Chemical Bonding & Molecular Structure

$\mathrm{SO}_2$ and $\mathrm{O}_3$ has $s p^2$ hybridisation with bent geometry.

Statement given in III and IV are correct. While the statements given in I and II are incorrect. Their correct forms are

I. When $\mathrm{O}_2$ is converted to $\mathrm{O}_2^{2+}$, the bond order increases from 2 to 3. II. In conversion of $\mathrm{O}_2$ to $\mathrm{O}_2^{2+}$, the magnetic property changes from paramagnetic to diamagnetic.

The correct order is,

$ \mathrm{H}-\mathrm{H}<\mathrm{O}-\mathrm{H}<\mathrm{C}-\mathrm{H}<\mathrm{C}-\mathrm{C} $

$\mathrm{H}-\mathrm{H}$ is smallest due to small size of H atom.

$\mathrm{C}-\mathrm{C}$ is larger than both H and O thus this single bond is largest among the given options.

$\mathrm{O}_2^{2+}=14$ electron,

B.O. $\left(\mathrm{O}_2^{2-}\right)=\frac{1}{2}(10-8)=1$;

B.O. $\left(\mathrm{O}_2^{+}\right)=\frac{1}{2}(10-5)=2.5$

B.O. $\left(\mathrm{O}_2^{-}\right)=\frac{1}{2}(10-7)=1.5$;

B.O. $\left(\mathrm{O}_2\right)=\frac{1}{2}(10-6)=2$

Unpaired electron

$ \mathrm{O}_2^{2+}, \mathrm{O}_2^{2-}=\mathrm{O} ; \mathrm{O}_2^{+}, \mathrm{O}_2^{-}=\mathrm{I} ; \mathrm{O}_2=2 $

Sum of B.O. and unpaired electron

$ 3+1+2.5+1.5+2=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ 0+0+1+1+2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

The correctly matched geometry and hybridisation is given in, I and II. While III and IV are incorrect. Their correct form is :

$\mathrm{SF}_4=s p^3 d-$ Trigonal bipyramidal

$\mathrm{PbCl}_2 \rightarrow s p^2-\mathrm{V}$ shaped geometry.

The correct order of dipole moment is

$ \mathrm{B}<\mathrm{C}<\mathrm{A} $

$\mathrm{H}_2 \mathrm{O}$ has bent shape with two lone pairs on oxygen is highly electronegative, creating strong $\mathrm{O}-\mathrm{H}$ bond dipoles followed by $\mathrm{NH}_3$ and $\mathrm{CHCl}_3$.

Let's match each molecule with its approximate dipole moment:

1. HCl (Hydrogen Chloride): HCl is a diatomic molecule with a polar covalent bond. Chlorine is significantly more electronegative than hydrogen. This creates a net dipole moment. The experimentally observed dipole moment for HCl is approximately 1.08 D.

   • Matching with List-II, (II) 1.07 D is the closest and most accurate value. So, A - II.

2. NH₃ (Ammonia): NH₃ has a trigonal pyramidal geometry due to the presence of a lone pair on the nitrogen atom and three N-H bonds. Nitrogen is more electronegative than hydrogen, so the N-H bond dipoles point towards nitrogen. The lone pair also contributes significantly to the dipole moment, pointing away from the base of the pyramid. All these contributions add up constructively, resulting in a substantial net dipole moment. The experimentally observed dipole moment for NH₃ is approximately 1.47 D.

   • Matching with List-II, (IV) 1.47 D is a perfect match. So, B - IV.

3. H₂O (Water): H₂O has a bent (V-shaped) geometry due to two lone pairs on the oxygen atom and two O-H bonds. Oxygen is much more electronegative than hydrogen. The O-H bond dipoles point towards oxygen, and these dipoles, along with the lone pair contributions, add up constructively because of the bent shape, resulting in a very large net dipole moment. The experimentally observed dipole moment for H₂O is approximately 1.85 D.

   • Matching with List-II, (I) 1.85 D is a perfect match. So, C - I.

4. NF₃ (Nitrogen Trifluoride): NF₃ also has a trigonal pyramidal geometry, similar to NH₃. However, in NF₃, fluorine is more electronegative than nitrogen. Therefore, the N-F bond dipoles point towards the fluorine atoms, i.e., away from the nitrogen atom. The lone pair on nitrogen contributes a dipole moment pointing away from the base, similar to NH₃. However, the N-F bond dipoles oppose the dipole moment from the lone pair. This leads to a partial cancellation of dipoles, resulting in a much smaller net dipole moment compared to NH₃. The experimentally observed dipole moment for NF₃ is approximately 0.23 D.

   • Matching with List-II, (III) 0.23 D is a perfect match. So, D - III.

Combining the matches:

• (A) HCl - (II) 1.07 D

• (B) NH₃ - (IV) 1.47 D

• (C) H₂O - (I) 1.85 D

• (D) NF₃ - (III) 0.23 D

Now, let's check the given options:

Option A: A-II, B-IV, C-I, D-III. This matches our derived connections.

The final answer is $\boxed{\text{A}}$

Only $\mathrm{PCl}_3$ and $\mathrm{ClF}_3$ are correctly matched. While $\mathrm{SO}_2$ and $\mathrm{SF}_4$ are incorrect. Their correct match is

$ \begin{aligned} & \mathrm{SO}_2=1 \text { lone pair }=s p^2 \\ & \mathrm{SF}_4=1 \text { lone pair }=s p^3 d \end{aligned} $

The number of molecules having lone pair of electrons on central atoms are 5 (five).

They are

$\mathrm{SF}_4, \mathrm{NCl}_3, \mathrm{XeF}_6, \mathrm{XeF}_4, \mathrm{SnCl}_2$

Whereas $\mathrm{BF}_3, \mathrm{SiCl}_4, \mathrm{HgCl}_2$ and $\mathrm{PCl}_5=0$ lone pair

There are 4 Lewis bases in the given molecules.

They are $\mathrm{NH}_3, \mathrm{OH}, \mathrm{CO}, \mathrm{C}_2 \mathrm{H}_4$

Lewis bases are molecules or ions that can donate an electron pair.

$\mathrm{ClF}_3$ and $\mathrm{SF}_4$ has hybridisation as that of xenon (II) flouride. Both has $s p^3 d$ hybridisation.

$\mathrm{N}_2, \mathrm{CO}, \mathrm{NO}^{+}$are isoelectronic species.

Isoelectronic species are those species that have same number of electrons. $\mathrm{N}_2$, $\mathrm{Co}, \mathrm{No}^{+}, \mathrm{Au}$ has 14 electrons.

A molecule with $T$-shape has total-five electrons pairs in the valence shell of its central atom. This consists of three bond pairs and two lone pairs. This is in accordance with VSEPR theory.

Bond order is given by

B.O. $=\frac{1}{2}$ [Number of bonding electrons- number of antibonding electrons]

B.O. of $\mathrm{C}_2=2$

B.O. of $\mathrm{O}_2^{2+}=2.5, x=2+3=5$

B.O. $(a+b+c)=4.5$

$ a=\mathrm{B}_2, b=\mathrm{N}_2, C=\mathrm{F}_2 $

B.O. $=1+3+1=5$

Three molecules has two lone pairs of electrons on central atoms. These are $\mathrm{ClF}_3, \mathrm{XeF}_4$ and $\mathrm{H}_2 \mathrm{O}$ while $\mathrm{SF}_6, \mathrm{BF}_3, \mathrm{PCl}_5$ has 0 lone pair and $\mathrm{BrF}_5, \mathrm{SF}_4$ has 1 lone pair.

The pair of molecules that have the same bond order are $\mathrm{O}_2$ and $\mathrm{C}_2$.

Bond order shows how many chemical bonds are between two atoms.

To find bond order, use this formula:

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

For $\mathrm{O}_2$:

Number of bonding electrons = 10,
Number of antibonding electrons = 6

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

For $\mathrm{C}_2$:

Number of bonding electrons = 8,
Number of antibonding electrons = 4

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

So, both $\mathrm{O}_2$ and $\mathrm{C}_2$ have a bond order of 2.

Among the given options, only two molecules/ions has trigonal planar structure.

$\left[\mathrm{BO}_3\right]^{3-}, \mathrm{BCl}_3$ : Trigonal planar $\mathrm{NH}_3, \mathrm{PCl}_3, \mathrm{XeO}_3$ : Trigonal pyramidal $\mathrm{ClF}_3$ : T shape

Both $(\mathrm{A})$ and $(\mathrm{R})$ are correct and $(R)$ is the correct explanation of $(A)$.

$ \text { The reaction } $

$ \mathrm{O}_2^{+} \underset{\mathrm{I}}{\stackrel{ }{\longleftarrow}} \mathrm{O}_2 \underset{\mathrm{II}}{\longrightarrow} \mathrm{O}_2^{-} $

$ \begin{aligned} & \mathrm{O}_2: 16 \text { electrons: Paramagnetic } \\ & \mathrm{O}_2^{+}: 15 \text { electrons (1 removed); } \\ & \text { Paramagnetic } \\ & \mathrm{O}_2^{-}: 17 \text { electrons (1 added) : Paramagnetic } \\ & \mathrm{O}_2^{+}=\sigma \mathrm{ls}^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^0\right) \\ & \qquad \text { B.O. }=\frac{10-5}{2}=2.5 \\ & \mathrm{O}_2^{+}=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma * 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^1\right) \\ & \text { B.O. }=\frac{10-6}{2}=2 \\ & \mathrm{O}_2^{-}=\sigma l s^2 \sigma * 1 s^2 \sigma 2 s^2 \sigma * 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^2=\pi * 2 p_y^1\right) \end{aligned} $

$ \begin{aligned} &\text { B.O. }=\frac{10-7}{2}=1.5\\ &\text { Hence, statement } A \text { and } C \text { are correct. } \end{aligned} $

The correct order increasing of lone pair of electron on central atoms are

$ \mathrm{SiH}_4<\mathrm{SF}_4<\mathrm{ClF}_3<\mathrm{XeF}_2 $

zero lone pair one lone pairs two lone pairs 3 lone pair

$ \text { IV }<\text { III }<\text { I }<\text { II } $

$\mathrm{CH}_4, \mathrm{BF}_3, \mathrm{CO}_2$ all has zero dipole moments.

Dipole moments is given by $\mu=\sqrt{n(n+2)}$ B.M. where, $n$ is the number of unpaired electrons on central atom, in $\mathrm{CH}_4$, C has 0 unpaired electrons,

Thus $\mu=\sqrt{0(0+2)}=0$

Same is true for $\mathrm{BF}_3$ and $\mathrm{CO}_2$.

Order of bond angle is

$\mathrm{ClF}_3<\mathrm{PF}_3<\mathrm{BF}_3$

have central atom with sp$^2$ hybridisation NH$_2^-$ and H$_2$O have sp$^3$ hybridised central atom.

Carbocation is $s p^2$ hybridised hence it's shape is trigonal planar

In Molecular Orbital Theory, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). The wave functions of atomic orbitals $\psi_A$ and $\psi_B$ can combine in two ways:

1. Constructive Interference: This leads to the formation of a bonding molecular orbital ($\sigma$):

$ \sigma = \psi_A + \psi_B $

1. Destructive Interference: This leads to the formation of an antibonding molecular orbital ($\sigma^*$):

$ \sigma^* = \psi_A - \psi_B $

Conclusion

The antibonding molecular orbital $\sigma^*$ is represented by:

$ \sigma^* = \psi_A - \psi_B $

Therefore, the correct answer is:

Option B: $\psi_{\mathrm{A}} - \psi_{\mathrm{B}}$

To match the molecules in List I with their corresponding shapes in List II, we need to consider the electronic geometry and the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here’s the detailed analysis:

$\mathrm{NH_3}$ (Ammonia): Ammonia has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons. This leads to a trigonal pyramidal shape.

$\mathrm{BrF_5}$ (Bromine pentafluoride): Bromine pentafluoride has a central bromine atom bonded to five fluorine atoms and has one lone pair of electrons. This leads to a square pyramidal shape.

$\mathrm{PCl_5}$ (Phosphorus pentachloride): Phosphorus pentachloride has a central phosphorus atom bonded to five chlorine atoms and no lone pairs. This results in a trigonal bipyramidal shape.

$\mathrm{CH_4}$ (Methane): Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs, forming a tetrahedral structure.

Using the above analyses, we can match the molecules and shapes as follows:

• A ($\mathrm{NH_3}$) - III (Trigonal pyramidal)
• B ($\mathrm{BrF_5}$) - I (Square pyramid)
• C ($\mathrm{PCl_5}$) - IV (Trigonal bipyramidal)
• D ($\mathrm{CH_4}$) - II (Tetrahedral)

Therefore, the correct matching is option B:

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

$\begin{aligned} & \text { A } \rightarrow \text { Sea-saw (III) } \\ & \text { B } \rightarrow \text { Bent T-shape (IV) } \\ & \text { C } \rightarrow \text { Pyramidal (II) } \\ & \text { D } \rightarrow \text { Tetrahedral (I) } \end{aligned}$

$\begin{aligned} & \mathrm{A} \rightarrow \text { Tetrahedral (III) } \\ & \mathrm{B} \rightarrow \text { Paramagnetic (I) } \\ & \mathrm{C} \rightarrow \text { Diamagnetic (II) } \\ & \mathrm{D} \rightarrow \text { Linear (IV) } \end{aligned}$

$\begin{aligned} & \mathrm{ICl} \rightarrow s p^3 \rightarrow 1 \mathrm{bp}+3 \mathrm{lp} \rightarrow \text { Linear } \\ & \mathrm{ICl}_3 \rightarrow s p^3 d \rightarrow 3 \mathrm{bp}+2 \mathrm{lp} \rightarrow \mathrm{T} \text {-shape } \\ & \mathrm{CIF}_5 \rightarrow s p^3 d^2 \rightarrow 5 \mathrm{bp}+1 \mathrm{lp} \rightarrow \text { Square Pyramidal } \\ & \mathrm{IF}_7 \rightarrow s p^3 d^3 \rightarrow 7 \text { bp only } \rightarrow \text { Pentagonal bipyramidal } \\ & \text { (A)-(IV), (B)-(I), (C)-(II), D-(III) } \end{aligned}$

Dipole moment of $\mathrm{NH}_3$ is higher than $\mathrm{NF}_3$ as $\mu$ of lone-pair is in same direction as the resultant dipole moment of $\mathrm{N}-\mathrm{H}$ bonds.

So, both (A) & (R) are true, & (R) is the correct explanation of (A)

Let's analyze the bond structure in an ethylene molecule (C₂H₄) to determine the count of $\sigma$ and $\pi$ bonds.

Ethylene consists of two carbon atoms double-bonded to each other, with each carbon atom also bonded to two hydrogen atoms. Each single bond is a $\sigma$ bond. The double bond between the carbon atoms is made up of one $\sigma$ bond and one $\pi$ bond. Here's how the bonds break down:

1. Each carbon-hydrogen bond is a $\sigma$ bond. Since there are four C-H bonds, we have 4 $\sigma$ bonds from C-H bonding.
2. The carbon-carbon double bond consists of one $\sigma$ bond and one $\pi$ bond. Thus, we have 1 additional $\sigma$ bond from the C=C double bond.
3. Total $\sigma$ bonds = 4 (from C-H) + 1 (from C=C) = 5.
4. There is 1 $\pi$ bond present in the carbon-carbon double bond.

Therefore, the correct option is:

Option B: 5 and 1

In o-nitrophenol intra molecular hydrogen bonding is present.

In HF intermolecular hydrogen bonding is present.

Other statements are correct except B and C.

$\mathrm{SO}_3^{2-}$ is the only species with pyramidal geometry.

To determine the number of molecules/ions in which the central atom is involved in $\mathrm{sp}^3$ hybridization, we must analyze the hybridization state for each central atom. Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.

Let's evaluate each molecule/ion:

1. $\mathrm{NO}_3^{-}$:

The central atom is nitrogen (N). In $\mathrm{NO}_3^{-}$, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs. Using the formula for hybridization:

Number of hybrid orbitals = number of sigma bonds + number of lone pairs

For $\mathrm{NO}_3^{-}$: $3 + 0 = 3$ hybrid orbitals. Therefore, nitrogen is $\mathrm{sp}^2$ hybridized.

2. $\mathrm{BCl}_3$:

The central atom is boron (B). In $\mathrm{BCl}_3$, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula:

For $\mathrm{BCl}_3$: $3 + 0 = 3$ hybrid orbitals. Therefore, boron is $\mathrm{sp}^2$ hybridized.

3. $\mathrm{ClO}_2^{-}$:

The central atom is chlorine (Cl). In $\mathrm{ClO}_2^{-}$, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,:

For $\mathrm{ClO}_2^{-}$: $2 + 2 = 4$ hybrid orbitals. Thus, chlorine is $\mathrm{sp}^3$ hybridized.

4. $\mathrm{ClO}_3^{-}$:

The central atom is chlorine (Cl). In $\mathrm{ClO}_3^{-}$, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,:

For $\mathrm{ClO}_3^{-}$: $3 + 1 = 4$ hybrid orbitals. Thus, chlorine is $\mathrm{sp}^3$ hybridized.

Based on the analysis, the molecules/ions $\mathrm{ClO}_2^{-}$ and $\mathrm{ClO}_3^{-}$ have their central atoms involved in $\mathrm{sp}^3$ hybridization.

Therefore, the total number is 2.

Thus, the correct option is Option A.

$\mathrm{NH}_3 \text { have more dipole moment than } \mathrm{NF}_3 \text {. }$

Let's analyze both statements:

Statement (I): "A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis."

This statement is false. In molecular orbital (MO) theory, a pi bond ($\pi$ bond) is formed by the sideways overlap of p-orbitals from two adjacent atoms. The characteristic electron density of a $\pi$ bonding molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of electron density are where the p-orbitals overlap and electrons are likely to be found. This concentration of electron density above and below the inter-nuclear axis is what allows the $\pi$ bond to provide additional stability to the molecule, alongside any sigma ($\sigma$) bonds that may be present.

Statement (II): "The $\pi^*$ antibonding MO has a node between the nuclei."

This statement is true. In a $\pi^*$ (pi-star) antibonding molecular orbital, there is indeed a nodal plane between the nuclei where the probability of finding electrons is essentially zero. This node arises from the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive interference and a region of zero electron density in the bonding region. The presence of this nodal plane is what makes the $\pi^*$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.

Therefore, the correct answer is:

Option D: Statement I is false but Statement II is true.

Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another electronegative atom within the same molecule. This phenomenon tends to happen when the molecule can form a six-membered or five-membered ring as a result of this bonding, which offers a stable configuration.

Let's consider each of the given options:

• Option A : The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit intramolecular hydrogen bonding.


• Option B : Water ($\mathrm{H}_2 \mathrm{O}$) is capable of intermolecular hydrogen bonding, but it cannot form intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to form a stable ring structure within a single molecule.


• Option C : Ethanol ($\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$) can form intermolecular hydrogen bonds through its hydroxyl group with other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen bonds because it does not have the correct geometry for the hydroxyl group to bond within the same molecule.


• Option D : Ammonia ($\mathrm{NH}_3$) is known for intermolecular hydrogen bonding with other ammonia molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that would allow for intramolecular hydrogen bonding.

Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-nitrophenol molecule.

The correct answer is Option C: (A) is correct but (R) is not correct.

Assertion (A): $\mathrm{PH}_3$ has lower boiling point than $\mathrm{NH}_3$.

This assertion is true. The boiling point of ammonia ($\mathrm{NH}_3$) is higher than that of phosphine ($\mathrm{PH}_3$). Ammonia has a boiling point of about -33.34°C, whereas phosphine has a boiling point of about -87.7°C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the strength and type of intermolecular forces present in the substances.

Reason (R) : In liquid state $\mathrm{NH}_3$ molecules are associated through Vander Waals' forces, but $\mathrm{PH}_3$ molecules are associated through hydrogen bonding.

This reason is incorrect. Ammonia ($\mathrm{NH}_3$) can form hydrogen bonds due to the presence of a highly electronegative nitrogen atom bonded to hydrogen atoms. This allows $\mathrm{NH}_3$ molecules to strongly associate with each other through hydrogen bonding, which is a much stronger intermolecular force than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of ammonia.

On the other hand, phosphine ($\mathrm{PH}_3$) does not form hydrogen bonds because the electronegativity difference between phosphorus and hydrogen is not significant enough to enable the formation of hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals forces, leading to a lower boiling point when compared to ammonia.

In conclusion, the assertion that $\mathrm{PH}_3$ has a lower boiling point than $\mathrm{NH}_3$ is correct, due to the hydrogen bonding present in $\mathrm{NH}_3$ and absent in $\mathrm{PH}_3$. However, the reason provided is incorrect because it incorrectly states that $\mathrm{PH}_3$ forms hydrogen bonds and that $\mathrm{NH}_3$ is associated through Van der Waals forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false.

We can qualitatively assess the ionic character of the bonds in each molecule as follows :

Given these considerations, the correct order from least to most ionic character is as follows :

Therefore, the correct option is :

Option A : $\mathrm{N_2}<\mathrm{SO_2}<\mathrm{ClF_3}<\mathrm{K_2O}<\mathrm{LiF}$

$\mathrm{AgCl}<\mathrm{CoCl}_2<\mathrm{BaCl}_2<\mathrm{KCl}$ (ionic character)

Reason: $\mathrm{Ag}^{+}$ has pseudo inert gas configuration.

* Molecular orbital should have maximum overlap

* Symmetry about the molecular axis should be similar

$\mathrm{BrF_5}$

$\mathrm{AlCl}_3$ in acidified aqueous solution forms octahedral geometry $[\mathrm{Al}(\mathrm{H}_2 \mathrm{O})_6]^{3+}$

According to NCERT,

Statement-I : Factual data,

Statement-II is true.

But correct explanation is presence of completely filled $d$ and $f$-orbitals of heavier members

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

$\mu \neq 0$

$\mathrm{CHCl}_3$ is polar molecule and rest all molecules are non-polar.