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:

Related Formula

Core Logic

Step 1: Compare dipole moments:

• H₂O = 1.85 D$\text{H}_2\text{O} = 1.85\text{ D}$
• NH₃ = 1.47 D$\text{NH}_3 = 1.47\text{ D}$
• CHCl₃ = 1.04 D$\text{CHCl}_3 = 1.04\text{ D}$
• H₂S = 0.95 D$\text{H}_2\text{S} = 0.95\text{ D}$
• NF₃ = 0.23 D$\text{NF}_3 = 0.23\text{ D}$ (lowest dipole moment)

Step 2: Identify molecule X = NF₃$X = \text{NF}_3$.

Step 3: Central atom is N (2s² 2p³$2s^2 2p^3$). It forms 3 single bonds with F atoms and retains 1 lone pair of electrons.

Pattern Recognition

Sees: Comparison between NH₃$\text{NH}_3$ and NF₃$\text{NF}_3$ dipoles. Shortcut: In NF₃$\text{NF}_3$, fluorine's high electronegativity pulls electron density away from the lone pair direction, resulting in an exceptionally low dipole moment (0.23 D).

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Draw Lewis structures for all given molecules and sum the total lone pairs on all atoms. HNO₃$HNO_3$: 7 lone pairs total. H₂SO₄$H_2SO_4$: 8 lone pairs total. O₃$O_3$: 6 lone pairs total. NF₃$NF_3$: Nitrogen has 1 lone pair, and each of the three fluorine atoms has 3 lone pairs. Total = 1 + (3 × 3) = 10$1 + (3 \times 3) = 10$ lone pairs.

Step 1: Identifying Molecule X

Molecule X is NF₃$NF_3$ due to possessing the maximum number of lone pairs (10).

Step 2: Bond Angle Analysis

In NF₃$NF_3$, Nitrogen is sp³$sp^3$ hybridized. Due to the high electronegativity of Fluorine, the bond pair electron density shifts towards fluorine. This reduces bond pair-bond pair repulsion around the central nitrogen compared to ammonia (NH₃$NH_3$). As a result, the lone pair compresses the F-N-F bond angle more severely than in NH₃$NH_3$ (107^°$107^\circ$). The resulting F-N-F bond angle is approximately 102^°$102^\circ$.

Pattern Recognition

Electronegativity rules: If the surrounding atoms are more electronegative than the central atom, bond angles decrease because bonding electrons are pulled away from the central atom, allowing the lone pair to expand further and crush the angle.

Chapter Mix

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

Core Logic

(A) Dipole moment of NF₃$\mathrm{NF}_3$ is lower than that of NH₃$\mathrm{NH}_3$ because in NF₃$\mathrm{NF}_3$, the orbital dipole due to the lone pair and the bond dipoles (N-F) are in opposite directions, whereas in NH₃$\mathrm{NH}_3$ they reinforce each other. (Statement A is false).

(B) BeH₂$\mathrm{BeH}_2$ is sp$sp$ hybridized, a linear molecule. The two Be-H bond dipoles cancel out, resulting in a zero net dipole moment. (Statement B is true).

(C) Bond order of O₂²⁻$\mathrm{O}_2^{2-}$ (peroxide ion, 18 e^-$18 e^-$) = 1$= 1$. Bond order of F₂$\mathrm{F}_2$ (18 e^-$18 e^-$) = 1$= 1$. They are isoelectronic and have the same bond order. (Statement C is true).

(D) The formal charge on the central oxygen atom in ozone (O₃$\mathrm{O}_3$) is +1$+1$, not -1$-1$. (Statement D is false).

(E) In NO₂$\mathrm{NO}_2$, nitrogen is an odd-electron species (7+16 = 23 e^-$7+16 = 23 e^-$), thus it does not follow the octet rule. (Statement E is false).

Step 1: Final Conclusion

Only statements B and C are correct.

Pattern Recognition

Ammonia vs Nitrogen trifluoride is a classic NCERT dipole exception. Central atoms with odd electrons (like N in NO$\mathrm{NO}$, NO₂$\mathrm{NO}_2$, ClO₂$\mathrm{ClO}_2$) inherently violate the octet rule.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

(A) IF₃$\mathrm{IF}_3$: Iodine has 7 valence electrons. 3$3$ are shared with F, leaving 2$2$ lone pairs. Total electron pairs = 3 (bond) + 2 (lone) = 5$3 \text{ (bond)} + 2 \text{ (lone)} = 5$. Hybridization is sp³d$sp^3d$. Geometry is trigonal bipyramidal, shape is T-shaped. (Correct)

(B) IF₅$\mathrm{IF}_5$: Iodine has 7 valence electrons. 5$5$ are shared with F, leaving 1$1$ lone pair. Total electron pairs = 5 (bond) + 1 (lone) = 6$5 \text{ (bond)} + 1 \text{ (lone)} = 6$. Hybridization is sp³d²$sp^3d^2$. Shape is Square pyramidal. (Correct)

(C) IF₇$\mathrm{IF}_7$: Iodine has 7 valence electrons, all 7$7$ shared with F. Zero lone pairs. Total pairs = 7. Hybridization is sp³d³$sp^3d^3$. Shape is Pentagonal bipyramidal. (Correct)

(D) ClO₄^-$\mathrm{ClO}_4^-$: Chlorine forms 4 bonds (mostly double) with Oxygen, plus one extra electron overall. 4 bond regions + 0 lone pairs = 4$4 \text{ bond regions} + 0 \text{ lone pairs} = 4$. Hybridization is sp³$sp^3$. Shape is Tetrahedral. (Incorrect)

Step 1: Final Conclusion

Only combinations A, B, and C are correct.

Pattern Recognition

Steric Number (SN) = (Valence e^-$e^-$ on central atom + Monovalent atoms - Cation charge + Anion charge) / 2. Use SN to quickly deduce hybridization: 4 arrow sp³$\rightarrow sp^3$, 5 arrow sp³d$\rightarrow sp^3d$, 6 arrow sp³d²$\rightarrow sp^3d^2$, 7 arrow sp³d³$\rightarrow sp^3d^3$.

Chapter Mix

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

Core Logic

We need to determine the bond order and magnetic nature of the given diatomic species using Molecular Orbital Theory.

As seen, both O₂^+$O_2^+$ and N₂^-$N_2^-$ have a bond order of 2.5 and contain unpaired electrons making them paramagnetic.

Pattern Recognition

Isoelectronic species or species with equivalent valence electrons (e.g., both having 15 electrons) will exhibit identical bond orders (2.5) and have 1 unpaired electron, ensuring paramagnetism.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure