Chemical Bonding & Molecular Structure

NO$_2^-$ ion

Structure (Lewis Structure)

Nitrogen (N) has 2 non-bonding electrons. The total valence electrons of N is 5. Out of 5, 3 electrons are bonding electrons and 2 electrons are non-bonding electrons.

Double bonded oxygen has 4 non-bonding electrons. Oxygen valence electrons : 6

Out of 6, 2 are bonding electrons are 4 are non-bonding electrons.

Negatively charged oxygen has 6 non-bonding electrons. The negative charge is due to the extra electron present. So, the total number of non-bonded electrons present in NO$_2^-$ ion based on Lewis theory is

$2(of\,N) + 4(of\,O) + 6(of\,{O^ - })$

$ = 2 + 4 + 6 = 12$

Linear species are

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

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

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

Number of $\mathrm{e}^{-}$ present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{-}=3$

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

Species	Unpaired e
$ \mathrm{N}_2 $	0
$ \mathrm{O}_2 $	2
$ \mathrm{C}_2^{-} $	1
$ \mathrm{O}_2^{-} $	1
$ \mathrm{O}_2^{2-} $	0
$ \mathrm{H}_2^{+} $	1
$ \mathrm{CN}^{-} $	0
$ \mathrm{He}_2^{+} $	1

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).

Species	Shape (VSEPR)	Dipole moment?	Reason
HF	linear	nonzero	polar bond, no cancellation
$\mathrm{H}_2$	linear	zero	homonuclear, nonpolar
$\mathrm{H}_2\mathrm{S}$	bent	nonzero	bent shape → net dipole
$\mathrm{CO}_2$	linear	zero	symmetric linear cancellation
$\mathrm{NH}_3$	trigonal pyramidal	nonzero	lone pair breaks symmetry
$\mathrm{BF}_3$	trigonal planar	zero	symmetric planar cancellation
$\mathrm{CH}_4$	tetrahedral	zero	symmetric tetrahedral cancellation
$\mathrm{CHCl}_3$	tetrahedral	nonzero	not symmetric ($3\ \mathrm{Cl}$ and $1\ \mathrm{H}$)
$\mathrm{SiF}_4$	tetrahedral	zero	symmetric tetrahedral cancellation
$\mathrm{H}_2\mathrm{O}$	bent	nonzero	bent shape → net dipole
$\mathrm{BeF}_2$	linear	zero	symmetric linear cancellation

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.

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

$\mathrm{A}_2 \mathrm{~B} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{B}^{-2}$

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 $=4$

Sol.	Magnetic behaviour	Bond order
$\mathrm{H}_2$	Diamagnetic	$1$
$\mathrm{He_2^+}$	Paramagnetic	$0.5$
$\mathrm{O_2^+}$	Paramagnetic	$2.5$
$\mathrm{N_2^{2-}}$	Paramagnetic	$2$
$\mathrm{O_2^{2-}}$	Diamagnetic	$1$
$\mathrm{F_2}$	Diamagnetic	$1$
$\mathrm{Ne_2^+}$	Paramagnetic	$0.5$
$\mathrm{B_2}$	Paramagnetic	$1$

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.

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

The number of molecules having only 2 lone pair of electrons $=3$

Which are $\mathrm{H}_2 \mathrm{O} , \mathrm{N}_2$ and $\mathrm{CO}$

Note:-

$\mathrm{XeF}_4$ have 2 lps on central atom, but we are asked about lone pair in molecule

A bent-shaped molecule has a molecular geometry with a central atom bonded to two other atoms and one or two pairs of non-bonding electrons. The VSEPR (Valence Shell Electron Pair Repulsion) theory helps us predict the shapes of molecules.

Let's consider the given molecules:

1. N$_3^-$: The azide ion has a linear shape, not bent. The nitrogen in the middle is connected to two other nitrogens and there are no lone pairs on the central atom.




2. NO$_2^-$: The nitrite ion has a bent shape. The central nitrogen atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.




3. I$_3^-$: The triiodide ion has a linear shape, not bent. The central iodine atom is connected to two other iodine atoms and there are no lone pairs on the central atom.




4. O$_3$: Ozone has a bent shape. The central oxygen atom is connected to two other oxygen atoms and has one lone pair of electrons, giving it a bent geometry.




5. SO$_2$: Sulfur dioxide has a bent shape. The central sulfur atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.

Therefore, there are 3 bent-shaped molecules: NO$_2^-$, O$_3$, and SO$_2$.

Interhalogen compounds are the substances that consist of two different halogens. The most common type of interhalogen compounds are binary, containing only two different elements.

In IF$_5$, iodine (I) is the central atom. Iodine has 7 valence electrons, 5 of which are used for bonding with the 5 fluorine (F) atoms. This leaves 2 electrons, or 1 lone pair on the iodine atom.

In IF$_7$, iodine (I) is the central atom again. Iodine has 7 valence electrons, all of which are used for bonding with the 7 fluorine (F) atoms. So, there are no lone pairs on the iodine atom.

Therefore, the sum of lone pairs present on the central atom of the interhalogens IF$_5$ and IF$_7$ is 1.

So, Answer is 4

The covalent character of an ionic bond is largely determined by the polarization of the ions involved in the bond. Polarization refers to the distortion of the electron cloud of an anion by a cation. This distortion leads to a shift in electron density towards the cation, thereby increasing the covalent character of the bond.

Here's a breakdown of the options:

(A) Polarising power of cation: The greater the polarising power of a cation, the greater the distortion of the anion, and therefore, the greater the covalent character of the bond. Thus, this statement is correct.

(B) Extent of distortion of anion: The more an anion is distorted, the more the electron density is shifted towards the cation, and the greater the covalent character of the bond. This statement is also correct.

(C) Polarisability of the anion: The greater the polarisability of an anion, the more it can be distorted, and therefore, the greater the covalent character of the bond. Thus, this statement is also correct.

(D) Polarising power of anion: This statement is incorrect. It is not the polarising power of the anion, but the polarisability of the anion and the polarising power of the cation that influence the covalent character of the bond.

Therefore, three of the factors listed (A, B, C) affect the percent covalent character of the ionic bond.

$\mathrm{XeF}_4 \rightarrow$ Square planar

$\mathrm{SF}_4 \rightarrow$ See saw

$\mathrm{SiF}_4 \rightarrow$ Tetrahedral

$\mathrm{BF}_4^{-} \rightarrow$ Tetrahedral

$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \rightarrow$ Square planar

$\left[\mathrm{FeCl}_4\right]^{2-} \rightarrow$ Tetrahedral

$\left[\mathrm{PtCl}_4\right]^{2-} \rightarrow$ Square planar

$\mathrm{BrF}_4^{-} \rightarrow$ Square planar

So, 4 square planer shape compounds are present.

In an ice crystal, each water molecule is hydrogen bonded to four neighbouring molecules.

A square pyramidal structure has five bonds and one lone pair, making a total of six electron pairs around the central atom. The geometry of such a molecule can be analyzed using the VSEPR theory.

1. $\mathrm{PF}_{5}$: 5 bond pairs and 0 lone pairs, so its geometry is trigonal bipyramidal, not square pyramidal.


2. $\mathrm{BrF}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.


3. $\mathrm{IF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.


4. $\mathrm{BrF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.


5. $\mathrm{XeOF}_{4}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.


6. $\mathrm{ICl}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.

So the number of species from the given list that have a square pyramidal structure is 3.

Only $\mathrm{ICl}_4^{-}, \mathrm{BrF}_3$ and $\mathrm{NO}_2^{+}$have even number of electrons.

$ \begin{aligned} & \mathrm{NO}_2 \Rightarrow 23 e^{-} ; \\\\ & \mathrm{ICl}_4^{-} \Rightarrow 122 e^{-} ; \\\\ & \mathrm{BrF}_3 \Rightarrow 62 e^{-} ; \\\\ & \mathrm{ClO}_2 \Rightarrow 33 e^{-} ; \\\\ & \mathrm{NO}_2^{+} \Rightarrow 22 e^{-} ; \\\\ & \mathrm{NO} \Rightarrow 15 e^{-} \end{aligned} $

Total no, of lone pairs on oxygen atoms = 6

Number of atoms with zero oxidation state $=0$


Number of atom with zero oxidation state $=0$


Number of atoms where zero oxidation state $=2$


Number of atoms where zero oxidation state $=3$


Number of atoms with zero oxidation state $=1$

$ \therefore $ Total atom with zero oxidation number state are 6.

Those species which have unpaired electrons are called paramagnetic species.

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

B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

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

Here two unpaired electrons present. So it is paramagnetic.

$O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ 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^2}^ * \, = \,\pi _{2p_y^2}^ * $

Here is no unpaired electron so it is diamagnetic.

$O_2^{+}$ has 15 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^0}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

$C_2$ 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.

$C_2^{ - }$ has 13 electrons.

Moleculer orbital configuration of $C_2^{ - }$ is

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

Here 1 unpaired electron present, so it is paramagnetic.

Li₂ has 6 electrons.

Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$

Here no unpaired electron present, so it is diamagnetic.

Configuration of $He_2^ + $ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

$\mathrm{CIF}_{3} \rightarrow 3 \sigma$ bond $+2$ lone pair

$\mathrm{IF}_{7} \rightarrow 7 \sigma$ bond $+0$ lone pair

$\mathrm{BrF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal

$\mathrm{BrF}_{3} \rightarrow 3 \sigma$ bond $+2$ lone pair

$\mathrm{I}_{2} \mathrm{Cl}_{6} \rightarrow 4 \sigma$ bond $+2$ lone pair

$\mathrm{IF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal

$\mathrm{CIF} \rightarrow 1 \sigma$ bond $+3$ lone pair

$\mathrm{CIF}_{5} \rightarrow 5 \sigma$ bond $+1$ lone pair $\rightarrow$ Square pyramidal

$\mathrm{NO}_{3}^{\ominus} \rightarrow$ Trigonal planar (Planar)

$\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow$ Open book (Non-planar)

$\mathrm{BF}_{3} \rightarrow$ Trigonal planar (Planar)

$\mathrm{PCl}_{3} \rightarrow$ Pyramidal (Non-planar)

$\mathrm{XeF}_{4} \rightarrow$ Square planar (Planar)

$\mathrm{SF}_{4} \rightarrow$ See-Saw (Non-planar) $\mathrm{XeO}_{3} \rightarrow$ Pyramidal (Non-planar)

$\mathrm{PH}_{4}^{\oplus} \rightarrow$ Tetrahedral (Non-planar)

$\mathrm{SO}_{3} \rightarrow$ Trigonal planar (Planar)

$\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-} \rightarrow$ Tetrahedral (Non-planar)

Paramagnetic species: $\mathrm{KO}_{2}, \mathrm{NO}_{2}, \mathrm{ClO}_{2}, \mathrm{NO}$

Diamagnetic species are: $\mathrm{Na}_{2} \mathrm{O}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{Cl}_{2} \mathrm{O}$

$\mathrm{CN}^{-}, \mathrm{NO}^{+}$and $\mathrm{O}_{2}^{2+}$ have bond order of 3

$\mathrm{O}_{2}$ has bond order of 2

$\mathrm{O}_{2}^{+}$ has bond order of $2.5$

$\therefore 3$ species have similar bond order.

From structure, it is clear that it has five bond pairs and one lone pair.

Those species which have unpaired electrons are called paramagnetic species.

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

(1)   $N_2$ 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.

(2) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)

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

Here in $N_2^{ + }$, 1 unpaired electron present, so it is paramagnetic.

(3)   $\mathrm{N}_{2}^{2-}$ has 16 electrons.

Moleculer orbital configuration of $\mathrm{N}_{2}^{2-}$ 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 electron present, so it is paramagnetic.

(4)   $\mathrm{N}_{2}^{-}$ has 15 electrons.

Moleculer orbital configuration of $\mathrm{N}_{2}^{-}$ 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^0}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

(a)    $O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ 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^2}^ * \, = \,\pi _{2p_y^2}^ * $

Here is no unpaired electron so it is diamagnetic.

(b)    $O_2^{−}$ has 17 electrons.

Moleculer orbital configuration of $O_2^{2−}$ 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^2}^ * \, = \,\pi _{2p_y^1}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

(c)   $O_2$ 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 electron present, so it is paramagnetic.

(d)   $O_2^{+}$ has 15 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^0}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

Symmetrical molecules have zero dipole moment like $\mathrm{BF}_{3}$, $\mathrm{BeF}_{2}, \mathrm{SiF}_{4}, \mathrm{CCl}_{4}$ and $\mathrm{PF}_{5}$.

Unsymmetrical molecules have net diploe moment like $-\mathrm{NF}_3$, $\mathrm{CHCl}_3$ and $\mathrm{H}_2 \mathrm{S}$

$\mathrm{PF}_5 \Rightarrow \mathrm{sp}^3 \mathrm{~d}$ hybridisation

(5 sigma bonds, zero lone pair on central atom)

Value of $y=1$

$\mathrm{BeF}_2, \mathrm{BF}_3$ and $\mathrm{CCl}_4 \Rightarrow \mu_{\mathrm{net}}=0$

$\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{HCl} \Rightarrow \mu_{\mathrm{net}} \neq 0$