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The element which can displace three other halogens from their compound is : (a) \(\mathrm{F}\) (b) \(\mathrm{Cl}\) (c) \(\mathrm{Br}\) (d) I

Short Answer

Expert verified
The element that can displace three other halogens from their compounds is Fluorine (F), option (a).

Step by step solution

01

Understanding the Halogen Displacement Series

In the halogen group of the periodic table, the reactivity of the elements decreases as you move down the group. This means a more reactive halogen can displace a less reactive one from its compounds. The order of reactivity, from most to least, is Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).
02

Determining the Most Reactive Halogen

Since the question asks which halogen can displace three other halogens from their compounds, we are looking for the most reactive halogen. This is because only the most reactive halogen would be capable of displacing all the other three that come after it in the reactivity series.
03

Identifying the Correct Halogen

From the reactivity series established, we determine that Fluorine (F) is the most reactive halogen. Therefore, it has the ability to displace Chlorine (Cl), Bromine (Br), and Iodine (I) from their respective compounds.

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Key Concepts

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

Chemical Reactivity in Halogens
Halogens, found in Group 17 of the periodic table, include elements like fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). Understanding their chemical reactivity is key to many applications in chemistry.

Reactivity in halogens is largely dictated by their ability to gain electrons and form negative ions called halides. Due to their position on the periodic table, halogens have seven electrons in their outermost shell and are just one electron short of achieving a stable, 'full' electronic configuration. This makes them highly eager to gain an electron through reactions.

Electron Affinity and Reactivity

Fluorine, being the smallest halogen atom, exhibits the greatest attraction for electrons – a property known as electron affinity. This strong pull makes fluorine the most reactive halogen. As we move down the group (from F to I), the atoms increase in size, and the outer electrons experience a weaker attraction from the nucleus. Consequently, the reactivity decreases down the group.

Factors Influencing Reactivity

Several factors contribute to this trend of decreasing reactivity:
  • Electron shielding effect – inner shells of electrons block the outer electrons from the nucleus' pull, reducing the atom's ability to gain an electron.
  • Atomic radius – larger atomic size dilutes the nucleus' ability to pull on electrons from other atoms.
  • Electronegativity – the measure of an atom's ability to attract and hold onto electrons decreases down the group.
Periodic Table Trends
The periodic table is a systematic arrangement of the elements, allowing us to predict an element's behavior based on its position. As we explore the periodic table, we notice trends in properties, such as atomic size, electronegativity, ionization energy, and metallic character.

From Top to Bottom

Moving from top to bottom within a group, there is an increase in atomic radius and a general decrease in ionization energy, meaning it becomes easier for an atom to release an electron. Electronegativity also decreases, explaining why elements like fluorine are far more reactive than those like iodine. The opposite trends are observed moving from left to right across a period, exhibiting a decrease in atomic radius and an increase in ionization energy and electronegativity.

The periodic trends are essential in predicting an element's chemical reactivity and its ability to interact with other elements. These properties help chemists understand why certain reactions occur and forecast new ones. For the halogens, these trends explain their varying capacities to displace one another in chemical compounds.
Reactivity Series
A reactivity series is a ranking of elements according to their reactivity. In the context of halogens, the reactivity series orders them by their ability to displace each other from compounds. This series is pivotal in understanding reactions, such as single-replacement reactions, where one element replaces another in a compound.

The reactivity series for halogens, starting with the most reactive, is: fluorine, chlorine, bromine, and iodine. The sequence helps predict outcomes when these elements come into contact with each other in compounds. For instance, a more reactive halogen can displace a less reactive halogen from its salts.

Practical Implications

Knowledge of the halogen reactivity series is beneficial in many chemical processes, including water purification, where chlorine is used to displace less reactive impurities. Moreover, understanding this series allows us to predict which halogens can bond with particular metals, aiding in the formation of specific compounds used in pharmaceuticals and industrial applications.
  • Fluorine's superior reactivity lets it react with almost all other elements.
  • Chlorine is less reactive than fluorine but can displace bromine and iodine.
  • Bromine, being less reactive than both fluorine and chlorine, can only displace iodine.
  • Iodine is the least reactive and ordinarily does not displace other halogens.

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Most popular questions from this chapter

For the electrochemical cell \(\operatorname{Pt}(s) \mid \begin{aligned}&\mathrm{H}_{2}(g)\left|\mathrm{H}^{+}(1 M) \| \mathrm{Cu}^{2+}(1 M)\right| \mathrm{Cu}(s), \text { which one of the } \\\&1 \text { atm }\end{aligned}\) following statements is true ? (a) \(\mathrm{H}_{2}\) liberated at anode and \(\mathrm{Cu}\) is deposite at cathode. (b) \(\mathrm{H}_{2}\) liberated at cathode and \(\mathrm{Cu}\) is deposite at anode. (c) Oxidation occurs at cathode. (d) Reduction occurs at anode.

Resistance of a decimolar solution between two electrodes \(0.02\) meter apart and \(0.0004 \mathrm{~m}^{2}\) in \(\cdots\) area was found to be 50 ohm. Specific conductance \((\kappa)\) is : (a) \(0.1 \mathrm{Sm}^{-1}\) (b) \(1 \mathrm{~S} \mathrm{~m}^{-1}\) (c) \(10 \mathrm{~S} \mathrm{~m}^{-1}\) (d) \(4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{-1}\)

The \(E^{\circ}\) for the following cell is \(+0.34 \mathrm{~V} . \operatorname{In}(s)\left|\ln (\mathrm{OH})_{3}(a q) \| \mathrm{SbO}_{2}^{-}(a q)\right| \mathrm{Sb}(s)\). Using \(E^{\circ}=-1.0 \mathrm{~V}\) for the \(\operatorname{In}(\mathrm{OH})_{3} \mid\) In couple, calculate \(E^{\circ}\) for the \(\mathrm{SbO}_{2}^{-} \mid \mathrm{Sb}\) half-reaction : (a) \(-1.34\) (b) \(+0.66\) (c) \(+0.82\) (d) \(-0.66\)

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