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.

First, let us recall the autoionization of bromine trifluoride:

$ 2\,\mathrm{BrF_3} \rightleftharpoons \mathrm{BrF_2^+} + \mathrm{BrF_4^-} $

We have to determine the shapes of the cation and the anion.

• Central atom: Br

• Valence electrons of Br = 7

• Each F contributes one bond → 2 Br–F bonds = 2 pairs of bonding electrons

So, total electrons around Br for shape determination:

$ 7 - 2(\text{by bonding}) - 1(\text{positive charge}) = 4 $

These 4 electrons form 2 lone pairs on Br.

Hence, the total number of electron pairs around Br =

2 (bonding) + 2 (lone pairs) = 4 electron domains.

According to VSEPR theory:

• 4 electron pairs → tetrahedral electron geometry

• With 2 bonding and 2 lone pairs → bent shape

Thus, $\mathrm{BrF_2^+}$ is bent.

• Central atom: Br

• Valence electrons of Br = 7

• Each F contributes one bond → 4 Br–F bonds = 4 pairs of bonding electrons

• Additional one electron due to negative charge → +1

Hence total electrons on central bromine:

$ 7 + 1 - 4 = 4 $

which means 2 lone pairs on Br.

Thus, the total number of electron pairs =

4 (bonding) + 2 (lone pairs) = 6 electron domains.

According to VSEPR theory:

• 6 electron domains → octahedral electron geometry

• With 4 bonding and 2 lone pairs → square planar shape

Shapes are:

• $\mathrm{BrF_2^+}$ → bent

• $\mathrm{BrF_4^-}$ → square planar

Correct Option: A) bent, square planar

Hence, 4 compounds does not follow octet rule for sulphur.

$ \text { Only } 1 \text { set has all molecules with one Ip. } $

Order: $\quad \theta_1>\theta_3>\theta_2$

$SiF_4$ is covalent while $SnF_4$ and $PbF_4$ are ionic due to large electronegative difference between them.

$ \begin{aligned}(E)\,\, & \operatorname{MnO4}^{-2}=7+4 \times 2+2=4+10=17 e^{-} \\ & \operatorname{CrO4} ^{-2}=6+4 \times 2+2=6+10=16 e^{-} \end{aligned} $

since $e^{-}$count is notequal, leawis dot structure will be different.

1. Finding the Covalent Bond Length:

In chemistry, the covalent bond length is defined as the average distance between the centers of the nuclei of two bonded atoms in a molecule.

When atoms $A$ and $B$ form a single covalent bond, we can approximate the distance between their nuclei by simply adding their individual covalent radii.

Therefore, if the covalent radius of atom $A$ is $r_A$ and the covalent radius of atom $B$ is $r_B$, the internuclear distance, which is the bond length of the $AB$ molecule, is:

$ \text{Bond Length} = r_A + r_B $

2. Finding the Total Length of the AB Molecule:

To find the total length of the entire molecule, we need to consider the full physical extent of both atoms. Imagine the two atoms as spheres placed side by side on a straight axis.

• The center of atom $A$ to its outermost left edge is equal to its radius, $r_A$.

• The distance from the center of atom $A$ to the center of atom $B$ is the bond length, which we already established is $(r_A + r_B)$.

• The center of atom $B$ to its outermost right edge is equal to its radius, $r_B$.

The total length of the molecule is the distance from the outermost left edge of atom $A$ to the outermost right edge of atom $B$. We can find this by adding the three segments together:

$ \text{Total Length} = (\text{radius of } A) + (\text{internuclear distance}) + (\text{radius of } B) $

$ \text{Total Length} = r_A + (r_A + r_B) + r_B $

$ \text{Total Length} = 2r_A + 2r_B $

$ \text{Total Length} = 2(r_A + r_B) $

So, the covalent bond length is $(r_A + r_B)$ and the total length of the full $AB$ molecule is $2(r_A + r_B)$.

Comparing our findings with the given options, this corresponds perfectly to Option A.

Correct Answer:

Option A

$\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right), 2\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right)$

constructive Interferance due to same phase overlapping gives $\pi-B M O$.

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 - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (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_a = 6

N_b = 10

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

Here 2 unpaired electron present.

Molecular orbital configuration of O$_2^ + $ (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

Here 1 unpaired electron present.

Molecular orbital configuration of $O_2^ - $ (17 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^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Here 1 unpaired electron present.

Molecular orbital configuration of O $_2^{2 - }$ (18 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^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

Here 0 unpaired electron present.

As Bond strength $ \propto $ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

So, Correct order of Bond Order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So, Correct order of Bond Length is

$\ce{O2^{+}} < \ce{O2} < \ce{O2^{-}} < \ce{O2^{2-}}$

$\therefore $ The correct sequence of number of unpaired electrons is :

$\ce{O2} > \ce{O2^{+}} = \ce{O2^{-}} > \ce{O2^{2-}}$

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 $\mathrm{e}^{-}$

(A)	$\mathrm{O}_2$	2
(B)	$\mathrm{N}_2$	0
(C)	$\mathrm{F}_2$	0
(D)	$\mathrm{S}_2$	2
(E)	$\mathrm{Cl}_2$	0

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

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

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