### Related Formula textSteric Number = frac12 left( V + M - C + A right)$\text{Steric Number} = \frac{1}{2} \left( V + M - C + A \right)$ where, V$V$ = valence electrons of central atom M$M$ = number of monovalent surrounding atoms C$C$ = cationic charge, A$A$ = anionic charge ### Core Logic Let's calculate the hybridization, shape, and lone pairs for each option: 1. mathrmPF_5$\mathrm{PF}_5$:

- Central atom: Phosphorus (V=5$V=5$). - Steric Number = frac12(5 + 5) = 5 implies mathrmsp^3d$= \frac{1}{2}(5 + 5) = 5 \implies \mathrm{sp^3d}$ hybridization. - Lone pairs = 5 - 5 = 0$= 5 - 5 = 0$. - Geometry: Trigonal bipyramidal. It has different axial and equatorial bond lengths, but no lone pair on the central atom. 2. mathrmXeF_4$\mathrm{XeF}_4$: - Central atom: Xenon (V=8$V=8$). - Steric Number = frac12(8 + 4) = 6 implies mathrmsp^3d^2$= \frac{1}{2}(8 + 4) = 6 \implies \mathrm{sp^3d^2}$ hybridization (fails condition a). 3. mathrmSF_4$\mathrm{SF}_4$:

- Central atom: Sulfur (V=6$V=6$). - Steric Number = frac12(6 + 4) = 5 implies mathrmsp^3d$= \frac{1}{2}(6 + 4) = 5 \implies \mathrm{sp^3d}$ hybridization. - Lone pairs = 5 - 4 = 1$= 5 - 4 = 1$ lone pair on sulfur. - Shape: See-saw. It contains axial and equatorial bonds which have distinct lengths (1.64~mathrmmathringA$1.64~\mathrm{\mathring{A}}$ vs 1.54~mathrmmathringA$1.54~\mathrm{\mathring{A}}$ due to lone pair-bond pair repulsion). This satisfies all three conditions. 4. mathrmXeF_2$\mathrm{XeF}_2$: - Central atom: Xenon (V=8$V=8$). - Steric Number = frac12(8 + 2) = 5 implies mathrmsp^3d$= \frac{1}{2}(8 + 2) = 5 \implies \mathrm{sp^3d}$ hybridization. - Lone pairs = 5 - 2 = 3$= 5 - 2 = 3$ lone pairs on Xe. - Shape: Linear. Both Xe-F bonds are identical in length (fails condition b).

### Step 1: Conclusion Hence, only mathrmSF_4$\mathrm{SF}_4$ satisfies all the given parameters. ### Pattern Recognition For any trigonal bipyramidal molecular geometry (steric number 5), the axial bonds suffer more repulsion (from 3 equatorial bonds at 90^circ$90^\circ$) than the equatorial bonds (which have only 2 axial neighbors at 90^circ$90^\circ$). Consequently, the axial bonds are always longer and weaker than equatorial bonds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Related Formula Total valence shell electron pair system equation: textValence Pairs = textBond Pairs (BP) + textLone Pairs (LP)$\text{Valence Pairs} = \text{Bond Pairs (BP)} + \text{Lone Pairs (LP)}$ ### Core Logic Let's decode individual identities based on the descriptive properties: * The elements with the first and second highest electronegativity values across the entire periodic table are Fluorine (mathrmF$\mathrm{F}$) and Oxygen (mathrmO$\mathrm{O}$), matching labels mathrmX$\mathrm{X}$ and mathrmY$\mathrm{Y}$. * Element mathrmA$\mathrm{A}$ is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (mathrmXe$\mathrm{Xe}$). * Substituting these components into the target layout formula yields mathrmXeOF_4$\mathrm{XeOF_4}$: - Xenon brings 8 valence electrons. It forms 4 single bonds with F and 1 double bond with O, consuming 6 electrons and leaving 1 lone pair on the central atom. - Steric Number = 5 text bond regions + 1 text lone pair = 6$= 5 \text{ bond regions} + 1 \text{ lone pair} = 6$ (Octahedral electronic arrangement). ### Step 1: Geometry Determination Placing the double-bonded oxygen and lone pair along vertical spatial axes yields a stable square pyramidal molecular shape layout:

### Pattern Recognition In mathrmsp^3d^2$\mathrm{sp^3d^2}$ architectures containing an explicit lone pair along with an asymmetric double bond (like mathrmXeOF_4$\mathrm{XeOF_4}$), the lone pair always sits directly opposite the double bond to minimize electron repulsion, leaving a clean square pyramidal shape. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The p-Block Elements

### Related Formula Steric Number system equation for identifying electronic configurations: textSteric Number (SN) = frac12[mathrmV + mathrmM - mathrmC + mathrmA]$\text{Steric Number (SN)} = \frac{1}{2}[\mathrm{V} + \mathrm{M} - \mathrm{C} + \mathrm{A}]$ ### Core Logic Let's list parameters using a detailed structural grid: Comparing items: mathrmXeF_2$\mathrm{XeF}_2$ has 3 lone pairs but its linear architecture enforces mu = 0$\mu = 0$. Therefore, mathrmClF_3$\mathrm{ClF}_3$ has the highest count of lone pairs (2) with a net non-zero asymmetric dipole configuration. ### Step 1: Selection The hybridization of mathrmClF_3$\mathrm{ClF}_3$ is mathrmsp^3mathrmd$\mathrm{sp}^3\mathrm{d}$. ### Pattern Recognition Watch out for symmetry traps! mathrmXeF_2$\mathrm{XeF}_2$ contains the absolute maximum lone pairs, but its symmetric planar positioning perfectly cancels out the dipole vectors. Thus, the correct candidate slips down to mathrmClF_3$\mathrm{ClF}_3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Let's find the number of bond pairs (sigma$\sigma$-bonds or regions) and lone pairs on the central atom of each species: 1. **mathrmICl_2^-$\mathrm{ICl}_2^-$**: - Central atom Iodine has 7$7$ valence electrons + 1$+ 1$ negative charge = 8$= 8$ electrons. - Forms 2$2$ single bonds (bond pairs = 2$= 2$). - Remaining 6$6$ electrons form 3$3$ lone pairs. - Ratio is 2 : 3$2 : 3$ (Matches LIST-II, III).

2. **mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$**: - Oxygen has 6$6$ valence electrons. - Forms 2$2$ bond pairs with Hydrogens. - Remaining 4$4$ electrons form 2$2$ lone pairs. - Ratio is 2 : 2$2 : 2$ (Matches LIST-II, IV).

3. **mathrmSO_2$\mathrm{SO}_2$**: - Sulfur has 6$6$ valence electrons. - Forms 2$2$ double bonds (which are counted as 4$4$ bonding pairs of electrons/bond pairs in typical VSEPR representations here). - Remaining 2$2$ electrons form 1$1$ lone pair. - Ratio is 4 : 1$4 : 1$ (Matches LIST-II, II).

4. **mathrmXeF_4$\mathrm{XeF}_4$**: - Xenon has 8$8$ valence electrons. - Forms 4$4$ bond pairs with Fluorines. - Remaining 4$4$ electrons form 2$2$ lone pairs. - Ratio is 4 : 2$4 : 2$ (Matches LIST-II, I).

Thus, the correct mapping is: A-III, B-IV, C-II, D-I. ### Pattern Recognition For match-the-column with VSEPR structures: - Always find steric number: textSteric Number = frac12(V + M - C + A)$\text{Steric Number} = \frac{1}{2}(V + M - C + A)$. - Water is 2$2$ bond pairs, 2$2$ lone pairs (sp^3$sp^3$) rightarrow$\rightarrow$ B-IV. This alone helps eliminate multiple incorrect options immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### 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