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Hint: Coplanar objects are the objects which lie on the same plane whereas non-coplanar objects are the objects which do not lie in the same plane.
Complete step-by-step answer:
>In this compound, C=C is in-plane and the cyanide group attached to it will also be in the same plane but the ${ CH }_{ 2 }$ the group is ${ sp }^{ 3 }$ hybridized and the hydrogens are present above and below the plane. Hence, this compound will be non-planer.
>In this compound, all the atoms are ${ sp }^{ 3 }$ hybridized, so they have tetrahedral geometry. We know that in tetrahedral geometry, all the atoms are not in one plane. So, this compound is non-planer.
>In Biphenyl, all the carbon atoms are ${ sp }^{ 2 }$ hybridized. It undergoes nuclear substitution reaction, as one of the phenyl rings acts as electron-acceptor while the other will act as an electron-releasing group. Since all the hydrogens present are in one plane, this compound is coplanar.
>In this compound, all the atoms are ${ sp }^{ 3 }$ hybridized, so they have tetrahedral geometry. We know that in tetrahedral geometry, all the atoms are not in one plane. So, this compound is non-planer.
Hence, the correct option is C.
Note: The possibility to make a mistake is that you may choose option B. But in that compound, the carbon atoms are ${ sp }^{ 3 }$ hybridized, not ${ sp }^{ 2 }$ so they don’t lie in the same plane.
Complete step-by-step answer:
>In this compound, C=C is in-plane and the cyanide group attached to it will also be in the same plane but the ${ CH }_{ 2 }$ the group is ${ sp }^{ 3 }$ hybridized and the hydrogens are present above and below the plane. Hence, this compound will be non-planer.
>In this compound, all the atoms are ${ sp }^{ 3 }$ hybridized, so they have tetrahedral geometry. We know that in tetrahedral geometry, all the atoms are not in one plane. So, this compound is non-planer.
>In Biphenyl, all the carbon atoms are ${ sp }^{ 2 }$ hybridized. It undergoes nuclear substitution reaction, as one of the phenyl rings acts as electron-acceptor while the other will act as an electron-releasing group. Since all the hydrogens present are in one plane, this compound is coplanar.
>In this compound, all the atoms are ${ sp }^{ 3 }$ hybridized, so they have tetrahedral geometry. We know that in tetrahedral geometry, all the atoms are not in one plane. So, this compound is non-planer.
Hence, the correct option is C.
Note: The possibility to make a mistake is that you may choose option B. But in that compound, the carbon atoms are ${ sp }^{ 3 }$ hybridized, not ${ sp }^{ 2 }$ so they don’t lie in the same plane.
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