Match the LIST-I with LIST-II.
LIST-I (Molecules/ion) LIST-II (Hybridisation of central atom) A. $PF5$ $PF{5}$ I. $dsp^2$ $dsp^{2}$ B. $SF6$ $SF{6}$ II. $sp^3d$ $sp^{3}d$ C. $Ni(CO)4$ $Ni(CO){4}$ III. $sp^3d^2$ $sp^{3}d^{2}$ D. $[PtCl4]^2-$ $[PtCl{4}]^{2-}$ IV. $sp^3$ $sp^{3}$
Choose the correct answer from the options given below:

LIST-I (Molecules/ion)	LIST-II (Hybridisation of central atom)	A. $PF5$ $PF{5}$	I. $dsp^2$ $dsp^{2}$	B. $SF6$ $SF{6}$	II. $sp^3d$ $sp^{3}d$	C. $Ni(CO)4$ $Ni(CO){4}$	III. $sp^3d^2$ $sp^{3}d^{2}$	D. $[PtCl4]^2-$ $[PtCl{4}]^{2-}$	IV. $sp^3$ $sp^{3}$

A. A-II, B-III, C-IV, D-I
B. A-IV, B-I, C-II, D-III
C. A-I, B-II, C-III, D-IV
D. A-III, B-I, C-IV, D-II

Solution & Explanation

### Core Logic Let us evaluate each central atom configuration systematically: * A.

PF_5

: Phosphorus has 5 valence electrons, forming

5sigma

bonds with zero lone pairs. Steric number

= 5 implies sp^3d

hybridisation. * B.

SF_6

: Sulfur has 6 valence electrons, forming

6sigma

bonds with zero lone pairs. Steric number

= 6 implies sp^3d^2

hybridisation. * C.

Ni(CO)_4

: Nickel is in a 0 oxidation state (

3d^8 4s^2

). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled

3d^10

state. The vacant

4s

and three

4p

orbitals hybridise to give an

sp^3

configuration.

* D.

[PtCl_4]^2-

: Platinum is in the

+2

oxidation state (

5d^8

). Since it belongs to the

5d

transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar

dsp^2

hybridisation state.

### Pattern Recognition Shortcut: Match structural main elements first:

PF_5 ightarrow sp^3d

(II),

SF_6 ightarrow sp^3d^2

(III). This isolates option (1) instantly without checking d-block coordinate fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Reference Study Guides

Chemical Bonding And Molecular Structure

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 2

Q31 2025 VSEPR Theory and d-Electron Configurations

Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of

mathrmClF_3

. The ions from the following with 'n' number of unpaired electrons are : A.

mathrmV^3 +

mathrmTi^3+

mathrmCu^2 +

mathrmNi^2+

mathrmTi^2+

Choose the correct answer from the options given below:

A. $textA and C only$ $\text{A and C only}$
B. $textA, D and E only$ $\text{A, D and E only}$
C. $textB and C only$ $\text{B and C only}$
D. $textB and D only$ $\text{B and D only}$

Solution

### Step 1: Determine 'n'

mathrmClF_3

has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus,

n = 2

. ### Step 2: Find ions with 2 unpaired electrons Let us compute the number of unpaired electrons for each configuration: - **A.

mathrmV^3+

:**

[mathrmAr] 3d^2 rightarrow 2

unpaired electrons. - **B.

mathrmTi^3+

:**

[mathrmAr] 3d^1 rightarrow 1

unpaired electron. - **C.

mathrmCu^2+

:**

[mathrmAr] 3d^9 rightarrow 1

unpaired electron. - **D.

mathrmNi^2+

:**

[mathrmAr] 3d^8 rightarrow 2

unpaired electrons. - **E.

mathrmTi^2+

:**

[mathrmAr] 3d^2 rightarrow 2

unpaired electrons. Thus, A, D, and E have exactly

n=2

unpaired electrons. ### Pattern Recognition Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember

mathrmClF_3

is T-shaped with 2 equatorial lone pairs. Look for

d^2

d^8

configurations among the transition metal ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

Q35 2025 Molecular Geometry and VSEPR

The molecules having square pyramidal geometry are

A. $mathrmBrF_5 text \& mathrmXeOF_4$
B. $mathrmSbF_5 text \& mathrmXeOF_4$
C. $mathrmSbF_5 text \& mathrmPCl_5$
D. $mathrmBrF_5 text \& mathrmPCl_5$

Solution

### Core Logic Let us check the steric details using VSEPR theory: - **

mathrmBrF_5

:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (

sp^3d^2

), geometry is square pyramidal. - **

mathrmXeOF_4

:** Xenon has 8 valence electrons. It forms 1 double bond with Oxygen, 4 single bonds with Fluorine, and retains 1 lone pair. Steric number = 6 (

sp^3d^2

), geometry is square pyramidal. - **

mathrmSbF_5

mathrmPCl_5

:** Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 (

sp^3d

), geometry is trigonal bipyramidal. Visual representations of geometries:

Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning

### Pattern Recognition Sees: Steric count 6 with 5 bonded segments + 1 lone pair

rightarrow

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

Q44 2025 VSEPR Theory

Given below are two statements: Statement (I) : for

mathrmCell mathrmF_3

, all three possible structures may be drawn as follows.

ClF3 structure variant I for Q44 — The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.

Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the

ell mathfrakp-

bp repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is incorrect but statement II is correct.
B. Statement I is correct but statement II is incorrect.
C. Both Statement I and statement II are correct.
D. Both Statement I and statement II are incorrect.

Solution

### Related Formula

textSteric Number for ClF_3 = frac7+32 = 5 implies sp^3d text hybridization (Trigonal Bipyramidal geometry)

### Core Logic - **Statement I is correct:** The three structural arrangements represent the different ways to place three bond pairs and two lone pairs within a trigonal bipyramidal grid. - **Statement II is incorrect:** According to VSEPR theory and Bent's rule, in

sp^3d

hybridization, **lone pairs must occupy equatorial positions** to minimize strong

90^circ

lone pair-bond pair (

ell p-bp

) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable. ### Pattern Recognition For

sp^3d

configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience

120^circ

interactions, minimizing severe

90^circ

structural strains. This results in the classic stable T-shaped configuration for

ClF_3

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

Q28 2025 Molecular Orbital Theory

Which of the following molecules(s) show/s paramagnetic behavior? (A)

O_2

(B)

N_2

(C)

F_2

(D)

S_2

(E)

Cl_2

Choose the correct answer from the options given below:

A. $textB only$ $\text{B only}$
B. $textA \& C only$ $\text{A \& C only}$
C. $textA \& E only$ $\text{A \& E only}$
D. $textA \& D only$ $\text{A \& D only}$

Solution

### Related Formula Paramagnetism

implies

Presence of at least one unpaired electron in the molecular orbitals. ### Core Logic According to Molecular Orbital Theory (MOT): *

O_2

has 16 electrons. Its outer configuration contains two unpaired electrons in the anti-bonding orbitals:

pi^*_2p_x = pi^*_2p_y

. Thus, it is paramagnetic. *

S_2

belongs to the same oxygen family group and shares an analogous valence configuration with two unpaired electrons in its anti-bonding

pi^*

orbitals. Hence, it is also paramagnetic. *

N_2

(14e-),

F_2

(18e-), and

Cl_2

(34e-) have completely paired electronic systems and behave diamagnetically. ### Pattern Recognition Both

O_2

and

S_2

contain 2 unpaired electrons in their highest occupied molecular orbitals, making them classic examples of paramagnetic diatomic species. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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