
Among the fluorides below, the one which does not exist is:
A) \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\]
B) \[{\rm{He}}{{\rm{F}}_{\rm{4}}}\]
C) \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\]
D) \[{\rm{C}}{{\rm{F}}_{\rm{4}}}\]
Answer
163.8k+ views
Hint: We know that fluorine is a halogen. All halogens are positioned in the periodic table’s 17th group. They have one less valence electron in their outermost shell. Therefore, they require only one electron to complete its octet and attain stability.
Complete Step by Step Solution:
As stated above, due to the need for only one electron by the halogen atom to attain stability, they form covalent bonds with other atoms easily. Here, we have to identify the compound of fluorine that does not exist. Let’s discuss all the options one by one.
Option A is \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\]. Among the noble gases, xenon forms compounds with oxygen and fluorine. This is because of the larger size of the xenon atom. And, due to the large size, there is a weak attraction of the electrons to the nucleus. Therefore, existence of \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\] is possible.
Option C is \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\]. The Sulphur atom has six valence electrons. So, four fluorine atoms can form fir covalent bonds with the sulphur atom. So, the existence of \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\] is also possible.
Option D is \[{\rm{C}}{{\rm{F}}_{\rm{4}}}\]. The carbon atom has four numbers of valence electrons. So, four fluorine atoms can form four covalent bonds with the carbon atom. Therefore, the existence of \[{\rm{C}}{{\rm{F}}_{\rm{4}}}\] is also possible.
Option B is \[{\rm{He}}{{\rm{F}}_{\rm{4}}}\]. We know that helium has two electrons in the valence shell which is completely filled. Therefore, it cannot form compounds. Therefore, existence of \[{\rm{He}}{{\rm{F}}_{\rm{4}}}\] is not possible.
Hence, option B is right.
Note: Noble gas are elements that are positioned in the periodic table's 18th group. They are also named inter gases. Some examples as helium, argon, neon, xenon, Krypton, etc. are noble gases. Among the halogens, only xenon forms a group of compounds and others exist as monatomic molecules.
Complete Step by Step Solution:
As stated above, due to the need for only one electron by the halogen atom to attain stability, they form covalent bonds with other atoms easily. Here, we have to identify the compound of fluorine that does not exist. Let’s discuss all the options one by one.
Option A is \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\]. Among the noble gases, xenon forms compounds with oxygen and fluorine. This is because of the larger size of the xenon atom. And, due to the large size, there is a weak attraction of the electrons to the nucleus. Therefore, existence of \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\] is possible.
Option C is \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\]. The Sulphur atom has six valence electrons. So, four fluorine atoms can form fir covalent bonds with the sulphur atom. So, the existence of \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\] is also possible.
Option D is \[{\rm{C}}{{\rm{F}}_{\rm{4}}}\]. The carbon atom has four numbers of valence electrons. So, four fluorine atoms can form four covalent bonds with the carbon atom. Therefore, the existence of \[{\rm{C}}{{\rm{F}}_{\rm{4}}}\] is also possible.
Option B is \[{\rm{He}}{{\rm{F}}_{\rm{4}}}\]. We know that helium has two electrons in the valence shell which is completely filled. Therefore, it cannot form compounds. Therefore, existence of \[{\rm{He}}{{\rm{F}}_{\rm{4}}}\] is not possible.
Hence, option B is right.
Note: Noble gas are elements that are positioned in the periodic table's 18th group. They are also named inter gases. Some examples as helium, argon, neon, xenon, Krypton, etc. are noble gases. Among the halogens, only xenon forms a group of compounds and others exist as monatomic molecules.
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