Homework Answers
Number of valence electrons in Br is 7
5 of them are used in forming 5 bonds.
Remaining 2 electrons are present as 1 lone pair.
So, there are 6 electron group present (5 bond pair and 1 lone
pair)
Central atom has 1 lone pair and 5 bond pair
So, the shape is Square pyramidal
Answer: option 2
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Question 4 2.5 pts Give the molecular geometry and number of electron groups for BrF5. square...
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Provide the correct molecular geometry for (a), given the number of lone pairs and bonding groups on the central atom. trigonal planar octahedral trigonal bipyramidal trigonal pyramidal square pyramidal tetrahedral seesaw bent linearProvide the correct molecular geometry for (b), given the number of lone pairs and bonding groups on the central atom. trigonal bipyramidal square pyramidal bent tetrahedral trigonal planar seesaw trigonal pyramidal octahedral linear Provide the correct molecular geometry for (c), given the number of lone pairs and bonding groups on the...
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Number 1-6 1. What is the molecular geometry of SBr4? A) seesaw B) square planar C)...
Number 1-6
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Number 1-6
1. What is the molecular geometry of SBr4? A) seesaw B) square planar C) square pyramidal D) tetrahedral 2. Using the VSEPR model, the molecular geometry of the central atom in CF4 is A) linear B) trigonal planar C) tetrahedral D) bent E) trigonal pyramidal 3. Using the VSEPR model, the molecular geometry of the central atom in NC13 is A) linear B) trigonal planar C) tetrahedral D) bent E) trigonal pyramidal 4. Using the VSEPR model, the...
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olecular Geometry Saved Help Save & E Ch. Ex. 6 - Predict the Geometry Using VSEPR 2 attempts left Check my work Be sure to answer all parts. Report problem Hint Predict the geometry of the following molecules and ion using the VSEPR model. Enter the number corresponding to the correct geometry in the box next to the formula Electron Domain Molecular CHI 1 - bent 2 - linear 3-octahedral 4 - seesaw-shaped 5-square planar 6 - square pyramidal 7-tetrahedral...
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