Among mathrmSO_2$\mathrm{SO}_2$, mathrmNF_3$\mathrm{NF}_3$, mathrmNH_3$\mathrm{NH}_3$, mathrmXeF_2$\mathrm{XeF}_2$, mathrmClF_3$\mathrm{ClF}_3$ and mathrmSF_4$\mathrm{SF}_4$, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is

### Core Logic Let's compute the total valence electrons for the nitrite ion (NO_2^-$NO_2^-$): textValence electrons = 5text (from N) + 2 times 6text (from O) + 1text (negative charge) = 18text electrons$$\text{Valence electrons} = 5\text{ (from N)} + 2 \times 6\text{ (from O)} + 1\text{ (negative charge)} = 18\text{ electrons}$$ In the valid Lewis structural representation: * The central nitrogen atom forms one single bond and one double bond with the terminal oxygens, consuming 2 + 4 = 6$2 + 4 = 6$ bonding electrons. * Remaining non-bonded valence electrons = 18 - 6 = 12$18 - 6 = 12$ electrons. ### Step 1: Account for Lone Pairs Distribution of non-bonded electrons across the individual atoms: * Central Nitrogen atom has 1$1$ lone pair (2$2$ electrons). * Single-bonded Oxygen atom has 3$3$ lone pairs (6$6$ electrons). * Double-bonded Oxygen atom has 2$2$ lone pairs (4$4$ electrons). textTotal non-bonded electrons = 2 + 6 + 4 = 12$$\text{Total non-bonded electrons} = 2 + 6 + 4 = 12$$ ### Pattern Recognition Non-bonded electrons can always be obtained directly by subtracting total bonding electrons from total valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Step 1: Determine 'n' mathrmClF_3$\mathrm{ClF}_3$ has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus, n = 2$n = 2$. ### Step 2: Find ions with 2 unpaired electrons Let us compute the number of unpaired electrons for each configuration: - **A. mathrmV^3+$\mathrm{V}^{3+}$:** [mathrmAr] 3d^2 rightarrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons. - **B. mathrmTi^3+$\mathrm{Ti}^{3+}$:** [mathrmAr] 3d^1 rightarrow 1$[\mathrm{Ar}] 3d^1 \rightarrow 1$ unpaired electron. - **C. mathrmCu^2+$\mathrm{Cu}^{2+}$:** [mathrmAr] 3d^9 rightarrow 1$[\mathrm{Ar}] 3d^9 \rightarrow 1$ unpaired electron. - **D. mathrmNi^2+$\mathrm{Ni}^{2+}$:** [mathrmAr] 3d^8 rightarrow 2$[\mathrm{Ar}] 3d^8 \rightarrow 2$ unpaired electrons. - **E. mathrmTi^2+$\mathrm{Ti}^{2+}$:** [mathrmAr] 3d^2 rightarrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons. Thus, A, D, and E have exactly n=2$n=2$ unpaired electrons. ### Pattern Recognition Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember mathrmClF_3$\mathrm{ClF}_3$ is T-shaped with 2 equatorial lone pairs. Look for d^2$d^2$ or d^8$d^8$ configurations among the transition metal ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

### Core Logic Let us check the steric details using VSEPR theory: - **mathrmBrF_5$\mathrm{BrF}_5$:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (sp^3d^2$sp^3d^2$), geometry is square pyramidal. - **mathrmXeOF_4$\mathrm{XeOF}_4$:** Xenon has 8 valence electrons. It forms 1 double bond with Oxygen, 4 single bonds with Fluorine, and retains 1 lone pair. Steric number = 6 (sp^3d^2$sp^3d^2$), geometry is square pyramidal. - **mathrmSbF_5$\mathrm{SbF}_5$ & mathrmPCl_5$\mathrm{PCl}_5$:** Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 (sp^3d$sp^3d$), geometry is trigonal bipyramidal. Visual representations of geometries:

### Pattern Recognition Sees: Steric count 6 with 5 bonded segments + 1 lone pair rightarrow$\rightarrow$ always square pyramidal geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Let us evaluate each central atom configuration systematically: * A. PF_5$PF_5$: Phosphorus has 5 valence electrons, forming 5sigma$5\sigma$ bonds with zero lone pairs. Steric number = 5 implies sp^3d$= 5 implies sp^3d$ hybridisation. * B. SF_6$SF_6$: Sulfur has 6 valence electrons, forming 6sigma$6\sigma$ bonds with zero lone pairs. Steric number = 6 implies sp^3d^2$= 6 implies sp^3d^2$ hybridisation. * C. Ni(CO)_4$Ni(CO)_4$: Nickel is in a 0 oxidation state (3d^8 4s^2$3d^8 4s^2$). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled 3d^10$3d^{10}$ state. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridise to give an sp^3$sp^3$ configuration.

* D. [PtCl_4]^2-$[PtCl_4]^{2-}$: Platinum is in the +2$+2$ oxidation state (5d^8$5d^8$). Since it belongs to the 5d$5d$ transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar dsp^2$dsp^2$ hybridisation state.

### Pattern Recognition Shortcut: Match structural main elements first: PF_5 ightarrow sp^3d$PF_5 ightarrow sp^3d$ (II), SF_6 ightarrow sp^3d^2$SF_6 ightarrow sp^3d^2$ (III). This isolates option (1) instantly without checking d-block coordinate fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

### Related Formula textSteric Number for ClF_3 = frac7+32 = 5 implies sp^3d text hybridization (Trigonal Bipyramidal geometry)$$\text{Steric Number for } ClF_3 = \frac{7+3}{2} = 5 \implies sp^3d \text{ hybridization (Trigonal Bipyramidal geometry)}$$ ### Core Logic - **Statement I is correct:** The three structural arrangements represent the different ways to place three bond pairs and two lone pairs within a trigonal bipyramidal grid. - **Statement II is incorrect:** According to VSEPR theory and Bent's rule, in sp^3d$sp^3d$ hybridization, **lone pairs must occupy equatorial positions** to minimize strong 90^circ$90^\circ$ lone pair-bond pair (ell p-bp$\ell p-bp$) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable. ### Pattern Recognition For sp^3d$sp^3d$ configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience 120^circ$120^circ$ interactions, minimizing severe 90^circ$90^circ$ structural strains. This results in the classic stable T-shaped configuration for ClF_3$ClF_3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure