Question

question 6 (1 point)
the bond angle in co₃²⁻ is expected to be approximately
90 degrees
109.5 degrees
120 degrees
145 degrees
180 degrees
question 7 (1 point)
which of the following molecules demonstrates a square pyramidal molecular structure?
xeo₃
xef₄
ccl₄
ch₄

Explanation:

Question 6

• For \(CO_{3}^{2 -}\), the central carbon atom has \(3\) bonding pairs and \(0\) lone pairs.
• According to VSEPR (Valence - Shell Electron - Pair Repulsion) theory, when the steric number (sum of bonding and non - bonding electron pairs around the central atom) is \(3\) ( \(SN=3\) ), the molecular geometry is trigonal planar.
• In a trigonal planar geometry, the bond angles are \(120^{\circ}\).

• For \(XeO_{3}\): The central \(Xe\) atom has \(3\) bonding pairs (\(Xe - O\) bonds) and \(1\) lone pair. Using VSEPR, the molecular geometry is trigonal pyramidal.
• For \(XeF_{4}\): The central \(Xe\) atom has \(4\) bonding pairs (\(Xe - F\) bonds) and \(2\) lone pairs. According to VSEPR, the molecular geometry is square planar.
• For \(CCl_{4}\): The central \(C\) atom has \(4\) bonding pairs (\(C - Cl\) bonds) and \(0\) lone pairs. By VSEPR, the molecular geometry is tetrahedral.
• For \(CH_{4}\): The central \(C\) atom has \(4\) bonding pairs (\(C - H\) bonds) and \(0\) lone pairs. Using VSEPR, the molecular geometry is tetrahedral.
• None of the given options (\(XeO_{3}\), \(XeF_{4}\), \(CCl_{4}\), \(CH_{4}\)) have a square - pyramidal structure. But if we assume a mis - print and consider \(IF_{5}\) (not in the options, just for knowledge, central \(I\) has \(5\) bonding pairs and \(1\) lone pair giving square - pyramidal), among the given options, there is an error as none of them have square - pyramidal. But if we go by the closest VSEPR - related analysis of the given options:
• \(XeO_{3}\): \(SN=3 + 1=4\) (trigonal pyramidal)
• \(XeF_{4}\): \(SN = 4+2 = 6\) (square planar)
• \(CCl_{4}\) and \(CH_{4}\): \(SN=4\) (tetrahedral)

Answer:

C. 120 degrees

Question 7