The number of species from the following which are paramagnetic and with bond order equal to one is

Year	2026	2025	2024	Total
Questions	12	14	16	42

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

Q jee_main_2024_30_january_evening Dipole Moment

Given below are two statements: Statement-I: Since fluorine is more electronegative than nitrogen, the net dipole moment of

NF₃

is greater than

NH₃

. Statement-II: In

NH₃

, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite direction, but in

NF₃

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. $Statement I is true but Statement II is false.$
B. $Both Statement I and Statement II are false.$
C. $Both statement I and Statement II is are true.$
D. $Statement I is false but Statement II is are true.$

Solution

Core Logic

Statement I: The net dipole moment of $NH₃$ $NH_3$ ( $1.47 D$ $1.47\, D$) is actually greater than that of $NF₃$ $NF_3$ ( $0.23 D$ $0.23\, D$). Therefore, Statement I is false.

Statement II: In $NH₃$ $NH_3$, the $N-H$ $N-H$ bond dipole moments (pointing towards the more electronegative N) reinforce the orbital dipole moment of the lone pair. In $NF₃$ $NF_3$, 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 $NF₃$ $NF_3$ 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 Evening

Step 1: Final Conclusion

Since both statements assert the opposite of established facts regarding $NH₃$ $NH_3$ and $NF₃$ $NF_3$, both are false.

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2024_30_jan_morning VSEPR Theory

Match List-I with List-II.

List-I (Molecule)	List-II (Shape)
(A) $BrF₅$ $BrF_5$	(I) T-shape
(B) $H₂O$ $H_2O$	(II) See saw
(C) $ClF₃$ $ClF_3$	(III) Bent
(D) $SF₄$ $SF_4$	(IV) Square pyramidal

A. $(A)-(I), (B)-(II), (C)-(IV), (D)-(III)$
B. $(A)-(II), (B)-(I), (C)-(III), (D)-(IV)$
C. $(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$
D. $(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$

Solution

Core Logic

Using VSEPR theory: (A) $BrF₅$ $BrF_5$: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) $arrow$ $\rightarrow$ $sp³d²$ $sp^3d^2$ hybridization $arrow$ $\rightarrow$ Square pyramidal shape. (B) $H₂O$ $H_2O$: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) $arrow$ $\rightarrow$ $sp³$ $sp^3$ hybridization $arrow$ $\rightarrow$ Bent shape. (C) $ClF₃$ $ClF_3$: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) $arrow$ $\rightarrow$ $sp³d$ $sp^3d$ hybridization $arrow$ $\rightarrow$ T-shape. (D) $SF₄$ $SF_4$: S has 6 valence electrons. 4 form bonds with F, leaving 1 lone pair. (4 bp + 1 lp) $arrow$ $\rightarrow$ $sp³d$ $sp^3d$ hybridization $arrow$ $\rightarrow$ See-saw shape.

Step 1: Matching

(A) - (IV) (B) - (III) (C) - (I) (D) - (II)

VSEPR Theory solution diagram for Q71 - JEE Main 2024 Morning

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_30_jan_morning Molecular Orbital Theory

The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule

Numerical Answer. Answer: 8 to 8

Solution

Core Logic

According to Molecular Orbital Theory (MOT), the number of molecular orbitals (MOs) formed is equal to the total number of atomic orbitals (AOs) combined.

Step 1: Counting atomic orbitals

For a single atom in the 2nd period, the valence shell has: One 2s orbital Three 2p orbitals ( $2pₓ, 2py, 2pz$ $2p_x, 2p_y, 2p_z$) Total = 4 atomic orbitals per atom. For a diatomic molecule, two such atoms combine. Total atomic orbitals = $4 \times 2 = 8$ $4 \times 2 = 8$.

Step 2: Forming molecular orbitals

Combining these 8 atomic orbitals yields 8 molecular orbitals:

From 2s: $σ₂ₛ$ $\sigma_{2s}$ and $σ^*₂ₛ$ $\sigma^*_{2s}$ (2 MOs)
From 2p: $σ2pz, π2pₓ, π2py, π^2pₓ, π^2py, σ^*2pz$ (6 MOs)

Total MOs = $2 + 6 = 8$ $2 + 6 = 8$.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2024_31_jan_evening Ionic Bond and Lattice Energy

Which of the following is least ionic?

A. $BaCl₂$ $\text{}BaCl_2$
B. $AgCl$ $\text{}AgCl$
C. $KCl$ $\text{}KCl$
D. $CoCl₂$ $\text{}CoCl_2$

Solution

Core Logic

According to Fajan's rules, covalent character is favored by high charge and small size of the cation, and by cations with a pseudo-noble gas configuration. $Ag^+$ $Ag^+$ has a pseudo-noble gas configuration ( $ns²np⁶nd¹⁰$ $ns^2np^6nd^{10}$), which results in high polarizing power compared to s-block and typical transition elements. Therefore, $AgCl$ $AgCl$ has the maximum covalent character and is the least ionic among the given options.

Ionic character order: $AgCl < CoCl₂ < BaCl₂ < KCl$ $AgCl < CoCl_2 < BaCl_2 < KCl$

Step 1: Final Selection

Because $AgCl$ $AgCl$ is the most covalent, it is the least ionic. Hence, option (2) is correct.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q81 jee_main_2024_31_jan_evening Dipole Moment and Fractional Charge

A diatomic molecule has a dipole moment of

1.2 D

. If the bond distance is

1AA

, then fractional charge on each atom is _________

\times 10⁻¹⁰ esu

. (Given:

1 D = 10⁻¹⁸ esu cm

)

Numerical Answer. Answer: 1.2 to 1.2

Solution

Related Formula

μ = q \times d

Core Logic

Given dipole moment, $μ = 1.2 D = 1.2 \times 10⁻¹⁸ esu cm$ . Bond distance, $d = 1AA = 10⁻⁸ cm$ $d = 1\mathrm{\AA} = 10^{-8}\text{ cm}$.

We need to find the fractional charge $q$ $q$.

Step 1: Calculation

q = (μ)/(d)

q = 1.2 \times 10⁻¹⁸ esu cm10⁻⁸ cm

q = 1.2 \times 10⁻¹⁰ esu

Step 2: Final Formatting

The question asks for the fractional charge in the form $x \times 10⁻¹⁰ esu$ $x \times 10^{-10}\text{ esu}$. Therefore, the value is $1.2$ $1.2$.

Note: Based on NTA officially accepting $12$ $12$ (if asked for $x \times 10⁻¹¹$ $x \times 10^{-11}$) or $1.2$ $1.2$. We will format it exactly as calculated.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

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$$\mathrm {H}_2, \mathrm{He}_2^+, \mathrm{O}_2^+, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^+, \mathrm{B}_2$$

Solution & Explanation

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Final Conclusion

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Solution

Core Logic

Step 1: Matching

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Solution

Core Logic

Step 1: Counting atomic orbitals

Step 2: Forming molecular orbitals

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Solution

Core Logic

Step 1: Final Selection

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Solution

Related Formula

Core Logic

Step 1: Calculation

Step 2: Final Formatting

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