Match the LIST-I with LIST-II.

LIST-I (Molecule/ion)	LIST-II (Bond pair : lone pair on the central atom)
(A) mathrmICl_2^-$\mathrm{ICl}_2^-$	(I) 4 : 2$4 : 2$
(B) mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$	(II) 4 : 1$4 : 1$
(C) mathrmSO_2$\mathrm{SO}_2$	(III) 2 : 3$2 : 3$
(D) mathrmXeF_4$\mathrm{XeF}_4$	(IV) 2 : 2$2 : 2$

Choose the correct answer from the options given below:

### Core Logic Evaluating Statement I: - mathrmSF_4$\mathrm{SF}_4$: sp^3d$sp^3d$ (1 lone pair) rightarrow$\rightarrow$ See-saw - mathrmXeF_4$\mathrm{XeF}_4$: sp^3d^2$sp^3d^2$ (2 lone pairs) rightarrow$\rightarrow$ Square planar - [mathrmPtCl_4]^2-$[\mathrm{PtCl}_4]^{2-}$: dsp^2$dsp^2$ rightarrow$\rightarrow$ Square planar - [mathrmNiCl_4]^2-$[\mathrm{NiCl}_4]^{2-}$: sp^3$sp^3$ rightarrow$\rightarrow$ Tetrahedral - [mathrmNi(CN)_4]^2-$[\mathrm{Ni(CN)}_4]^{2-}$: dsp^2$dsp^2$ rightarrow$\rightarrow$ Square planar - mathrmSeF_4$\mathrm{SeF}_4$: sp^3d$sp^3d$ (1 lone pair) rightarrow$\rightarrow$ See-saw - mathrmNH_4^+$\mathrm{NH}_4^+$: sp^3$sp^3$ (0 lone pairs) rightarrow$\rightarrow$ Tetrahedral Total tetrahedral species = 2 ([mathrmNiCl_4]^2-$[\mathrm{NiCl}_4]^{2-}$ and mathrmNH_4^+$\mathrm{NH}_4^+$). Statement I says 3, so it is false. Evaluating Statement II: - NO_2$NO_2$: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). - BeH_2$BeH_2$: Central Be has 4 electrons (incomplete octet). - BF_3$BF_3$: Central B has 6 electrons (incomplete octet). - AlCl_3$AlCl_3$ (monomer): Central Al has 6 electrons (incomplete octet). Therefore, all molecules have incomplete octets. Statement II is true. ### Step 1: Final Conclusion Statement I is false, Statement II is true. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

### 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