### Core Logic Using VSEPR theory to find the hybridization and shape: 1. PF_5$PF_5$: P has 5 valence electrons, forms 5 single bonds with F. Steric number = 5 (sp3d). 0 lone pairs. Shape = Trigonal bipyramidal. 2. BrF_5$BrF_5$: Br has 7 valence electrons, forms 5 single bonds, 1 lone pair. Steric number = 6 (sp3d2). Shape = Square pyramidal. 3. PCl_5$PCl_5$: P has 5 valence electrons, 5 bonds, 0 lone pairs. Steric number = 5 (sp3d). Shape = Trigonal bipyramidal. 4. [PtCl_4]^2-$[PtCl_4]^{2-}$: Pt^2+$Pt^{2+}$ is a d^8$d^8$ system. With Cl^-$Cl^-$ (but 4d/5d transition metals always form low spin square planar complexes), it's dsp^2$dsp^2$ hybridized. Shape = Square planar. 5. BF_3$BF_3$: B has 3 valence electrons, 3 bonds, 0 lone pairs. Steric number = 3 (sp2). Shape = Trigonal planar. 6. Fe(CO)_5$Fe(CO)_5$: Fe (d6s2 -> d8 under strong field CO$CO$). Carbonyls strongly prefer 5-coordinate trigonal bipyramidal geometry for d^8$d^8$ (dsp^3$dsp^3$ hybridization). Shape = Trigonal bipyramidal. ### Step 1: Count Trigonal Bipyramidal Molecules Molecules with trigonal bipyramidal shape: - PF_5$PF_5$ - PCl_5$PCl_5$ - Fe(CO)_5$Fe(CO)_5$ Total count = 3. ### Pattern Recognition Steric Number = 5 with 0 lone pairs ALWAYS yields Trigonal Bipyramidal geometry. Watch out for BrF_5$BrF_5$ which has 5 bonds but 1 lone pair (SN = 6, Square Pyramidal). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

### Related Formula textTotal Atomic Orbitals Combinations = textBonding MOs + textAntibonding MOs$\text{Total Atomic Orbitals Combinations} = \text{Bonding MOs} + \text{Antibonding MOs}$ ### Core Logic When atomic orbitals combine, they form an equal number of molecular orbitals: * Two 2s$2s$ atomic orbitals combine to form **1** bonding orbital (sigma_2s$\sigma_{2s}$) and **1** antibonding orbital (sigma^*_2s$\sigma^*_{2s}$). * Six 2p$2p$ atomic orbitals combine to form **3** bonding orbitals (sigma_2p_z, pi_2p_x, pi_2p_y$\sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}$) and **3** antibonding orbitals (sigma^*_2p_z, pi^*_2p_x, pi^*_2p_y$\sigma^*_{2p_z}, \pi^*_{2p_x}, \pi^*_{2p_y}$). ### Step 1: Total Summation Summing the antibonding orbitals from both subshells: textTotal Antibonding Molecular Orbitals = 1 text (from 2s) + 3 text (from 2p) = 4$\text{Total Antibonding Molecular Orbitals} = 1 \text{ (from 2s)} + 3 \text{ (from 2p)} = 4$ ### Pattern Recognition The linear combination of N$N$ atomic orbitals always yields exactly fracN, 2$\frac{N}, {2}$ antibonding molecular orbitals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Related Formula vecmu_textnet = sum vecmu_i = 0 quad text(For perfectly symmetrical geometry configurations)$\vec{\mu}_{\text{net}} = \sum \vec{\mu}_i = 0 \quad \text{(For perfectly symmetrical geometry configurations)}$ ### Core Logic Analyze the molecular geometry and symmetry of each molecule: 1. textCH_4$\text{CH}_4$: Symmetrical tetrahedral geometry implies mu = 0$\implies \mu = 0$. 2. textBF_3$\text{BF}_3$: Symmetrical trigonal planar geometry implies mu = 0$\implies \mu = 0$. 3. textH_2textO$\text{H}_2\text{O}$: Bent shape due to lone pairs implies mu neq 0$\implies \mu \neq 0$. 4. textHF$\text{HF}$: Linear asymmetric diatomic molecule implies mu neq 0$\implies \mu \neq 0$. 5. textNH_3$\text{NH}_3$: Trigonal pyramidal shape due to a lone pair implies mu neq 0$\implies \mu \neq 0$. 6. textCO_2$\text{CO}_2$: Symmetrical linear structure (O=C=O$O=C=O$) where dipoles cancel out implies mu = 0$\implies \mu = 0$. 7. textSO_2$\text{SO}_2$: Bent angular geometry due to a lone pair implies mu neq 0$\implies \mu \neq 0$. ### Step 1: Final Counting The molecules with a net zero dipole moment are textCH_4$\text{CH}_4$, textBF_3$\text{BF}_3$, and textCO_2$\text{CO}_2$. This gives a total count of **3**. ### Pattern Recognition Molecules with a symmetrical arrangement of identical bonds and no lone pairs on the central atom (e.g., tetrahedral textCH_4$\text{CH}_4$, trigonal planar textBF_3$\text{BF}_3$, linear textCO_2$\text{CO}_2$) always have a net dipole moment of zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic CCl_4 rightarrow mu = 0 quad text(Symmetrical tetrahedral)$CCl_4 \rightarrow \mu = 0 \quad \text{(Symmetrical tetrahedral)}$ CO_2 rightarrow mu = 0 quad text(Linear structure)$CO_2 \rightarrow \mu = 0 \quad \text{(Linear structure)}$ CH_2=CH_2 rightarrow mu = 0 quad text(Planar symmetrical structure)$CH_2=CH_2 \rightarrow \mu = 0 \quad \text{(Planar symmetrical structure)}$ For CHCl_3$CHCl_3$, the individual dipole vectors do not cancel due to differing electronegativities of H and Cl, leading to a permanent non-zero dipole moment (mu neq 0$\mu \neq 0$).

### Pattern Recognition Symmetry yields vector cancellation rightarrow mu=0$\rightarrow \mu=0$. Asymmetry in CHCl_3$CHCl_3$ prevents cancellation rightarrow$\rightarrow$ polar. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### 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