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Question: Describe the shapes of \[B{F_3}\] and \(B{H_4}^ - \). Assign the hybridization of boron in these spe...

Describe the shapes of BF3B{F_3} and BH4B{H_4}^ - . Assign the hybridization of boron in these species.

Explanation

Solution

For finding out the hybridization of the given molecules, we can use the hybridization formula, and by referring to the arrangement of the spdf orbitals we can conclude the structure of the given molecular formulas.

Complete step by step answer:
The general formula of hybridization is given as
H=(V+X+C+A)/2H = (V + X + C + A)/2
Where,
VV is the number of valence electrons of the central atom.
XX is the number of monovalent atoms attached to the central atom.
CC is the total positive charge on the molecule.
AA is the total negative charge on the molecule.
Now, for BF3B{F_3} since we know that central atom is boron and there are 33 fluorine atom surrounding it, and the number of valence electron of boron is 33 , then writing all the values;
Total number of valence electrons of boron; V=3V = 3
The number of fluorine attached to boron; X=3X = 3
Total positive charge on the molecule; C=0C = 0
Total negative charge on the molecule; A=0A = 0
Therefore,
H=(3+3+0+0)/2H = (3 + 3 + 0 + 0)/2
H=6/2H = 6/2,
H=3H = 3.
And for BF4B{F_4}^ - since we know that central atom is boron and there are 44 fluorine atom surrounding it, and the number of valence electron of boron is 33 , then writing all the values;
Total number of valence electrons of boron; V=3V = 3
The number of fluorine attached to boron; X=4X = 4
Total positive charge on the molecule; C=0C = 0
Total negative charge on the molecule; A=1A = 1
Therefore,
H=(3+4+0+1)/2H = (3 + 4 + 0 + 1)/2
H=8/2H = 8/2,
H=4H = 4,
Now referring to the table given below,

VALUES OF HHYBRIDIZATIONSTRUCTURE
2spspLinear
3sp2s{p^2}Trigonal planar
4sp3s{p^3}Tetrahedral
5sp3ds{p^3}dTrigonal bipyramidal
6sp3d2s{p^3}{d^2}Octahedral
7sp3d3s{p^3}{d^3}Pentagonal bipyramidal

By considering the above table we get to know that since our HH is 33 for BF3B{F_3} and for BH4B{H_4}^ - it is 44 , therefore, the structure of BF3B{F_3} is trigonal planar and BH4B{H_4}^ - is tetrahedral.

Note:
In questions related to structures, always remember to apply hybridization formula, and refer to the table given above, it applies to all compounds which have hybridization seven or less than it.