Question
The $$Cl - C - Cl$$ angle in 1, 1, 2, 2 - tetrachloroethene and tetrachloromethane respectively will be about
A.
$${120^ \circ }\,\,{\text{and}}\,\,{109.5^ \circ }$$
B.
$${90^ \circ }\,\,{\text{and}}\,\,{109.5^ \circ }$$
C.
$${109.5^ \circ }\,\,{\text{and}}\,\,{90^ \circ }$$
D.
$${109.5^ \circ }\,\,{\text{and}}\,\,{120^ \circ }$$
Answer :
$${120^ \circ }\,\,{\text{and}}\,\,{109.5^ \circ }$$
Solution :
The bond angle in $$s{p^3},s{p^2}$$ and $$sp$$ hybridization is respectively $$109.28',{120^ \circ }$$ and $${180^ \circ }.$$
Tetrachloroethene being an alkene has $$s{p^2}$$ hybridised $$C $$ - atoms and hence the $$Cl - C - Cl$$ angle is $${120^ \circ },$$ whereas in tetrachloromethane, carbon is $$s{p^3}$$ hybridised, so the angle is $${109^ \circ }.28'.$$
The bond angle in $$s{p^3},s{p^2}$$ and $$sp$$ hybridization is respectively $$109.28',{120^ \circ }$$ and $${180^ \circ }.$$
Tetrachloroethene being an alkene has $$s{p^2}$$ hybridised $$C $$ - atoms and hence the $$Cl - C - Cl$$ angle is $${120^ \circ },$$ whereas in tetrachloromethane, carbon is $$s{p^3}$$ hybridised, so the angle is $${109^ \circ }.28'.$$