
The smallest bond angle is found in
A. \[IF_7\]
B. \[CH_4\]
C. \[BeF_2\]
D. \[BF_3\]
Answer
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Hint: \[IF_7\] has a pentagonal bipyramidal structure.
\[CH_4\], \[BeF_2\], and \[BF_3\] have tetrahedral, linear, and trigonal planar geometry respectively.
Linear molecules have a bond angle of \[180^o\] which is the highest bond angle.
Complete step by step solution:Here in this question, we have to find out which of the given molecules has the smallest bond angle.
The shape is predicted by VSEPR theory.
A. \[IF_7\]
Here Iodine is the central atom. It is an interhalogen compound.
Iodine has seven valence electrons all of which make a bond with each electron of seven F atoms.
It contains seven bond pairs and zero lone pairs.
So, its structure is pentagonal bipyramidal.
Five bond pairs are in the same plane at an angle of \[72^o\], and the remaining two bond pairs are perpendicular to the plane creating an angle of \[90^o\] with the plane.

Image: Structure of \[IF_7\]
So, A is correct.
B. \[CH_4\]
Here carbon is the central atom.
It has four valence electrons in its valence shell all of which are involved in bond formation.
So, there are four bond pairs.
So, it has a tetrahedral structure.
The bond angle is \[109^o\].
So, B is incorrect.
C. \[BeF_2\]
Here beryllium is the central atom.
It has two valence electrons in its valence shell all of which are involved in bond formation.
So, there are two bond pairs.
So, it has a tetrahedral structure.
The bond angle is \[180^o\].
So, C is incorrect.
D. \[BF_3\]
Here boron is the central atom.
It has three valence electrons in its valence shell all of which are involved in bond formation.
So, there are three bond pairs.
So, it has a trigonal planar structure.
The bond angle is \[120^o\].
So, D is incorrect.
So, \[IF_7\] has a pentagonal pyramidal structure and has the smallest bond angle.
So, option A is correct.
Note: An interhalogen compound is a molecule that includes two or more distinct halogen atoms and no atoms of elements from any other group.
Most interhalogen compounds are binary i.e., composed of two distinct elements.
\[CH_4\], \[BeF_2\], and \[BF_3\] have tetrahedral, linear, and trigonal planar geometry respectively.
Linear molecules have a bond angle of \[180^o\] which is the highest bond angle.
Complete step by step solution:Here in this question, we have to find out which of the given molecules has the smallest bond angle.
The shape is predicted by VSEPR theory.
A. \[IF_7\]
Here Iodine is the central atom. It is an interhalogen compound.
Iodine has seven valence electrons all of which make a bond with each electron of seven F atoms.
It contains seven bond pairs and zero lone pairs.
So, its structure is pentagonal bipyramidal.
Five bond pairs are in the same plane at an angle of \[72^o\], and the remaining two bond pairs are perpendicular to the plane creating an angle of \[90^o\] with the plane.

Image: Structure of \[IF_7\]
So, A is correct.
B. \[CH_4\]
Here carbon is the central atom.
It has four valence electrons in its valence shell all of which are involved in bond formation.
So, there are four bond pairs.
So, it has a tetrahedral structure.
The bond angle is \[109^o\].
So, B is incorrect.
C. \[BeF_2\]
Here beryllium is the central atom.
It has two valence electrons in its valence shell all of which are involved in bond formation.
So, there are two bond pairs.
So, it has a tetrahedral structure.
The bond angle is \[180^o\].
So, C is incorrect.
D. \[BF_3\]
Here boron is the central atom.
It has three valence electrons in its valence shell all of which are involved in bond formation.
So, there are three bond pairs.
So, it has a trigonal planar structure.
The bond angle is \[120^o\].
So, D is incorrect.
So, \[IF_7\] has a pentagonal pyramidal structure and has the smallest bond angle.
So, option A is correct.
Note: An interhalogen compound is a molecule that includes two or more distinct halogen atoms and no atoms of elements from any other group.
Most interhalogen compounds are binary i.e., composed of two distinct elements.
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