The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:

### 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

### Step 1: Determine 'n' mathrmClF_3$\mathrm{ClF}_3$ has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus, n = 2$n = 2$. ### Step 2: Find ions with 2 unpaired electrons Let us compute the number of unpaired electrons for each configuration: - **A. mathrmV^3+$\mathrm{V}^{3+}$:** [mathrmAr] 3d^2 rightarrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons. - **B. mathrmTi^3+$\mathrm{Ti}^{3+}$:** [mathrmAr] 3d^1 rightarrow 1$[\mathrm{Ar}] 3d^1 \rightarrow 1$ unpaired electron. - **C. mathrmCu^2+$\mathrm{Cu}^{2+}$:** [mathrmAr] 3d^9 rightarrow 1$[\mathrm{Ar}] 3d^9 \rightarrow 1$ unpaired electron. - **D. mathrmNi^2+$\mathrm{Ni}^{2+}$:** [mathrmAr] 3d^8 rightarrow 2$[\mathrm{Ar}] 3d^8 \rightarrow 2$ unpaired electrons. - **E. mathrmTi^2+$\mathrm{Ti}^{2+}$:** [mathrmAr] 3d^2 rightarrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons. Thus, A, D, and E have exactly n=2$n=2$ unpaired electrons. ### Pattern Recognition Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember mathrmClF_3$\mathrm{ClF}_3$ is T-shaped with 2 equatorial lone pairs. Look for d^2$d^2$ or d^8$d^8$ configurations among the transition metal ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

### Core Logic Let us check the steric details using VSEPR theory: - **mathrmBrF_5$\mathrm{BrF}_5$:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (sp^3d^2$sp^3d^2$), geometry is square pyramidal. - **mathrmXeOF_4$\mathrm{XeOF}_4$:** Xenon has 8 valence electrons. It forms 1 double bond with Oxygen, 4 single bonds with Fluorine, and retains 1 lone pair. Steric number = 6 (sp^3d^2$sp^3d^2$), geometry is square pyramidal. - **mathrmSbF_5$\mathrm{SbF}_5$ & mathrmPCl_5$\mathrm{PCl}_5$:** Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 (sp^3d$sp^3d$), geometry is trigonal bipyramidal. Visual representations of geometries:

### Pattern Recognition Sees: Steric count 6 with 5 bonded segments + 1 lone pair rightarrow$\rightarrow$ always square pyramidal geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Let us evaluate each central atom configuration systematically: * A. PF_5$PF_5$: Phosphorus has 5 valence electrons, forming 5sigma$5\sigma$ bonds with zero lone pairs. Steric number = 5 implies sp^3d$= 5 implies sp^3d$ hybridisation. * B. SF_6$SF_6$: Sulfur has 6 valence electrons, forming 6sigma$6\sigma$ bonds with zero lone pairs. Steric number = 6 implies sp^3d^2$= 6 implies sp^3d^2$ hybridisation. * C. Ni(CO)_4$Ni(CO)_4$: Nickel is in a 0 oxidation state (3d^8 4s^2$3d^8 4s^2$). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled 3d^10$3d^{10}$ state. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridise to give an sp^3$sp^3$ configuration.

* D. [PtCl_4]^2-$[PtCl_4]^{2-}$: Platinum is in the +2$+2$ oxidation state (5d^8$5d^8$). Since it belongs to the 5d$5d$ transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar dsp^2$dsp^2$ hybridisation state.

### Pattern Recognition Shortcut: Match structural main elements first: PF_5 ightarrow sp^3d$PF_5 ightarrow sp^3d$ (II), SF_6 ightarrow sp^3d^2$SF_6 ightarrow sp^3d^2$ (III). This isolates option (1) instantly without checking d-block coordinate fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

### Related Formula textSteric Number for ClF_3 = frac7+32 = 5 implies sp^3d text hybridization (Trigonal Bipyramidal geometry)$$\text{Steric Number for } ClF_3 = \frac{7+3}{2} = 5 \implies sp^3d \text{ hybridization (Trigonal Bipyramidal geometry)}$$ ### Core Logic - **Statement I is correct:** The three structural arrangements represent the different ways to place three bond pairs and two lone pairs within a trigonal bipyramidal grid. - **Statement II is incorrect:** According to VSEPR theory and Bent's rule, in sp^3d$sp^3d$ hybridization, **lone pairs must occupy equatorial positions** to minimize strong 90^circ$90^\circ$ lone pair-bond pair (ell p-bp$\ell p-bp$) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable. ### Pattern Recognition For sp^3d$sp^3d$ configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience 120^circ$120^circ$ interactions, minimizing severe 90^circ$90^circ$ structural strains. This results in the classic stable T-shaped configuration for ClF_3$ClF_3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure