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What is the shape of BF₄⁻ ? A. Tetrahedra

What is the shape of BF₄⁻ ?

A. Tetrahedral

B. Pyramidal

C. Trigonal Planar

D. Bent

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1 Year agoGrade
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1 Answer

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Askiitians Tutor Team

ApprovedApproved Tutor Answer1 Year ago

The shape of the tetrafluoroborate ion, BF₄⁻, is best described as tetrahedral. So, the correct answer is A. Tetrahedral. Let’s break down why this is the case.

Understanding Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. To determine the shape of BF₄⁻, we can use the Valence Shell Electron Pair Repulsion (VSEPR) theory, which helps predict the geometry based on the repulsion between electron pairs around a central atom.

Analyzing BF₄⁻

In the case of BF₄⁻, we have the following components:

  • Boron (B): This is the central atom.
  • Fluorine (F): There are four fluorine atoms bonded to the boron atom.
  • Charge: The ion carries a negative charge, which means there is an extra electron contributing to the overall electron count.

Counting Electron Pairs

Boron has three valence electrons, and each fluorine contributes one electron through its bond. Therefore, in BF₄⁻, we have:

  • 3 (from B) + 4 (from 4 F) + 1 (extra electron due to the negative charge) = 8 total valence electrons.

These 8 electrons will form 4 bonding pairs (one for each B-F bond) and no lone pairs on the boron atom.

Applying VSEPR Theory

According to VSEPR theory, when there are four bonding pairs and no lone pairs around the central atom, the electron pairs will arrange themselves to minimize repulsion. This arrangement leads to a tetrahedral geometry.

Visualizing the Tetrahedral Shape

Imagine a pyramid with a triangular base. The boron atom is at the center, while the four fluorine atoms are positioned at the corners of a tetrahedron. This spatial arrangement allows the bonds to be as far apart as possible, which is crucial for stability.

Real-World Analogy

Think of a tetrahedron like a pyramid with a triangular base, where each corner of the base and the apex represent the positions of the fluorine atoms. Just as the corners of a pyramid are evenly spaced from each other, the fluorine atoms in BF₄⁻ are symmetrically arranged around the boron atom.

Conclusion

In summary, the tetrafluoroborate ion, BF₄⁻, exhibits a tetrahedral shape due to the arrangement of its four bonding pairs around the central boron atom. This geometry is a direct result of the principles outlined in VSEPR theory, making it a classic example of molecular geometry in chemistry.


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is Pradeep Chemistry+Ncert sufficient for all of the theory of Jee Mains Chemistry?


Hi, I’m a self-study JEE 2027 aspirant with a large 11th backlog. Since I don’t think 6–8 hour one-shots are sufficient to cover the full theory + breadth of problem types of a JEE chapter, I’ve shifted from lecture-based preparation to book-based self-study.


I have a problem with using NCERT as my primary chemistry source. In my experience, it is concise and information-heavy, but often lacks detailed explanations, systematic organisation of concepts, and categorisation of problem types. This seems especially problematic for Physical and Organic Chemistry, where I feel that simply reading NCERT may not give enough depth or problem-solving preparation.


So I’m considering this approach:


Pradeep Chemistry → complete the chapter thoroughly from Pradeep → then read NCERT for that chapter → solve JEE Main PYQs/problems.


My main question is:


Is Pradeep + NCERT sufficient as the complete theory source for JEE Main-level Chemistry?


In other words, after thoroughly completing a chapter from Pradeep and then NCERT, can I consider the theory part of that chapter complete for JEE Main, with only PYQs/problem practice remaining?


I’m asking specifically about JEE Main level, not Advanced.


Also, if Pradeep is not sufficient, what specific gap does it leave, and what would you recommend as a better book-based alternative for self-study?

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