
Why $ \text{Cl}{{\text{F}}_{\text{3}}} $ exists but $ \text{FC}{{\text{l}}_{\text{3}}} $ does not?
Answer
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Hint: A molecule with two or more distinct halogen atoms (fluorine, chlorine, bromine, iodine, or astatine) and no atoms from any other group is called an interhalogen compound. The majority of known interhalogen chemicals are binary. In general, their formulas are $ X{{Y}_{n}} $ , where n = 1, 3, 5, or 7, and X is the less electronegative of the two halogens. Because halogens have odd valances, the value of n in interhalogens is always odd. They're all susceptible to hydrolysis and ionisation, resulting in polyhalogen ions. Because astatine is very radioactive, those produced with it have a very short half-life.
Complete Step By Step Answer:
$ \text{Cl}{{\text{F}}_{\text{3}}} $ is the formula for chlorine trifluoride, an interhalogen chemical. This colourless, toxic, corrosive, and highly reactive gas condenses into a pale-greenish yellow liquid, which is the most common form in which it is marketed (pressurized at room temperature). The chemical is mainly used in rocket fuels, plasma less cleaning and etching procedures in the semiconductor sector, nuclear reactor fuel processing, and other industrial processes.
There are two main reasons for this: Cl is a large atom, while F is a tiny one. As a result, whereas Cl can contain three F atoms, F cannot.
Cl does not have d orbitals, whereas F does.
Because all halogens have a $ {{p}^{5}} $ configuration, each atom must unpair its paired electrons and promote it to the d orbital for increased valency. In the absence of this, F cannot have a valency greater than one, preventing the creation of $ \text{FC}{{\text{l}}_{\text{3}}} $ .
Chlorine possesses vacant d-orbitals, which get excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.
Due to the lack of vacant d-orbitals in the 2nd energy shell, fluorine cannot extend its octet.
As a result, $ \text{Cl}{{\text{F}}_{\text{3}}} $ exists but $ \text{FC}{{\text{l}}_{\text{3}}} $ does not.
Note:
Chlorine trifluoride is used to clean chemical vapour deposition chambers in the semiconductor sector. It has the benefit of not requiring the chamber to be dismantled in order to remove semiconductor material from the walls. Unlike most other compounds employed in this capacity, it does not require plasma activation since the heat of the chamber is sufficient to cause it to break down and react with the semiconductor material.
Complete Step By Step Answer:
$ \text{Cl}{{\text{F}}_{\text{3}}} $ is the formula for chlorine trifluoride, an interhalogen chemical. This colourless, toxic, corrosive, and highly reactive gas condenses into a pale-greenish yellow liquid, which is the most common form in which it is marketed (pressurized at room temperature). The chemical is mainly used in rocket fuels, plasma less cleaning and etching procedures in the semiconductor sector, nuclear reactor fuel processing, and other industrial processes.
There are two main reasons for this: Cl is a large atom, while F is a tiny one. As a result, whereas Cl can contain three F atoms, F cannot.
Cl does not have d orbitals, whereas F does.
Because all halogens have a $ {{p}^{5}} $ configuration, each atom must unpair its paired electrons and promote it to the d orbital for increased valency. In the absence of this, F cannot have a valency greater than one, preventing the creation of $ \text{FC}{{\text{l}}_{\text{3}}} $ .
Chlorine possesses vacant d-orbitals, which get excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.
Due to the lack of vacant d-orbitals in the 2nd energy shell, fluorine cannot extend its octet.
As a result, $ \text{Cl}{{\text{F}}_{\text{3}}} $ exists but $ \text{FC}{{\text{l}}_{\text{3}}} $ does not.
Note:
Chlorine trifluoride is used to clean chemical vapour deposition chambers in the semiconductor sector. It has the benefit of not requiring the chamber to be dismantled in order to remove semiconductor material from the walls. Unlike most other compounds employed in this capacity, it does not require plasma activation since the heat of the chamber is sufficient to cause it to break down and react with the semiconductor material.
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