Question

question 7 (1 point)
which of the following molecules demonstrates a square pyramidal molecular structure?
xeo₃
xef₄
ccl₄
ch₄

Explanation:

• For \(XeO_3\): The central atom \(Xe\) has 8 valence electrons. In \(XeO_3\), \(Xe\) forms 3 double bonds with \(O\) atoms. The steric number \(=3 + \frac{8 - 6}{2}=4\). The electron - pair geometry is tetrahedral and the molecular geometry is trigonal pyramidal.
• For \(XeF_4\): The central atom \(Xe\) has 8 valence electrons. \(Xe\) forms 4 single bonds with \(F\) atoms and has 2 lone pairs. The steric number \(=4+\frac{8 - 4}{2}=6\). The electron - pair geometry is octahedral. Due to the presence of 2 lone pairs, the molecular geometry is square planar.
• For \(CCl_4\): The central atom \(C\) has 4 valence electrons. \(C\) forms 4 single bonds with \(Cl\) atoms. The steric number \(=4+\frac{4 - 4}{2}=4\). The electron - pair geometry and molecular geometry are tetrahedral.
• For \(CH_4\): The central atom \(C\) has 4 valence electrons. \(C\) forms 4 single bonds with \(H\) atoms. The steric number \(=4+\frac{4 - 4}{2}=4\). The electron - pair geometry and molecular geometry are tetrahedral.

Answer:

None of the given options (\(XeO_3\), \(XeF_4\), \(CCl_4\), \(CH_4\)) have a square - pyramidal molecular structure. If we assume there is a mistake in the options and consider \(XeOF_4\) (not in the given list), for \(XeOF_4\): The central atom \(Xe\) has 8 valence electrons. \(Xe\) forms 1 double bond with \(O\) and 4 single bonds with \(F\) atoms and has 1 lone pair. The steric number \(=5+\frac{8-(2 + 4)}{2}=6\). The electron - pair geometry is octahedral and the molecular geometry is square - pyramidal. But based on the given options, there is no correct answer.