Final answer:
In the reaction BF3 + NaF → NaBF4, the hybridization of boron changes from sp2 to sp3. In ethene reacting with H2 to form CH2Br-CH2Br, carbon atoms change from sp2 to sp3 hybridization. Other reactions mentioned are not relevant to the specific inquiry about hybridization changes.
Step-by-step explanation:
In the reaction BF3 + NaF → NaBF4, the hybridization of boron in BF3 is sp2 before the reaction, due to its three bonded fluorine atoms and no lone pairs, resulting in a trigonal planar shape. When boron accepts an additional fluoride ion to form NaBF4, the hybridization of boron changes to sp3 to accommodate the additional bond and lone pairs, resulting in a tetrahedral shape.
As for the reaction starting with ethene (CH2=CH2) and resulting in CH2Br-CH2Br, the hybridization of carbon atoms in ethene is initially sp2 due to the double bond (one σ bond and one π bond). After the addition of H2, the resulting dibromomethane has carbon atoms that are sp3 hybridized, with four σ bonds and no π bonds, again leading to a tetrahedral geometry.
The SO3 molecule in the reaction S2 + 1/2O2 → SO3 does not pertain to the question asked, as it is not related to a change in hybridization. The same applies to the examples provided in the typos such as CS2 + SH → HCS3- or F + SO3 → SFO3. These do not directly answer the question regarding the hybridization change for the underlined atom in the specified equations. It is important to stick to the original question for accurate and relevant answers.