Chemical Bonding & Molecular Structure

$\mathrm{XeF}_2$ and $\mathrm{I}_3^{-}$ have $2$ bond pairs and $3$ lone pairs on the central atom (total $5$ electron pairs). In this case, the molecular shape becomes linear.

$\mathrm{XeOF}_4$ and $\mathrm{BrF}_5$ have $5$ bond pairs and $1$ lone pair on the central atom (total $6$ electron pairs). This gives a square pyramidal shape.

$\mathrm{XeO}_2 \mathrm{~F}_2$ and $\mathrm{SF}_4$ have $4$ bond pairs and $1$ lone pair on the central atom (total $5$ electron pairs). This leads to a see-saw shape.

$\mathrm{XeO}_3$ and $\mathrm{NH}_3$ have $3$ bond pairs and $1$ lone pair on the central atom (total $4$ electron pairs). So, the shape is pyramidal.

Statement I

1. $\mathrm{BF_4^-}$

   Central atom B has 4 bond pairs and 0 lone pairs $\Rightarrow$ tetrahedral.

   All four $\mathrm{B-F}$ bonds are equivalent $\Rightarrow$ equal bond lengths.

2. $\mathrm{SiF_4}$

   Central atom Si has 4 bond pairs and 0 lone pairs $\Rightarrow$ tetrahedral.

   All four $\mathrm{Si-F}$ bonds equivalent $\Rightarrow$ equal bond lengths.

3. $\mathrm{XeF_4}$

   Xe has 4 bond pairs and 2 lone pairs ($\mathrm{AX_4E_2}$) $\Rightarrow$ square planar.

   All four $\mathrm{Xe-F}$ bonds equivalent by symmetry $\Rightarrow$ equal bond lengths.

4. $\mathrm{SF_4}$

   S has 4 bond pairs and 1 lone pair ($\mathrm{AX_4E}$) $\Rightarrow$ see-saw shape.

   There are two axial and two equatorial $\mathrm{S-F}$ bonds, and they are not equivalent (axial bonds are longer).

   $\Rightarrow$ unequal bond lengths.

So, number of species with unequal $\mathrm{E-F}$ bond lengths $= 1$ (only $\mathrm{SF_4}$), not 2.

$\Rightarrow$ Statement I is false.

Statement II

Bond order (NCERT MO theory):

$ \text{Bond order}=\frac{N_b-N_a}{2} $

Using known MO results for these diatomic species:

• $\mathrm{O_2^+}$ has bond order $2.5$

• $\mathrm{O_2}$ has bond order $2$

• $\mathrm{O_2^-}$ has bond order $1.5$

• $\mathrm{O_2^{2-}}$ has bond order $1$

• $\mathrm{F_2}$ has bond order $1$

Highest bond order among $\mathrm{O_2^-}, \mathrm{O_2^{2-}}, \mathrm{F_2}, \mathrm{O_2^+}$ is $\mathrm{O_2^+}$, not $\mathrm{O_2^-}$.

$\Rightarrow$ Statement II is false.

Correct Option

Both Statement I and Statement II are false.

Option A

$ \begin{array}{|l|l|l|} \hline \text { Species } & \begin{array}{l} \text { Bond } \\ \text { order } \end{array} & \begin{array}{l} \text { Magnetic } \\ \text { Nature } \end{array} \\ \hline \mathrm{O}_2^{+} & 2.5 & \text { Paramagnetic } \\ \hline \mathrm{O}_2^{-} & 1.5 & \text { Paramagnetic } \\ \hline \mathrm{O}_2^{+} & 2.5 & \text { Paramagnetic } \\ \hline \mathrm{N}_2^{-} & 2.5 & \text { Paramagnetic } \\ \hline \mathrm{N}_2^{2-} & 2 & \text { Paramagnetic } \\ \hline \end{array} $

Check each statement:

A. $\mu(\mathrm{NF}_3) > \mu(\mathrm{NH}_3)$

Both are trigonal pyramidal. In $\mathrm{NH}_3$, the bond dipoles and lone pair dipole add to give a larger net dipole moment. In $\mathrm{NF}_3$, the $\mathrm{N-F}$ bond dipoles oppose the lone pair direction, so net dipole is small.

So, $\mu(\mathrm{NF}_3) < \mu(\mathrm{NH}_3)$.

A is incorrect.

B. Dipole moment of $\mathrm{BeH}_2$ is zero

$\mathrm{BeH}_2$ is linear: $\mathrm{H-Be-H}$ (NCERT: $sp$ hybridised). The two $\mathrm{Be-H}$ bond dipoles are equal and opposite, hence cancel.

So, $\mu = 0$.

B is correct.

C. Bond order of $\mathrm{O}_2^{2-}$ and $\mathrm{F}_2$ is same

Using MO theory:

• $\mathrm{O}_2^{2-}$ has 14 valence electrons (like $\mathrm{F}_2$). For peroxide ion, bond order becomes

  $\text{B.O.} = 1$

• For $\mathrm{F}_2$, bond order is also

  $\text{B.O.} = 1$

So they are the same.

C is correct.

D. Formal charge on central oxygen of ozone is $-1$

In one resonance form: $\mathrm{O=O-O}$ (central O has one lone pair, one double bond, one single bond)

Formal charge on central O:

$\text{F.C.} = 6 - \left(2 + \frac{6}{2}\right) = 6 - (2+3) = +1$

So it is $+1$, not $-1$.

D is incorrect.

E. In $\mathrm{NO}_2$, all atoms satisfy octet rule, hence very stable

$\mathrm{NO}_2$ has an odd number of electrons (17 valence electrons), so it is an odd-electron molecule and nitrogen does not complete its octet. Hence it is quite reactive (dimerises to $\mathrm{N}_2\mathrm{O}_4$).

E is incorrect.

Correct statements: B and C only

✅ Option A

(B) This statement is not true because Xe has 7 electron pairs in its valence shell in this molecule.

(C) This statement is not true because the ${O}-{Xe}-{O}$ bond angle is close to $120^{\circ}$, not $180^{\circ}$.

(D) This statement is true because the F-Xe-F bond angle is close to $180^{\circ}$.

(E) This statement is not true because Xe has 14 valence electrons in ${XeO}_2 {F}_2$, not 16.

Hybridization : $\mathrm{sp}^3$

Number of lone pair in $\mathrm{NF}_3=10$

Bond angle in $\mathrm{NF}_3 \approx 102^{\circ}$

$ \begin{array}{ll} \text { Molecule } & \text { Dipole moment } \\ \mathrm{H}_2 \mathrm{~S} & 0.95 \\ \mathrm{H}_2 \mathrm{O} & 1.85 \\ \mathrm{NF}_3 & 0.23 \text { (minimum) } \\ \mathrm{NH}_3 & 1.47 \\ \mathrm{CHCl}_3 & 1.04 \end{array} $

From the given values, the dipole moment is the smallest for $\mathrm{NF}_3$ (it is $0.23$), so molecule $(X)$ is $\mathrm{NF}_3$.

In $\mathrm{NF}_3$, the central atom is nitrogen $(\mathrm{N})$.

Nitrogen has $5$ valence electrons. It makes $3$ $\mathrm{N{-}F}$ bonds, so $3$ electrons are used for bonding.

The remaining $2$ valence electrons stay on nitrogen as one lone pair.

So, the number of lone pairs on the central atom in $(X)=\mathrm{NF}_3$ is $1$.

Number of lone pair on central atom = 1

$\mathrm{XF}_3$ acts as a Lewis acid, so it must have an incomplete octet (it can accept an electron pair).

A common p-block fluoride with formula $\mathrm{EF}_3$ that is a Lewis acid is $\mathrm{BF}_3$.

In $\mathrm{BF}_3$, boron forms three $\sigma$ bonds with fluorine and has no lone pair, so it has only 6 electrons around it. Therefore, it is electron-deficient and behaves as a Lewis acid.

With three bond pairs around boron, the hybridization is $\mathrm{sp}^2$.

$\mathrm{YF}_3$ acts as a Lewis base, so it must have a lone pair to donate.

A common p-block fluoride with formula $\mathrm{EF}_3$ that is a Lewis base is $\mathrm{NF}_3$.

In $\mathrm{NF}_3$, nitrogen forms three $\sigma$ bonds with fluorine and has one lone pair, so it can donate that lone pair and behave as a Lewis base.

With four electron pairs (3 bond pairs + 1 lone pair) around nitrogen, the hybridization is $\mathrm{sp}^3$.

So, $\mathrm{XF}_3=\mathrm{BF}_3$ has hybridization $\mathrm{sp}^2$ and $\mathrm{YF}_3=\mathrm{NF}_3$ has hybridization $\mathrm{sp}^3$.

$\begin{aligned} & \mathrm{XF}_3=\mathrm{BF}_3 ; \mathrm{sp}^2 \\ & \mathrm{YF}_3=\mathrm{NF}_3 ; \mathrm{sp}^3\end{aligned}$

Consider the structure of $\mathrm{HNO}_3$

$\begin{aligned} & 1:-0 \\ & 2:-(+1) \\ & 3:-0 \\ & 4:-(-1) \\ & \text { Formal charge }=\text { valence } \mathrm{e} \text { 's }- \text { non bonding e's } - (\frac{\text { bonding electrons }}{2})\end{aligned}$

Bond length is inversely related to bond order (as per NCERT idea):

triple bond $<$ double bond $<$ single bond (shorter to longer).

Also, for single bonds, bond length increases when the bonded atom is larger (so $C{-}O$ is longer than $C{-}H$).

Now compare:

• $C{-}H$ (A): single bond, but H is very small $\Rightarrow$ short

• $C{\equiv}N$ (D): triple bond $\Rightarrow$ very short

• $C{=}O$ (C): double bond $\Rightarrow$ short, but longer than triple

• $C{-}O$ (B): single bond with larger O $\Rightarrow$ longest

So increasing order of bond length is:

$A < D < C < B$

Correct option: D

Statement I: True

Bond dissociation enthalpy generally decreases down the group because bond length increases (bond becomes weaker). But $F_2$ is an exception: its $F-F$ bond is unusually weak due to strong lone pair–lone pair repulsions on the small fluorine atoms. Hence,

$D(\mathrm{Cl_2}) > D(\mathrm{Br_2}) > D(\mathrm{F_2}) > D(\mathrm{I_2})$

So Statement I is correct.

Statement II: False

Using Fajans’ rule (NCERT): covalent character increases when the cation has higher charge and smaller size (greater polarising power).

• $\mathrm{SnCl_4}$ vs $\mathrm{SnCl_2}$: $Sn^{4+}$ has higher charge and smaller size than $Sn^{2+}$

  $\Rightarrow \mathrm{SnCl_4} \text{ is more covalent than } \mathrm{SnCl_2} \ (\text{True})$

• $\mathrm{PbCl_4}$ vs $\mathrm{PbCl_2}$: similarly, $Pb^{4+}$ polarises more than $Pb^{2+}$

  $\Rightarrow \mathrm{PbCl_4} > \mathrm{PbCl_2} \text{ in covalent character (True)}$

• $\mathrm{UF_4}$ vs $\mathrm{UF_6}$: here $U^{6+}$ has higher charge than $U^{4+}$, so it has greater polarising power

  $\Rightarrow \mathrm{UF_6} \text{ should be more covalent than } \mathrm{UF_4}$

But the statement claims $\mathrm{UF_4 > UF_6}$, which is incorrect.

So Statement II is false.

Correct Option: D

$\boxed{\text{Statement I is true but Statement II is false}}$

Statement I: Count species with tetrahedral molecular geometry:

• SF₄ → seesaw (AX₄E), not tetrahedral

• NH₄⁺ → tetrahedral

• [NiCl₄]²⁻ (Ni²⁺, d⁸ with weak-field Cl⁻) → tetrahedral

• XeF₄ → square planar, not tetrahedral

• [PtCl₄]²⁻ (Pt²⁺, d⁸) → square planar, not tetrahedral

• SeF₄ → seesaw (AX₄E), not tetrahedral

• [Ni(CN)₄]²⁻ (Ni²⁺, d⁸ with strong-field CN⁻) → square planar, not tetrahedral

So tetrahedral = 2, not 3 ⇒ Statement I is false.

Statement II: Check incomplete octet around central atom:

• NO₂: odd-electron molecule; N does not complete octet

• BeH₂: Be has only 4 electrons around it

• BF₃: B has only 6 electrons

• AlCl₃: Al has only 6 electrons in monomeric form (though it dimerizes to Al₂Cl₆ to complete octet)

Thus the statement is taken as true ⇒ Statement II is true.

Correct option: C — Statement I is false but Statement II is true.

Statement 1 is correct.

Statement 2 is incorrect since in $\mathrm{sp}^3 \mathrm{~d}$ hybridization; lone pair cannot occupy axial position.

Statement - I :

$\mu=\mathrm{q} \times \mathrm{d}$

More charges and more distance between charges than other compound so more dipole moment.

Statement-I is true.

Statement-II :

$\mathrm{C}_1-\mathrm{C}_2$ bond has partial double bond character that means lesser bond length than $C_1-C_2$ bond of other compound.

Statement-II is false.

No. of unpaired $\mathrm{e}^{-}$

If species contain unpaired electron than it is paramagnetic.

So A & D are paramagnetic.

The correct choice is Option D: Statement I is true but Statement II is false.

Statement I (True)

Cotton is made from cellulose, which is a kind of carbohydrate. Cellulose has many –OH groups (hydroxyl groups) which can make strong hydrogen bonds with water. Because of this, cotton holds more water and takes more time to dry than nylon.

Statement II (False)

Nylon is made of polymer chains with only a few special groups (amide –NH and C=O) that can form hydrogen bonds with water. Nylon forms fewer hydrogen bonds with water compared to cotton. So, cotton actually makes more hydrogen bonds with water than nylon does.

Given A is rarest, monoatomic, non-radioactive p-block element and form $\mathrm{AX}_4 \mathrm{Y}$ type of molecule.

$\therefore$ It is concluded that it is Xe

It is given the electronegativity of A is less than X & Y

It is given the electronegativity of $\mathrm{X} \& \mathrm{Y}$ is highest and second highest respectively among all element.

$\therefore \mathrm{X}$ & Y are F & O

$\therefore$ Compound is consider as $\mathrm{XeOF}_4$ with square pyramidal shape.

1. Molecules with non-zero dipole moment

• SO₂ (bent)

• NF₃ (pyramidal)

• NH₃ (pyramidal)

• ClF₃ (T-shaped)

• SF₄ (seesaw)

  (XeF₂ is linear → zero dipole)

1. Lone-pair count on central atom

• SO₂: 1

• NF₃: 1

• NH₃: 1

• SF₄: 1

• ClF₃: 2

  → ClF₃ has the highest number (2) of lone pairs among those with nonzero dipole.

1. Hybridization of Cl in ClF₃

   • Total electron domains = 3 bonding pairs + 2 lone pairs = 5

   • For 5 domains → $sp^3d$ hybridization

Answer: Option D ($sp^3d$).

Rate of hydrolysis $\propto$ Leaving group ability

$\mathrm{n}=2($ No of lone pair present in equitorial plane)

$\begin{array}{lc} & \text { (Unpaired } \mathrm{e}^{-} \text {) } \\ \text { (A) } \mathrm{V}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \\ \text { (B) } \mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^1 & 1 \\ \text { (C) } \mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9 & 1 \\ \text { (D) } \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 & 2 \\ \text { (E) } \mathrm{Ti}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \end{array}$

$\mathrm{BrF}_5$ : Square pyramedal

$\mathrm{XeOF}_4$ : Square pyramedal

$\mathrm{SbF}_5$ : Trigonal bipyramidal

$\mathrm{PCl}_5$ : Trigonal bipyramidal

Due to resonance bond length is identical in ozone. Therefore statement I is true and statement II is false.

Dipole moment $\quad \mathrm{H}_2 \mathrm{O}>\mathrm{NH}_3>\mathrm{CH}_4$

$\mathrm{H}_2 \mathrm{O} \& \mathrm{NH}_3$ are Lewis Bases

$\mathrm{NH}_3$ act as Lowry- Bronsted base

Hence, A, B & C are correct

(A) $\mathrm{A} \rightarrow$ IV

(B) B $\rightarrow$ II

(C) $\mathrm{C} \rightarrow \mathrm{I}$

(D) $\mathrm{D} \rightarrow$ III

We need to compare the experimental (or well‐established) dipole moments of the given molecules: $\mathrm{HBr}, \mathrm{H_2S}, \mathrm{NF_3},$ and $\mathrm{CHCl_3}$.

1. $\mathbf{NF_3}$

Structure: Trigonal pyramidal (like $\mathrm{NH_3}$), but each bond is more polar toward fluorine.

Net dipole moment: Quite small, about $0.23\text{–}0.24\,D$ (the bond dipoles partly oppose the lone‐pair contribution).

2. $\mathbf{HBr}$

Structure: Simple diatomic.

Dipole moment: About $0.78\,D$.

3. $\mathbf{H_2S}$

Structure: Bent (like $\mathrm{H_2O}$) but S is less electronegative than O, and the S–H bond angle is wider than the O–H angle in water.

Dipole moment: About $0.95\,D$.

4. $\mathbf{CHCl_3}$ (chloroform)

Structure: Tetrahedral around carbon with three Cl and one H.

Dipole moment: About $1.01\,D$.

Ordering from smallest to largest dipole moment

$ \boxed{\mathrm{NF_3} \;<\; \mathrm{HBr} \;<\; \mathrm{H_2S} \;<\; \mathrm{CHCl_3}.} $

Hence, the correct choice is:

$ \boxed{\text{Option (C)}\quad NF_3 < HBr < H_2S < CHCl_3.} $

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.