Chemical Bonding and Molecular Structure

Number of $\sigma$ bonds $=5$

Number of $\pi$ bond $=1$

$\mathrm{C}_2 \mathrm{H}_2 \Rightarrow \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$

Number of $\sigma$-bonds $=3$

Number of $\pi$-bonds $=2$

$ \text { Number of } \sigma \text {-bonds }=4 $

Number of $\sigma$-bonds $=3$

Number of lone pair = one A-IV, B-I, C-III, D-II

$\mathrm{ClF}_3$ has two lone pairs of electrons on Cl atom with bent T -shape structure (geometry)

Formal charge on $2^{\text {nd }}$ oxygen atom $=6-\frac{1}{2}(4)-4=0$

Formal charge on $1^{\text {st }}$ oxygen atom $=6-\frac{1}{2}(6)-2=+1$

Formal charge on $3^{\text {rd }}$ oxygen atom $=6-\frac{1}{2}(2)-6=-1$

$\bullet$ A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.

$\bullet$ When bond order increases, the bond length decreases.

A.

B. $\mathrm{XeF}_4: 2$ lone pairs of electron

$\mathrm{XeO}_3: 1$ lone pair of electron

$\mathrm{XeF}_2$ : 3 lone pairs of electron

C. Order of Bond length :- $\mathrm{N}-\mathrm{O}>\mathrm{C}-\mathrm{H}>\mathrm{O}-\mathrm{H}$

D. $\quad \mathrm{N}_2$ Bond order is 3

$\mathrm{H}_2$ Bond order is 1

$\mathrm{O}_2$ Bond order is 2

Dipole moment of a molecule depends both on shape and bond dipole.

Molecule	Shape	$\mu$(Debye)
$ \mathrm{CCl}_4 $		0
$\mathrm{HI}$	$\mathrm{H-I}$	0.38
$\mathrm{CO_2}$	$\mathrm{O=C=O}$	0
$\mathrm{BF_3}$		0

Since $\mathrm{CHCl}_3$ is an organic solvent so, covalent (non-polar) compounds will be more soluble in it. As the dipole moment of solute increases, solubility in chloroform decreases. Hence increasing order of solubility.

$\mathrm{NaCl}<\mathrm{CH}_3 \mathrm{CN}<\mathrm{CH}_3 \mathrm{OH}<$ Cyclohexane

It is $s p^3 d$ hybridised with axial to equatorial angle of $90^{\circ}$ and equatorial bond angles of $120^{\circ}$. It has five $\mathrm{P}-\mathrm{Cl}$ sigma bonds.

Axial bonds are longer than equatorial bonds.

The given question involves understanding the boiling points of Group 16 hydrides and the effect of molecular mass and hydrogen bonding on these boiling points.

Statement I claims that the boiling points of Group 16 hydrides follow this order:

$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{S}.$

This statement is indeed true. Water ($\mathrm{H}_2 \mathrm{O}$) has a much higher boiling point compared to the other hydrides in Group 16. This anomaly primarily arises due to the extensive hydrogen bonding present in water, which greatly increases its boiling point. In the absence of such strong hydrogen bonding, heavier molecules like $\mathrm{H}_2 \mathrm{Te}$, $\mathrm{H}_2 \mathrm{Se}$, and $\mathrm{H}_2 \mathrm{S}$ would typically have higher boiling points due to increased van der Waals forces (attributable to larger molecular mass and size). But in this case, $\mathrm{H}_2 \mathrm{O}$ surpasses them because hydrogen bonds are much stronger than van der Waals forces.

Statement II states that based on molecular mass alone, $\mathrm{H}_2 \mathrm{O}$ should have a lower boiling point than the other Group 16 hydrides but has a higher boiling point due to extensive hydrogen bonding. This statement is also true. Water, having a lower molecular mass, would typically have a lower boiling point if we consider only molecular mass. However, the strong hydrogen bonds between water molecules contribute to a much higher boiling point. Hydrogen bonding significantly impacts physical properties like boiling point because it requires more energy to break these bonds during the phase change from liquid to gas.

In summary, based on the analysis of the two statements:

Option A: Both Statement I and Statement II are true is the correct answer. Each statement is correct, and both relate accurately to the anomalous behavior of water among the hydrides of Group 16 elements.

In o-nitrophenol intramolecular H-bonding is present.

$\mathrm{NH}_3 \Rightarrow s p^3$ hybridised with 1 lone pair.

Structure will be Trigonal Pyramidal.

$\mathrm{BrF}_5 \Rightarrow s p^3 d^2$ hybridised with 1 lone pair.

Structure will be Square Pyramidal.

$\mathrm{XeF}_4 \Rightarrow s p^3 d^2$ with two lone pairs.

Structure will be Square Planar.

$\mathrm{SF}_6 \Rightarrow s p^3 d^2$ with no lone pair.

Structure will be Octahedral.

A-I, B-IV, C-II, D-III

(1) In ozone; there are two resonating structures.

$\mathrm{HF}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{S}>\mathrm{CH}_4$ (Non-polar)

Total numbers electrons are same

$\mathrm{Ca}^{+2}, \mathrm{Ar}, \mathrm{K}^{+}, \mathrm{Cl}^{-} \rightarrow 20$ electrons

In the formation of $\mathrm{BMO}$, the two electron waves of the bonding atoms reinforce each other due to constructive interference. Molecular orbitals obtained from $2 \mathrm{P}_{\mathrm{x}}$ and $2 \mathrm{P}_{\mathrm{y}}$ orbitals are 'unsymmetrical' around bond axis.

Hydrated chlorides and Bromides of $\mathrm{Ca}, \mathrm{Sr}$ and $\mathrm{Ba}$ are Ionic so undergo dehydration after heating. Hydrated chlorides and Bromides of $\mathrm{Be}$ and $\mathrm{Mg}$ are covalent so undergo hydrolysis on Heating.

Total number of species = 3

Molecular orbital (energy) diagram / sequence of N$_2$

T$l^+$ & I$^-$ > T$l^{+3}$ & 3I$^-$

due to inert pair effect T$l^{+}$ is more stable than T$l^{+3}$.

Intermolecular forces means force of attraction between two or more molecules

dipole-dipole (attraction between two or more polar molecules).

Dipole induced dipole (attraction between polar and non polar molecules)

Hydrogen bonding (it is a special type of dipole-dipole and ion-dipole attraction)

Dispersion forces (mainly acts between non polar molecules).

Covalent bonding (acts between atom not between molecules )

(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

CO₂ $\Rightarrow$ sp² hybridisation, bond angle = 180$^\circ$

NH$_4^ + $ $\Rightarrow$ sp³ hybridisation, bond angle = 109$^\circ$ 28'

NH₃ $\Rightarrow$ sp³ hybridisation with one lone pair on central atom, bond angle $ \simeq $ 107$^\circ$

H₂O $\Rightarrow$ sp³ hybridisation with two lone pairs on central atom, bond angle $\simeq$ 104.5$^\circ$

In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because X - X' bond in interhalogens is weaker than X - X bond in halogens excepts F - F bond. Therefore I-Cl is more reactive than I₂ because of weaker I - Cl bond then I - I bond.

XeF₂ having maximum lone pairs, so, it has maximum 'lone pair-lone pair' electron repulsions.

Due to one unpaired electron in $\pi$ * 2p molecular orbital, O$_2^ + $ is a paramagnetic ion.

Structure of $\mathrm{XeO}_4$ and $\mathrm{XeO}_3 \mathrm{F}_2$ do not have any lone pair at central-atom. i.e.

Thus, option (a) is the correct. While in $\mathrm{XeO}_3, \mathrm{XeO}_2 \mathrm{~F}_2$ and $\mathrm{XeF}_6$ one lone pair of electrons is present on central atom of each compound.

More the covalent nature of compounds lesser will be its solubility.

Also, according to Fajan's rule, lesser be the size of cation or more be size of anion, species show more covalent nature.

As, $\mathrm{Mg}$ show +2 oxidation state and $\mathrm{Na}$ is in +1 oxidation state the size of $\mathrm{Mg}^{2+}$ is less than that of $\mathrm{Na}^{+}$. On the other hand, sulphur is in $-2$ oxidation state while chlorine is in $-1$ oxidation state, therefore sulphur anion is larger in size. Thus, among the given species, the combination of $\mathrm{Mg}$ and sulphur ($\mathrm{S}$) show maximum covalent nature and therefore is least soluble.

Hence, option (b) is the correct.

Isostructural species have same structure and same number of lone pair of electrons over central atom.

The structure of BO$^{-3}_3$ is

Thus, these are not isostructural because sulphur has one lone pair in its structure. Thus, both assertion and reason are correct and given reason is the correct answer of the given assertion. Hence, option (a) is the correct.

Due to hydration, addition of $\mathrm{H}_2 \mathrm{O}$ is by Markownikoff's rule and by hydroboration oxidation, addition of $\mathrm{H}_2 \mathrm{O}$ is by anti-Markownikoff's rule.

The structure of $\mathrm{IF}_7$ is pentagonal bipyramidal.

It can be depicted as follow

($Z=53$) (Ground state) $=$

I (Excited state) =

Here, $X$ denotes electrons of $\mathrm{F}$-atoms. The structure of $\mathrm{IF}_7$ is shown below :

The bond enthalpies of the two $\mathrm{O}-\mathrm{H}$ bonds in $\mathrm{H}-\mathrm{O}-\mathrm{H}$ are not equal. This is because electronic environment around $\mathrm{O}$ is not same after breakage of one $\mathrm{O}-\mathrm{H}$ bond. Therefore, both assertion and reason are incorrect.