/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Hydrolysis of \(\mathrm{NCl}_{3}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Hydrolysis of \(\mathrm{NCl}_{3}\) gives \(\mathrm{NH}_{3}\) and \(\mathrm{X}\). Which of the following is \(\mathrm{X} ?\) (a) \(\mathrm{HClO}_{4}\) (b) \(\mathrm{HClO}_{3}\) (c) \(\mathrm{HOCl}\) (d) \(\mathrm{HClO}_{2}\)

Short Answer

Expert verified
The product \(\mathrm{X}\) is \(\mathrm{HOCl}\), option (c).

Step by step solution

01

Write the Hydrolysis Reaction

First, write the balanced chemical equation for the hydrolysis of nitrogen trichloride (\(\mathrm{NCl}_3\)). The hydrolysis of \(\mathrm{NCl}_3\) involves its reaction with water to form ammonia (\(\mathrm{NH}_3\)) and another product \(\mathrm{X}\). The chemical equation is: \[\mathrm{NCl}_3 + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NH}_3 + 3\mathrm{HOCl}\] In this equation, it's clear that \(\mathrm{X} = \mathrm{HOCl}\).
02

Determine the Product X

Based on the hydrolysis reaction written in Step 1, compare the options given in the problem with the products of the reaction. The product \(\mathrm{X}\) matches the compound \(\mathrm{HOCl}\) which is hypochlorous acid. Thus, the answer can be identified as option (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hydrolysis Reaction
A hydrolysis reaction involves breaking down a compound by reacting it with water. This type of reaction is crucial in many chemical processes and in the breakdown of different compounds. In the context of our exercise, we see hydrolysis applied to nitrogen trichloride, a molecule that reacts with water to form new products.
This reaction typically occurs when the water causes a cleaving of bonds within the compound, leading to the formation of new, generally simpler compounds. The main reactants in a hydrolysis reaction include:
  • A compound (like nitrogen trichloride in our case)
  • Water, which facilitates the breakdown
Through hydrolysis, hydrogen atoms (H) from water often attach to one section of the compound, while hydroxide groups (OH) do the same with another section. This results in the formation of discrete compounds, such as ammonia and hypochlorous acid in our particular instance of hydrolysis.
Nitrogen Trichloride
Nitrogen trichloride ( ext{NCl}_3 ) is a yellow, oily chemical compound known for being highly reactive and explosive. It is formed by treating ammonium salts with chlorine gas. Due to its reactivity, it easily engages in hydrolysis reactions.
When nitrogen trichloride undergoes hydrolysis, it splits, forming ammonia ( ext{NH}_3 ) and hypochlorous acid ( ext{HOCl} ) as the notable products. Key properties of nitrogen trichloride include:
  • Unstable, particularly when subjected to shock or applied pressure
  • Has a strong chlorine smell
  • Dissolves in organic solvents easily
This compound presents hazards due to its explosiveness, requiring careful handling, especially when dealing with it in a laboratory or industrial setting.
Hypochlorous Acid
Hypochlorous acid ( ext{HOCl} ) is an important chemical compound, known for its strong disinfectant properties. It's the active compound in many household bleaches and disinfectants used to sanitize surfaces.
In the exercise, hypochlorous acid emerges as the product of the hydrolysis of nitrogen trichloride. This formation can be understood by its tendency to form when chlorine-containing compounds react with water. Some features of hypochlorous acid include:
  • Effective at killing bacteria, viruses, and fungi
  • Formed in a weak acid solution, showcasing acidity property
  • Used extensively in water treatment facilities and sanitation
Hypochlorous acid is considered safe in diluted forms and is often used for disinfecting drinking water and in healthcare settings to ensure microbial safety.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Of the following sets, which one does not contain isoelectronic species? (a) \(\mathrm{CN}^{-}, \mathrm{N}_{2}, \mathrm{C}_{2}{ }^{2-}\) (b) \(\mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}\) (c) \(\mathrm{SO}_{3}{ }^{2-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\) (d) \(\mathrm{BO}_{3}^{3-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\)

In analogy to \(\mathrm{O}_{2}^{+}\left[\mathrm{PtF}_{6}\right]^{-}\)a compound \(\mathrm{N}_{2}^{+}\left[\mathrm{PtF}_{6}\right] \mathrm{Z}^{-}\), will not be formed because: (a) The ionization of \(\mathrm{N}_{2}\) gas is higher than that of \(\mathrm{N}\) atom (b) The ionization of \(\mathrm{N}_{2}\) gas is lower than that of \(\mathrm{O}_{2}\) gas (c) The ionization of \(\mathrm{N}_{2}^{2}\) gas is higher than that of \(\mathrm{O}_{2}\) gas (d) none of these

Which of the following statement is not correct? (1) the hybridization of \(\mathrm{Xe}\) in \(\mathrm{XeF}_{4}\) is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) (2) \(\mathrm{XeO}_{3}\) has four \(\sigma\) and four \(\pi\) bonds (3) liquid helium is used as cryogenic liquid (4) among noble gases, the occurrence (\% by weight) of argon is higher in air. (a) 1 and 2 (b) 2 and 3 (c) 2 only (d) 3 only

Match the following Match the following Column-I (a) \(\mathrm{XeF}_{2}\) (b) \(\mathrm{XeF}_{4}\) (c) \(\mathrm{XeF}_{6}\) (d) \(\mathrm{XeOF}_{4}\) Column-II (p) One lone pair electron on Xe-atom (q) \(\mathrm{sp}^{3} \mathrm{~d}\) (r) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) (s) \(\mathrm{sp}^{3} \mathrm{~d}^{3}\) (t) \(+6\) oxidation state of xenon

The bond dissociation energy of \(\mathrm{B}-\mathrm{F}\) in \(\mathrm{BF}_{3}\) is 646 \(\mathrm{kJ} \mathrm{mol}^{-1}\) whereas that of \(\mathrm{C}-\mathrm{F}\) in \(\mathrm{CF}_{4}\) is \(515 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The correct reason for higher B - F bond dissociation energy as compared to that of \(\mathrm{C}-\mathrm{F}\) is: (a) stronger bond between \(\mathrm{B}\) and \(\mathrm{F}\) in \(\mathrm{BF}_{3}\) as compared to that between \(\mathrm{C}\) and \(\mathrm{F}\) is \(\mathrm{CF}_{4}\) (b) significant \(\mathrm{p}-\mathrm{p}\) interaction between \(\mathrm{B}\) and \(\mathrm{F}\) in \(\mathrm{BF}_{3}\) whereas there is no possibility of such interaction between \(\mathrm{C}\) and \(\mathrm{F}\) in \(\mathrm{CF}_{4}\) (c) lower degree of \(\mathrm{p}-\mathrm{p}\) interaction between \(\mathrm{B}\) and \(\mathrm{F}\) \(\mathrm{BF}_{3}\) than that between \(\mathrm{C}\) and \(\mathrm{F}\) in \(\mathrm{CF}_{4}\) (d) smaller size of \(\mathrm{B}\) - atom as compared to that of C- atom

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.