and iodine. Iodine pentoxide is a white crystalline solid in which the oxidation state of iodine is \[\left( { + 5} \right)\] and it decomposes from a temperature between 300 – 500 $ ^{\text{0}}{\text{C}} $ . Lastly the tetraiodine nonoxide decomposes at 75 $ ^{\text{0}}{\text{C}} $ to form hydroiodic acid and iodine.