Respuesta :
Correct Answer: CH2CCI2 and CH2CH2.
Reason:
Structure of compound depends on the hybridization of central atom. In both of the above molecules (i.e. CH2CCL2 and CH2CH2). Central atom i.e. carbon is sp^2 hybridized. Thus 3 hybridized orbitals of 'C' is involved in sigma bonding. Further, these orbitals are placed spatially at an angle of [tex] 120^{0} [/tex]. The un-hybridized 2pz orbital of 'C' is involved in pi-bonding. Due to this, both of the above molecules possess a triangular planner geometry.
Reason:
Structure of compound depends on the hybridization of central atom. In both of the above molecules (i.e. CH2CCL2 and CH2CH2). Central atom i.e. carbon is sp^2 hybridized. Thus 3 hybridized orbitals of 'C' is involved in sigma bonding. Further, these orbitals are placed spatially at an angle of [tex] 120^{0} [/tex]. The un-hybridized 2pz orbital of 'C' is involved in pi-bonding. Due to this, both of the above molecules possess a triangular planner geometry.
Answer:
1,1-Dichloroethene (CH₂CCl₂) and Ethene (H₂CCH₂) have same molecular geometries.
Explanation:
The molecular geometries of all molecules are drawn below. Among these molecules four molecules contains four atoms which has same geometry. These are Boron in BF₃, Carbon in CH₂O, both carbon atoms CH₂CCl₂ and both carbon atoms H₂CCH₂. All specified elements in these molecules have same geometry (Trigonal Planar).
While, the overall geometry of only two molecules is same. Those two are CH₂CCl₂ and H₂CCH₂ because in these molecules all carbon atoms have Trigonal Planar Geometry. Hence, we can confirm that among given compounds this pair has same geometry.
![Ver imagen transitionstate](https://us-static.z-dn.net/files/dad/b64a8b8ae4d96dcf2f855e3736d56094.png)