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

The oxidation state of sodium in sodium amalgam is (a) \(-1\) (b) \(+1\) (c) \(+2\) (d) 0

Short Answer

Expert verified
(d) 0

Step by step solution

01

Understanding Sodium Amalgam

Sodium amalgam is an alloy of sodium and mercury. In alloys, the elements are mixed together physically, not chemically bonded in a way that typically changes oxidation states.
02

Determining Oxidation State

In elemental form or in simple mixtures such as alloys, the elements retain their elemental oxidation state. Sodium, in its elemental form, has an oxidation state of 0.
03

Final Conclusion

Given that sodium in sodium amalgam remains in its elemental form and does not form a specific compound altering its oxidation state, sodium in sodium amalgam retains its elemental oxidation state, which is 0.

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.

Sodium Amalgam
Sodium amalgam is an interesting combination of metals, specifically sodium and mercury. It's important to understand that when these two elements come together to form an amalgam, they don't create a new chemical compound. Instead, they form a physical mixture. Think of it as a blend, where both elements maintain their identity while exhibiting a uniform composition.
  • Being an alloy, sodium amalgam is utilized in a variety of applications, including as a reducing agent in chemical synthesis.
  • The properties of sodium amalgam can be quite different from each of its components, but its fundamental chemistry is reflected in its composition.
Understanding the concept of sodium amalgam is essential when determining how elements behave within this mixture.
Elemental Form
The concept of the "elemental form" is key to understanding how oxidation states work in mixtures like sodium amalgam. When elements are in their pure, uncombined state, we refer to them as being in their "elemental form." This is crucial when considering their oxidation states.
In the case of sodium amalgam, sodium doesn't chemically bind with mercury. Instead, it stays in its elemental form within the alloy.
  • An element in its elemental form naturally has an oxidation state of 0.
  • This rule helps simplify many chemistry problems involving metals and other non-chemically bound mixtures.
This principle allows us to easily determine the oxidation state of sodium when in an amalgam with mercury.
Alloys
Alloys are fascinating materials that result from mixing two or more metals, like sodium and mercury in sodium amalgam. They are widely used due to their enhanced characteristics compared to their individual metal components.
Alloys can exhibit improved properties such as increased strength, ductility, and corrosion resistance, depending on their composition.
  • They inherently retain the chemical properties of the individual metals, especially in terms of oxidation states.
  • When determining oxidation states within alloys, it is important to recognize that the individual metals maintain their elemental states, making it easier to analyze their chemical behavior.
The unique nature of alloys highlights the distinct relationship between physical mixing and chemical bonding.
Oxidation State Determination
Oxidation state determination is a fundamental concept in understanding how elements interact within compounds and mixtures. The oxidation state is a figure that indicates the degree of oxidation or reduction an element has undergone.
For simple mixtures like alloys—where elements don't form new compounds but exist side by side—the elements retain their original elemental oxidation states, typically 0.
  • This is particularly useful because it shortens the process of determining oxidation states in complex mixtures.
  • In sodium amalgam, since sodium remains in its elemental form, its oxidation state remains 0, aligning with fundamental oxidation state rules.
Grasping these principles equips students with the tools to accurately resolve typical oxidation state challenges in chemistry.

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

The increase in boiling points of noble gases from He to Xe is due to the (a) increase in atomic volume (b) increase in electron affinity (c) increase in polarizability (d) decrease in ionization energy

Noble gases can act as (a) reducing agents (b) oxidizing agents (c) both of these (d) none of these

Among the following, the correct statement is (a) Between \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\) is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional (b) Between \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}, \mathrm{PH}_{3}\) is a better electron donor because the lone pair of electrons occupies \(s p^{3}\) orbital and is more directional (c) Between \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\) is a better electron donor because the lone pair of electrons occupies \(s p^{3}\) orbital and is less directional

The van der Waals forces in halogen decrease in the order (a) \(\mathrm{I}_{2}>\mathrm{Br}_{2}>\mathrm{Cl}_{2}>\mathrm{F}_{2}\) (b) \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) (c) \(\mathrm{I}_{2}>\mathrm{F}_{2}>\mathrm{Br}_{2}>\mathrm{Cl}_{2}\) (d) \(\mathrm{Cl}_{2}>\mathrm{I}_{2}>\mathrm{Br}_{2}>\mathrm{F}_{2}^{2}\)

Which of the following are arranged in correct increasing order of solubilities? (a) \(\mathrm{KHCO}_{3}<\mathrm{NaHCO}_{3}<\mathrm{CaCO}_{3}\) (b) \(\mathrm{CaCO}_{3}<\mathrm{KHCO}_{3}<\mathrm{NaHCO}_{3}\) (c) \(\mathrm{CaCO}_{3}<\mathrm{NaHCO}_{3}<\mathrm{KHCO}_{3}\) (d) \(\mathrm{NaHCO}_{3}<\mathrm{KHCO}_{3}<\mathrm{CaCO}_{3}\)

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