
Which of the following has a linear structure?
A) \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]
B) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]
C) \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\]
D) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_4}\]
Answer
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Hint: To check if the molecular structure of the given compounds is linear or not, first we have to check the hybridization of the compound. If the hybridization of the compound is sp, then the molecule has a linear shape.
Complete step by step solution:Let's understand how to check hybridization of an atom. First, we have to check the number of surrounding atoms. The surrounding groups are atoms bonded to the atom and the lone pairs of the atom.
Let's discuss all the options one by one.
Option A is \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]. Here, four chlorine atoms are bonded to the carbon atom and there is no lone pair present on the carbon atom. So, hybridization of \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\] is \[s{p^3}\] . So, its structure is tetrahedral.
Option B is \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]. Its structure is \[{\rm{HC}} \equiv {\rm{CH}}\] . C1 and C2 both are bonded to two atoms. So, the hybridization of C atoms in ethyne is \[sp\] . So, the molecular shape of \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]is linear.
Option C is \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\] . Here, two oxygen atoms are bonded to the S atom. And the count of lone pairs on the s-atom is two. So, the total count of groups surrounding the sulfur atom is 4. So, hybridization of S is \[s{p^3}\] .
Option D is \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_4}\]. Its structure is \[{{\rm{H}}_{\rm{2}}}{\rm{C}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\] . The hybridization of both the carbon atoms is 3. So, its hybridization is \[s{p^2}\] . So, its shape is trigonal planar.
Hence, option B is right.
Note: It is to be noted that, if the count of surrounding groups is 4, then the hybridization is \[s{p^3}\] . If the count of the surrounding groups is 3, hybridization is \[s{p^2}\]and if the count of surrounding groups is 2, the hybridization is \[sp\].
Complete step by step solution:Let's understand how to check hybridization of an atom. First, we have to check the number of surrounding atoms. The surrounding groups are atoms bonded to the atom and the lone pairs of the atom.
Let's discuss all the options one by one.
Option A is \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]. Here, four chlorine atoms are bonded to the carbon atom and there is no lone pair present on the carbon atom. So, hybridization of \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\] is \[s{p^3}\] . So, its structure is tetrahedral.
Option B is \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]. Its structure is \[{\rm{HC}} \equiv {\rm{CH}}\] . C1 and C2 both are bonded to two atoms. So, the hybridization of C atoms in ethyne is \[sp\] . So, the molecular shape of \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]is linear.
Option C is \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\] . Here, two oxygen atoms are bonded to the S atom. And the count of lone pairs on the s-atom is two. So, the total count of groups surrounding the sulfur atom is 4. So, hybridization of S is \[s{p^3}\] .
Option D is \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_4}\]. Its structure is \[{{\rm{H}}_{\rm{2}}}{\rm{C}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\] . The hybridization of both the carbon atoms is 3. So, its hybridization is \[s{p^2}\] . So, its shape is trigonal planar.
Hence, option B is right.
Note: It is to be noted that, if the count of surrounding groups is 4, then the hybridization is \[s{p^3}\] . If the count of the surrounding groups is 3, hybridization is \[s{p^2}\]and if the count of surrounding groups is 2, the hybridization is \[sp\].
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