The incorrect hybridization is the first compound.
Be brings in 2 electrons (2nd group) and Cl brings 7 electrons each, totalling 14. When you do the electron dot configuration, the Be is double bonded to each Cl. This makes the SN (how many atoms the central is attached to, plus the number of PAIRS of electrons) equal to 2.
The superscripts of the hybridizations have to add up to the hybridization (ie. SN=2 equals S1P1 or just SP)
Another example of this is if the SN=4, it would be SP3 (the s has a hidden superscript of 1, plus the 3 on the P). It is NOT sp2.
The BF3 is a tricky example. It is commonly known that Boron doesn't follow the octet rule. (Octet rule means that the central atom has to have 8 electons) The B brings 3 electrons, and F brings 7 electrons each totalling 21. Overall there are 24 electrons in the drawing. After drawing, you see that the B is attached to three with no pairs. (SN=3)
So, the hybridization has to add up to 3...meaning it is SP2.
The C2H2 is a little difficult to figure out as well. Overall there are 10 electrons to draw on the figure, which doesnt allow the central Carbons to fulfill the Octet rule. In order to do this, they have to share some of the electrons (a triple bond). This means that the central ones now have NO electron pairs that are 'loose'. The central atoms are bonded to two atoms each, with no electron pairs, so SN=2. This helps determine hybridization which is SP.
Finally, H2O, the central oxygen is bonded to the two hydrogens. Overall, there are 8 electrons to draw. Hydrogens need no more than the bond they share with the oxygen, and this is true for any atom they are bonded to. The remaining electrons (4) are placed around the oxygen. So, we add how many atoms are attached to the central, 2 (Hydrogens) and then add how many PAIRS of electrons around the central atom (2) which means SN=4. This has to be the sum of the superscripts. So it has SP3 hybridization.
Source(s): I'm a chemistry major.
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