Shape of $BF_{4}^{-}$ is:
(a)- Tetrahedral
(b)- Pyramidal
(c)- Trigonal planar
(d)- Bent
Answer
645k+ views
Hint: 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:
$X=\dfrac{1}{2}\left[ \begin{align}
& \{\text{no}\text{. of valence electrons of central atom }\!\!\}\!\!\text{ + }\!\!\{\!\!\text{ no}\text{. of monovalent atoms }\!\!\}\!\!\text{ } \\
& \text{ - }\!\!\{\!\!\text{ charge on cation }\!\!\}\!\!\text{ + }\!\!\{\!\!\text{ charge on the anion }\!\!\}\!\!\text{ } \\
\end{align} \right]$
$X=\dfrac{1}{2}\left[ VE+MA-c+a \right]$
Given a molecule is $BF_{4}^{-}$, 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 $BF_{4}^{-}$ and -1 anionic charge.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 3+4-0+1 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in$C{{H}_{3}}^{+}$. So, $s{{p}^{3}}$ 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.
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:
$X=\dfrac{1}{2}\left[ \begin{align}
& \{\text{no}\text{. of valence electrons of central atom }\!\!\}\!\!\text{ + }\!\!\{\!\!\text{ no}\text{. of monovalent atoms }\!\!\}\!\!\text{ } \\
& \text{ - }\!\!\{\!\!\text{ charge on cation }\!\!\}\!\!\text{ + }\!\!\{\!\!\text{ charge on the anion }\!\!\}\!\!\text{ } \\
\end{align} \right]$
$X=\dfrac{1}{2}\left[ VE+MA-c+a \right]$
Given a molecule is $BF_{4}^{-}$, 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 $BF_{4}^{-}$ and -1 anionic charge.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 3+4-0+1 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in$C{{H}_{3}}^{+}$. So, $s{{p}^{3}}$ 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.
Recently Updated Pages
Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

