Chemical Bonding & Molecular Structure

Dipole moment $=1.5 \times 10^{-29} \mathrm{C} . \mathrm{m}$ (given)

Bond length $=150 \mathrm{~pm}=150 \times 10^{-12} \mathrm{~m}$

Charge $=1.602 \times 10^{-19} \mathrm{C}$

$\begin{aligned} \text { Ionic character } & =\frac{\text { Dipole moment }}{\text { Bond length } \times \text { Charge }} \\ & =\frac{1.5 \times 10^{-29}}{150 \times 10^{-12} \times 1.602 \times 10^{-19}} \\ & =0.625 \end{aligned}$

Percentage ionic character

$\begin{aligned} & =0.625 \times 100 \\ & =62.5 \% \end{aligned}$

$\mathrm{SeF}_4$

$\mathrm{Se}=[\mathrm{Ar}] 3 d^{10} 4 s^2 4 p^4 $

Selenium (Se) undergoes $s p^3 d$ hybridisation and form five $s p^3 d$ hybridised orbitals.

Out of five $s p^3 d$ orbitals one is completely filled and four are half filled.

Therefore, geometry of H$_2$O and NH$_3$ are different.

Element with maximum bond energy is carbon because $\mathrm{C}-\mathrm{C}$ bond is strongest as carbon atoms are smaller in size.

Correct order of electronegativity of hybrid orbitals of carbon is

$sp > sp^2 > sp^3$

Electronegativity of carbon increases as the s-character of hybrid orbital increases.

Hybridisation	% of s
$sp$	50%
$sp^2$	33%
$sp^3$	25%

Higher percentage of s-character indicates that hybrid orbital is more closer to the nucleus and that’s why it has higher electronegativity.

Bond length is inversely proportional to the bond order. Higher bond order results in stronger bond, which are accompanied by stronger forces of attraction holding the atoms together.

Since, electronegativity difference is highest in case of N $-$ N bonds, therefore NH$_3$ has highest dipole moment,

As dipole moment is vector quantity. Due to electron density in opposite direction, net dipole moment cancel out each other.

Average bond order = 6/4 = 1.5

Average bond order = 6/3 = 2

Average bond order = 4/3 = 1.33

Correct sequence of bond order is

SO$_3$ > SO$_4^{2-}$ > S$_2$O$_3^{2-}$

The sulphur dioxide molecule have dipole moment value of 161 debye. It is a bent molecule with a sulphur atom placed in the centre and two atoms of oxygen on the side, one oxygen atom is connected by a double bond and the oxygen ion by a single bond.

Hence, among all SO$_2$, have highest dipole moment.

Bond order (BO) = $\mathrm{{{(Number\,of\,electron\,in\,bonding\,MO) - (Number\,of\,electron\,in\,anti - bonding\,MO)} \over 2}}$

S. No.	Molecule	BO
1.	H$_2^+$	$1-0/2=0.5$
2.	He$_2^ + $	$2-1/2=0.5$
3.	He$_2^ - $	$2-1/2=0.5$
4.	B$_2^ + $	$3-2/2=0.5$
5.	F$_2^ - $	$4-2/2=1$
6.	Be$_2^{2 - }$	$3-1/2=1$

Hence, among all of these only $\mathrm{H_2^ + ,He_2^ + ,He_2^ - ,B_2^ +}$ give 0.5 bond order.

S. No.	Molecule	Steric no.	Hybridisation	Structure	Shape
(A)	$\mathrm{SnCl_2}$	3	$sp^2$		Angular or bent (1)
(B)	$\mathrm{XeF_4}$	6	$sp^3d^2$		Square planar (5)
(C)	$\mathrm{CIF_3}$	5	$sp^3d$		T-shape (4)
(D)	$\mathrm{IF_5}$	6	$sp^3d^2$		Square pyramidal (3)

Diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ is one of the simplest boron hydride.

The structure of diborane molecule consist four $\mathrm{H}$-atom and two boron atoms coming on the same plane. The boron atom present in $s p^3$ hybridised state and from these four hybrid orbitals, three of orbitals have one electron each, and one is an empty orbital. It will form $\mathrm{B}-\mathrm{H}-\mathrm{B}$ bonds called banana bond.

(a) $\mathrm{NCl}_5$ does not exist while $\mathrm{PCl}_5$ does. This is correct due to $\mathrm{N}$ does not have vacant $d$-orbitals and it cannot expand its oxidation state but $\mathrm{P}$ has vacant $d$-orbitals.

(b) $\mathrm{Pb}$ prefer to form tetravalent compound. This is wrong as to lead prefers to form bivalent compounds due to inert pair effect.

(c) The three $\mathrm{C}-\mathrm{O}$ bond are equal in carbonate ion. This is because the bonds are in resonance.

(d) Both O$_2^+$ and NO are paramagnetic as they contain unpaired electrons.

O$_2^+$ $\Rightarrow$ 15e$^-$ (odd) paramagnetic

NO $\Rightarrow$ 15e$^-$ (odd) paramagnetic

Hence, only (b) option is incorrect.

$p \pi-p \pi$ bonding is weak bonding. Due to larger atomic size of phosphorus. It is unstable to form $\pi-\pi$ bond and so, it is tetra-atomic in which each $p$-atom is linked with 3 others P-atom by $3 \sigma$-bonds. In nitrogen, due to smaller atomic size $\mathrm{N}$ forms $1 \sigma$ and $2 \pi$-bonds i.e. triple bonds with other $\mathrm{N}$ atom and exists as diatomic molecules. Hence, the bonding is responsible for that. Phosphorus from $\mathrm{P}_4$ because in $\mathrm{P}_2, p \pi$-$p \pi$ bond is weaker.

In $\mathrm{SF}_6$, the sulphur atom is shielded by 6 fluorine atoms which hinders kinetically any reaction with water, alkali hydroxide, ammonia or strong acid, as a result, it remains inert to these reagents.

$\mathrm{SF}_6$ contain 6 steric number and $s p^3 d^2$ hybridisation. $\mathrm{S}_2 \mathrm{O}_3^{2-}$ is similar look like ammonia shape i.e. tetrahedral.

There are two $\pi$-bond present in $\mathrm{SO}_4^{2-} \cdot \mathrm{SO}_4^{2-}$ is a polyatomic ion as well as a non-polar covalent compound.

According to Mulliken equation,

$\mathrm{EN}=\frac{\mathrm{IE}+\mathrm{EA}}{5.6},$ ..... (i)

when both IE and EA are taken in $\mathrm{eV}$.

(Here, IE = Ionisation energy

EA = Electron gain enthalpy

EN = Electros negativity)

IE for chlorine $=13 \mathrm{eV}$

EA for chlorine $=4 \mathrm{eV}$

Put value in Eq. (i)

$\begin{aligned} \mathrm{EN} & =\frac{13+4}{5.6}=\frac{17}{5.6} \\ & =3.03 \approx 3.0 \mathrm{eV} \end{aligned}$

Dipole moment is a measure of polarity of a bond. It is the product of charges and the distance between partial charge. Dipole moment of polar molecule containing lone pairs is vector sum of dipole of lone pair and net dipole moments of bonds.

Electronegativity of F is more than that of N, thus the direction of dipole moment of N-F bond will be form F to N. So, net dipole moment of NF$_3$ found be 0.24 D.

Due to less electronegativity of H in ammonia, it has maximum dipole moment. Dipole moment order :

$\matrix{ {N{H_3}} & < & {NB{r_3}} & < & {NC{l_3}} & < & {N{F_3}} \cr {(1.4D)} & {} & {(0.8D)} & {} & {(0.6D)} & {} & {(0.24D)} \cr } $

Here, D = Debye or coulomb metre.

S. No.	Molecule	Steric number	Hybridisation	Shape	Structure
1	$\mathrm{BF_3}$	$\frac{3+3}{2}=3$	$sp^2$	Trigonal planar
2.	$\mathrm{NO_2^-}$	$\frac{5+1}{2}=3$	$sp^2$	Bent shape
3.	$\mathrm{NH_2^-}$	$\frac{5+2+1}{2}=4$	$sp^3$	Angular or V-shape
4.	$\mathrm{H_2O}$	$\frac{6+2}{2}=4$	$sp^3$	V-shape or bent

Hence, BR$_3$ and NO$_2^-$ are sp$^2$ hybridised.

For p-dichlorobenzene and p-cyanobenzene, the dipole moment of individual bonds cancel each other as they are equal in magnitude and opposite in direction. Hence, resultant dipole moment of molecules is zero.

For p-hydroquinone and p-benzenedithiol, the O$-$H and S$-$H bond dipoles in these molecules do not cancel each other as they are not in opposite directions due to existence in different conformation.

o-nitrophenol forms intramolecular H-bonding whereas molecules of p-nitrophenol get associated through intermolecular H-bond.

$\therefore$ o-nitrophenol is the most volatile compound.

The shape of CIF₃ is as follows

It has two lone pair at equatorial position.

Molecular electronic configuration for given species are

(a) ${O_2} = \sigma \,1{s^2}\,\sigma *1{s^2}\,\sigma 2{s^2}\,\sigma *2{s^2}\,\sigma 2p_x^2\,\pi \,2p_y^2\,\pi \,2p_z^2\,\pi *2p_y^1\,\pi *2p_z^1$

It has two unpaired electrons.

(b) $O_2^{2 - }$ : It has no unpaired electron.

(c) $N_2^{2 + }$ : It has one unpaired electron.

(d) B₂ : It has two unpaired electrons.

$\therefore$ $O_2^{2 - }$ is diamagnetic in nature.

Bond order for the various compounds are as follows

for SO₂,

Number of bonds = 4

Number of (SO) groups = 2

$\therefore$ Bond order $ = {4 \over 2} = 2$

For SO₃,

Number of bonds = 6

Number of (SO) groups = 3

$\therefore$ Bond order $ = {6 \over 3} = 2$

For SO$_4^{2 - }$,

Number of bonds = 6

Number of (SO) groups = 4

$\therefore$ Bond order $ = {6 \over 4} = 1.5$

For SO$_3^{2 - }$,

Number of bonds = 4

Number of (SO) groups = 3

$\therefore$ Bond order $ = {4 \over 3} = 1.33$

Hence, correct order is $S{O_2} = S{O_3} > SO_4^{2 - } > SO_3^{2 - }$.

CH₄ - Bond angle = 109^o28'

H₂S - Bond angle = 92^o

NH₃ - Bond angle = 107^o

H₂O - Bond angle = 104^o5'

Potential energy curve for H₂ molecules is

In solid state PCl₅ exist in Ionpair i.e. [PCl₄] ⁺ and [PCl₆] ^–

[PCl₄]⁺ is tetrahedral.
[PCl₆]^– is octahedral.

To determine the option in which the hybridisation of the underlined atom is affected, we need to look at each reaction and predict the changes in hybridisation that might be occuring with the reactants and products.

Option A:

In the reaction $ \underline {Xe}F_4 + SbF_5 \to $, xenon tetrafluoride (XeF₄) reacts with antimony pentafluoride (SbF₅). XeF₄ is a square planar molecule, and the central Xe atom is $sp^3d^2$ hybridised. The reaction with SbF₅ can lead to the formation of a complex where the hybridisation of Xe changes if it coordinates with SbF₅. Thus, the hybridisation of Xe may indeed be affected in this case.

$\underline {Xe} $F4	+	SbF5	$ \to $	[XeF3]+[SbF6]-
sp3d2				sp3d   sp3d2

Option B:

In the reaction $ H_2\underline{S}O_4 + NaCl \buildrel {420K} \over \longrightarrow $, sulfuric acid (H₂SO₄) reacts with sodium chloride (NaCl). The underlined atom here is sulfur (S). In sulfuric acid, sulfur is hybridised as $sp^3$. Upon heating, there is typically no change in the hybridisation of the sulfur atom within sulfuric acid as part of this reaction. Therefore, the hybridisation of S is likely not affected.

Option C:

In the reaction $ H_3\underline{P}O_2 \buildrel {Disproportionation} \over \longrightarrow $, phosphorous acid (H₃PO₂) is undergoing a disproportionation reaction. The underlined atom is phosphorus (P). In H₃PO₂, the hybridisation of the phosphorus atom is $sp^3$. A typical disproportionation reaction of phosphorous acid would result in phosphine (PH₃) and orthophosphoric acid (H₃PO₄). The phosphorus atom in PH₃ will have $sp^3$ hybridisation while in H₃PO₄ the phosphorus atom is $sp^3$ hybridised. Given this information, the hybridisation of phosphorus may or may not change, depending on the resulting state of the phosphorus atom in the products. It's not a clear-cut case without knowing the specific products.

Option D:

In the reaction $ \underline{N}H_3 \buildrel {{H^+}} \over \longrightarrow $, ammonia (NH₃) is reacting with a proton (H⁺). The underlined atom is nitrogen (N). In NH₃, the nitrogen is $sp^3$ hybridised. When ammonia accepts a proton, it forms ammonium ion (NH₄⁺), which is also $sp^3$ hybridised. Therefore, there is no change in the hybridisation of nitrogen in this case.

Given this information, the option in which the hybridisation of the underlined atom is most clearly affected is Option A, where the reaction can lead to coordination changes to the Xe atom, potentially affecting its hybridisation.

From the given graph, potential energy of A-B molecule is minimum.

Thus A-B bond is most stable as it requires more energy to dissociate and have strongest bond amongst these.

B = Most electronegative

D = Least electronegative

A - B = Shortest bond length

A - B = Largest bond enthalpy

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 - - -



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

Similarly Molecular orbital configuration of NO²⁺ (13 electrons) is

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

$\therefore\,\,\,\,$ N_b = 9

N_a = 4

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

Molecular orbital configuration of NO^- (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_b = 10

N_a = 6

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

As Bond strength $ \propto $ Bond order

$ \therefore $ NO^– will have minimum bond strength.

Ion-ion interaction energy $ \propto $ ${1 \over r}$

Dipole-dipole interaction energy $ \propto $ ${1 \over {{r^3}}}$

London dispersion $ \propto $ ${1 \over {{r^6}}}$

[XeF₅]^– $ \to $ 5BP + 2LP $ \to $ sp³d³ hybridisation $ \to $ Pentagonal planar

XeO₃F₂ $ \to $ 5BP + 0LP $ \to $ sp³d hybridisation $ \to $ Trigonal bipyramidal

Here 4$\sigma $ bonds +1 lone pair

Electronic geometry is Trigonal bipyramidal. Moleculer Geometry or Moleculer Shape is see-saw.

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Compound	Electronic Geometry	Moleculer Geometry or Moleculer Shape	Lone Pair	Polarity
AB4	Tetrahedral	Tetrahedral	0	Nonpolar
AB4	Trigonal Bipyramidal	see-saw	1	Polar
AB4	Octahedral (Square Bipyramidal)	Square Planar	2	Nonpolar

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Here in this question it is not clear which geometry they are talking about, correct answer should be either Trigonal Bipyramidal or see-saw.

Each carbon atom is sp² hybrid

$ \therefore $ 3 sp² hybrid orbitals are formed by each carbon atom

Total sp² orbitals = 6 × 3 = 18

Magnetic moment = 1.73 BM

$ \therefore $ Unpaired electron = 1

(1) $O_2$ 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.

(2)   $O_2^{+}$ 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.

(3)   $O_2^{-}$ has 17 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^2}^ * = \,\,\pi _{2p_y^1}^ * $

Here 1 unpaired electron present.

Hence $O_2^{-}$ & $O_2^{+}$ have one unpaired electron.

Bond length order in carbon halogen bonds are

in the order of C – F $<$ C – Cl $<$ C – Br $<$ C – I

Hence, Bond energy order

C – F $>$ C – Cl $>$ C – Br $>$ C – I

    CN^- has 14 electrons.

Moleculer orbital configuration of CN^- is

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

Here is no unpaired electron so it is diamagnetic.
N_b = 10

N_a = 4

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

Ionic interactions are stronger as compared to van der waal interactions.

So, correct order is

ion-ion $>$ ion-dipole $>$ dipole-dipole