Chemical Bonding & Molecular Structure

Option A

A charge–charge interaction falls off as

$U_{qq}\propto\frac1r$

while a charge–dipole interaction goes as

$U_{q\mu}\propto\frac1{r^2}\,. $

So the charge–dipole potential actually decays faster.

⇒ A is false.

Option B

A fixed dipole–dipole energy is

$U_{\mu\mu}\propto\frac1{r^3}\,, $

but once both molecules tumble freely, the thermal (Keesom) average gives

$\langle U\rangle_{\rm Keesom}\propto -\frac{\mu^4}{kT\,(4\pi\varepsilon_0)^2\,r^6}\,. $

That falls off as $1/r^6$, not $1/r^3$.

⇒ B is false.

Option C

The permanent–induced dipole (Debye) interaction is

$U_{\rm Debye}=-\tfrac12\,\alpha\,E^2\propto -\frac{\alpha\,\mu^2}{(4\pi\varepsilon_0)^2\,r^6}\,, $

and when you average over random orientations (assuming the interaction is weak compared with $kT$), no $T$ appears in the result.

⇒ C is true.

Option D

Even totally nonpolar species experience instantaneous‐induced (London) dispersion forces that attract as

$U_{\rm disp}\propto -\frac1{r^6}\,. $

⇒ D is true.

Answer: C and D.

Molecular orbital (MO) energy levels for homonuclear diatomic molecules: For Molecular orbital theory, all the valence atomic orbitals are included. The total number of atomic orbitals must always equal the total number of molecular orbitals. Bonding molecular orbitals are more stable than the atomic orbitals and increase the electron density between the nuclei Antibonding molecular orbitals are. less stable than the atomic orbitals and decrease the electron density between the nuclei. (A) Bond older can be calculated using the formula:
Bond order $=\frac{1}{2}\left(N_b-N_a\right)$
$N_b$ - Number of bonding electrons
$\mathrm{N_a}^{\text { }}$ - Number of antibonding electrons
For $\mathrm{Ne}_2$, total number of electrons $=20 \quad\left\{\begin{array}{l}\text { Number of electrons } \\ \text { for } \mathrm{Ne}=10\end{array}\right.$

The total electrons are filled in the molecular orbitals as,${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * \sigma _{2{p_z^2}}^ *$$ \begin{aligned} N_b=10, & N_a=10 \\ \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-10) \\ & =\frac{1}{2}(0) \\ & =0 \end{aligned} $Bond order of $\mathrm{Ne}_2$ is zero. It is correct. B) $H O M O$ of $F_2$ The number of electrons in $F$ is 9 . So, the total number of electrons in $F_2$ molecule is $18(9+9)$ The 18 electrons are filled in the molecular orbitals as,${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $
Homo is the highest occupied molecular orbital. For $F_2$, the last electron is occupied in the $\pi^*$ orbital. So, the highest occupied molecular orbital of $F_2$ is $\pi$ - type, not $\sigma$-type. It is not correct. c) Bond energy is directly proportional to bond order. The strength of a chemical bond is directly proportional to the amount of energy required to break it. To compare the bond energy of $\mathrm{O}_2^{+}$and $\mathrm{O}_2$, calculate the bond older. The molecule with higher bond order has higher bond energy$\mathrm{O}_2$ Number of elections of oxygen $(0)=8$ number of elections of $\mathrm{O}_2$ molecule $=16$ The 16 elections are filled in the molecular orbitals as: ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * $

$ \begin{aligned} & N_b=10, N_a=6 \end{aligned} $ $ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-6) \\ & =\frac{1}{2} \cdot(4) \\ & =2 \end{aligned} $ $\mathrm{O}_2^{+}$ Total number of elections for $0_2$ molecule $=16$ For $\mathrm{O}_2^{+}$, one electron is removed from $\mathrm{O}_2$. So, the total number of electrons $=16-1=15$ The 15 electrons are filled in the molecular orbitals as, ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * $$ \begin{aligned} N_b=10, & N_a=5 \\ \text { Bond order } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-5) \\ & =\frac{1}{2}(5) \\ & =2.5 \end{aligned} $Bond order of $\mathrm{O}_2^{+}$is higher than the bond order of $\mathrm{O}_2$. Bond order and bond energy are directly proportional so, the higher the bond oder, the higher the bond energy. Therefore, $\mathrm{O}_2{ }^{+}$molecule which has higher bond order and also has higher bond energy.So, bond energy of $\mathrm{O}_2^{+}$is higher than the bond energy of $\mathrm{O}_2$.$ \text { It is not correct: } $ D) Bond length of $Li_2$ and $B_2$ Bond length and bond older are inversely proportional to each other so; higher the bond order, shorter the bond length. To get the bond length of $L_2$ and $B_2$, first calculate the bond order of $Li_2$ and $B_2$. $\mathrm{Li}_2$ Number of electrons of $\mathrm{Li}=3$ : total number of electrons in $\mathrm{Li}_2$ molecule $=6$ The total electrons are filled in the molecular orbitals as Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$$ N_b=4, N_a=2 $$ \begin{aligned} \text { Bond loeder } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(4-2) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $ $B_2$ Number of electrons of Balon $(B)=5$ Total number of electrons of $B_2$ molecule $=10$ The total electrons are filled in the molecular orbitals as ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$ N_b=6 \quad N_a=4 $$ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b+N_a\right) \\ & =\frac{1}{2}(6-4) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $Bond order is equal for $Li_{2}$ and $B_2$ so, atomic size is consider Comparing the size of lithium and boron, lithium is a larger atom than boron. Bond length generally increases with the size of atom so, bond length of $\mathrm{Li}_2$ is larger than the bond length of $\mathrm{B_2}$ Size of $Li>B$ So, Bond length of $Li_2>B_2$ This statement is correct: Answer: The incorrect statements are (B) and (C). (B) and (C).

The octet rule states that atoms tend to combine in such a way that they each have eight electrons in their valence shell, giving them the same electronic configuration as a noble gas. Let's examine each option to determine which molecules follow the octet rule.

Option A: $\mathrm{CO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{NO}$ and $\mathrm{HCl}$

1. $\mathrm{CO}_2$: Carbon dioxide follows the octet rule. Carbon forms double bonds with two oxygen atoms, allowing it to have 8 electrons in its valence shell.

2. $\mathrm{C}_2 \mathrm{H}_4$: Ethylene follows the octet rule. Each carbon forms a double bond with the other carbon and single bonds with two hydrogen atoms, resulting in 8 electrons in the valence shell of the carbons.

3. $\mathrm{NO}$: Nitric oxide does not follow the octet rule. It has an odd number of electrons, making it a free radical.

4. $\mathrm{HCl}$: Hydrogen chloride follows the octet rule. Chlorine has 8 electrons in its valence shell when bonded with hydrogen.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{C}_2 \mathrm{H}_4$, and $\mathrm{HCl}$ follow the octet rule here.

Option B: $\mathrm{NO}_2, \mathrm{O}_3, \mathrm{HCl}$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{NO}_2$: Nitrogen dioxide does not follow the octet rule. It has an odd number of electrons.

2. $\mathrm{O}_3$: Ozone follows the octet rule. It has a resonance structure that allows each oxygen atom to have 8 electrons in its valence shell.

3. $\mathrm{HCl}$: As mentioned, hydrogen chloride follows the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule. All atoms achieve stable configurations through bonding.

Thus, three molecules $\mathrm{O}_3$, $\mathrm{HCl}$, and $\mathrm{H}_2 \mathrm{SO}_4$ follow the octet rule in this option.

Option C: $\mathrm{BCl}_3, \mathrm{NO}, \mathrm{NO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{BCl}_3$: Boron trichloride does not follow the octet rule. Boron has only 6 electrons in its valence shell.

2. $\mathrm{NO}$: As mentioned, nitric oxide does not follow the octet rule.

3. $\mathrm{NO}_2$: As mentioned, nitrogen dioxide does not follow the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule.

Thus, only one molecule, $\mathrm{H}_2 \mathrm{SO}_4$, follows the octet rule in this option.

Option D: $\mathrm{CO}_2, \mathrm{BCl}_3, \mathrm{O}_3$ and $\mathrm{C}_2 \mathrm{H}_4$

1. $\mathrm{CO}_2$: As mentioned, carbon dioxide follows the octet rule.

2. $\mathrm{BCl}_3$: As mentioned, boron trichloride does not follow the octet rule.

3. $\mathrm{O}_3$: As mentioned, ozone follows the octet rule.

4. $\mathrm{C}_2 \mathrm{H}_4$: As mentioned, ethylene follows the octet rule.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{O}_3$, and $\mathrm{C}_2 \mathrm{H}_4$ follow the octet rule in this option.

Conclusion: The correct options are Option A and Option D, both having at least three molecules that follow the octet rule.

The molecules which gives permanent dipole moment are polar in nature.





Thus, options (a, c) are correct.

Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :

For $C_2^{2 - }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $O_2^{2 + }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $

Thus the bond order = (10 $-$ 4)/2 = 3;

Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

So O$_2$ has 2 unpaired electrons. for O₂, bond order = 2.

As bond order is inversely proportional to bond length, thus the bond length of $O_2^{2 + }$ is less than the bond length of O₂.

Option (C) : For $N_2^ + $, the total number of electrons is 13. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Bond order of $N_2^ + $ = (9 $-$ 4)/2 = 2.5

   $N_2^{ - }$ has 15 electrons.

Moleculer orbital configuration of $N_2^{ - }$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$

$\therefore\,\,\,\,$N_a = 5

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right] = 2.5$

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $He_2^ + $ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.

Structure of each of the compound is represented as :

(A) BrF₅



Structure : Square pyramidal

(i) One lone pair

(ii) Four bond pair

(B) ClF₃



Structure : T-Shaped

(i) Two lone pair

(ii) Three bond pair

(C) XeF₄



Structure : Square planar

(i) Two lone pair

(ii) Four bond pair

(D) SF₄



Structure : See-saw

(i) One lone pair

(ii) Four bond pair

(a) Ice floats in water due to the low density of ice as compare to water which is due to open cage like structure (formed by intermolecular H-bonding).

(b) Basic strength of RNH₂ > R₃N. It is also explained by hydrogen bonding.


(c)
(d) Dimerisation of acetic acid in benzene is due to intermolecular hydrogen bonding.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A)   $N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here no unpaired electron present, so it is diamagnetic.

(B)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $

Here 2 unpaired electrons present, so it is paramagnetic.

(C)    $S_2$ has 32 electrons.

It also has 2 unpaired electrons like $O_2$, so it is paramagnetic.

(D)   $C_2$ has 12 electrons.

Moleculer orbital configuration of $C_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$

Here no unpaired electron present, so it is diamagnetic.

2, 2-dimethylpropane and 2, 2, 3, 3 - tetramethylbutane are symmetrical molecules so their net dipole moment is zero. (b) and (c) have resultant dipole moment.

To determine which molecule or ion assumes a linear structure, we need to consider the molecular geometry, which is often predicted by the VSEPR (Valence Shell Electron Pair Repulsion) theory.

Let’s analyze each option:

Option A: SnCl₂

Tin(II) chloride, SnCl₂, has a bent structure rather than a linear one due to the presence of a lone pair of electrons on the tin atom, which repels the bonds between Sn and Cl.

Option B: NCO^-

The cyanate ion, NCO^-, has a linear structure because the central atom carbon forms a double bond with nitrogen and a triple bond with oxygen, aligning all atoms in a straight line.

Option C: CS₂

Carbon disulfide, CS₂, has a linear structure. The central carbon atom forms double bonds with each sulfur atom, allowing the molecule to maintain a straight line.

Option D: $NO_2^+$

The nitronium ion, $NO_2^+$, also has a linear structure. It has a central nitrogen atom bonded to two oxygen atoms with no lone pairs on nitrogen, thus the molecule assumes a linear shape.

So, the linear structures are assumed by Options B, C, and D.

Isostructural compounds are those that have the same shape and similar bond angles. To determine which compound is isostructural with carbon dioxide (CO₂), we need to analyze the molecular geometry of each of the options.

1. CO₂ has a linear structure with the carbon atom in the center, double-bonded to two oxygen atoms, giving it a bond angle of 180°. The structure can be represented as:

$ \text{O} = \text{C} = \text{O} $

2. HgCl₂ also has a linear structure. The mercury atom in the center is single-bonded to two chlorine atoms, giving it a bond angle of 180°. The structure can be represented as:

$ \text{Cl} - \text{Hg} - \text{Cl} $

3. SnCl₂ has a bent (or V-shaped) structure due to lone pairs on the tin atom. Hence, it is not isostructural with CO₂.

4. C₂H₂ (acetylene) has a linear structure with a carbon-carbon triple bond and hydrogen atoms at each end, giving it a bond angle of 180°. The structure can be represented as:

$ \text{H} - \text{C} \equiv \text{C} - \text{H} $

5. NO₂ has a bent structure due to the presence of a lone electron on the nitrogen atom, which causes repulsion and bends the molecule. Therefore, it is not isostructural with CO₂.

Given this analysis, CO₂ is isostructural with both HgCl₂ and C₂H₂.