Chemical Bonding & Molecular Structure

$\begin{aligned} & \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\ & \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right] \end{aligned}$

Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2=2$

Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{+}=1$

So total $\mathrm{e}^{-}$ in $\left(\pi^*\right)=2+1+3=6$

To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:

$\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}$

Let's evaluate each molecule individually:

1. $\mathrm{C}_2$:

For $\mathrm{C}_2$, the total number of electrons is 12. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 8 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{8 - 4}{2} = 2$

2. $\mathrm{O}_2$:

For $\mathrm{O}_2$, the total number of electrons is 16. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^1 \left( \pi_{2p_y}^* \right)^1$. There are 10 bonding electrons and 6 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 6}{2} = 2$

3. $\mathrm{Be}_2$:

For $\mathrm{Be}_2$, the total number of electrons is 8. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2$. There are 4 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 4}{2} = 0$

4. $\mathrm{Li}_2$:

For $\mathrm{Li}_2$, the total number of electrons is 6. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2$. There are 4 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 2}{2} = 1$

5. $\mathrm{Ne}_2$:

For $\mathrm{Ne}_2$, the total number of electrons is 20. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^2 \left( \pi_{2p_y}^* \right)^2 \left( \sigma_{2p_z}^* \right)^2$. There are 10 bonding electrons and 10 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 10}{2} = 0$

6. $\mathrm{N}_2$:

For $\mathrm{N}_2$, the total number of electrons is 14. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 10 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 4}{2} = 3$

7. $\mathrm{He}_2$:

For $\mathrm{He}_2$, the total number of electrons is 4. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2$. There are 2 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{2 - 2}{2} = 0$

Thus, the molecules with a bond order of 2 from the given list are $\mathrm{C}_2$ and $\mathrm{O}_2$. Therefore, the number of molecules having bond order 2 is 2.

$\mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2$

These are exception of octet rule

$\mathrm{NiS + HN{O_3} + HCl\buildrel {} \over \longrightarrow \mathop {NiC{l_2}}\limits_{(A)} + S + NO + {H_2}O}$

$\mathrm{\mathop {NiC{l_2}}\limits_{(A)} + N{H_4}OH +}$ Dimethylgyoxime $\mathrm{\buildrel {} \over \longrightarrow \mathop {Ni{{(dmg)}_2}}\limits_{(B)} + N{H_4}Cl + {H_2}O}$

Total 12 proton do not involve in H-bond

$\mathrm{SO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3$ are $s p^2$ hybridised central atom

compounds

$\mathrm{CH_3OH}$

$\mathrm{H_2O}$

$\mathrm{HF}$

$\mathrm{NH}_3$

Can show $\mathrm{H}$-Bonding.

A molecule has zero dipole moment when its bond dipoles cancel by symmetry (or when bonds are nonpolar).

Compounds with zero dipole moment:
$\mathrm{H}_2,\ \mathrm{CO}_2,\ \mathrm{BF}_3,\ \mathrm{CH}_4,\ \mathrm{SiF}_4,\ \mathrm{BeF}_2$

Number = $6$.

$\text { Lewis dot structure of } \mathrm{NO}_2^{-} \text {is: }$

$\text { Total number of valence } \mathrm{e}^{\ominus} \text { around } \mathrm{N}=8$

$\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{S}, \mathrm{HBr}, \mathrm{HCl}$ have non zero dipole moments.

$\mathrm{O}_2^{-} \text {and } \mathrm{NO} \text { have } 1 \text { unpaired electron. }$

When an atom has the lowest oxidation number of -2 in a compound, it means the atom has gained two electrons beyond its neutral state to achieve this oxidation state. This is because gaining electrons makes the oxidation number more negative. In a neutral atom, the electrons in the outermost shell are known as valence electrons, which play a crucial role in chemical reactions and bonding.

In the given compound $\mathrm{A}_2\mathrm{B}$, atom B has an oxidation state of -2, indicating it has gained two electrons. For an atom to gain two electrons to complete its octet implies that its neutral state had six valence electrons. Therefore, before gaining electrons, the atom B would have had six electrons in its valence shell to begin with. As it gains two more electrons, it reaches an oxidation number of -2, implying it now has a complete octet or eight electrons in its valence shell, but the original count of valence electrons in its neutral state was six.

To find the fractional charge on each atom in a diatomic molecule, we can use the relationship between dipole moment (μ), charge (q), and distance (d):

$ \mu = q \times d $

Where:

• μ is the dipole moment.


• q is the magnitude of the charge on each atom.


• d is the distance between the charges.

First, convert units to be consistent with esu:

• Bond distance in cm: $ 1 \mathrm{~A}^{\circ} = 1 \times 10^{-8} \mathrm{~cm} $


• Dipole moment in esu·cm: $ 1.2 \mathrm{~D} = 1.2 \times 10^{-18} \mathrm{~esucm} $

We are given that the bond distance ($ d $) is $ 1 \mathrm{~A}^{\circ} $ and that the dipole moment $( \mu $) is $ 1.2 \mathrm{~D} $. Using these values, we can calculate the charge ($ q $) as follows:

$ q = \frac{\mu}{d} $

Now plug in the given values:

$ q = \frac{1.2 \times 10^{-18} \mathrm{~esucm}}{1 \times 10^{-8} \mathrm{~cm}} $

$ q = 1.2 \times 10^{-10} \mathrm{~esu} $

$ q = 1.2 \times 10^{-9} \times 10^{-1} \mathrm{~esu} $

$ q = 0.000000012 \times 10^{-1} \mathrm{~esu} $

To determine the number of species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.

1. $\mathrm{NH}_3$ (Ammonia):

The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

2. $\mathrm{SO}_2$ (Sulfur dioxide):

The central atom is sulfur. Sulfur dioxide has 2 sigma bonds and 1 lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.

3. $\mathrm{SiO}_2$ (Silicon dioxide):

The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.

4. $\mathrm{BeCl}_2$ (Beryllium chloride):

The central atom is beryllium. Beryllium chloride has 2 sigma bonds and no lone pair. Thus, the steric number is 2, which corresponds to $\mathrm{sp}$ hybridization.

5. $\mathrm{CO}_2$ (Carbon dioxide):

The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.

6. $\mathrm{H}_2\mathrm{O}$ (Water):

The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

7. $\mathrm{CH}_4$ (Methane):

The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

8. $\mathrm{BF}_3$ (Boron trifluoride):

The central atom is boron. Boron trifluoride has 3 sigma bonds and no lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.

From the above analysis, the species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals are:

• $\mathrm{NH}_3$
• $\mathrm{H}_2\mathrm{O}$
• $\mathrm{CH}_4$

Therefore, the number of species where the central atom uses $\mathrm{sp}^3$ hybrid orbitals is 3.

Two molecular orbitals $\sigma 2 \mathrm{s}$ and $\sigma * 2 \mathrm{s}$.

Six molecular orbitals $\sigma 2 \mathrm{p}_z$ and $\sigma * 2 \mathrm{p}_{\mathrm{z}}$.

$\pi 2 \mathrm{p}_{\mathrm{x}}, \pi 2 \mathrm{p}_{\mathrm{y}}$ and $\pi * 2 \mathrm{p}_{\mathrm{x}}, \pi^* 2 \mathrm{p}_{\mathrm{y}}$

To determine which molecules have a zero dipole moment from the list provided, we need to consider their molecular geometry and symmetry. Molecules that are symmetrical often have zero dipole moment because the bond dipoles cancel each other out.

CO₂ (Carbon Dioxide): This molecule is linear in shape. The two opposing oxygen atoms pull equally on the carbon, canceling out the dipoles, resulting in a net dipole moment of zero.

CH₄ (Methane): Methane is tetrahedral and symmetrical, with all four hydrogen atoms equally spaced around the carbon atom. This symmetry leads to the cancellation of dipoles, giving a net dipole moment of zero.

BF₃ (Boron Trifluoride): This molecule is trigonal planar and symmetrical. The three fluorine atoms are arranged evenly around the central boron atom, causing the dipoles to cancel each other out, resulting in a zero dipole moment.

Therefore, the molecules with zero dipole moment among the list provided are CO₂, CH₄, and BF₃.

Antibonding molecular orbital from $2 \mathrm{~s}=1$

Antibonding molecular orbital from $2 p=3$

Total $=4$

To determine whether a molecule is polar or non-polar, we must consider the difference in electronegativity between the atoms and the symmetry of the molecule. Polar molecules occur when there is an electronegativity difference between the bonded atoms. Non-polar molecules either do not have any polar bonds or the polarities cancel each other out because of a symmetrical arrangement. Let's consider each molecule individually:

• HF (Hydrogen fluoride): This molecule is polar due to the high electronegativity difference between hydrogen (H) and fluorine (F).
• H₂O (Water): It is polar because of its bent shape and the difference in electronegativity between oxygen and hydrogen.
• SO₂ (Sulfur dioxide): The molecule is non-linear and has polar bonds, making it a polar molecule.
• H₂ (Hydrogen): This molecule is non-polar because it consists of two identical atoms which share electrons evenly.
• CO₂ (Carbon dioxide): Even though C=O bonds are polar, the molecule is linear and the polarities cancel out, making CO₂ non-polar.
• CH₄ (Methane): It is non-polar because the C-H bonds are evenly distributed in a tetrahedral shape, canceling out any dipole moments.
• NH₃ (Ammonia): This molecule is polar; it has a trigonal pyramidal shape with a lone pair on nitrogen, creating a dipole moment.
• HCl (Hydrogen chloride): The molecule is polar due to the electronegativity difference between hydrogen and chlorine.
• CHCl₃ (Chloroform): It has polar C-Cl bonds and is not symmetrical, resulting in a polar molecule.
• BF₃ (Boron trifluoride): This molecule is non-polar despite having polar bonds, because the shape of the molecule is trigonal planar and the dipoles cancel out.

Given this information, the non-polar molecules from the list are: $\mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{BF}_3$.

Therefore, there are 4 non-polar molecules in the list given.

$\begin{array}{lcl} \mathrm{CO} \Rightarrow & \overline{\mathrm{C}} \equiv \stackrel{+}{\mathrm{O}} & : \mathrm{BO}=3 \\ \mathrm{NO}^{+} \Rightarrow & \mathrm{N} \equiv \mathrm{O}^{+} & : \mathrm{BO}=3 \end{array}$