
Which of the following will be octahedral?
A) \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]
B) \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]
C) \[{\rm{PC}}{{\rm{l}}_{\rm{5}}}\]
D) \[{\rm{B}}{{\rm{O}}_{\rm{3}}}^{3 - }\]
Answer
162.3k+ views
Hint: An octahedral molecule has \[s{p^3}{d^2}\] hybridization. An octahedral molecule is to be identified by counting the groups surrounding the central atom. Also, we have to count the lone pairs. If the total count of the group is 6, then the molecule is \[s{p^3}{d^2}\]hybridized.
Complete step by step solution:Let’s discuss all the options one by one.
Option A is \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]. The Sulphur atom is bonded to six fluorine atoms. And there is no lone pair on the atom of Sulphur. Therefore, \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]is \[s{p^3}{d^2}\]hybridized and the molecular shape is octahedral.
Option B is\[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]. We can calculate the hybridization of B atom by the formula of \[\dfrac{{V + X - C + A}}{2}\] , where, V denotes the valence electrons, X is for the monovalent atoms, C is for the cationic charge and A is for the anionic charge.
For \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \],
\[H = \dfrac{{3 + 4 + 1}}{2} = 4\], so, \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]is \[s{p^3}\] hybridized and the molecular shape is tetrahedral.
Option C is \[{\rm{PC}}{{\rm{l}}_{\rm{5}}}\]. Here, the molecule is \[s{p^3}d\] hybridized and the molecular shape is trigonal pyramidal.
Option D is\[{\rm{B}}{{\rm{O}}_{\rm{3}}}^{3 - }\]. We have to use the above formula to find the hybridization of Boron atoms. As there are no monovalent atoms present. Therefore,
\[H = \dfrac{{3 + 3}}{2} = 3\]
As H=3, the hybridization is \[s{p^2}\] and the molecular shape is trigonal planar.
Hence, the option A is right.
Note: The theory used above to determine the molecular shape is VSEPR theory. This theory guides to predict the molecular shape considering the lone pairs and bond pairs around the central atom. This theory also gives the molecular geometry.
Complete step by step solution:Let’s discuss all the options one by one.
Option A is \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]. The Sulphur atom is bonded to six fluorine atoms. And there is no lone pair on the atom of Sulphur. Therefore, \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]is \[s{p^3}{d^2}\]hybridized and the molecular shape is octahedral.
Option B is\[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]. We can calculate the hybridization of B atom by the formula of \[\dfrac{{V + X - C + A}}{2}\] , where, V denotes the valence electrons, X is for the monovalent atoms, C is for the cationic charge and A is for the anionic charge.
For \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \],
\[H = \dfrac{{3 + 4 + 1}}{2} = 4\], so, \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]is \[s{p^3}\] hybridized and the molecular shape is tetrahedral.
Option C is \[{\rm{PC}}{{\rm{l}}_{\rm{5}}}\]. Here, the molecule is \[s{p^3}d\] hybridized and the molecular shape is trigonal pyramidal.
Option D is\[{\rm{B}}{{\rm{O}}_{\rm{3}}}^{3 - }\]. We have to use the above formula to find the hybridization of Boron atoms. As there are no monovalent atoms present. Therefore,
\[H = \dfrac{{3 + 3}}{2} = 3\]
As H=3, the hybridization is \[s{p^2}\] and the molecular shape is trigonal planar.
Hence, the option A is right.
Note: The theory used above to determine the molecular shape is VSEPR theory. This theory guides to predict the molecular shape considering the lone pairs and bond pairs around the central atom. This theory also gives the molecular geometry.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
