The molecules having square pyramidal geometry are

Core Logic

Statement I: The net dipole moment of NH₃$NH_3$ (1.47 D$1.47\, D$) is actually greater than that of NF₃$NF_3$ (0.23 D$0.23\, D$). Therefore, Statement I is false.

Statement II: In NH₃$NH_3$, the N-H$N-H$ bond dipole moments (pointing towards the more electronegative N) reinforce the orbital dipole moment of the lone pair. In NF₃$NF_3$, the N-F$N-F$ bond dipole moments point away from N (towards the more electronegative F), opposing the orbital dipole moment of the lone pair. This partial cancellation in NF₃$NF_3$ makes its net dipole moment lower. Therefore, Statement II is also false, as it reverses the correct orientations.

Step 1: Final Conclusion

Since both statements assert the opposite of established facts regarding NH₃$NH_3$ and NF₃$NF_3$, both are false.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Using VSEPR theory: (A) BrF₅$BrF_5$: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) arrow$\rightarrow$ sp³d²$sp^3d^2$ hybridization arrow$\rightarrow$ Square pyramidal shape. (B) H₂O$H_2O$: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) arrow$\rightarrow$ sp³$sp^3$ hybridization arrow$\rightarrow$ Bent shape. (C) ClF₃$ClF_3$: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) arrow$\rightarrow$ sp³d$sp^3d$ hybridization arrow$\rightarrow$ T-shape. (D) SF₄$SF_4$: S has 6 valence electrons. 4 form bonds with F, leaving 1 lone pair. (4 bp + 1 lp) arrow$\rightarrow$ sp³d$sp^3d$ hybridization arrow$\rightarrow$ See-saw shape.

Step 1: Matching

(A) - (IV) (B) - (III) (C) - (I) (D) - (II)

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

According to Molecular Orbital Theory (MOT), the number of molecular orbitals (MOs) formed is equal to the total number of atomic orbitals (AOs) combined.

Step 1: Counting atomic orbitals

For a single atom in the 2nd period, the valence shell has: One 2s orbital Three 2p orbitals (2pₓ, 2py, 2pz$2p_x, 2p_y, 2p_z$) Total = 4 atomic orbitals per atom. For a diatomic molecule, two such atoms combine. Total atomic orbitals = 4 × 2 = 8$4 \times 2 = 8$.

Step 2: Forming molecular orbitals

Combining these 8 atomic orbitals yields 8 molecular orbitals:

• From 2s: σ₂ₛ$\sigma_{2s}$ and σ^*₂ₛ$\sigma^*_{2s}$ (2 MOs)
• From 2p: σ2pz, π2pₓ, π2py, π^2pₓ, π^2py, σ^*2pz$\sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}, \pi^_{2p_x}, \pi^_{2p_y}, \sigma^*_{2p_z}$ (6 MOs)

Total MOs = 2 + 6 = 8$2 + 6 = 8$.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

According to Fajan's rules, covalent character is favored by high charge and small size of the cation, and by cations with a pseudo-noble gas configuration. Ag^+$Ag^+$ has a pseudo-noble gas configuration (ns²np⁶nd¹⁰$ns^2np^6nd^{10}$), which results in high polarizing power compared to s-block and typical transition elements. Therefore, AgCl$AgCl$ has the maximum covalent character and is the least ionic among the given options.

Ionic character order: AgCl < CoCl₂ < BaCl₂ < KCl$AgCl < CoCl_2 < BaCl_2 < KCl$

Step 1: Final Selection

Because AgCl$AgCl$ is the most covalent, it is the least ionic. Hence, option (2) is correct.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Related Formula

Core Logic

Given dipole moment, μ = 1.2 D = 1.2 × 10⁻¹⁸ esu cm$\mu = 1.2\text{ D} = 1.2 \times 10^{-18}\text{ esu cm}$. Bond distance, d = 1AA = 10⁻⁸ cm$d = 1\mathrm{\AA} = 10^{-8}\text{ cm}$.

We need to find the fractional charge q$q$.

Step 1: Calculation

Step 2: Final Formatting

The question asks for the fractional charge in the form x × 10⁻¹⁰ esu$x \times 10^{-10}\text{ esu}$. Therefore, the value is 1.2$1.2$.

Note: Based on NTA officially accepting 12$12$ (if asked for x × 10⁻¹¹$x \times 10^{-11}$) or 1.2$1.2$. We will format it exactly as calculated.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure