Problem 3
Question
When chlorine is passed through concentrated hot solution of \(\mathrm{KOH}\), the compound formed is (a) \(\mathrm{KClO}\) (b) \(\mathrm{KClO}_{2}\) (c) \(\mathrm{KClO}_{3}\) (d) \(\mathrm{KClO}_{4}\)
Step-by-Step Solution
Verified Answer
The compound formed is KClO₃, corresponding to option (c).
1Step 1: Identify the Reaction Conditions
We are given that chlorine is passed through a concentrated hot solution of KOH. This suggests that the reaction conditions are both concentrated and hot, which will influence the type of reaction that occurs.
2Step 2: Recall Relevant Chemical Reaction
Recall the chemical reaction for chlorine gas reacting with hot, concentrated \ \ \( \text{KOH} \ \): \[\mathrm{3Cl_2 + 6KOH \rightarrow 5KCl + KClO_3 + 3H_2O} \]This reaction states that \ \(\text{KClO}_3\) (potassium chlorate) is formed as one of the products.
3Step 3: Analyze the Products
The main products of the reaction include potassium chloride (KCl) and potassium chlorate (KClO₃). We are primarily interested in the oxychloride product, which in this case is KClO₃.
4Step 4: Choose the Correct Option
From the given options:
(a) KClO
(b) KClO₂
(c) KClO₃
(d) KClO₄,
the correct answer is (c) KClO₃, as this is the product formed when chlorine reacts with hot, concentrated KOH.
Key Concepts
Chlorine and KOH ReactionOxychlorides of ChlorineChemical Reactions in Inorganic Chemistry
Chlorine and KOH Reaction
When chlorine gas is bubbled into a concentrated hot solution of potassium hydroxide, a unique chemical reaction unfolds. This specific interaction is influenced by the conditions both in the concentration and temperature of KOH. Under heated, concentrated conditions, chlorine reacts in a manner that typically leads to a rearrangement of the elements involved.
In simple terms, during this reaction:
In simple terms, during this reaction:
- Chlorine, initially a two-atom molecule, dissociates and engages with potassium and hydroxide ions present.
- This leads to the formation of potassium chloride (KCl) and potassium chlorate (KClO₃) along with water (H₂O).
Oxychlorides of Chlorine
The term "oxychlorides" refers to compounds where chlorine is bonded to oxygen. In this reaction with potassium hydroxide, one of the products is potassium chlorate (
KClO₃), which is a type of oxychloride.
To comprehend why KClO₃ forms during the reaction:
This highlights how versatile chlorine can be when interacting with other elements, especially under varied chemical environments.
To comprehend why KClO₃ forms during the reaction:
- We need to look at chlorine's ability to bond not only with metals but also with oxygen.
- This dual binding capacity allows it to form a structure containing both chlorine and oxygen, known as an oxychloride.
This highlights how versatile chlorine can be when interacting with other elements, especially under varied chemical environments.
Chemical Reactions in Inorganic Chemistry
Inorganic chemistry explores reactions involving non-organic compounds. The reaction between chlorine and potassium hydroxide is a quintessential example of an inorganic chemical reaction. Such reactions emphasize the importance of conditions like temperature and concentration in determining the products formed.
To understand these reactions:
To understand these reactions:
- It's vital to grasp how elemental substances like chlorine react under different conditions.
- Inorganic reactions often involve shifts in oxidation states, as seen when chlorine turns into oxychlorides like KClO₃.
Other exercises in this chapter
Problem 1
A radioactive halogen is (a) Polonium (b) Radon (c) Astatine (d) lodine
View solution Problem 4
Which of the following is planar? (a) \(\mathrm{XeF}_{4}\) (b) \(\mathrm{XeO}_{4}\) (c) \(\mathrm{XeO}_{2} \mathrm{~F}_{2}\) (d) \(\mathrm{XeOF}_{4}\)
View solution Problem 5
In Kroll and 1.C.I. process of the production of titanium, the inert gas used is (b) Ar (c) \(\mathrm{Kr}\) (d) \(\mathrm{Xe}\) (a) \(\mathrm{Ne}\)
View solution Problem 6
Which one of the following acts as an antichlor ? (a) \(\mathrm{MnO}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{K}_{2} \mathrm{C
View solution