Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the properties of interstitial hydrides is correct? (a) They give rise to metals fit for fabrication (b) They generally form non-stoichiometric species (c) They can be used as hydrogenation catalysts (d) The hydrogen dissolved in titanium improves its mechanical properties

Short Answer

Expert verified
The correct property is that interstitial hydrides generally form non-stoichiometric species (option b).

Step by step solution

01

Understanding Interstitial Hydrides

Interstitial hydrides, also known as metallic hydrides, are hydrogen compounds where hydrogen atoms occupy the interstitial spaces (gaps) within the metal lattice. These hydrides are typically formed by transition metals.
02

Analyzing Option (a)

Option (a) states that interstitial hydrides give rise to metals fit for fabrication. While some hydrides might alter metallic properties, interstitial hydrides are generally not directly used for fabrication due to their variable composition.
03

Analyzing Option (b)

Option (b) states that interstitial hydrides generally form non-stoichiometric species. This is correct because the number of hydrogen atoms in the lattice can vary, leading to non-stoichiometric compounds where the hydrogen-to-metal ratio is not fixed.
04

Analyzing Option (c)

Option (c) suggests that interstitial hydrides can be used as hydrogenation catalysts. While some transition metals are used as catalysts, this does not specifically apply to interstitial hydrides themselves.
05

Analyzing Option (d)

Option (d) states that the hydrogen dissolved in titanium improves its mechanical properties. In fact, hydrogenation often embrittles metals so this is generally incorrect.
06

Conclusion

After evaluating all options, option (b) is the correct property of interstitial hydrides as they are known to be non-stoichiometric.

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 Vaia!

Key Concepts

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

Non-stoichiometric Species
Non-stoichiometric species are compounds where the ratio of the constituent elements does not adhere to simple whole numbers. This often happens in materials like interstitial hydrides. In these compounds, hydrogen atoms add to a metal lattice in varied amounts. As a result, the hydrogen-to-metal ratios can fluctuate. This variability makes them non-stoichiometric. Non-stoichiometric compounds offer unique properties due to this irregularity. For example, their electrical or magnetic behaviors can be fine-tuned by changing the concentration of one of their components. In industrial and scientific realms, this flexibility makes them very useful. Not sticking to fixed ratios allows for greater versatility in their applications.
Hydrogen Compounds
Hydrogen compounds are chemical compounds where hydrogen bonds with other elements. These can range from simple molecules like water ( H_2O ) to more complex structures like hydrocarbons ( C_nH_m ). An interesting subset of hydrogen compounds is interstitial hydrides. Here, hydrogen atoms squeeze into the tiny gaps within metal lattices.
  • These hydrogen compounds can alter the metal’s properties significantly.
  • They allow the metal to absorb and release hydrogen easily.
Interstitial hydrides are noted for their ability to store large quantities of hydrogen, making them important in energy storage technologies. They also play a pivotal role in various industrial processes. Thus, understanding hydrogen compounds is crucial to harnessing the benefits of these material properties.
Transition Metals
Transition metals are elements found in the middle portion of the periodic table. They are known for their ability to form compounds with variable oxidation states and for their colorful appearance. Transition metals play a central role in forming interstitial hydrides. The loosely packed atoms in their metallic lattice permit hydrogen atoms to occupy spaces. This feature makes them suitable for the creation of non-stoichiometric hydrides. Some key properties of transition metals include:
  • High melting and boiling points, due to strong metallic bonding.
  • Excellent conductors of electricity, as their electrons can move freely.
  • Act as catalysts in many chemical reactions, thanks to their complex electron configurations.
The ability to form such active compounds makes transition metals critical in many modern technologies and industrial applications.

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

Which of the following is not correct? (a) Nitric oxide in solid state exhibits diamagnetic property. (b) \(\mathrm{NH}_{3}\) is a weak reducing agent compared to \(\mathrm{PH}_{3}\) (c) Hydrolysis of \(\mathrm{NCl}_{3}\) gives \(\mathrm{NH}_{3}\) and \(\mathrm{HOCl}\) (d) \(\mathrm{NH}_{3}\) is less stable than \(\mathrm{PH}_{3}\)

In context with the industrial preparation of hydrogen from water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\), which of the following is the correct statement? (a) \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) are fractionally separated using differences in their densities (b) \(\mathrm{CO}\) is removed by absorption in aqueous \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\) solution (c) \(\mathrm{H}_{2}\) is removed through occlusion with \(\mathrm{Pd}\) (d) \(\mathrm{CO}\) is oxidized to \(\mathrm{CO}_{2}\) with steam in the presence of a catalyst followed by absorption of \(\mathrm{CO}_{2}\) in alkali.

Find \([\mathrm{A}],[\mathrm{B}]\) and \([\mathrm{C}]\) respectively in the reaction given below: \([\mathrm{A}]+\mathrm{Na}\left[\mathrm{Cr}(\mathrm{OH})_{4}\right] \stackrel{\mathrm{H}_{2} \mathrm{O}_{2}, \text { boil }}{\longrightarrow}[\mathrm{B}] \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{\text { Yellow }}-[\mathrm{C}]\) compound (a) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{Na}_{2} \mathrm{CrO}_{4}\) (b) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{NaOH}, \mathrm{K}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

The percentage by weight of hydrogen in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is (a) \(6.5\) (b) \(5.88\) (c) 25 (d) 50

Sugarcane on reaction with nitric acid gives (a) \(2 \mathrm{HCOOH}\) (two moles) (b) \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) (c) no reaction (d) \((\mathrm{COOH})_{2}\)

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.

Sign-up for free