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

### Step 1: Determine the bond order of textCO$\text{CO}$ Carbon monoxide (textCO$\text{CO}$) contains 6 + 8 = 14$6 + 8 = 14$ total electrons. Its structural representation is textCequivtextO$\text{C}\equiv\text{O}$, matching a bond order value of 3. ### Step 2: Determine the bond order of textNO^+$\text{NO}^+$ The nitrosonium ion (textNO^+$\text{NO}^+$) contains 7 + 8 - 1 = 14$7 + 8 - 1 = 14$ total electrons. Since it is isoelectronic with textN_2$\text{N}_2$ and textCO$\text{CO}$ (14text electrons$14\text{ electrons}$), its corresponding bond order value is also 3. ### Step 3: Sum the results textSum = 3 + 3 = 6$$\text{Sum} = 3 + 3 = 6$$ ### Pattern Recognition Isoelectronic species possessing 14 total electrons consistently demonstrate a bond order value of 3. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Let us determine the steric number (Z$Z$) and number of lone pairs (LP$LP$) for the central atom in each given molecule. Formula: Z = frac12 (V + M - C + A)$Z = \frac{1}{2} (V + M - C + A)$ Where V$V$ = valence electrons on central atom, M$M$ = number of monovalent atoms, C$C$ = cationic charge, A$A$ = anionic charge. LP = Z - textBond Pairs (B.P.)$LP = Z - \text{Bond Pairs (B.P.)}$ 1. **O_3$O_3$**: Central atom O (V=6$V=6$). It forms one double bond and one dative bond. It has 1 lone pair remaining. 2. **H_2O$H_2O$**: Central atom O (V=6$V=6$). Z = frac12(6 + 2) = 4$Z = \frac{1}{2}(6 + 2) = 4$. LP = 4 - 2 = 2$LP = 4 - 2 = 2$. 3. **SF_4$SF_4$**: Central atom S (V=6$V=6$). Z = frac12(6 + 4) = 5$Z = \frac{1}{2}(6 + 4) = 5$. LP = 5 - 4 = 1$LP = 5 - 4 = 1$ (See-saw shape). 4. **ClF_3$ClF_3$**: Central atom Cl (V=7$V=7$). Z = frac12(7 + 3) = 5$Z = \frac{1}{2}(7 + 3) = 5$. LP = 5 - 3 = 2$LP = 5 - 3 = 2$ (T-shape). 5. **NH_3$NH_3$**: Central atom N (V=5$V=5$). Z = frac12(5 + 3) = 4$Z = \frac{1}{2}(5 + 3) = 4$. LP = 4 - 3 = 1$LP = 4 - 3 = 1$ (Pyramidal). 6. **BrF_5$BrF_5$**: Central atom Br (V=7$V=7$). Z = frac12(7 + 5) = 6$Z = \frac{1}{2}(7 + 5) = 6$. LP = 6 - 5 = 1$LP = 6 - 5 = 1$ (Square Pyramidal). 7. **XeF_4$XeF_4$**: Central atom Xe (V=8$V=8$). Z = frac12(8 + 4) = 6$Z = \frac{1}{2}(8 + 4) = 6$. LP = 6 - 4 = 2$LP = 6 - 4 = 2$ (Square Planar). ### Step 1: Final Counting

The compounds containing exactly ONE lone pair on the central atom are O_3$O_3$, SF_4$SF_4$, NH_3$NH_3$, and BrF_5$BrF_5$. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Using Molecular Orbital (MO) Theory, we evaluate the bond order (BO = fracN_b - N_a2$BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species: ### Step 1: Final Selection We need the species that satisfies BOTH conditions: 1. Paramagnetic 2. Bond Order = 1 Looking at the table, B_2$B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate pi_2p$\pi_{2p}$ orbitals) and has a bond order of 1. Total number of such species = 1. ### Pattern Recognition B_2$B_2$ (10 electrons) and O_2$O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. B_2$B_2$ has BO = 1, and O_2$O_2$ has BO = 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic According to VSEPR theory: 1. [Ni(CN)_4]^2-$[Ni(CN)_4]^{2-}$: dsp^2$dsp^2$ hybridization rightarrow$\rightarrow$ Square Planar. 2. PCl_5$PCl_5$: sp^3d$sp^3d$ hybridization with 0 lone pairs rightarrow$\rightarrow$ Trigonal Bipyramidal. 3. BrF_5$BrF_5$: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine, leaving 1 lone pair. sp^3d^2$sp^3d^2$ hybridization rightarrow$\rightarrow$ geometry is octahedral, but shape is Square Pyramidal. 4. PF_5$PF_5$: sp^3d$sp^3d$ hybridization with 0 lone pairs rightarrow$\rightarrow$ Trigonal Bipyramidal.

### Pattern Recognition AX_5E_1$AX_5E_1$ configuration (5 bond pairs + 1 lone pair) typically yields a Square Pyramidal shape. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

### Core Logic Statement I: The net dipole moment of NH_3$NH_3$ (1.47\, D$1.47\, D$) is actually greater than that of NF_3$NF_3$ (0.23\, D$0.23\, D$). Therefore, Statement I is false. Statement II: In NH_3$NH_3$, the N-H$N-H$ bond dipole moments (pointing towards the more electronegative N) reinforce the orbital dipole moment of the lone pair. In NF_3$NF_3$, 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 NF_3$NF_3$ 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 NH_3$NH_3$ and NF_3$NF_3$, both are false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure