Which among the following molecules is
(a) involved in mathrmsp^3d$\mathrm{sp^3d}$ hybridization,
(b) has different bond lengths and
(c) has lone pair of electrons on the central atom?
A.mathrmPF_5$\mathrm{PF}_5$
B.mathrmXeF_4$\mathrm{XeF}_4$
C.mathrmSF_4$\mathrm{SF}_4$
D.mathrmXeF_2$\mathrm{XeF}_2$
Solution & Explanation
### Related Formula
textSteric Number = frac12 left( V + M - C + A right)$$\text{Steric Number} = \frac{1}{2} \left( V + M - C + A \right)$$
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:
1. mathrmPF_5$\mathrm{PF}_5$:VSEPR Theory and Hybridization
- Central atom: Phosphorus (V=5$V=5$).
- Steric Number = frac12(5 + 5) = 5 implies mathrmsp^3d$= \frac{1}{2}(5 + 5) = 5 \implies \mathrm{sp^3d}$ 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.
2. mathrmXeF_4$\mathrm{XeF}_4$:
- Central atom: Xenon (V=8$V=8$).
- Steric Number = frac12(8 + 4) = 6 implies mathrmsp^3d^2$= \frac{1}{2}(8 + 4) = 6 \implies \mathrm{sp^3d^2}$ hybridization (fails condition a).
3. mathrmSF_4$\mathrm{SF}_4$:VSEPR Theory and Hybridization
- Central atom: Sulfur (V=6$V=6$).
- Steric Number = frac12(6 + 4) = 5 implies mathrmsp^3d$= \frac{1}{2}(6 + 4) = 5 \implies \mathrm{sp^3d}$ 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~mathrmmathringA$1.64~\mathrm{\mathring{A}}$ vs 1.54~mathrmmathringA$1.54~\mathrm{\mathring{A}}$ due to lone pair-bond pair repulsion). This satisfies all three conditions.
4. mathrmXeF_2$\mathrm{XeF}_2$:
- Central atom: Xenon (V=8$V=8$).
- Steric Number = frac12(8 + 2) = 5 implies mathrmsp^3d$= \frac{1}{2}(8 + 2) = 5 \implies \mathrm{sp^3d}$ 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).
VSEPR Theory and Hybridization
### Step 1: Conclusion
Hence, only mathrmSF_4$\mathrm{SF}_4$ 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^circ$90^\circ$) than the equatorial bonds (which have only 2 axial neighbors at 90^circ$90^\circ$). Consequently, the axial bonds are always longer and weaker than equatorial bonds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 5
Q89jee_main_2024_27_jan_morningMolecular Orbital Theory
Sum of bond order of textCO$\text{CO}$ and textNO^+$\text{NO}^+$ is textquadquad$\text{\quad\quad}$.
Numerical Answer.Answer: 6 to 6
Solution
### 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
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
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