1- Molecular geometry for CO₂ is nonlinear with net dipole.

Here are the answers to the True/False questions:

1- Molecular geometry for $\text{CO}_2$ is nonlinear with net dipole.

• $\text{CO}_2$ has a central carbon atom double-bonded to two oxygen atoms ($\text{O=C=O}$). There are two electron domains around the central carbon, resulting in a linear electronic and molecular geometry.
• Due to its linear and symmetrical structure, the bond dipoles cancel out, meaning $\text{CO}_2$ has no net dipole.
• Therefore, the statement is (X) False.

2- Electronic geometry for molecules with formula $\text{AB}_3\text{U}$ is tetrahedral.

• The formula $\text{AB}_3\text{U}$ indicates a central atom (A) bonded to three other atoms (B) and having one lone pair (U).
• The total number of electron domains is $3 \text{ (bonding pairs)} + 1 \text{ (lone pair)} = 4$.
• For 4 electron domains, the electronic geometry is always tetrahedral.
• Therefore, the statement is (V) True.

3- Electronic geometry equal to molecular geometry for $\text{POCl}_3$ molecule.

• In $\text{POCl}_3$, Phosphorus (P) is the central atom. It forms a double bond with Oxygen (O) and single bonds with three Chlorine (Cl) atoms.
• The central P atom has 4 bonding domains (1 double bond to O, 3 single bonds to Cl) and no lone pairs.
• With 4 electron domains and no lone pairs, both the electronic geometry and the molecular geometry are tetrahedral.
• Therefore, the statement is (V) True.

4- Molecular geometry for $\text{sp}^2$ hybridization molecules is trigonal planar.

• $\text{sp}^2$ hybridization corresponds to 3 electron domains.
• If all 3 domains are bonding pairs (e.g., $\text{BF}_3$), the molecular geometry is trigonal planar.
• However, if there are 2 bonding pairs and 1 lone pair (e.g., $\text{SO}_2$), the molecular geometry is bent.
• Since not all $\text{sp}^2$ hybridized molecules have a trigonal planar molecular geometry, the statement is too general.
• Therefore, the statement is (X) False.

5- V.B.T. predicts $\text{sp}^2$ hybridization at one oxygen atom for $\text{O}_2$ molecule and diamagnetic.

• In the $\text{O}_2$ molecule, each oxygen atom forms one sigma bond and one pi bond, and each has two lone pairs. This arrangement is consistent with $\text{sp}^2$ hybridization for the sigma framework and lone pairs.
• However, Valence Bond Theory (VBT) in its simplest form typically predicts $\text{O}_2$ to be diamagnetic (all electrons paired). Experimentally, $\text{O}_2$ is known to be paramagnetic (has unpaired electrons), which is correctly predicted by Molecular Orbital Theory (MOT).
• Since the prediction of diamagnetism is incorrect, the entire statement is false.
• Therefore, the statement is (X) False.

6- The energy of nonbonding orbitals higher than anti bonding molecular orbitals.

• Nonbonding orbitals are typically atomic orbitals that do not participate in bonding and have energies similar to the original atomic orbitals.
• Antibonding molecular orbitals are formed from the destructive interference of atomic orbitals and are always higher in energy than the corresponding bonding molecular orbitals and the original atomic orbitals.
• Therefore, antibonding orbitals are generally higher in energy than nonbonding orbitals.
• Therefore, the statement is (X) False.

7- All hetero triatomic molecules in linear structure have net dipole.

• A hetero triatomic molecule consists of three atoms, where at least two are different.
• Consider $\text{CO}_2$ ($\text{O=C=O}$). It is a hetero triatomic molecule (carbon is different from oxygen) and has a linear structure. However, the two equal and opposite bond dipoles cancel out, resulting in no net dipole.
• Since $\text{CO}_2$ is a counterexample, the statement is false.
• Therefore, the statement is (X) False.

8- Repulsion force between bonded pair and lone pair of electrons is higher than that between lone pair and lone pair.

• According to VSEPR theory, the order of electron pair repulsion is: Lone pair-Lone pair (LP-LP) > Lone pair-Bonded pair (LP-BP) > Bonded pair-Bonded pair (BP-BP).
• The statement claims LP-BP repulsion is higher than LP-LP repulsion, which is incorrect.
• Therefore, the statement is (X) False.

9- Each lone pair of electron on central atom is counted as one region of high electron density.

• In VSEPR theory, electron domains (regions of high electron density) include single bonds, multiple bonds (double or triple bonds count as one domain), and lone pairs.
• Each lone pair on the central atom is indeed considered one electron domain.
• Therefore, the statement is (V) True.

10- The $\text{SO}_2$ molecule has bent shape.

• In $\text{SO}_2$, Sulfur (S) is the central atom. It forms two bonds with oxygen atoms and has one lone pair.
• The central S atom has 2 bonding domains (one double bond, one single bond, or two resonance structures with double bonds) and 1 lone pair.
• Total electron domains = 3.
• Electronic geometry: Trigonal planar.
• Molecular geometry: With 2 bonding domains and 1 lone pair, the molecular geometry is bent (or V-shaped).
• Therefore, the statement is (V) True.