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

Regarding \(\mathrm{F}^{-}\)and \(\mathrm{Cl}^{-}\)which of the following statements is correct? 1\. Cl can give up an electron more easily than \(\mathrm{F}^{-}\) 2\. \(\mathrm{Cl}^{-}\)is a better reducing agent than \(\mathrm{F}^{-}\) 3\. \(\mathrm{Cl}^{-}\)is smaller in size than \(\mathrm{F}^{-}\) 4\. \(\mathrm{F}^{-}\)can be oxidized more readily than \(\mathrm{Cl}^{-}\) (a) 1,2 and 4 (b) 1 only (c) 1 and 2 (d) 3 and 4

Short Answer

Expert verified
Correct options are 1 and 2 (Choice C).

Step by step solution

01

Analyze Oxidation Trends

Analyzing the tendency for oxidation, we know that the ability of a species to lose electrons is linked to being oxidized. Both \( \text{F}^- \) and \( \text{Cl}^- \) are anions, negatively charged atoms, meaning they have extra electrons compared to their neutral atoms. \( \text{F}^- \) is smaller and holds its electrons more tightly due to higher electronegativity, making \( \text{Cl}^- \) the better candidate to give up an electron and thus to be oxidized more easily.
02

Evaluate Reducing Agent Ability

A reducing agent is a species capable of donating electrons to another species. \( \text{Cl}^- \), with its weaker hold on its outer electron compared to \( \text{F}^- \), can more readily donate an electron, making \( \text{Cl}^- \) a better reducing agent than \( \text{F}^- \).
03

Consider Atomic Size

When considering the size of ions, \( \text{F}^- \) is generally smaller in size than \( \text{Cl}^- \). This trend fits with periodic trends where the size increases down the group (with \( \text{Cl}^- \) being below \( \text{F}^- \) in the same group on the periodic table).
04

Final Assessment of Statements

Based on steps 1-3, recall the correct comparisons: \( \text{Cl}^- \) can give up an electron more easily and is a better reducing agent than \( \text{F}^- \), and \( \text{F}^- \) is indeed smaller than \( \text{Cl}^- \). Hence, statements 1 and 2 are correct, while statement 4, which suggests \( \text{F}^- \) can be oxidized more readily than \( \text{Cl}^- \), is incorrect.

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.

Oxidation Trends
Oxidation refers to the process where an atom or ion loses electrons. In the world of chemistry, this is an essential concept, especially when evaluating anions like \( \text{F}^- \) and \( \text{Cl}^- \). These ions have gained an electron compared to their neutral forms, making them negatively charged.

To determine which ion is more likely to undergo oxidation, we need to look at their electronegativity and size. Fluoride ions (\( \text{F}^- \)) are smaller and have a higher electronegativity compared to chloride ions (\( \text{Cl}^- \)). This means \( \text{F}^- \) holds its additional electron more tightly and is less likely to give it up. Hence, \( \text{Cl}^- \) is more prone to oxidation as it can lose its electron more easily than \( \text{F}^- \).
  • \( \text{Oxidation:} \) Loss of electrons.
  • \( \text{High electronegativity:} \) Provides stronger electron hold
  • \( \text{\( \text{Cl}^- \) is more oxidizable than \( \text{F}^- \):} \) \( \text{Cl}^- \) is larger, holds its electron less tightly.
Reducing Agents
A reducing agent is a substance that donates electrons, causing another substance to be reduced. This donating capability makes them crucial players in redox reactions.

In comparing \( \text{F}^- \) and \( \text{Cl}^- \), the ability to act as reducing agents is determined by how readily they can give away electrons. Due to its larger size and weaker electron holding capacity, \( \text{Cl}^- \) acts as a better reducing agent than \( \text{F}^- \). The extra space in \( \text{Cl}^- \)'s electron cloud allows easier electron donation.
  • \( \text{Reducing Agent:} \) Donates electrons.
  • \( \text{\( \text{Cl}^- \) as a reducing agent:} \) Larger, with a weaker electron hold.
  • \( \text{Electron donation capability:} \) Higher in \( \text{Cl}^- \) than \( \text{F}^- \).
Atomic Size
Atomic size, or ionic radius, is another crucial factor when discussing chemical behavior of anions like \( \text{F}^- \) and \( \text{Cl}^- \). This characteristic influences many properties, including oxidation and reducing potential.

Within the periodic table, atomic size increases as you move down a group. This puts \( \text{Cl}^- \) at a larger size compared to \( \text{F}^- \) since chlorine is located below fluorine. The extra electron shells in \( \text{Cl}^- \) result in a larger overall ion size.
  • \( \text{Atomic size increases down a group:} \) More electron shells added.
  • \( \text{\( \text{Cl}^- \) is larger than \( \text{F}^-\):} \) Due to additional electron shell.
  • \( \text{Size influences reactivity:} \) Larger ions often have more significant reaction tendencies in redox processes.
Periodic Trends
Understanding periodic trends is indispensable for grasping chemistry's big picture. These trends help explain why elements behave differently, including their oxidation and size characteristics.

Elements in the same group of the periodic table show consistent trends in properties. For instance, as you progress down a group:
  • \( \text{Atomic size increases:} \) Each element has an additional electron shell.
  • \( \text{Electronegativity decreases:} \) Larger atoms have less electromagnetic pull on their electrons.
  • \( \text{Ionization energy decreases:} \) Easier to remove electrons from larger atoms.
For \( \text{F}^- \) and \( \text{Cl}^- \):
  • \( \text{\( \text{F}^- \) is smaller and more electronegative than \( \text{Cl}^- \)} \)
  • \( \text{\( \text{Cl}^- \) is larger, has lower ionization energy, thus is better at electron donation.} \)
These periodic trends shed light on their oxidation capabilities, where \( \text{Cl}^- \) is more likely to lose an electron compared to the tightly held extra electron in \( \text{F}^- \). Knowing these patterns helps us predict and understand chemical behaviors across the periodic table.

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 represents the correct order of increasing first ionization enthalpy for \(\mathrm{Ca}, \mathrm{Ba}, \mathrm{S}, \mathrm{Se}\) and Ar? (a) \(\mathrm{Ba}<\mathrm{Ca}<\mathrm{Se}<\mathrm{S}<\mathrm{Ar}\) (b) \(\mathrm{Ca}<\mathrm{Ba}<\mathrm{S}<\mathrm{Sr}<\mathrm{Ar}\) (c) \(\mathrm{Ca}<\mathrm{S}<\mathrm{Ba}<\mathrm{Se}<\mathrm{Ar}\) (d) \(\mathrm{S}<\mathrm{Se}<\mathrm{Ca}<\mathrm{Ba}<\mathrm{Ar}\)

Which one of the following ion has the highest value of ionic radius? (a) \(\mathrm{F}^{-}\) (b) \(\mathrm{O}^{2-}\) (c) \(\mathrm{B}^{3+}\) (d) \(\mathrm{Li}^{+}\)

What is the order of basic nature of hydrides of group \(\mathrm{V}\) elements? (a) \(\mathrm{AsH}_{3}>\mathrm{SbH}_{3}>\mathrm{PH}_{3}>\mathrm{NH}_{3}\) (b) \(\mathrm{NH}_{3}>\mathrm{SbH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}\) (c) \(\mathrm{NH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3}\) (d) \(\mathrm{PH}_{3}>\mathrm{NH}_{3}>\mathrm{SbH}_{3}>\mathrm{AsH}_{3}\)

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) \(\mathrm{NaHCO}_{3}<\mathrm{KHCO}_{3}<\mathrm{CaCO}_{3}\)

Which of the following is the correct relation with regard to the electron afinity of halogen? (a) \(\mathrm{F}>\mathrm{Cl}\) (b) \(\mathrm{F}>\mathrm{I}\) (c) \(\mathrm{Br}>\mathrm{F}\) (d) \(\mathrm{Br}>\mathrm{Cl}\)

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