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.

$\begin{array}{cccc}\mathrm{NO}_2^{+} & \mathrm{NO}_2 & \mathrm{NO}_3^{-} & \mathrm{NO}_2^{-} \\ s p & s p^2 & s p^2 & s p^2\end{array}$

$\mathrm{NO}_2^{-}<\mathrm{NO}_3^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}$bond angle order.

$\therefore $ Order of dipole moment : $\mathrm{BF}_3=\mathrm{NH}_4^{+}<\mathrm{NF}_3<\mathrm{NH}_3$

See-saw → 1

T-shape → 2

Trigonal pyramidal → zero

Square pyramidal → 3

$(P) \rightarrow(1) ;(Q) \rightarrow(2) ;(R) \rightarrow(5) ;(S) \rightarrow(3)$

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 $\mathrm{CHCl_3}$.

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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.} $

Option A

A charge–charge interaction falls off as

$U_{qq}\propto\frac1r$

while a charge–dipole interaction goes as

$U_{q\mu}\propto\frac1{r^2}\,. $

So the charge–dipole potential actually decays faster.

⇒ A is false.

Option B

A fixed dipole–dipole energy is

$U_{\mu\mu}\propto\frac1{r^3}\,, $

but once both molecules tumble freely, the thermal (Keesom) average gives

$\langle U\rangle_{\rm Keesom}\propto -\frac{\mu^4}{kT\,(4\pi\varepsilon_0)^2\,r^6}\,. $

That falls off as $1/r^6$, not $1/r^3$.

⇒ B is false.

Option C

The permanent–induced dipole (Debye) interaction is

$U_{\rm Debye}=-\tfrac12\,\alpha\,E^2\propto -\frac{\alpha\,\mu^2}{(4\pi\varepsilon_0)^2\,r^6}\,, $

and when you average over random orientations (assuming the interaction is weak compared with $kT$), no $T$ appears in the result.

⇒ C is true.

Option D

Even totally nonpolar species experience instantaneous‐induced (London) dispersion forces that attract as

$U_{\rm disp}\propto -\frac1{r^6}\,. $

⇒ D is true.

Answer: C and D.

Molecular orbital (MO) energy levels for homonuclear diatomic molecules: For Molecular orbital theory, all the valence atomic orbitals are included. The total number of atomic orbitals must always equal the total number of molecular orbitals. Bonding molecular orbitals are more stable than the atomic orbitals and increase the electron density between the nuclei Antibonding molecular orbitals are. less stable than the atomic orbitals and decrease the electron density between the nuclei. (A) Bond older can be calculated using the formula:
Bond order $=\frac{1}{2}\left(N_b-N_a\right)$
$N_b$ - Number of bonding electrons
$\mathrm{N_a}^{\text { }}$ - Number of antibonding electrons
For $\mathrm{Ne}_2$, total number of electrons $=20 \quad\left\{\begin{array}{l}\text { Number of electrons } \\ \text { for } \mathrm{Ne}=10\end{array}\right.$

The total electrons are filled in the molecular orbitals as,${\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}^ * \sigma _{2{p_z^2}}^ *$$ \begin{aligned} N_b=10, & N_a=10 \\ \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-10) \\ & =\frac{1}{2}(0) \\ & =0 \end{aligned} $Bond order of $\mathrm{Ne}_2$ is zero. It is correct. B) $H O M O$ of $F_2$ The number of electrons in $F$ is 9 . So, the total number of electrons in $F_2$ molecule is $18(9+9)$ The 18 electrons are filled in the molecular orbitals as,${\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}^ * $
Homo is the highest occupied molecular orbital. For $F_2$, the last electron is occupied in the $\pi^*$ orbital. So, the highest occupied molecular orbital of $F_2$ is $\pi$ - type, not $\sigma$-type. It is not correct. c) Bond energy is directly proportional to bond order. The strength of a chemical bond is directly proportional to the amount of energy required to break it. To compare the bond energy of $\mathrm{O}_2^{+}$and $\mathrm{O}_2$, calculate the bond older. The molecule with higher bond order has higher bond energy$\mathrm{O}_2$ Number of elections of oxygen $(0)=8$ number of elections of $\mathrm{O}_2$ molecule $=16$ The 16 elections are filled in the molecular orbitals as: ${\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}^ * $

$ \begin{aligned} & N_b=10, N_a=6 \end{aligned} $ $ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-6) \\ & =\frac{1}{2} \cdot(4) \\ & =2 \end{aligned} $ $\mathrm{O}_2^{+}$ Total number of elections for $0_2$ molecule $=16$ For $\mathrm{O}_2^{+}$, one electron is removed from $\mathrm{O}_2$. So, the total number of electrons $=16-1=15$ The 15 electrons are filled in the molecular orbitals as, ${\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}^ * $$ \begin{aligned} N_b=10, & N_a=5 \\ \text { Bond order } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-5) \\ & =\frac{1}{2}(5) \\ & =2.5 \end{aligned} $Bond order of $\mathrm{O}_2^{+}$is higher than the bond order of $\mathrm{O}_2$. Bond order and bond energy are directly proportional so, the higher the bond oder, the higher the bond energy. Therefore, $\mathrm{O}_2{ }^{+}$molecule which has higher bond order and also has higher bond energy.So, bond energy of $\mathrm{O}_2^{+}$is higher than the bond energy of $\mathrm{O}_2$.$ \text { It is not correct: } $ D) Bond length of $Li_2$ and $B_2$ Bond length and bond older are inversely proportional to each other so; higher the bond order, shorter the bond length. To get the bond length of $L_2$ and $B_2$, first calculate the bond order of $Li_2$ and $B_2$. $\mathrm{Li}_2$ Number of electrons of $\mathrm{Li}=3$ : total number of electrons in $\mathrm{Li}_2$ molecule $=6$ The total electrons are filled in the molecular orbitals as Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$$ N_b=4, N_a=2 $$ \begin{aligned} \text { Bond loeder } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(4-2) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $ $B_2$ Number of electrons of Balon $(B)=5$ Total number of electrons of $B_2$ molecule $=10$ The total electrons are filled in the molecular orbitals as ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$ N_b=6 \quad N_a=4 $$ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b+N_a\right) \\ & =\frac{1}{2}(6-4) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $Bond order is equal for $Li_{2}$ and $B_2$ so, atomic size is consider Comparing the size of lithium and boron, lithium is a larger atom than boron. Bond length generally increases with the size of atom so, bond length of $\mathrm{Li}_2$ is larger than the bond length of $\mathrm{B_2}$ Size of $Li>B$ So, Bond length of $Li_2>B_2$ This statement is correct: Answer: The incorrect statements are (B) and (C). (B) and (C).

Since, $Z$ is an alkali metal with atomic number $a+2$

$Y$ has atomic number $a+1$, so it must be a noble gas.$X$ has atomic number a, so it must be a halogen.

$X$ is a halogen which is non-metal. Z is an alkali metal, which is a metal. Compound formed between metal and non-metal is ionic.

Set of II and III i.e.

$\mathrm{HgCl}_2, \mathrm{SO}_3, \mathrm{SF}_6$ and $\mathrm{BeCl}_2, \mathrm{BF}_3, \mathrm{PCl}_5$ has no lone pair of electrons present on central atom

The structure of the given molecules with their bond angles are given below.

Hence, the increasing order of bond angle is

$ \underset{(\approx 104.5)}{\mathrm{H}_2 \mathrm{O}}<\underset{(\approx 107)}{\mathrm{NH}_3} \underset{(\approx 116.8)}{<\mathrm{O}_3} \underset{\substack{\approx 1199 \\ \approx \mathrm{SO}_2}}{\stackrel{\mathrm{SO}_2}{\sim}} $

The boiling point order is primarily determined by hydrogen bonding strength and network. $\mathbf{H}_2 \mathbf{O}$ forms the most extensive hydrogen bond network due to two lone pairs.

HF possesses very strong individual hydrogen bonds, but a less extensive network.

$\mathbf{N H}_{\mathbf{3}}$ exhibits weaker hydrogen bonding compared to $\mathrm{H}_2 \mathrm{O}$ and HF .

$\mathbf{P H}_3$ lacks hydrogen bonding, having only weaker dipole-dipole forces.

Hence, the correct order is

$ \mathrm{H}_2 \mathrm{O}>\mathrm{HF}>\mathrm{NH}_3>\mathrm{PH}_3 $

The structure of $\mathrm{S}_8, \mathrm{P}_4, \mathrm{~S}_6$ and $\mathrm{O}_3$ are given below.

So, the increasing order of their bond angles is $\mathrm{P}_4<\mathrm{S}_6<\mathrm{S}_8<\mathrm{O}_3$.

Step 1: Find bond orders using the formula

Bond order = $\frac{1}{2}$(number of bonding electrons - number of antibonding electrons)

Bond order of $\mathrm{O}_2^{-}$ (17 electrons):

$ =\frac{1}{2}(10-7)=1.5 $

Bond order of $\mathrm{O}_2^{2-}$ (18 electrons):

$ =\frac{1}{2}(10-8)=1.0 $

Bond order of $\mathrm{O}_2^{2+}$ (14 electrons):

$ =\frac{1}{2}(10-4)=3.0 $

Step 2: Add the bond orders for $\mathrm{O}_2^{-}$ and $\mathrm{O}_2^{2-}$

The question says $x$ is their sum.

$ x=1.5+1.0=2.5 $

Step 3: Express the bond order of $\mathrm{O}_2^{2+}$ in terms of $x$

We already found bond order of $\mathrm{O}_2^{2+}$ is 3.0.

We want to write 3.0 as a multiple of $x$:

$ \begin{aligned} 3.0 & =k \cdot x \\ 3.0 & =k \cdot 2.5 \\ k & =\frac{3.0}{2.5}=1.20 \end{aligned} $

So, bond order of $\mathrm{O}_2^{2+}$ is $1.20x$.

The structure of $\mathrm{CH}_2(\mathrm{CN})_2$ is as follows

So, number of $\sigma$ bonds $=6$

Number of $\pi$ bonds $=4$

So, the ratio of $\sigma: \pi$ bonds $=3: 2$

Bond order

$ =\frac{\text {Bonding electrons - Non-bonding electrons }}{2} $

B.O. of $\mathrm{O}_2=\frac{1}{2}[10-6]=2$

$ \begin{aligned} & \mathrm{O}_2^{+}=\frac{1}{2}[10-5]=2.5 \\ & \mathrm{O}_2^{-}=\frac{1}{2}(10-7)=1.5 \\ & \mathrm{O}_2^{2+}=\frac{1}{2}(10-4)=3 \end{aligned} $

Sum of B.O. of $\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}$and $\mathrm{O}_2^{2+}$

$ =2+2.5+1.5+3=9 $

Statement I is not correct but statement II is correct. The correct form of statement I is,

Both $\mathrm{SF}_6$ and $\mathrm{BrF}_5$ has same hybridisation i.e., $s p^3 d^2$.

Among the given molecules, total five molecules are polar in nature

$ \mathrm{NH}_3, \mathrm{NF}_3, \mathrm{H}_2 \mathrm{~S}, \mathrm{CHCl}_3, \mathrm{H}_2 \mathrm{O} $

Total three, are non-polar molecules

$ \mathrm{BF}_3, \mathrm{CO}_2, \mathrm{CH}_4 $

$ \begin{aligned} &\text { }\\ &\begin{aligned} \mathrm{ClF}_3 & =T \text {-shape } \\ & =s p^3 d \text { hybridisation } \\ \mathrm{XeF}_4 & =\text { square planar }=s p^3 d^2 \text { hybridisation } \\ \mathrm{SF}_4 & =s e e \text {-saw }=s p^3 d \text { hybridisation } \\ \mathrm{BrF}_5 & =\text { square pyramidal }=s p^3 d^2 \end{aligned} \end{aligned} $