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Chemical Bonding and Molecular Structure appeared 42 times across 3 years — 4.9% of Chemistry. This question is from Molecular Orbital Theory.

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The number of species from the following which are paramagnetic and with bond order equal to one is $H₂, He₂^+, O₂^+, N₂²⁻, O₂²⁻, F₂, Ne₂^+, B₂$

Numerical Answer Type:

Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

Core Logic

Using Molecular Orbital (MO) Theory, we evaluate the bond order ( $BO = (Nb - Nₐ)/(2)$ $BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Species	Magnetic behaviour	Bond order
$H₂$ $H_2$	Diamagnetic	1
$He₂^+$ $He_2^+$	Paramagnetic	0.5
$O₂^+$ $O_2^+$	Paramagnetic	2.5
$N₂²⁻$ $N_2^{2-}$	Paramagnetic	2
$O₂²⁻$ $O_2^{2-}$	Diamagnetic	1
$F₂$ $F_2$	Diamagnetic	1
$Ne₂^+$ $Ne_2^+$	Paramagnetic	0.5
$B₂$ $B_2$	Paramagnetic	1

Step 1: Final Selection

We need the species that satisfies BOTH conditions:

Paramagnetic
Bond Order = 1

Looking at the table, $B₂$ $B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate $π₂ₚ$ $\pi_{2p}$ orbitals) and has a bond order of 1.

Total number of such species = 1.

Pattern Recognition

$B₂$ $B_2$ (10 electrons) and $O₂$ $O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. $B₂$ $B_2$ has BO = 1, and $O₂$ $O_2$ has BO = 2.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

Chemical Bonding And Molecular Structure

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 6

Q43 jee_main_2025_24_jan_morning Hybridization and Molecular Geometry

Which of the following statement is true with respect to

H₂O, NH₃

and

CH₄

? A. The central atoms of all the molecules are

sp³

hybridized. B. The H-O-H, H-N-H and H-C-H angles in the above molecules are

104.5°

,

107.5°

and

109.5°

respectively. C. The increasing order of dipole moment is

CH₄ < NH₃ < H₂O

. D. Both

H₂O

and

NH₃

are Lewis acids and

CH₄

is a Lewis base E. A solution of

NH₃

in

H₂O

is basic. In this solution

NH₃

and

H₂O

act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below:

A. A, B and C only
B. C, D and E only
C. A, D and E only
D. A, B, C and E only

Solution

Core Logic

Analyzing each statement individually:

Statement A is true: The central atoms ( $O, N, C$ $O, N, C$) all possess an electron steric number equal to 4, indicating $sp³$ $\mathfrak{sp}^3$ hybridization state pathways.
Statement B is true: Due to valence shell electron pair repulsions, the bond angles decrease from the ideal tetrahedral angle ( $109.5°$ $109.5^{\circ}$ in $CH₄$ $CH_4$,
Water molecule structural bond configuration shape representation
) as lone pairs are added ( $107.5°$ $107.5^{\circ}$ in $NH₃$ $NH_3$ with 1 lone pair,
Water molecule structural bond configuration shape representation
; $104.5°$ $104.5^{\circ}$ in $H₂O$ $H_2O$ with 2 lone pairs,
Water molecule structural bond configuration shape representation
).
Statement C is true: The dipole moment increases alongside central atom electronegativity and asymmetric lone pair configurations, following the sequence $CH₄ (0 D) < NH₃ (1.47 D) < H₂O (1.85 D)$ .

Pattern Recognition

Lone pairs repel bonding electron pairs more strongly than bonding pairs repel each other, systematically compressing adjacent bond angles.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q74 jee_main_2024_01_february_morning Ionic Character

Arrange the bonds in order of increasing ionic character in the molecules.

LiF

,

K₂O

,

N₂

,

SO₂

and

ClF₃

.

A. $ClF₃ < N₂ < SO₂ < K₂O < LiF$ $ClF_3 < N_2 < SO_2 < K_2O < LiF$
B. $LiF < K₂O < ClF₃ < SO₂ < N₂$ $LiF < K_2O < ClF_3 < SO_2 < N_2$
C. $N₂ < SO₂ < ClF₃ < K₂O < LiF$ $N_2 < SO_2 < ClF_3 < K_2O < LiF$
D. $N₂ < ClF₃ < SO₂ < K₂O < LiF$ $N_2 < ClF_3 < SO_2 < K_2O < LiF$

Solution

Core Logic

The ionic character of a bond is directly proportional to the electronegativity difference ( $Δ EN$ $\Delta EN$) between the two bonded atoms. Larger $Δ EN$ $\Delta EN \implies$ higher ionic character.

Step 1: Assess Electronegativity Differences

$N₂$ $N_2$: Both atoms are Nitrogen. $Δ EN = 0$ $\Delta EN = 0$. Purely covalent. (Lowest ionic character)
$SO₂$ $SO_2$: Bond between S and O. Moderate $Δ EN$ $\Delta EN$. Covalent with some polarity.
$ClF₃$ $ClF_3$: Bond between Cl and F. $Δ EN$ $\Delta EN$ is higher than S-O as F is the most electronegative element.
$K₂O$ $K_2O$: Bond between K (alkali metal, very low EN) and O. Very high $Δ EN$ $\Delta EN$. Ionic.
$LiF$ $LiF$: Bond between Li (alkali metal) and F (highest EN). Maximum $Δ EN$ $\Delta EN$ possible among these options. Most ionic.

Step 2: Order Derivation

Increasing order of ionic character (or $Δ EN$ $\Delta EN$): $N₂ < SO₂ < ClF₃ < K₂O < LiF$ $N_2 < SO_2 < ClF_3 < K_2O < LiF$

Pattern Recognition

Homodiatomic ( $N₂$ $N_2$) 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.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_01_february_morning VSEPR Theory

The number of molecules/ion/s having trigonal bipyramidal shape is ....

PF₅

,

BrF₅

,

PCl₅

,

[PtCl₄]²⁻

,

BF₃

,

Fe(CO)₅

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

Using VSEPR theory to find the hybridization and shape:

$PF₅$ $PF_5$: P has 5 valence electrons, forms 5 single bonds with F. Steric number = 5 (sp3d). 0 lone pairs. Shape = Trigonal bipyramidal.
$BrF₅$ $BrF_5$: Br has 7 valence electrons, forms 5 single bonds, 1 lone pair. Steric number = 6 (sp3d2). Shape = Square pyramidal.
$PCl₅$ $PCl_5$: P has 5 valence electrons, 5 bonds, 0 lone pairs. Steric number = 5 (sp3d). Shape = Trigonal bipyramidal.
$[PtCl₄]²⁻$ $[PtCl_4]^{2-}$: $Pt²⁺$ $Pt^{2+}$ is a $d⁸$ $d^8$ system. With $Cl^-$ $Cl^-$ (but 4d/5d transition metals always form low spin square planar complexes), it's $dsp²$ $dsp^2$ hybridized. Shape = Square planar.
$BF₃$ $BF_3$: B has 3 valence electrons, 3 bonds, 0 lone pairs. Steric number = 3 (sp2). Shape = Trigonal planar.

6. $Fe(CO)₅$ $Fe(CO)_5$: Fe (d6s2 $arrow$ $\rightarrow$ d8 under strong field $CO$ $CO$). Carbonyls strongly prefer 5-coordinate trigonal bipyramidal geometry for $d⁸$ $d^8$ ( $dsp³$ $dsp^3$ hybridization). Shape = Trigonal bipyramidal.

Step 1: Count Trigonal Bipyramidal Molecules

Molecules with trigonal bipyramidal shape:

$PF₅$ $PF_5$
$PCl₅$ $PCl_5$
$Fe(CO)₅$ $Fe(CO)_5$

Total count = 3.

Pattern Recognition

Steric Number = 5 with 0 lone pairs ALWAYS yields Trigonal Bipyramidal geometry. Watch out for $BrF₅$ $BrF_5$ which has 5 bonds but 1 lone pair (SN = 6, Square Pyramidal).

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Q jee_main_2024_29_january_evening Molecular Orbital Theory

The total number of anti bonding molecular orbitals, formed from 2s and 2p atomic orbitals in a diatomic molecule is ________.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Total Atomic Orbitals Combinations = Bonding MOs + Antibonding MOs

Core Logic

When atomic orbitals combine, they form an equal number of molecular orbitals:

Two $2s$ $2s$ atomic orbitals combine to form 1 bonding orbital ( $σ₂ₛ$ $\sigma_{2s}$) and 1 antibonding orbital ( $σ^*₂ₛ$ $\sigma^*_{2s}$).
Six $2p$ $2p$ atomic orbitals combine to form 3 bonding orbitals ( $σ2pz, π2pₓ, π2py$ $\sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}$) and 3 antibonding orbitals ( $σ^2pz, π^2pₓ, π^*2py$ ).

Step 1: Total Summation

Summing the antibonding orbitals from both subshells:

Total Antibonding Molecular Orbitals = 1 (from 2s) + 3 (from 2p) = 4

Pattern Recognition

The linear combination of $N$ $N$ atomic orbitals always yields exactly $(N)/(2)$ $\frac{N}{2}$ antibonding molecular orbitals.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2024_29_january_evening Dipole Moment

The total number of molecules with zero dipole moment among

CH₄

,

BF₃

,

H₂O

,

HF

,

NH₃

,

CO₂

, and

SO₂

is ________.

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

μₙₑₜ = Σ μᵢ = 0 (For perfectly symmetrical geometry configurations)

Core Logic

Analyze the molecular geometry and symmetry of each molecule:

$CH₄$ $\text{CH}_4$: Symmetrical tetrahedral geometry $μ = 0$ $\implies \mu = 0$.
$BF₃$ $\text{BF}_3$: Symmetrical trigonal planar geometry $μ = 0$ $\implies \mu = 0$.
$H₂O$ $\text{H}_2\text{O}$: Bent shape due to lone pairs $μ \neq 0$ $\implies \mu \neq 0$.
$HF$ $\text{HF}$: Linear asymmetric diatomic molecule $μ \neq 0$ $\implies \mu \neq 0$.
$NH₃$ $\text{NH}_3$: Trigonal pyramidal shape due to a lone pair $μ \neq 0$ $\implies \mu \neq 0$.
$CO₂$ $\text{CO}_2$: Symmetrical linear structure ( $O=C=O$ $\mathrm{O}=\mathrm{C}=\mathrm{O}$) where dipoles cancel out $μ = 0$ $\implies \mu = 0$.
$SO₂$ $\text{SO}_2$: Bent angular geometry due to a lone pair $μ \neq 0$ $\implies \mu \neq 0$.

Step 1: Final Counting

The molecules with a net zero dipole moment are $CH₄$ $\text{CH}_4$, $BF₃$ $\text{BF}_3$, and $CO₂$ $\text{CO}_2$. This gives a total count of 3.

Pattern Recognition

Molecules with a symmetrical arrangement of identical bonds and no lone pairs on the central atom (e.g., tetrahedral $CH₄$ $\text{CH}_4$, trigonal planar $BF₃$ $\text{BF}_3$, linear $CO₂$ $\text{CO}_2$) always have a net dipole moment of zero.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Questions — jee_main_2024_29_jan_morning

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