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Question: Shape of \(BF_{4}^{-}\) is: (a)- Tetrahedral (b)- Pyramidal (c)- Trigonal planar (d)- Bent...

Shape of BF4−BF_{4}^{-} is:
(a)- Tetrahedral
(b)- Pyramidal
(c)- Trigonal planar
(d)- Bent

Explanation

Solution

The shape or structure of the compound can be predicted by calculating the hybridization of the compound. The hybridization can be calculated with the number of valence electrons of the central atom, the number of monovalent atoms/ groups surrounding the central atom, charge on the cation, and charge on the anion.

Complete step by step answer:
To predict the shape of the molecule by:
If the number of hybrid orbitals is equal to the number of surrounding groups, it has a regular geometry and the shape is the same as that predicted by hybridization. If the number of surrounding groups is less than the hybrid orbitals, the difference gives the number of lone pairs present. So, the molecule will have irregular geometry and the shape is predicted by leaving the hybrid orbitals containing the lone pairs.

For calculating the number of hybrid orbital or hybridization of the central atom we can use the formula:
X=12[no. of valence electrons of central atom !!!! + !!!! no. of monovalent atoms !!!!   - !!!! charge on cation !!!! + !!!! charge on the anion !!!!  ]X=\dfrac{1}{2}\left[ \begin{aligned} & \\{\text{no}\text{. of valence electrons of central atom }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ no}\text{. of monovalent atoms }\\!\\!\\}\\!\\!\text{ } \\\ & \text{ - }\\!\\!\\{\\!\\!\text{ charge on cation }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ charge on the anion }\\!\\!\\}\\!\\!\text{ } \\\ \end{aligned} \right]
X=12[VE+MA−c+a]X=\dfrac{1}{2}\left[ VE+MA-c+a \right]
Given a molecule is BF4−BF_{4}^{-}, the central atom is boron, and it has a monovalent atom, fluorine.
The central atom has 3 valence electrons. There are 4 monovalent atoms in BF4−BF_{4}^{-} and -1 anionic charge.

So, the hybridization will be,
X=12[3+4−0+1]=82=4X=\dfrac{1}{2}\left[ 3+4-0+1 \right]=\dfrac{8}{2}=4
The value of X is 4, therefore, the hybridization is sp3s{{p}^{3}}
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs inCH3+C{{H}_{3}}^{+}. So, sp3s{{p}^{3}} hybridization has a tetrahedral shape.
So, the correct answer is “Option A”.

Note: For predicting the shape of the molecule, the only number of atoms or groups is considered. Even, if they are a monovalent or divalent atom or molecule.