Answer:
1. sp = XY₂
2. sp² = XY₂Z, XY₃
3. sp3³ = XY₄, XY₂Z₂, XY₃Z
4. sp³d = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z
5. sp³d² = XY₆, XY₄Z₂, XY₅Z
Explanation:
this is quite dicey, so it should be looked into carefully.
we would classify each of the abbreviation according to their hybridization and it electron domain.
⇒ sp hybridization = XY₂
in this, we can see that the central atom X is bonded to two outer atoms Y.
this makes the no of hybrid orbitals and the no of sigma bonds both 2.
electron domain = 2.
⇒ sp² hybridization = XY₂Z, XY₃
Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.
electron domain = 3.
⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z
for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.
In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.
⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z
for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;
XY₂Z₃:
total electron domain = 2+3 = 5
XY₃Z₂:
total electron domain = 2+3 = 5
XY₄Z:
total electron domain = 1+4 = 5
⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z
Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.
XY4Z2:
total electron domain = 2+4 = 6
cheers, i hope this helps.
