Describe the shapes of \[B{F_3}\] and $B{H_4}^ - $. Assign the hybridization of boron in these species.
Answer
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Hint: 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
$H = (V + X + C + A)/2$
Where,
$V$ is the number of valence electrons of the central atom.
$X$ is the number of monovalent atoms attached to the central atom.
$C$ is the total positive charge on the molecule.
$A$ is the total negative charge on the molecule.
Now, for $B{F_3}$ since we know that central atom is boron and there are $3$ fluorine atom surrounding it, and the number of valence electron of boron is $3$ , then writing all the values;
Total number of valence electrons of boron; $V = 3$
The number of fluorine attached to boron; $X = 3$
Total positive charge on the molecule; $C = 0$
Total negative charge on the molecule; $A = 0$
Therefore,
$H = (3 + 3 + 0 + 0)/2$
$H = 6/2$,
$H = 3$.
And for $B{F_4}^ - $ since we know that central atom is boron and there are $4$ fluorine atom surrounding it, and the number of valence electron of boron is $3$ , then writing all the values;
Total number of valence electrons of boron; $V = 3$
The number of fluorine attached to boron; $X = 4$
Total positive charge on the molecule; $C = 0$
Total negative charge on the molecule; $A = 1$
Therefore,
$H = (3 + 4 + 0 + 1)/2$
$H = 8/2$,
$H = 4$,
Now referring to the table given below,
By considering the above table we get to know that since our $H$ is $3$ for \[B{F_3}\] and for $B{H_4}^ - $ it is $4$ , therefore, the structure of \[B{F_3}\] is trigonal planar and $B{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.
Complete step by step answer:
The general formula of hybridization is given as
$H = (V + X + C + A)/2$
Where,
$V$ is the number of valence electrons of the central atom.
$X$ is the number of monovalent atoms attached to the central atom.
$C$ is the total positive charge on the molecule.
$A$ is the total negative charge on the molecule.
Now, for $B{F_3}$ since we know that central atom is boron and there are $3$ fluorine atom surrounding it, and the number of valence electron of boron is $3$ , then writing all the values;
Total number of valence electrons of boron; $V = 3$
The number of fluorine attached to boron; $X = 3$
Total positive charge on the molecule; $C = 0$
Total negative charge on the molecule; $A = 0$
Therefore,
$H = (3 + 3 + 0 + 0)/2$
$H = 6/2$,
$H = 3$.
And for $B{F_4}^ - $ since we know that central atom is boron and there are $4$ fluorine atom surrounding it, and the number of valence electron of boron is $3$ , then writing all the values;
Total number of valence electrons of boron; $V = 3$
The number of fluorine attached to boron; $X = 4$
Total positive charge on the molecule; $C = 0$
Total negative charge on the molecule; $A = 1$
Therefore,
$H = (3 + 4 + 0 + 1)/2$
$H = 8/2$,
$H = 4$,
Now referring to the table given below,
| VALUES OF H | HYBRIDIZATION | STRUCTURE |
| 2 | $sp$ | Linear |
| 3 | $s{p^2}$ | Trigonal planar |
| 4 | $s{p^3}$ | Tetrahedral |
| 5 | $s{p^3}d$ | Trigonal bipyramidal |
| 6 | $s{p^3}{d^2}$ | Octahedral |
| 7 | $s{p^3}{d^3}$ | Pentagonal bipyramidal |
By considering the above table we get to know that since our $H$ is $3$ for \[B{F_3}\] and for $B{H_4}^ - $ it is $4$ , therefore, the structure of \[B{F_3}\] is trigonal planar and $B{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.
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