Question

question 4 (1 point)
a molecule exhibits the trigonal planar shape. how many electron groups exist in the valence shell for the central atom in this molecule?
two
three
four
five
more information is required to answer this question
question 5 (1 point)
which of the following molecules demonstrates a trigonal planar molecular shape?
no₃
sf₃
bf₃
h₃o⁺

Explanation:

• Question 4: According to VSEPR (Valence - Shell Electron - Pair Repulsion) theory, a trigonal planar shape is associated with \(AX_3\) (where \(A\) is the central atom and \(X\) is a surrounding atom). In the VSEPR model, the number of electron groups (bonding pairs) around the central atom for a trigonal planar geometry is three.
• Question 5:
• For \(NO_3\), it is not a neutral molecule (it should be \(NO_3^-\)). The \(NO_3^-\) ion has a trigonal planar shape (\(AX_3\) with no lone pairs on the central \(N\) atom). But if we consider the given option \(NO_3\) (assuming it is a typo for \(NO_3^-\)), it has a trigonal planar shape.
• \(SF_3\) is not a stable species. If we consider \(SF_3^+\), it has a trigonal pyramidal shape (\(AX_3E\) in VSEPR, where \(E\) is a lone pair).
• \(BF_3\) has a central \(B\) atom with three bonding pairs (\(AX_3\) in VSEPR) and a trigonal planar shape.
• \(H_3O^+\) has a central \(O\) atom with three bonding pairs and one lone pair (\(AX_3E\) in VSEPR), giving it a trigonal pyramidal shape.

Answer:

• Question 4: Three
• Question 5: \(NO_3\) (assuming it is \(NO_3^-\)) and \(BF_3\)