Chemical Bonding & Molecular Structure

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

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.

We want to find which of the given diatomic species (among $\mathrm{O}_2^+$, $\mathrm{O}_2^-$, $\mathrm{N}_2^+$, and $\mathrm{NO}^+$) has the same bond order and the same magnetic property (dia- or paramagnetic) as the acetylide ion, $\mathrm{C}_2^{2-}$.

1. Electron Count and MO Considerations

(a) Acetylide Ion, $\mathbf{C_2^{2-}}$

A neutral $\mathrm{C}_2$ molecule would have $2 \times 6 = 12$ electrons.

In $\mathrm{C}_2^{2-}$, there are 2 extra electrons, giving a total of $12 + 2 = 14$ electrons.

(b) Matching Total Electrons Among the Options

Let us see how many electrons are in each species:

$\mathrm{O}_2$ has $2 \times 8 = 16$ electrons

$\mathrm{O}_2^+$ has $16 - 1 = 15$ electrons

$\mathrm{O}_2^-$ has $16 + 1 = 17$ electrons

$\mathrm{N}_2$ has $2 \times 7 = 14$ electrons

$\mathrm{N}_2^+$ has $14 - 1 = 13$ electrons

$\mathrm{NO}$ has $7 + 8 = 15$ electrons

$\mathrm{NO}^+$ has $15 - 1 = 14$ electrons

Hence the only diatomic species in the list that has 14 electrons (like $\mathrm{C}_2^{2-}$) is $\mathrm{NO}^+$.

2. Bond Order and Magnetic Property

(a) $\mathrm{C}_2^{2-}$ (14 Electrons)

From molecular orbital (MO) theory for 14-electron diatomics (similar to $\mathrm{N}_2$ with 14e), we generally get:

Bond order = 3

All electrons paired in MOs → diamagnetic.

(b) $\mathrm{NO}^+$ (also 14 Electrons)

Similarly, $\mathrm{NO}^+$ also has 14 electrons and, by the same reasoning in MO theory, has:

Bond order = 3

Diamagnetic (no unpaired electrons).

3. Conclusion

Among the given choices:

$\mathrm{O}_2^+$ has 15 electrons (not 14)

$\mathrm{O}_2^-$ has 17 electrons

$\mathrm{N}_2^+$ has 13 electrons

$\mathrm{NO}^+$ has 14 electrons

Therefore, the species with the same bond order (3) and the same magnetic property (diamagnetic) as $\mathrm{C}_2^{2-}$ is:

$ \boxed{\mathrm{NO}^{+} \text{ (Option D).}} $

Let's consider each species in List I and determine their geometries/shapes.

A. $\mathrm{H_3O^+}$: This ion is formed by the addition of a proton to a water molecule. The central atom (O) is surrounded by three Hydrogen atoms and one lone pair, making its shape pyramidal.

B. Acetylide anion: The acetylide ion ($\mathrm{C_2H^-}$ or $\mathrm{C_2^{2-}}$) has a linear geometry as the three atoms are in a straight line.

C. $\mathrm{NH_4^+}$: The ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms, giving it a tetrahedral geometry.

D. $\mathrm{ClO_2^-}$: The chlorite ion has a central chlorine atom surrounded by two oxygen atoms and one lone pair, which gives it a bent or V-shaped geometry.

Matching these to List II, we get:

A. $\mathrm{H_3O^+}$ - III. Pyramidal

B. Acetylide anion - II. Linear

C. $\mathrm{NH_4^+}$ - I. Tetrahedral

D. $\mathrm{ClO_2^-}$ - IV. Bent