There are 5 basic VSEPR geometries, which vary by their steric number (# of bond
ID: 591076 • Letter: T
Question
There are 5 basic VSEPR geometries, which vary by their steric number (# of bonded atoms + lone pairs): 2) Linear (3) trigonal planar 4) tetrahedral (5) trigonal bipyramidal (6) octahedral What are the bond angles in a tetrahedral base geometry? How do the bond angles compare in CH4, NH3, and H20? a. s compare in BF3 No2, and NO2? What are the bond angles in a trigonal bipyramidal base geometry? How do the bond angles compare in PCl5, SF4, ClF3, and /3? Are the central atoms' lone pairs placed in the equatorial or axial positions? What are the bond angles in an octahedral base geometry? How do the bond angles compare in SFG, BrF5, and XeF4? c. Whatare the bo dnd is Are the d.Explanation / Answer
a. bond angle in tetrahedral = 109.5 degrees.
by increasing no .of lone pairs on central atom, bond angle decreases.
so that,
in CH4 - 0 lone pairs= 109.5 degrees
NH3 - 1 lone pairs = 107 degrees
h2o - 2 lone pairs = 104 degrees
b. bond angle in trigonalplanar = 120 degrees.
by increasing no .of lone pairs on central atom, bond angle decreases.
so that,
in BF3 - 0 lone pairs= 120 degrees
NO2 - 1 odd eelctron = 134 degrees
NO2 - - 1 lone pairs = 115.4 degrees
C. bond angle in trigonalbipyramidal = 120,90 degrees.
by increasing no .of lone pairs on central atom, bond angle decreases.
so that,
in Pcl5 - 0 lone pairs= 120,90 degrees
SF4 - 1 lone pairs = 102 degrees (see-saw)
ClF3 - 2 lone pairs = 87.5,185 degrees(T-shape)
I3- - 3 lone pairs = 180 degrees (lineat shape)
d.
bond angle in octahedralr = 90 degrees.
by increasing no .of lone pairs on central atom, bond angle decreases.
so that,
in SF6 - 0 lone pairs= 90 degrees
BrF5- 1 lone pair = 84.8 degrees (square pyramidal)
FeF4- - 2 lone pairs = 90 degrees (square planar)
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