Chemical Bonding & Molecular Structure

Those species which have unpaired electrons are called paramagnetic species.

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

(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)   NO has 15 electrons.

Moleculer orbital configuration of NO 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 is 1 unpaired electron, 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.

(A) Moleculer orbital configuration of $C_2$ (12 electrons)

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

$\therefore\,\,\,\,$N_a = 4

N_b = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right] = 2$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $C_2^+$ (11 electrons)

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

$\therefore\,\,\,\,$N_a = 4

N_b = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {7 - 4} \right] = 1.5$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $N_2$ (14 electrons)

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

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here no unpaired electron present, so it is diamagnetic.

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

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

$\therefore\,\,\,\,$N_a = 4

N_b = 9

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right] = 2.5$

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer orbital configuration of NO (15 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^0}^ * $

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

N_b = 10

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

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO⁺ (14 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^0}^ * = \,\,\pi _{2p_y^0}^ * $

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here is no unpaired electron, So it is diamagnetic.

(D) 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}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons present, so it is paramagnetic.

Molecular orbital configuration of O$_2^ + $ (15 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^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

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

Here 1 unpaired electrons present, so it is also paramagnetic.

Species having the same number of electrons will have the same bond order.

Number of electrons in CO are 14 (8 from oxygen and 6 from carbon).

NO$^-$ has 16 electrons (7 from nitrogen, 8 from oxygen, add 1 for negative charge).

NO$^+$ has 14 electrons (7 from nitrogen, 8 from oxygen, subtract 1 for positive charge).

CN$^-$ has 14 electrons (7 from nitrogen, 6 from carbon, add 1 for negative charge).

N$_2$ has 14 electrons (7 from each nitrogen).

$\begin{array}{ll}\text { (a) } \mathrm{Na}_2 \mathrm{O}_2 & \rightarrow \mathrm{Na}^{+} \quad 8 \mathrm{e}^{-} \text {all shells are filled. } \\ \mathrm{O}_2^{2-} & \left.\rightarrow \quad \begin{array}{lll}1 s^2 & 2 s^2 & 2 p^5 \\ 1 s^2 & 2 s^2 & 2 p^5\end{array}\right] 10 e^{-}\end{array}$

(c) $\mathrm{N}_2 \mathrm{O}$

$ : \mathrm{N}=\mathrm{N} \rightarrow \ddot{\mathrm{O}}: $

all electrons are paired.

(d) $\mathrm{KO}_2 \rightarrow \mathrm{~K}^{+} \mathrm{O}_2^{-}$

$ \left.\mathrm{O}_2^{-}=\begin{array}{ccc} 1 s^2 & 2 s^2 & 2 p^4 \\ 1 s^2 & 2 s^2 & 2 p^5 \end{array}\right] 9 e^{-} $

Statement 1: Boron always forms covalent bonds.

→ Boron has a small size and high ionization energy. It does not easily lose electrons to form $ B^{3+} $ ions, since removing three electrons requires a lot of energy.

→ Therefore, boron compounds (like BF₃, BCl₃, BH₃, etc.) are covalent in nature.

✅ Statement 1 is True.

Statement 2: The small size of $ B^{3+} $ favors formation of covalent bond.

→ According to Fajan’s rules, small cations with high charge density have strong polarizing power, which distorts the electron cloud of the anion, leading to covalency.

→ $ B^{3+} $ is extremely small and highly charged, therefore its compounds are predominantly covalent.

✅ Statement 2 is True.

Furthermore, Statement 2 gives the correct reason why boron compounds are covalent — due to the small size and high charge of $ B^{3+} $.

✅ Correct Option: A

Answer: Option A — Both statements are true, and Statement 2 is the correct explanation for Statement 1.

(a)   XeF₄  is sp³d² hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.



(b) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.



(c)   SF₄ is sp³d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120^o and 179^o instead of the expected angles of 120^o and 180^o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.



(d) SiF₄ is sp³ hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.

(A) Molecular orbital configuration of O $_2^{2 - }$ (18 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^2}^ * \, = \,\pi _{2p_y^2}^ * $

So O $_2^{2 - }$ has no unpaired electrons.

(B) Molecular orbital configuration of B₂ (10 electrons) 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 in B₂, 2 unpaired electrons present.

(C) 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.

(D) 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.

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

1. Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

Mathematically:

$U \propto \frac{q_1 \cdot q_2}{r}$

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

2. Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.

Considering these two factors, the correct choice is:

Option D: Charge on the ion and size of the ion

TIPS/Formulae :

A diamagnetic substance contains no unpaired electron.

${H_2}$ is diamagnetic as it contains all paired electrons

$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)} $

$\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)} $

To determine which species has the maximum number of lone pairs of electrons on the central atom, we'll examine the electron configuration of each molecule's central atom:

Option A: $[ClO_3]^-$

The central atom is chlorine (Cl).

Chlorine has 7 valence electrons. In the $[ClO_3]^-$ ion, chlorine forms three bonds with oxygen atoms and there is an additional negative charge (one extra electron). Thus, chlorine has:

$7 + 1 - 3 = 5$ valence electrons left, which form 2.5 lone pairs. However, due to the requirement to minimize formal charges, typically chlorine in $[ClO_3]^-$ is involved in resonance structures reducing the average lone pairs per resonance structure to 1.

So, $[ClO_3]^-$ has 1 lone pair on the central atom.

Option B: $XeF_4$

The central atom is xenon (Xe).

Xenon has 8 valence electrons. In $XeF_4$, xenon forms four bonds with four fluorine atoms. Thus, xenon has:

$8 - 4 = 4$ valence electrons left, which form 2 lone pairs.

So, $XeF_4$ has 2 lone pairs on the central atom.

Option C: $SF_4$

The central atom is sulfur (S).

Sulfur has 6 valence electrons. In $SF_4$, sulfur forms four bonds with four fluorine atoms. Thus, sulfur has:

$6 - 4 = 2$ valence electrons left, which form 1 lone pair.

So, $SF_4$ has 1 lone pair on the central atom.

Option D: $[I_3]^-$

The central atom is iodine (I).

Iodine has 7 valence electrons. In the $[I_3]^-$ ion, iodines form a linear structure where the central iodine forms two bonds with two other iodine atoms and there is an additional negative charge (one extra electron). Thus, the central iodine has:

$7 + 1 - 2 = 6$ valence electrons left, which form 3 lone pairs.

So, $[I_3]^-$ has 3 lone pairs on the central atom.

Hence, the species with the maximum number of lone pairs of electrons on the central atom is:

Option D: $[I_3]^-$

Here eight 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4.

Here twelve 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4, $\angle $6M1, $\angle $6M2, $\angle $6M3, $\angle $6M4.

Here four 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1.

Here six 90^o angles between bond pair and bond pair. Those angles are $\angle $1M3, $\angle $1M4, $\angle $1M5, $\angle $2M3, $\angle $2M4, $\angle $2M5.

Structure of  H₃BO₃  is



Here Boron has 3 sigma bond. So it is sp² hybridised.

And oxygen has two sigma bond and two lone pair. So it sp³ hybridised.

1. In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

2. In NH₃, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107^o.



3. $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



4.
Here Si is sp³ hybridised and shape is regular tetrahedral and bond angle 109^o 28'

Molecular orbital configuration of NO (15 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^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

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

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

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

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Note :
(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So, bond length in NO > NO⁺

Regular Tetrahedral structure is possible in sp³ hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF₄  is sp³d² hybridised and structure is square planar.



(b)   [Ni(CN)₄]^$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni⁺² is $=$ [Ar]3d⁸



But because of CN^$-$ ion which is a strong field ligand , it can perform pairing of electron.



And the structure of dsp² hybridization is square planar.



(C) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral.



(d)   

SF₄ is sp³d hybridised and structure is see-saw.

Here $\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $_total = $\mu $₁ $-$ $\mu $₂ = 0

$\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $₁ + $\mu $₂ + $\mu $₃ = $\mu $₄

   $\mu $_total = 0

In NO₂ and O₃ have permanent dipole moment as both of them are not symmetric molecule.

In CO₂ and SiF₄ each individual bond c = O and S- $-$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled.

(a)   H₂O is sp³ hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $-$ H bond is 104.5^o

(b)   In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(s) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

(c)   NH₃ has sp³ hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107^o

(d)   SO₂ is sp² hybridized and bond angle is 117^o.

(a) $\,\,\,\,$

(b) $\,\,\,\,$

(c) $\,\,\,\,$

(d) $\,\,\,\,$

So, option (B) is correct

Note :
If in carbon(C) all 4 bonds are sigma bond then it is sp³ hybridization and if there is 3 sigma bond then sp² and if there is only 2 sigma bond there sp hybridization.

(A) Electronegativity of F is 4 and Electronegativity of H is 2.1.

$ \therefore $ Electronegativity difference between F and H is = 4 - 2.1 = 1.9

Now Electronegativity of Br is 2.8 and Electronegativity of H is 2.1.

$ \therefore $ Electronegativity difference between Br and H is = 2.8 - 2.1 = 0.7

As in HF Electronegativity difference is more then it's polar nature is also more.

(B) At equilibrium pure H₂O produce very less H⁺ and OH^- ions.

(C) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion.

(D) In covalenrt bond, sharing of electons take place. Transfer of electrons take place in ionic bond.

Hybridization of square planar complex is dsp² .



In square planar complex all 4 surrounding atoms and central atom are in the same plane and let this plane is x $-$ y plane. As it is dsp² hybridized, so it has 1 d orbital, 1 s orbital and 2 p orbital.

s orbital is non-directional, so we just write it as s.



In p_x the two lobes are along x-axis and in p_y two lobes are along y-axis

One d orbital is d_{x² $-$ y²}, it means out of four lobes of d orbital two along x-axis and two along y-axis.

(a) $\,\,\,$ AlH₃ + H^$-$ $ \to $ AlH$_4^ - $

Steric number of AlH₃ is = ${1 \over 2}$ [3 + 3] = 3

$\therefore\,\,\,$ AlH₃ is sp² hybridized.

Steric number of AlH$_4^ - $ = ${1 \over 2}$ [3 + 4 +1] = 4

$\therefore\,\,\,$ AlH$_4^ - $ is sp³ hybridized.

(b) $\,\,\,$ H₂O + H⁺ $ \to $ H₃O⁺

Steric no of H₂O = ${1 \over 2}$ (6+ 2) = 4

$\therefore\,\,\,\,$ H₂O s sp³ hybridized.

Steric no of H₃O⁺ = ${1 \over 2}$ [ 8 + 3 $-$1] = 4

$\therefore\,\,\,$ H₃O⁺ s also sp³ hybridized.

(c) $\,\,\,$ NH₃ + H⁺ $ \to $ NH₄⁺

Steric no of NH₃ = ${1 \over 2}$ [5 + 3] = 4

$\therefore\,\,\,$ hybridization of NH₃ is sp³

Steric number of NH₄⁺ = ${1 \over 2}$ [5 + 4 $-$ ] = 4

$\therefore\,\,\,$ Hybridization of NH₄⁺ is sp³

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ 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}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 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^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

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

Molecular orbital configuration of $O_2^ - $ (17 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^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 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^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond strength $ \propto $ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

Bond angle 109^o 28' means Regular Tetrahedral geometry and hybridization is sp³.

Steric Number (SN) for sp³ hybridization is 4.

In sp³ molecules can have different shapes which is decided by SN number

(1) $\,\,\,\,$ 4 bond pair in a molecule then angle between bonds 109^o 28^'

(2) $\,\,\,\,$ 3 bond pair and 1 lone pair in a molecule then angle between bonds 107^o.

(3) $\,\,\,\,$ 2 bond pair and 2 lone pair in a molecule then angle between bonds 104.5^o

(4) $\,\,\,\,$ 1 bond pair and 3 lone pair in a molecule then bond angle is undefined as there is only one bond.

(1) $\,\,\,\,$ In NH₃, 3 Bond pair(BP) +1 lone pair (LP) present so angle between bond 107^o.



(2) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^'.



(3) In NH$_4^ + $, 4 bond pair present so angle between bond is 109^o 28^'



(4) $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



(5) $\,\,\,\,$ In NH$_2^ - $, 2 bond pair and 2 lone pair present, so bond angle is 104.5^o



So, $NH{}^ + $ and BF$_4^ - $ has bond angle 109^o 28^'.

To determine which molecular species have unpaired electrons, let's examine the electron configurations of each species using molecular orbital (MO) theory. The presence of unpaired electrons can be deduced from the electron filling in the molecular orbitals.

Option A: N₂

For nitrogen diatomic (N₂), each nitrogen atom has 7 electrons, so the total number of electrons is 14. According to the molecular orbital theory, the electrons fill the molecular orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$.

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2$

So, N₂ has no unpaired electrons.

Option B: F₂

For fluorine diatomic (F₂), each fluorine atom has 9 electrons, resulting in a total of 18 electrons. They fill the orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals also results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, F₂ has no unpaired electrons.

Option C: $O_2^-$

The dioxygen anion ($O_2^-$) has one extra electron compared to neutral O₂, which normally has 16 electrons. Hence, $O_2^-$ has 17 electrons. These electrons fill the orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals, the last electron will be placed in one of the degenerate $\pi_{2p_x}^{*}$ or $\pi_{2p_y}^{*}$ orbitals, resulting in one unpaired electron:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$

So, $O_2^-$ has one unpaired electron.

Option D: $O_2^{2-}$

The peroxide ion ($O_2^{2-}$) has two extra electrons compared to neutral O₂, giving it a total of 18 electrons. These electrons fill the orbitals as follows: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, $O_2^{2-}$ has no unpaired electrons.

Based on the molecular orbital theory, the molecular species that has unpaired electron(s) is Option C: $O_2^-$.

To answer the question about the common features among the species CN^-, CO, and NO⁺, let's first understand the electronic configurations and bond orders of each species.

1. Electronic Configurations:

- CN^-: The cyanide ion (CN^-) has an electronic configuration similar to N₂ because CN has 10 electrons (from C and N); adding one more electron for the negative charge brings it to 11 electrons.

- CO: Carbon monoxide (CO) has a total of 10 valence electrons (4 from Carbon and 6 from Oxygen).

- NO⁺: The nitrosonium ion (NO⁺) has a total of 10 electrons as well, considering the positive charge removes one electron from the original 11 valence electrons of NO.

Thus, all three species CN^-, CO, and NO⁺ are isoelectronic, meaning they all have the same number of electrons (10 valence electrons).

2. Bond Order: The bond order is the number of chemical bonds between a pair of atoms. It is calculated as half the difference between the number of bonding electrons and the number of antibonding electrons. All three species have a bond order of 3, as can be seen from their molecular orbital diagrams.

3. Ligand Field Strength:

CO and CN^- are generally considered strong field ligands due to their ability to partake in back-donation, where electrons from the metal center can be donated back into the empty π* orbitals of the ligand. NO⁺ is also a reasonably strong field ligand due to its ability to accept electrons.

Conclusion:

Based on the above analysis, the correct option is:

Option A: bond order three and isoelectronic

This option accurately describes the common features among CN^-, CO, and NO⁺.

To determine the order of increasing C-O bond lengths for CO, $CO_3^{2-}$ and CO₂, it's essential to understand the types of bonding in these molecules and ions.

1. Carbon Monoxide (CO):

Carbon monoxide has a triple bond between carbon and oxygen. Triple bonds are stronger and shorter compared to single and double bonds. Thus, CO has a very short bond length.

2. Carbon Dioxide (CO₂):

In CO₂, carbon is double-bonded to each oxygen atom. Double bonds are longer than triple bonds but shorter than single bonds. Therefore, the C-O bond length in CO₂ is longer than in CO but shorter than in a molecule with single bonds.

3. Carbonate Ion ($CO_3^{2-}$):

In the carbonate ion, the bond order for each C-O bond is 1.33. This is because the carbonate ion has a resonance structure where one carbon-oxygen bond is a double bond and the others are single bonds. Due to the resonance, the effective bond order between carbon and oxygen is between a single and a double bond, making its bond length longer than that in CO₂ and shorter than a pure single bond.

Therefore, the increasing order of C-O bond length is:

CO < CO₂ < $CO_3^{2-}$

The correct option is:

Option D

CO < CO₂ < $CO_3^{2-}$

$\mathrm{H}_2 \mathrm{~S}$ has $s p^3$ hybridised sulphur, therefore, angular in shape with non-zero dipole moment.

(Non-linear, polar molecule)

The correct answer is Option C: If assertion is correct but reason is incorrect.

Explanation:

• Assertion: LiCl (Lithium Chloride) is indeed predominantly a covalent compound. This is due to the high polarizing power of the small lithium ion (Li⁺), which distorts the electron cloud of the larger chloride ion (Cl⁻), resulting in a significant degree of electron sharing and hence covalency.


• Reason: The reason provided is incorrect. The electronegativity difference between lithium (Li) and chlorine (Cl) is substantial. It is this difference in electronegativity that creates the polarity in the Li-Cl bond, making it an ionic bond with some covalent character.

Key Point: The high polarizing power of the Li⁺ ion, due to its small size and high charge density, is the main reason for the covalent character of LiCl, not the small electronegativity difference.

The critical temperature of a substance is the temperature above which it cannot be liquefied, regardless of the pressure applied. It depends on the intermolecular forces within the substance. To understand why the critical temperature of water (H₂O) is higher than that of oxygen (O₂), we need to examine the characteristics and interactions of these molecules.

Let’s break down the options:

Option A: Fewer electrons than O₂

This option is incorrect. The number of electrons in a molecule does not directly determine the critical temperature. While electron count can affect molecular interactions to an extent, it is not the primary reason for the difference in critical temperatures between H₂O and O₂.

Option B: Two covalent bonds

Although H₂O does have two covalent bonds, this feature alone does not explain its higher critical temperature compared to O₂. The nature of the bonds plays a more crucial role, particularly the type of intermolecular forces these bonds enable.

Option C: V-shape

H₂O indeed has a V-shaped (or bent) molecular geometry, while O₂ is linear. This V-shape contributes to some properties of water, like its polarity, but it is not the most direct reason for the higher critical temperature.

Option D: Dipole moment

This is the correct answer. H₂O has a significant dipole moment due to its molecular geometry and the difference in electronegativity between hydrogen and oxygen atoms. This results in strong hydrogen bonding between water molecules. Hydrogen bonds are a type of dipole-dipole interaction, which are much stronger than the van der Waals forces (or London dispersion forces) experienced by O₂ molecules. These strong intermolecular hydrogen bonds in water require more energy (higher temperature) to break, resulting in a higher critical temperature.

Therefore, the primary reason the critical temperature of water is higher than that of oxygen is due to the strong dipole moment leading to significant hydrogen bonding in water. The correct answer is:

Option D: dipole moment

To determine which of the given compounds has sp² hybridization, we need to understand the structure of each molecule and the hybridization of the central atom in each case.

Option A: CO₂

In carbon dioxide (CO₂), the carbon atom is the central atom. Carbon forms two double bonds with two oxygen atoms.

In this configuration, carbon has two regions of electron density (each double bond counts as one region of electron density). According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, this would result in a linear structure with a bond angle of 180 degrees. The hybridization of the carbon atom, in this case, is sp, because it needs two hybrid orbitals to form the two double bonds.

So, CO₂ does not have sp² hybridization.

Option B: SO₂

In sulfur dioxide (SO₂), the sulfur atom is the central atom. Sulfur forms two double bonds with two oxygen atoms.

However, sulfur also has a lone pair of electrons. This results in three regions of electron density around the sulfur atom (two double bonds and one lone pair). According to the VSEPR theory, this configuration leads to a bent molecular shape.

With three regions of electron density, the hybridization of sulfur is sp². Therefore, SO₂ has sp² hybridization.

Option C: N₂O

In nitrous oxide (N₂O), one nitrogen atom is central. The molecule can be arranged as N=N=O, where the central nitrogen forms a double bond with each of the terminal nitrogen and oxygen atoms.

This gives the central nitrogen two regions of electron density, resulting in linear geometry. The hybridization of the central nitrogen atom here is sp, not sp².

So, N₂O does not have sp² hybridization.

Option D: CO

In carbon monoxide (CO), the carbon atom forms a triple bond with the oxygen atom. This involves one sigma bond and two pi bonds, with one lone pair on each atom.

This gives each atom two regions of electron density, resulting in linear geometry. The hybridization of carbon in CO is also sp.

So, CO does not have sp² hybridization.

Based on the analysis of each option, the correct answer is:

Option B: SO₂

To identify the isostructural pairs among the given species, we first need to understand their molecular geometries and the electron-pair arrangements.

1. NF₃: Nitrogen trifluoride has a trigonal pyramidal structure due to the presence of one lone pair of electrons on nitrogen. It follows the AX₃E steric number formula (where A is the central atom, X represents bonded atoms, and E represents lone pairs).

2. $NO_3^-$: The nitrate ion has a trigonal planar structure. This can be rationalized using VSEPR theory, where it follows the AX₃ steric number formula with no lone pairs on the central nitrogen atom.

3. BF₃: Boron trifluoride is also trigonal planar. It conforms to the AX₃ steric number formula with no lone pairs on boron.

4. H₃O⁺: The hydronium ion has a trigonal pyramidal structure, similar to NF₃, with one lone pair of electrons. It follows the AX₃E formula.

5. HN₃: Hydronitrenium ion has a linear or slightly bent structure, depending on how we look at the resonance structures. Overall, it does not share the same geometric configuration with any of the molecules listed above.

From the above analysis:

• Isostructural Pair 1: NF₃ and H₃O⁺ (both have a trigonal pyramidal structure)
• Isostructural Pair 2: $NO_3^-$ and BF₃ (both have a trigonal planar structure)

Therefore, the correct option is:

Option C:

[NF₃, H₃O⁺] and [$NO_3^-$, BF₃].

Calcium carbide ($CaC_2$) is an ionic compound consisting of ${Ca^{2+}}$ and carbide ions ${C_2^{2-}}$. In this compound, the two carbon atoms are bonded together in the carbide ion ${C_2^{2-}}$. The bonding in the carbide ion involves one sigma ($\sigma$) bond and two pi ($\pi$) bonds, which forms a triple bond between the two carbon atoms.

The reasoning is as follows:

Each carbon atom in the ${C_2^{2-}}$ ion is sp hybridized, having one sigma ($\sigma$) bond from the overlap of sp hybrid orbitals and two pi ($\pi$) bonds from the lateral overlap of p orbitals. Therefore, the carbide ion ${C_2^{2-}}$ has a carbon-carbon triple bond.

Thus, the correct option is:

Option B: one sigma ($\sigma$) and two pi ($\pi$) bonds