Which among the following molecules is (a) involved in sp3d hybridization, (b) has different bond lengths and (c) has…

Q: Which among the following molecules is (a) involved in sp3d hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom?

### Related Formula where, V = valence electrons of central atom M = number of monovalent surrounding atoms C = cationic charge, A = anionic charge ### Core Logic Let's calculate the hybridization, shape, and lone pairs for each option: 1. PF5: - Central atom: Phosphorus (V=5). - Steric Number = (1/2)(5 + 5) = 5 sp3d hybridization. - Lone pairs = 5 - 5 = 0. - Geometry: Trigonal bipyramidal. It has different axial and equatorial bond lengths, but no lone pair on the central atom. 2. XeF4: - Centr

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Q64 jee_main_2026_21_jan_morning VSEPR Theory

Given below are two statements: Statement I: The number of species among

mathrmSF_4

mathrmNH_4^+

[mathrmNiCl_4]^2-

mathrmXeF_4

[mathrmPtCl_4]^2-

mathrmSeF_4

and

[mathrmNi(CN)_4]^2-

, that have tetrahedral geometry is 3. Statement II: In the set

[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. $textStatement I is true but Statement II is false$
B. $textBoth Statement I and Statement II are false$
C. $textStatement I is false but Statement II is true$
D. $textBoth Statement I and Statement II are true$

Solution

### Core Logic Evaluating Statement I: -

mathrmSF_4

sp^3d

(1 lone pair)

rightarrow

See-saw -

mathrmXeF_4

sp^3d^2

(2 lone pairs)

rightarrow

Square planar -

[mathrmPtCl_4]^2-

dsp^2

rightarrow

Square planar -

[mathrmNiCl_4]^2-

sp^3

rightarrow

Tetrahedral -

[mathrmNi(CN)_4]^2-

dsp^2

rightarrow

Square planar -

mathrmSeF_4

sp^3d

(1 lone pair)

rightarrow

See-saw -

mathrmNH_4^+

sp^3

(0 lone pairs)

rightarrow

Tetrahedral Total tetrahedral species = 2 (

[mathrmNiCl_4]^2-

and

mathrmNH_4^+

). Statement I says 3, so it is false. Evaluating Statement II: -

NO_2

: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). -

BeH_2

: Central Be has 4 electrons (incomplete octet). -

BF_3

: Central B has 6 electrons (incomplete octet). -

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

Q jee_main_2025_02_april_morning Interhalogen Molecular Geometry

A molecule with the formula

mathrmAX_4mathrmY

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 :

A. $(1)\ textSquare pyramidal$ $(1)\ \text{Square pyramidal}$
B. $(2)\ textOctahedral$ $(2)\ \text{Octahedral}$
C. $(3)\ textPentagonal planar$ $(3)\ \text{Pentagonal planar}$
D. $(4)\ textTrigonal bipyramidal$ $(4)\ \text{Trigonal bipyramidal}$

Solution

### Related Formula Total valence shell electron pair system equation:

textValence Pairs = textBond Pairs (BP) + textLone 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

) and Oxygen (

mathrmO

), matching labels

mathrmX

and

mathrmY

. * Element

mathrmA

is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (

mathrmXe

). * Substituting these components into the target layout formula yields

mathrmXeOF_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

(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

architectures containing an explicit lone pair along with an asymmetric double bond (like

mathrmXeOF_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

Q35 jee_main_2025_02_april_morning VSEPR and Dipole Moments

Among

mathrmSO_2

mathrmNF_3

mathrmNH_3

mathrmXeF_2

mathrmClF_3

and

mathrmSF_4

, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is

A. $(1)\ mathrmsp^3$ $(1)\ \mathrm{sp}^3$
B. $(2)\ mathrmdsp^2$ $(2)\ \mathrm{dsp}^2$
C. $(3)\ mathrmsp^3mathrmd^2$ $(3)\ \mathrm{sp}^3\mathrm{d}^2$
D. $(4)\ mathrmsp^3mathrmd$ $(4)\ \mathrm{sp}^3\mathrm{d}$

Solution

### Related Formula Steric Number system equation for identifying electronic configurations:

textSteric Number (SN) = frac12[mathrmV + mathrmM - mathrmC + mathrmA]

### 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

has 3 lone pairs but its linear architecture enforces

mu = 0

. Therefore,

mathrmClF_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

mathrmsp^3mathrmd

. ### Pattern Recognition Watch out for symmetry traps!

mathrmXeF_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

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2025_07_april_morning VSEPR 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

-bonds or regions) and lone pairs on the central atom of each species: 1. **

mathrmICl_2^-

**: - Central atom Iodine has

7

valence electrons

+ 1

negative charge

= 8

electrons. - Forms

2

single bonds (bond pairs

= 2

). - Remaining

6

electrons form

3

lone pairs. - Ratio is

2 : 3

(Matches LIST-II, III).

2. **

mathrmH_2mathrmO

**: - Oxygen has

6

valence electrons. - Forms

2

bond pairs with Hydrogens. - Remaining

4

electrons form

2

lone pairs. - Ratio is

2 : 2

(Matches LIST-II, IV).

3. **

mathrmSO_2

**: - Sulfur has

6

valence electrons. - Forms

2

double bonds (which are counted as

4

bonding pairs of electrons/bond pairs in typical VSEPR representations here). - Remaining

2

electrons form

1

lone pair. - Ratio is

4 : 1

(Matches LIST-II, II).

4. **

mathrmXeF_4

**: - Xenon has

8

valence electrons. - Forms

4

bond pairs with Fluorines. - Remaining

4

electrons form

2

lone pairs. - Ratio is

4 : 2

(Matches LIST-II, I).

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)

. - Water is

2

bond pairs,

2

lone pairs (

sp^3

)

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

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?

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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?

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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?