Chemistry Exam > Chemistry Questions > The shape of [BrF4]- ion isa)Regular tetrahed...
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The shape of [BrF4]- ion is
- a)Regular tetrahedron
- b)Square planar
- c)Trigonal pyramidal
- d)See-saw OR irregular tetrahedron
Correct answer is option 'B'. Can you explain this answer?
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The shape of [BrF4]- ion isa)Regular tetrahedronb)Square planarc)Trigo...
The shape of the [BrF4]- ion is square planar. To understand why, we need to look at the electron configuration of the bromine atom and the number of bonding electrons involved in the ion.
Bromine has 7 valence electrons, and each of the four fluorine atoms contributes one electron for bonding. Additionally, there is one extra electron due to the negative charge of the ion, resulting in a total of 12 valence electrons involved in bonding.
According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the molecular shape is determined by the repulsion of electron pairs around the central atom. In the case of [BrF4]-, the central bromine atom is surrounded by 4 bonding pairs of electrons and 1 lone pair.
The number of electron pairs around the central atom determines the electron pair geometry. In this case, with 5 electron pairs, the electron pair geometry is trigonal bipyramidal. However, the molecular shape is determined by the arrangement of the bonded atoms around the central atom, not including the lone pairs.
In the case of [BrF4]-, the lone pair of electrons occupies one of the equatorial positions in the trigonal bipyramidal geometry, causing the four fluorine atoms to take up the remaining positions. These fluorine atoms form a square planar arrangement around the central bromine atom.
Therefore, the shape of the [BrF4]- ion is square plan
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The shape of [BrF4]- ion isa)Regular tetrahedronb)Square planarc)Trigo...
Shape of [BrF4]- ion
The shape of an ion or molecule is determined by its electronic geometry, which refers to the arrangement of its bonding and non-bonding electron pairs around the central atom. In the case of [BrF4]- ion, the central atom is bromine (Br) with a negative charge, and it is surrounded by four fluorine (F) atoms.
Electronic geometry of [BrF4]- ion
To determine the electronic geometry, we need to consider the total number of electron pairs around the central atom. In this case, we have one lone pair of electrons on the bromine atom and four bonding pairs (one with each fluorine atom). Therefore, the total number of electron pairs is 5.
Valence electron configuration
Before we determine the shape, let's first look at the valence electron configuration of bromine and fluorine to understand their bonding behavior.
- Bromine (Br): [Kr] 5s2 4d10 4p5
- Fluorine (F): [He] 2s2 2p5
VSEPR theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a useful model for predicting the shapes of molecules and ions based on minimizing the repulsion between electron pairs.
According to VSEPR theory, the electron pairs (both bonding and non-bonding) around the central atom will arrange themselves in a way that maximizes their distance from each other, resulting in a specific shape.
Determining the shape
Let's use the VSEPR theory to determine the shape of [BrF4]- ion.
1. Count the total number of electron pairs: 5
2. Determine the arrangement of electron pairs: The electron pairs will arrange themselves in a way that maximizes their distance from each other. The possible arrangements for 5 electron pairs are trigonal bipyramidal and square pyramidal.
3. Consider the lone pair: The presence of a lone pair will distort the shape from the ideal arrangement. A lone pair exerts greater repulsion compared to a bonding pair.
4. Determine the shape: In the case of [BrF4]- ion, the lone pair on the bromine atom will push the bonding pairs closer together, resulting in a square planar shape. Therefore, the correct answer is option 'B' - square planar.
Conclusion
The shape of the [BrF4]- ion is square planar due to the presence of a lone pair on the central bromine atom, which distorts the ideal trigonal bipyramidal arrangement. The repulsion between the lone pair and bonding pairs causes the bonding pairs to be pushed closer together, resulting in a square planar shape.
The shape of an ion or molecule is determined by its electronic geometry, which refers to the arrangement of its bonding and non-bonding electron pairs around the central atom. In the case of [BrF4]- ion, the central atom is bromine (Br) with a negative charge, and it is surrounded by four fluorine (F) atoms.
Electronic geometry of [BrF4]- ion
To determine the electronic geometry, we need to consider the total number of electron pairs around the central atom. In this case, we have one lone pair of electrons on the bromine atom and four bonding pairs (one with each fluorine atom). Therefore, the total number of electron pairs is 5.
Valence electron configuration
Before we determine the shape, let's first look at the valence electron configuration of bromine and fluorine to understand their bonding behavior.
- Bromine (Br): [Kr] 5s2 4d10 4p5
- Fluorine (F): [He] 2s2 2p5
VSEPR theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a useful model for predicting the shapes of molecules and ions based on minimizing the repulsion between electron pairs.
According to VSEPR theory, the electron pairs (both bonding and non-bonding) around the central atom will arrange themselves in a way that maximizes their distance from each other, resulting in a specific shape.
Determining the shape
Let's use the VSEPR theory to determine the shape of [BrF4]- ion.
1. Count the total number of electron pairs: 5
2. Determine the arrangement of electron pairs: The electron pairs will arrange themselves in a way that maximizes their distance from each other. The possible arrangements for 5 electron pairs are trigonal bipyramidal and square pyramidal.
3. Consider the lone pair: The presence of a lone pair will distort the shape from the ideal arrangement. A lone pair exerts greater repulsion compared to a bonding pair.
4. Determine the shape: In the case of [BrF4]- ion, the lone pair on the bromine atom will push the bonding pairs closer together, resulting in a square planar shape. Therefore, the correct answer is option 'B' - square planar.
Conclusion
The shape of the [BrF4]- ion is square planar due to the presence of a lone pair on the central bromine atom, which distorts the ideal trigonal bipyramidal arrangement. The repulsion between the lone pair and bonding pairs causes the bonding pairs to be pushed closer together, resulting in a square planar shape.
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