Given below are two statements:
Statement I: The number of species among mathrmSF_4$\mathrm{SF}_4$, mathrmNH_4^+$\mathrm{NH}_4^+$, [mathrmNiCl_4]^2-$[\mathrm{NiCl}_4]^{2-}$, mathrmXeF_4$\mathrm{XeF}_4$, [mathrmPtCl_4]^2-$[\mathrm{PtCl}_4]^{2-}$, mathrmSeF_4$\mathrm{SeF}_4$ and [mathrmNi(CN)_4]^2-$[\mathrm{Ni(CN)}_4]^{2-}$, that have tetrahedral geometry is 3.
Statement II: In the set [NO_2, BeH_2, BF_3, AlCl_3]$[NO_{2}, BeH_{2}, BF_{3}, AlCl_{3}]$, all the molecules have incomplete octet around central atom.
In the light of the above statements, choose the correct answer 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.textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
D.textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
Solution & Explanation
### Core Logic
Evaluating Statement I:
- mathrmSF_4$\mathrm{SF}_4$: sp^3d$sp^3d$ (1 lone pair) rightarrow$\rightarrow$ See-saw
- mathrmXeF_4$\mathrm{XeF}_4$: sp^3d^2$sp^3d^2$ (2 lone pairs) rightarrow$\rightarrow$ Square planar
- [mathrmPtCl_4]^2-$[\mathrm{PtCl}_4]^{2-}$: dsp^2$dsp^2$rightarrow$\rightarrow$ Square planar
- [mathrmNiCl_4]^2-$[\mathrm{NiCl}_4]^{2-}$: sp^3$sp^3$rightarrow$\rightarrow$ Tetrahedral
- [mathrmNi(CN)_4]^2-$[\mathrm{Ni(CN)}_4]^{2-}$: dsp^2$dsp^2$rightarrow$\rightarrow$ Square planar
- mathrmSeF_4$\mathrm{SeF}_4$: sp^3d$sp^3d$ (1 lone pair) rightarrow$\rightarrow$ See-saw
- mathrmNH_4^+$\mathrm{NH}_4^+$: sp^3$sp^3$ (0 lone pairs) rightarrow$\rightarrow$ Tetrahedral
Total tetrahedral species = 2 ([mathrmNiCl_4]^2-$[\mathrm{NiCl}_4]^{2-}$ and mathrmNH_4^+$\mathrm{NH}_4^+$). Statement I says 3, so it is false.
Evaluating Statement II:
- NO_2$NO_2$: Central N has 7 valence electrons (odd-electron molecule, incomplete octet).
- BeH_2$BeH_2$: Central Be has 4 electrons (incomplete octet).
- BF_3$BF_3$: Central B has 6 electrons (incomplete octet).
- AlCl_3$AlCl_3$ (monomer): Central Al has 6 electrons (incomplete octet).
Therefore, all molecules have incomplete octets. Statement II is true.
### Step 1: Final Conclusion
Statement I is false, Statement II is true.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Class 12 Chemistry: Coordination Compounds
Keywords:#tetrahedral geometry#incomplete octet#JEE Main 2026 Morning Q64#Chemical Bonding and Molecular Structure JEE Main 2026
More Chemical Bonding and Molecular Structure Previous-Year Questions
Qjee_main_2025_02_april_eveningVSEPR Theory and Hybridization
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
### 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
A molecule with the formula mathrmAX_4mathrmY$\mathrm{AX}_4\mathrm{Y}$ has all its elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is :
### Related Formula
Total valence shell electron pair system equation:
textValence Pairs = textBond Pairs (BP) + textLone Pairs (LP)$$\text{Valence Pairs} = \text{Bond Pairs (BP)} + \text{Lone Pairs (LP)}$$
### 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 (mathrmF$\mathrm{F}$) and Oxygen (mathrmO$\mathrm{O}$), matching labels mathrmX$\mathrm{X}$ and mathrmY$\mathrm{Y}$.
* Element mathrmA$\mathrm{A}$ is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (mathrmXe$\mathrm{Xe}$).
* Substituting these components into the target layout formula yields mathrmXeOF_4$\mathrm{XeOF_4}$:
- 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 text bond regions + 1 text 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:
XeOF4 square pyramidal spatial geometry diagram for Q45
### Pattern Recognition
In mathrmsp^3d^2$\mathrm{sp^3d^2}$ architectures containing an explicit lone pair along with an asymmetric double bond (like mathrmXeOF_4$\mathrm{XeOF_4}$), the lone pair always sits directly opposite the double bond to minimize electron repulsion, leaving a clean square pyramidal shape.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Class 12 Chemistry: The p-Block Elements
Q35jee_main_2025_02_april_morningVSEPR and Dipole Moments
Among mathrmSO_2$\mathrm{SO}_2$, mathrmNF_3$\mathrm{NF}_3$, mathrmNH_3$\mathrm{NH}_3$, mathrmXeF_2$\mathrm{XeF}_2$, mathrmClF_3$\mathrm{ClF}_3$ and mathrmSF_4$\mathrm{SF}_4$, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is
### Related Formula
Steric Number system equation for identifying electronic configurations:
textSteric Number (SN) = frac12[mathrmV + mathrmM - mathrmC + mathrmA]$$\text{Steric Number (SN)} = \frac{1}{2}[\mathrm{V} + \mathrm{M} - \mathrm{C} + \mathrm{A}]$$
### Core Logic
Let's list parameters using a detailed structural grid:
Molecule
Hybridisation
Dipole Moment
Lone pair on the central atom
mathrmSO_2$\mathrm{SO}_2$
mathrmsp^2$\mathrm{sp}^2$
Non-zero
1
mathrmNF_3$\mathrm{NF}_3$
mathrmsp^3$\mathrm{sp}^3$
Non-zero
1
mathrmNH_3$\mathrm{NH}_3$
mathrmsp^3$\mathrm{sp}^3$
Non-zero
1
mathrmXeF_2$\mathrm{XeF}_2$
mathrmsp^3mathrmd$\mathrm{sp}^3\mathrm{d}$
Zero
3
mathrmClF_3$\mathrm{ClF}_3$
mathrmsp^3mathrmd$\mathrm{sp}^3\mathrm{d}$
Non-zero
2
mathrmSF_4$\mathrm{SF}_4$
mathrmsp^3mathrmd$\mathrm{sp}^3\mathrm{d}$
Non-zero
1
Comparing items: mathrmXeF_2$\mathrm{XeF}_2$ has 3 lone pairs but its linear architecture enforces mu = 0$\mu = 0$. Therefore, mathrmClF_3$\mathrm{ClF}_3$ has the highest count of lone pairs (2) with a net non-zero asymmetric dipole configuration.
### Step 1: Selection
The hybridization of mathrmClF_3$\mathrm{ClF}_3$ is mathrmsp^3mathrmd$\mathrm{sp}^3\mathrm{d}$.
### Pattern Recognition
Watch out for symmetry traps! mathrmXeF_2$\mathrm{XeF}_2$ contains the absolute maximum lone pairs, but its symmetric planar positioning perfectly cancels out the dipole vectors. Thus, the correct candidate slips down to mathrmClF_3$\mathrm{ClF}_3$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Qjee_main_2025_07_april_morningVSEPR Theory
Match the LIST-I with LIST-II.
LIST-I (Molecule/ion)
LIST-II (Bond pair : lone pair on the central atom)
(A) mathrmICl_2^-$\mathrm{ICl}_2^-$
(I) 4 : 2$4 : 2$
(B) mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$
(II) 4 : 1$4 : 1$
(C) mathrmSO_2$\mathrm{SO}_2$
(III) 2 : 3$2 : 3$
(D) mathrmXeF_4$\mathrm{XeF}_4$
(IV) 2 : 2$2 : 2$
Choose the correct answer from the options given below:
A. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Solution
### Core Logic
Let's find the number of bond pairs (sigma$\sigma$-bonds or regions) and lone pairs on the central atom of each species:
1. **mathrmICl_2^-$\mathrm{ICl}_2^-$**:
- Central atom Iodine has 7$7$ valence electrons + 1$+ 1$ negative charge = 8$= 8$ electrons.
- Forms 2$2$ single bonds (bond pairs = 2$= 2$).
- Remaining 6$6$ electrons form 3$3$ lone pairs.
- Ratio is 2 : 3$2 : 3$ (Matches LIST-II, III).
VSEPR linear shape of ICl2-
2. **mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$**:
- Oxygen has 6$6$ valence electrons.
- Forms 2$2$ bond pairs with Hydrogens.
- Remaining 4$4$ electrons form 2$2$ lone pairs.
- Ratio is 2 : 2$2 : 2$ (Matches LIST-II, IV).
VSEPR linear shape of ICl2-
3. **mathrmSO_2$\mathrm{SO}_2$**:
- Sulfur has 6$6$ valence electrons.
- Forms 2$2$ double bonds (which are counted as 4$4$ bonding pairs of electrons/bond pairs in typical VSEPR representations here).
- Remaining 2$2$ electrons form 1$1$ lone pair.
- Ratio is 4 : 1$4 : 1$ (Matches LIST-II, II).
VSEPR linear shape of ICl2-
4. **mathrmXeF_4$\mathrm{XeF}_4$**:
- Xenon has 8$8$ valence electrons.
- Forms 4$4$ bond pairs with Fluorines.
- Remaining 4$4$ electrons form 2$2$ lone pairs.
- Ratio is 4 : 2$4 : 2$ (Matches LIST-II, I).
VSEPR linear shape of ICl2-
Thus, the correct mapping is: A-III, B-IV, C-II, D-I.
### Pattern Recognition
For match-the-column with VSEPR structures:
- Always find steric number: textSteric Number = frac12(V + M - C + A)$\text{Steric Number} = \frac{1}{2}(V + M - C + A)$.
- Water is 2$2$ bond pairs, 2$2$ lone pairs (sp^3$sp^3$) rightarrow$\rightarrow$ B-IV. This alone helps eliminate multiple incorrect options immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
More Chemical Bonding and Molecular Structure Questions — jee_main_2026_21_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
.svg)
