### Core Logic
The ionic character of a bond is directly proportional to the electronegativity difference (Delta EN$\Delta EN$) between the two bonded atoms.
Larger Delta EN implies$\Delta EN \implies$ higher ionic character.
### Step 1: Assess Electronegativity Differences
- N_2$N_2$: Both atoms are Nitrogen. Delta EN = 0$\Delta EN = 0$. Purely covalent. (Lowest ionic character)
- SO_2$SO_2$: Bond between S and O. Moderate Delta EN$\Delta EN$. Covalent with some polarity.
- ClF_3$ClF_3$: Bond between Cl and F. Delta EN$\Delta EN$ is higher than S-O as F is the most electronegative element.
- K_2O$K_2O$: Bond between K (alkali metal, very low EN) and O. Very high Delta EN$\Delta EN$. Ionic.
- LiF$LiF$: Bond between Li (alkali metal) and F (highest EN). Maximum Delta EN$\Delta EN$ possible among these options. Most ionic.
### Step 2: Order Derivation
Increasing order of ionic character (or Delta EN$\Delta EN$):
N_2 < SO_2 < ClF_3 < K_2O < LiF$N_2 < SO_2 < ClF_3 < K_2O < LiF$
### Pattern Recognition
Homodiatomic (N_2$N_2$) is always 0% ionic. Alkali metal + Halogen (LiF$LiF$) represents the extreme of the ionic spectrum. Sorting non-metals by group distance yields the middle ranks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Keywords:#ionic character order#JEE Main 2024 Morning Q74#Chemical Bonding and Molecular Structure JEE Main 2024
More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 5
Q81jee_main_2024_29_jan_morningVSEPR Theory
Number of compounds with one lone pair of electrons on central atom amongst following is
O_3$O_3$, H_2O$H_2O$, SF_4$SF_4$, ClF_3$ClF_3$, NH_3$NH_3$, BrF_5$BrF_5$, XeF_4$XeF_4$
Numerical Answer.Answer: 4 to 4
Solution
### 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
VSEPR Theory diagram for Q81 - JEE Main 2024 MorningVSEPR Theory diagram for Q81 - JEE Main 2024 MorningVSEPR Theory diagram for Q81 - JEE Main 2024 Morning
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
Q88jee_main_2024_29_jan_morningMolecular Orbital Theory
The number of species from the following which are paramagnetic and with bond order equal to one is
mathrm H_2, mathrmHe_2^+, mathrmO_2^+, mathrmN_2^2-, mathrmO_2^2-, mathrmF_2, mathrmNe_2^+, mathrmB_2$\mathrm {H}_2, \mathrm{He}_2^+, \mathrm{O}_2^+, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^+, \mathrm{B}_2$
Numerical Answer.Answer: 1 to 1
Solution
### 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:
Species
Magnetic behaviour
Bond order
H_2$H_2$
Diamagnetic
1
He_2^+$He_2^+$
Paramagnetic
0.5
O_2^+$O_2^+$
Paramagnetic
2.5
N_2^2-$N_2^{2-}$
Paramagnetic
2
O_2^2-$O_2^{2-}$
Diamagnetic
1
F_2$F_2$
Diamagnetic
1
Ne_2^+$Ne_2^+$
Paramagnetic
0.5
B_2$B_2$
Paramagnetic
1
### 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
Q69jee_main_2024_30_january_eveningVSEPR Theory and Molecular Shapes
### 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.
Square Pyramidal structure of BrF5 diagram for Q69 - JEE Main 2024 Evening
### 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
Q75jee_main_2024_30_january_eveningDipole Moment
Given below are two statements:
Statement-I: Since fluorine is more electronegative than nitrogen, the net dipole moment of NF_3$NF_3$ is greater than NH_3$NH_3$.
Statement-II: In NH_3$NH_3$, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite direction, but in NF_3$NF_3$ the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction.
In the light of the above statements. Choose the most appropriate from the options given below.
A.textStatement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$
B.textBoth Statement I and Statement II are false.$\text{Both Statement I and Statement II are false.}$
C.textBoth statement I and Statement II is are true.$\text{Both statement I and Statement II is are true.}$
D.textStatement I is false but Statement II is are true.$\text{Statement I is false but Statement II is are true.}$
Solution
### 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.
Dipole moments in NH3 and NF3 diagram for Q75 - JEE Main 2024 EveningDipole moments in NH3 and NF3 diagram for Q75 - JEE Main 2024 Evening
### 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
### Core Logic
Using VSEPR theory:
(A) BrF_5$BrF_5$: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) rightarrow$\rightarrow$sp^3d^2$sp^3d^2$ hybridization rightarrow$\rightarrow$ Square pyramidal shape.
(B) H_2O$H_2O$: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) rightarrow$\rightarrow$sp^3$sp^3$ hybridization rightarrow$\rightarrow$ Bent shape.
(C) ClF_3$ClF_3$: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) rightarrow$\rightarrow$sp^3d$sp^3d$ hybridization rightarrow$\rightarrow$ T-shape.
(D) SF_4$SF_4$: S has 6 valence electrons. 4 form bonds with F, leaving 1 lone pair. (4 bp + 1 lp) rightarrow$\rightarrow$sp^3d$sp^3d$ hybridization rightarrow$\rightarrow$ See-saw shape.
### Step 1: Matching
(A) - (IV)
(B) - (III)
(C) - (I)
(D) - (II)
VSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 Morning
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
More Chemical Bonding and Molecular Structure Questions — jee_main_2024_01_february_morning
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