Question

Asked Feb 20, 2020

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Consider the following AB_{3} molecules and ions:

PCl_{3}, SO_{3}, AlCl_{3}, SO_{3}^{2-} , and CH_{3}^{+}. How many of these

molecules and ions do you predict to have a trigonal-planar

molecular geometry? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

Step 1

to find the structure of the molecule or ion we need to find the hybridisation first.

Since number of hybridised orbitals = number of sigma bonds + number of lone pairs

1) PCl_{3} her we know that P have 5 e- in its valence shell and 3 of them are making 3 sigma bonds with 3 Cl. hence 2 e- is remaining as 1 lone pair.

Hence number of hybridised orbitals = 3+1 =4

=> hybridisation = sp^{3} => tetrahedral electron geometry and trigonal pyramidal molecular geometry.

2) SO_{3} her we know that S have 6 e- in its valence shell and all of them are making 3 sigma bonds with 3 O and 3 pi bonds. hence 0 e- is remaining Hence lone pair = 0

Hence number of hybridised orbitals = 3+0 =3

=> hybridisation = sp^{2} => trigonal planar electron geometry and trigonal planar molecular geometry.

Step 2

4) AlCl3 her we know that Al have 3 e- in its valence shell and all of them are making 3 sigma bonds with 3 Cl.

hence 0 e- is remaining

Hence lone pair = 0

Hence number of hybridised orbitals = 3+0 =3

=> hybridisation = sp2 => trigonal planar electron geometry and trigonal planar molecular geometry.

4) SO32- her we know that S have 6 e- in its valence shell and 2 extra because of 2- hence total 8.

Now 4 of them are maki...

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