Match List-I with List-II according to shape.

List-I	List-II
(A) XeO_3$XeO_{3}$	(I) BrF_5$BrF_{5}$
(B) XeF_2$XeF_{2}$	(II) NH_3$NH_{3}$
(C) XeO_2F_2$XeO_{2}F_{2}$	(III) [I_3]^-$[I_{3}]^{-}$
(D) XeOF_4$XeOF_{4}$	(IV) SF_4$SF_{4}$

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

### Core Logic Comparing bond lengths: - C–H (A): sim 107 text pm$\sim 107 \text{ pm}$ - C≡N (D): sim 116 text pm$\sim 116 \text{ pm}$ - C=O (C): sim 121 text pm$\sim 121 \text{ pm}$ - C–O (B): sim 143 text pm$\sim 143 \text{ pm}$ ### Step 1: Final Conclusion Thus, the increasing order is A < D < C < B$A < D < C < B$, corresponding to option (2). ### Pattern Recognition Sees: covalent bond length comparison across bond orders and atomic radii. Trap: Assuming triple bonds are always longer or shorter without accounting for smaller atomic radii like hydrogen. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic - Statement I: Bond dissociation energy order for halogens is textCl_2 > textBr_2 > textF_2 > textI_2$\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$ due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so textUF_6 > textUF_4$\text{UF}_6 > \text{UF}_4$ (higher oxidation state has greater covalent character), making the statement II claim regarding textUF_4 > textUF_6$\text{UF}_4 > \text{UF}_6$ false. ### Step 1: Final Conclusion Statement I is true but Statement II is false, corresponding to option (1). ### Pattern Recognition Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Related Formula textFormal charge = (textValence e^-) - (textNon-bonding e^-) - fractextBonding e^-2$$\text{Formal charge} = (\text{Valence } e^{-}) - (\text{Non-bonding } e^{-}) - \frac{\text{Bonding } e^{-}}{2}$$ ### Core Logic Evaluate the structure of HNO_3$HNO_{3}$ shown in the solution image:

Atom 1 (Oxygen with H and N): F.C. = 6 - 4 - frac42 = 0$F.C. = 6 - 4 - \frac{4}{2} = 0$ Atom 2 (Nitrogen): F.C. = 5 - 0 - frac82 = +1$F.C. = 5 - 0 - \frac{8}{2} = +1$ Atom 3 (Double bonded Oxygen): F.C. = 6 - 4 - frac42 = 0$F.C. = 6 - 4 - \frac{4}{2} = 0$ Atom 4 (Single bonded Oxygen): F.C. = 6 - 6 - frac22 = -1$F.C. = 6 - 6 - \frac{2}{2} = -1$ ### Step 1: Final Conclusion The formal charges on atoms (1), (2), (3), and (4) are respectively 0, +1, 0, -1. ### Pattern Recognition In nitro groups (-NO_2$-NO_2$), the central nitrogen is always +1, the single bonded oxygen is -1, and the double-bonded oxygen is 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic XF_3$XF_3$ acts as a Lewis acid, meaning it is an electron-deficient species capable of accepting an electron pair. A common example from the p-block is BF_3$BF_3$. The central Boron atom has 3 bond pairs and 0 lone pairs. Therefore, its steric number is 3, corresponding to sp^2$sp^2$ hybridization. YF_3$YF_3$ acts as a Lewis base, meaning it has an available lone pair to donate. A common example is NF_3$NF_3$. The central Nitrogen atom has 3 bond pairs and 1 lone pair. Its steric number is 4, corresponding to sp^3$sp^3$ hybridization. ### Step 1: Final Conclusion The hybridization of X and Y respectively are sp^2$sp^2$ and sp^3$sp^3$. ### Pattern Recognition Electron deficient central atoms (Group 13) form sp^2$sp^2$ planar molecules. Atoms with a lone pair (Group 15) form sp^3$sp^3$ pyramidal molecules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 11 Chemistry: p-Block Elements