Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of ClF₃$\mathrm{ClF}_3$ . The ions from the following with 'n' number of unpaired electrons are : A. V3 +$\mathrm{V}^{3 + }$ B. Ti³⁺$\mathrm{Ti}^{3+}$ C. Cu2 +$\mathrm{Cu}^{2 + }$ D. Ni²⁺$\mathrm{Ni}^{2+}$ E. Ti²⁺$\mathrm{Ti}^{2+}$ Choose the correct answer from the options given below:

Step 1: Evaluate Statement I

BF₄⁻$BF_{4}^{-}$ (Tetrahedral, all bonds identical)\nSiF₄$SiF_{4}$ (Tetrahedral, all bonds identical)\nXeF₄$XeF_{4}$ (Square planar, all bonds identical)\nSF₄$SF_{4}$ (See-saw geometry, axial and equatorial bond lengths are unequal).\nThus, only ONE species (SF₄$SF_4$) has unequal bond lengths. Statement I is false.

Step 2: Evaluate Statement II

Using MOT to find bond orders (B.O.):\nO₂⁺$O_2^{+}$: B.O. = 2.5$2.5$\nO₂⁻$O_2^{-}$: B.O. = 1.5$1.5$\nO₂²⁻$O_2^{2-}$: B.O. = 1$1$\nF₂$F_2$: B.O. = 1$1$\nThe species with the highest bond order is O₂⁺$O_2^{+}$ (2.5$2.5$), not O₂⁻$O_2^{-}$. Statement II is false.

Final Conclusion

Both statements are false.

Pattern Recognition

Axial bonds are structurally longer than equatorial bonds in trigonal bipyramidal derivatives (like see-saw SF₄$SF_4$). Bond order maps predictably: 14e^- = 3.0$14e^- = 3.0$, stepping down 0.5$0.5$ per electron added or removed.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Evaluate steric number (SN) = Bond Pairs (BP) + Lone Pairs (LP): (A) XeO₃$XeO_3$: 3 BP, 1 LP arrow$\rightarrow$ Pyramidal geometry. Matches NH₃$NH_3$ (3 BP, 1 LP). (B) XeF₂$XeF_2$: 2 BP, 3 LP arrow$\rightarrow$ Linear geometry. Matches [I₃]^-$[I_3]^-$ (2 BP, 3 LP). (C) XeO₂F₂$XeO_2F_2$: 4 BP, 1 LP arrow$\rightarrow$ See-saw geometry. Matches SF₄$SF_4$ (4 BP, 1 LP). (D) XeOF₄$XeOF_4$: 5 BP, 1 LP arrow$\rightarrow$ Square pyramidal geometry. Matches BrF₅$BrF_5$ (5 BP, 1 LP).

Step 1: Final Conclusion

Matching pairs: A-II, B-III, C-IV, D-I.

Pattern Recognition

VSEPR iso-structural matching. Always count valence electrons of central atom minus bonds to find lone pairs. 4 domains with 1 LP = See-saw. 5 domains with 3 LP = Linear. 4 domains with 1 LP (sp3) = Pyramidal.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The p-Block Elements

Related Formula

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:

• PF₅$\mathrm{PF}_5$:

  

• Central atom: Phosphorus (V=5$V=5$).
• Steric Number = (1)/(2)(5 + 5) = 5 sp³d$= \frac{1}{2}(5 + 5) = 5 \implies \mathrm{sp^3d}$ 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.
• XeF₄$\mathrm{XeF}_4$:
• Central atom: Xenon (V=8$V=8$).
• Steric Number = (1)/(2)(8 + 4) = 6 sp³d²$= \frac{1}{2}(8 + 4) = 6 \implies \mathrm{sp^3d^2}$ hybridization (fails condition a).
• SF₄$\mathrm{SF}_4$:

  

• Central atom: Sulfur (V=6$V=6$).
• Steric Number = (1)/(2)(6 + 4) = 5 sp³d$= \frac{1}{2}(6 + 4) = 5 \implies \mathrm{sp^3d}$ 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~ A$1.64~\mathrm{\mathring{A}}$ vs 1.54~ A$1.54~\mathrm{\mathring{A}}$ due to lone pair-bond pair repulsion). This satisfies all three conditions.
• XeF₂$\mathrm{XeF}_2$:
• Central atom: Xenon (V=8$V=8$).
• Steric Number = (1)/(2)(8 + 2) = 5 sp³d$= \frac{1}{2}(8 + 2) = 5 \implies \mathrm{sp^3d}$ 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 SF₄$\mathrm{SF}_4$ 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^°$90^\circ$) than the equatorial bonds (which have only 2 axial neighbors at 90^°$90^\circ$). Consequently, the axial bonds are always longer and weaker than equatorial bonds.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Related Formula

Total valence shell electron pair system equation:

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 (F$\mathrm{F}$) and Oxygen (O$\mathrm{O}$), matching labels X$\mathrm{X}$ and Y$\mathrm{Y}$.
• Element A$\mathrm{A}$ is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (Xe$\mathrm{Xe}$).
• Substituting these components into the target layout formula yields XeOF₄$\mathrm{XeOF_4}$:
• 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 bond regions + 1 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 sp³d²$\mathrm{sp^3d^2}$ architectures containing an explicit lone pair along with an asymmetric double bond (like XeOF₄$\mathrm{XeOF_4}$), the lone pair always sits directly opposite the double bond to minimize electron repulsion, leaving a clean square pyramidal shape.

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:

Core Logic

Let's list parameters using a detailed structural grid:

Comparing items: XeF₂$\mathrm{XeF}_2$ has 3 lone pairs but its linear architecture enforces μ = 0$\mu = 0$. Therefore, ClF₃$\mathrm{ClF}_3$ has the highest count of lone pairs (2) with a net non-zero asymmetric dipole configuration.

Step 1: Selection

The hybridization of ClF₃$\mathrm{ClF}_3$ is sp³d$\mathrm{sp}^3\mathrm{d}$.

Pattern Recognition

Watch out for symmetry traps! XeF₂$\mathrm{XeF}_2$ contains the absolute maximum lone pairs, but its symmetric planar positioning perfectly cancels out the dipole vectors. Thus, the correct candidate slips down to ClF₃$\mathrm{ClF}_3$.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure