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HI can be prepared by which of the following methods (a) \(\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{~S}\) (b) \(\mathrm{PI}_{3}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{KI}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{H}_{2}+\mathrm{I}_{2} \stackrel{2}{\mathrm{Pt}}\)

Short Answer

Expert verified
Methods (a), (b), and (d) can prepare HI.

Step by step solution

01

Analyze each reaction's capability

To determine the correct method for preparing HI, we need to analyze whether each chemical reaction can produce HI. Let's do this one by one: (a) \( \mathrm{I}_2 + \mathrm{H}_2 \mathrm{S} \): When iodine (\( \mathrm{I}_2 \)) reacts with hydrogen sulfide (\( \mathrm{H}_2 \mathrm{S} \)), sulfur and HI are produced. This reaction is viable: \( \mathrm{I}_2 + \mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{2HI} + \mathrm{S} \).(b) \( \mathrm{PI}_3 + \mathrm{H}_2 \mathrm{O} \): This reaction does not directly give HI. Instead, it primarily produces phosphorous acid and HI: \( \mathrm{PI}_3 + 3\mathrm{H}_2 \mathrm{O} \rightarrow 3\mathrm{HI} + \mathrm{H}_3\mathrm{PO}_3 \).(c) \( \mathrm{KI} + \mathrm{H}_2 \mathrm{SO}_4 \): This reaction produces iodine (\( \mathrm{I}_2 \)), water, and sulfur dioxide (\( \mathrm{SO}_2 \)), not HI. It actually oxidizes iodide ions (\( \mathrm{I}^- \)) to iodine: \( 2\mathrm{KI} + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{I}_2 + 2\mathrm{H}_2 \mathrm{O} + \mathrm{SO}_2 \).(d) \( \mathrm{H}_2 + \mathrm{I}_2 \) over platinum (Pt) catalyst: This reaction is a direct way to synthesize HI. The reaction is: \( \mathrm{H}_2 + \mathrm{I}_2 \rightarrow 2\mathrm{HI} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
In the realm of chemistry, every transformation of molecules involves a chemical reaction. Here, we dive into the specific reactions involved in the preparation of hydrogen iodide (HI). Each reaction has its unique characteristics that determine the substances produced. Understanding chemical reactions is crucial as it involves recognizing reactants and predicting products.
  • When iodine reacts with hydrogen sulfide, the chemical reaction results in the formation of HI and sulfur: \( \mathrm{I}_2 + \mathrm{H}_2 \mathrm{S} \rightarrow 2\mathrm{HI} + \mathrm{S} \).
  • Meanwhile, when phosphorus triiodide reacts with water, it leads to the formation of HI and phosphorous acid: \( \mathrm{PI}_3 + 3\mathrm{H}_2 \mathrm{O} \rightarrow 3\mathrm{HI} + \mathrm{H}_3\mathrm{PO}_3 \).
  • Reactions sometimes do not produce HI directly, such as when potassium iodide reacts with sulfuric acid, resulting in iodine, water, and sulfur dioxide: \( 2\mathrm{KI} + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{I}_2 + 2\mathrm{H}_2 \mathrm{O} + \mathrm{SO}_2 \).
  • The reaction between hydrogen and iodine over a catalyst like platinum provides a straightforward synthesis of HI: \( \mathrm{H}_2 + \mathrm{I}_2 \rightarrow 2\mathrm{HI} \).
Recognizing the conditions under which each chemical reaction occurs is essential for successfully preparing desired compounds like HI.
Inorganic Chemistry
Inorganic chemistry primarily deals with compounds that are not based on carbon-hydrogen bonds, which includes a vast array of molecules such as HI. Understanding how to manipulate and synthesize these compounds is key in inorganic chemistry.
In the context of HI preparation, each reaction brings its unique aspect of inorganic chemistry:
  • In reactions involving iodine and hydrogen sulfide, we see elements from different groups of the periodic table interacting: viably synthesizing acids like HI using these nonmetals.
  • Reactions involving phosphorus triiodide and water offer insights into the handling of compounds that contain multiple elements, showcasing the diversity in inorganic reactions.
  • The reaction of potassium iodide with sulfuric acid highlights oxidizing reactions—one of the many processes used in inorganic chemistry to manipulate substances by changing their oxidation states, although in this case, HI is not a product.
  • The interaction of hydrogen and iodine, especially through catalytic means, underscores the importance of elemental reactions in forming new compounds, a fundamental part of inorganic chemistry.
These examples emphasize the diverse and complex nature of inorganic chemistry, where elements are combined in myriad ways to produce useful compounds.
Synthesis Methods
Synthesis methods in chemistry are strategies used to construct chemical compounds from simpler substances. When preparing a compound like HI, choosing the right method of synthesis is essential.
  • One straightforward synthesis method is reacting iodine with hydrogen sulfide. This method efficiently produces HI while also generating byproducts such as sulfur, as demonstrated in \( \mathrm{I}_2 + \mathrm{H}_2 \mathrm{S} \rightarrow 2\mathrm{HI} + \mathrm{S} \).
  • Using phosphorus triiodide and water is another synthesis approach. It not only yields HI but also phosphorous acid, necessitating the consideration of managing or utilizing the byproducts: \( \mathrm{PI}_3 + 3\mathrm{H}_2 \mathrm{O} \rightarrow 3\mathrm{HI} + \mathrm{H}_3\mathrm{PO}_3 \).
  • Inorganic reactions like the reduction of iodine by hydrogen with a platinum catalyst offer a direct method to achieve pure product outcomes, perfect for industrial uses: \( \mathrm{H}_2 + \mathrm{I}_2 \rightarrow 2\mathrm{HI} \).
The efficiency, purity of the product, and handling of byproducts are all considerations in selecting the most suitable synthesis method. Understanding these methods aids not only in the practical production of substances like HI but also in appreciating their applications in various chemical industries.

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Most popular questions from this chapter

Match the following: List I List II 1\. \(\mathrm{XeF}_{4}\) (i) distorted octahedral 2\. \(\mathrm{XeF}_{6}\) (ii) tetrahedral 3\. \(\mathrm{XeO}_{3}\) (iii) square planar 4\. \(\mathrm{XeO}_{4}\) (iv) pyramidal The correct matching is 1 2 3 4 (a) (iii) (i) (iv) (ii) (b) (ii) (i) (iii) (iv) (c) (i) (iii) (ii) (iv) (d) (iii) (iv) (i) (ii)

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Which one of the following halogen liberates oxygen, when passed through hot concentrated KOH solution? (a) \(\mathrm{Cl}_{2}\) (b) \(\mathrm{F}_{2}\) (c) \(\mathrm{I}_{2}\) (d) \(\mathrm{Br}_{2}\)

Consider the following substances: 1\. \(\mathrm{OF}_{2}\) 2\. \(\mathrm{Cl}_{2} \mathrm{O} \quad\) 3. \(\mathrm{Br}_{2} \mathrm{O}\) The correct sequence of \(X-O=X\) bond angle is (a) \(3>2>1\) (b) \(1>2>3\) (c) \(2>1>3\) (d) \(1>3>2\)

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