$$ \text { What is the most appropriate Lewis structure of } \mathrm{NO}_2 \mathrm{~F} \text { ? } $...

With $\mathrm{NO}_2 \mathrm{~F}$ you'll need to form a double bond between one of the Oxygen atoms and a Nitrogen atom to fill the octets and still use only the 24 valence electrons available for the molecule.Odd-electron moleculesIn molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, $\mathrm{NO}_2$, the octet rule is not satisfied for all the atoms

Amongst $\mathrm{O}_2, \mathrm{~N}_2, \mathrm{~F}_2$, and $\mathrm{B}_2$, which molecules will be at...

To determine which molecules are attracted to an external magnetic field, we need to understand their magnetic properties, specifically if they are paramagnetic or diamagnetic. Paramagnetic substances, which have unpaired electrons, are attracted to an external magnetic field. Diamagnetic substances, with all electrons paired, are not attracted to the magnetic field.$\mathrm{O}_2$ (Oxygen molecule):$\mathrm{O}_2$ is paramagnetic due to the presence of two unpaired electrons in its molecular orbital configuration.$$ \left(\sigma_{1s}\right)^2\left(\sigma_{1s}^*\right)^2\left(\sigma_{2s}\right)^2\left(\sigma_{2s}^*\right)^2\left(\sigma_{2p_z}\right)^2\left(\pi_{2p_x}\right)^2\left(\pi_{2p_y}\right)^2\left(\pi_{2p_x}^*\right)^1\left(\pi_{2p_y}^*\right)^1 $$$\mathrm{N}_2$ (Nitrogen molecule):$\mathrm{N}_2$ is diamagnetic because all its electrons are paired in the molecular orbital configuration.$$ \left(\sigma_{1s}\right)^2\left(\sigma_{1s}^*\right)^2\left(\sigma_{2s}\right)^2\left(\sigma_{2s}^*\right)^2\left(\pi_{2p_x}\right)^2\left(\pi_{2p_y}\right)^2\left(\sigma_{2p_z}\right)^2 $$$\mathrm{F}_2$ (Fluorine molecule):$\mathrm{F}_2$ is diamagnetic as it has all paired electrons in its molecular orbital configuration.$$ \left(\sigma_{1s}\right)^2\left(\sigma_{1s}^*\right)^2\left(\sigma_{2s}\right)^2\left(\sigma_{2s}^*\right)^2\left(\sigma_{2p_z}\right)^2\left(\pi_{2p_x}\right)^2\left(\pi_{2p_y}\right)^2\left(\pi_{2p_x}^*\right)^2\left(\pi_{2p_y}^*\right)^2 $$$B_2$ (Boron molecule):$B_2$ is paramagnetic because it has two unpaired electrons in its molecular orbital configuration.$$ \left(\sigma_{1s}\right)^2\left(\sigma_{1s}^*\right)^2\left(\sigma_{2s}\right)^2\left(\sigma_{2s}^*\right)^2\left(\pi_{2p_x}\right)^1\left(\pi_{2p_y}\right)^1 $$Therefore, the molecules that will be attracted to an external magnetic field are $\mathrm{O}_2$ and $\mathrm{B}_2$.

Which one of the following statements is correct about $\mathrm{N}_2, \mathrm{CO}$, and $\mathrm{NO}...

Isoelectronic Series and Bond OrderIsoelectronic SeriesAnisoelectronic seriesrefers to a group of molecules or ions that share the same number of electrons. These entities generally possess identical electronic configurations, leading to similar chemical and physical properties.Bond OrderThebond orderof a molecule can be determined using Molecular Orbital Theory. It is calculated with the formula:$ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) $Application to $\mathrm{N}_2, \mathrm{CO}$, and $\mathrm{NO}^{+}$1. Molecular Orbital ConfigurationsNitrogen ($\mathrm{N}_2$):$ \left[\left(\sigma_{1 s}^2 \sigma_{1 s}^*{ }^2\right)\left(\sigma_{2 s}^2 \sigma_{2 s}^*\right)\left(\pi_{2 p_x}^2=\pi_{2 p_y}^2\right)\left(\sigma_{2 p_z}^2\right)\right] $Bonding electrons:10 (2 from $\sigma_{2 s}$, 6 from $\pi_{2 p}$, and 2 from $\sigma_{2 p_z}$)Antibonding electrons:4 (2 from $\sigma_{1 s}^*$ and 2 from $\sigma_{2 s}^*$)Bond order calculation:$ \text{Bond Order} = \frac{1}{2}(10-4) = 3 $Carbon Monoxide (CO):Total electrons: 6 (from C) + 8 (from O) = 14$ \left[\left(\sigma_{1 s}^2 \sigma_{1 s}^{*2}\right)\left(\sigma_{2 s}^2 \sigma_{2 s}^{*2}\right)\left(\pi_{2 p_x}^2=\pi_{2 p_y}^2\right)\left(\sigma_{2 p_z}^2\right)\right] $Bonding electrons:10 (2 from $\sigma_{2 s}$, 6 from $\pi_{2 p}$, and 2 from $\sigma_{2 p_z}$)Antibonding electrons:4 (2 from $\sigma_{1 s}^*$ and 2 from $\sigma_{2 s}^*$)Bond order calculation:$ \text{Bond Order} = \frac{1}{2}(10-4) = 3 $Nitrosyl Ion ($\mathrm{NO}^{+}$):Total electrons: 7 (from N) + 8 (from O) - 1 (due to positive charge) = 14$ \left[\left(\sigma_{1 s}^2 \sigma_{1 s}^*{ }^2\right)\left(\sigma_{2 s}^2 \sigma_{2 s}^*\right)\left(\pi_{2 p_x}^2=\pi_{2 p_y}^2\right)\left(\sigma_{2 p_z}^2\right)\right] $Bonding electrons:10 (2 from $\sigma_{2 s}$, 6 from $\pi_{2 p}$, and 2 from $\sigma_{2 p_z}$)Antibonding electrons:4 (2 from $\sigma_{1 s}^*$ and 2 from $\sigma_{2 s}^*$)Bond order calculation:$ \text{Bond Order} = \frac{1}{2}(10-4) = 3 $ConclusionAll three species, $\mathrm{N}_2$, $\mathrm{CO}$, and $\mathrm{NO}^{+}$, are isoelectronic as they each contain 14 electrons. Moreover, they share an identical bond order of 3.

According to the VSEPR theory, what are the most stable shapes of $X_4 F_4$ and $\mathrm{SF}_4$, res...

$\mathrm{XeF}_4$ : With 4 bonding pairs and 2 lone pairs around xenon, the most stable shape is square planar.$\mathbf{S F}_4$ : With 4 bonding pairs and 1 lone pair around sulfur, the most stable shape is see-saw.

The dipole moments of three $\mathrm{AB}_3$-type molecules $\mathbf{I}, \mathbf{I I}$, and $\mathbf{...

Given:Three AB₃-type molecules I, II, III have dipole momentsI =0.0 DII =0.2 DIII =1.5 DKey conceptsTrigonal planar (AX₃, no lone pair)→ perfectly symmetrical →μ = 0Examples:BF₃, BCl₃Trigonal pyramidal (AX₃E, one lone pair)→ polar →μ ≠ 0Dipole moment depends on bond polarityanddirection.NH₃ vs NF₃NH₃:Bond dipoles and lone-pair dipole act in thesame direction→large μ ≈ 1.47 DNF₃:Bond dipoles oppose lone-pair dipole →small μ ≈ 0.24 DIdentificationI (0.0 D)→ must benonpolar trigonal planar⇒BF₃II (0.2 D)→ very small dipole moment⇒NF₃III (1.5 D)→ large dipole moment⇒NH₃✅ Correct option:AI : BF₃, II : NF₃, III : NH₃

Chemical Bonding And Molecular Structure - IAT-IISER Chemistry PYQs

Q: Identify the correct order of the molecules with respect to the magnitude of their dipole moment:...

i) Dipole moment $=0$, (cancelled out)ii) $\sqrt{P_{C l}^2+P_{C l}^2+2 P_{C l}^2 \cos (120)}=P_{C l}$iii) $\sqrt{P_{C l}^2+P_{C l}^2+2 P_{C l}^2 \cos (60)}=\sqrt{3} P_{C l}$iv) $\sqrt{P_F^2+P_F^2+2 P_F^2 \cos (60)}=\sqrt{3} P_F$$\left|P_F\right|>\left|P_{C l}\right|$ as Fluorine is more electronegative

01

PYQ 2020

easy

chemistry ID: iat-iise

$ $

02

PYQ 2022

easy

chemistry ID: iat-iise

Identify the correct order of the molecules with respect to the magnitude of their dipole moment:

03

PYQ 2023

medium

chemistry ID: iat-iise

Amongst, and, which molecules will be attracted to an external magnetic field?

1

and

2

and

3

, and

4

, and

04

PYQ 2024

medium

chemistry ID: iat-iise

Which one of the following statements is correct about, and ?

1

These are isoelectronic and have identical bond order.

2

These are isoelectronic and have different bond orders.

3

These are not isoelectronic but have identical bond order.

4

These are neither isoelectronic nor have identical bond order.

05

PYQ 2024

easy

chemistry ID: iat-iise

According to the VSEPR theory, what are the most stable shapes of and, respectively?

1

See-saw and square planar

2

Both see-saw

3

Square planar and see-saw

4

Both square planar

06

PYQ 2025

medium

chemistry ID: iat-iise

Which among the following processes is/are associated with increasing bond order but no change in diamagnetic/paramagnetic behaviour?(i) (ii) (iii)

1

(ii) only

2

(i) and (ii)

3

(ii) and (iii)

4

(iii) only

07

PYQ 2025

medium

chemistry ID: iat-iise

The dipole moments of three -type molecules, and are measured to be, and 1.5 D , respectively. Which one of the following options is correct regarding the identity of, and ?

1

I:

2

I:

3

I:

4

I:

Chemical Bonding And Molecular Structure

High-Yield Trend

Questions
7 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Chemical Bonding And Molecular Structure

High-Yield Trend

Questions 7 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
7 MCQs