Chemical Bonding & Molecular Structure

$ \mathrm{BrF}_5 \longrightarrow s p^2 d^2 $

The decomposition of ammonium nitrate is,

$ \mathrm{NH}_4 \mathrm{NO}_3 \longrightarrow \mathrm{~N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} $

$\mathrm{N}_2 \mathrm{O}$ is colourless neutral gas.

The complete bond cleavage reaction is,

$ \begin{aligned} &\text { (Negative charge is more on } \mathrm{CH}_3 \mathrm{CH}_2 \text {.) }\\ &\therefore A=\mathrm{CH}_3 \mathrm{CH}_2^{+}, B=\mathrm{I}^{-}, \mathrm{C}=\mathrm{CH}_3 \mathrm{CH}_2^{-}, D=\mathrm{Cu}^{+} \end{aligned} $

Basicity of an acid is the number of replaceable hydrogen in the acid.

For the given reaction below, the hybridisaion of central atom does not change.

$ \begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & s p^3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s p^3 \\ & \underset{s p^2}{\mathrm{BF}_3}+\mathrm{F} \longrightarrow \underset{s p^3}{\mathrm{BF}_4^{-}} \\ & \mathrm{PCl}_3+\mathrm{Cl}_2 \longrightarrow \mathrm{PCl}_5 \\ & s p^3 \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s p^3 d \\ & \mathrm{ClF}_3+\mathrm{F}_2 \longrightarrow \mathrm{ClF}_5 \\ & s p^3 d \quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, s p^3 d^2 \end{aligned} $

Explanation of Bond Order Calculation

Bond Order (B.O.) is calculated using the formula:

$ \text{Bond Order} = \frac{1}{2} (\text{number of electrons in bonding orbitals} - \text{number of electrons in antibonding orbitals}) $

Let's analyze each set:

Set (a): $\mathrm{B}_2, \mathrm{CN}^-, \mathrm{O}_2^{2-}$

$\mathrm{B}_2$:

$ \text{Number of bonding electrons} = 4, \text{Number of antibonding electrons} = 2 $

$ \text{B.O.} = \frac{1}{2}(4-2) = 1 $

$\mathrm{CN}^-$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

$\mathrm{O}_2^{2-}$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $

$ \text{B.O.} = \frac{1}{2}(10-8) = 1 $

Total Bond Order for Set (a):

$ 1 + 3 + 1 = 5 $

Set (b): $\mathrm{O}_2, \mathrm{F}_2, \mathrm{O}_2^{2+}$

$\mathrm{O}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $

$ \text{B.O.} = \frac{1}{2}(10-6) = 2 $

$\mathrm{F}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $

$ \text{B.O.} = \frac{1}{2}(10-8) = 1 $

$\mathrm{O}_2^{2+}$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

Total Bond Order for Set (b):

$ 2 + 1 + 3 = 6 $

Set (c): $\mathrm{O}_2^-, \mathrm{N}_2, \mathrm{O}_2^+$

$\mathrm{O}_2^-$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 7 $

$ \text{B.O.} = \frac{1}{2}(10-7) = 1.5 $

$\mathrm{N}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

$\mathrm{O}_2^+$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 5 $

$ \text{B.O.} = \frac{1}{2}(10-5) = 2.5 $

Total Bond Order for Set (c):

$ 1.5 + 3 + 2.5 = 7 $

Set (d): $\mathrm{C}_2, \mathrm{O}_2, \mathrm{He}_2^{2+}$

$\mathrm{C}_2$:

$ \text{Number of bonding electrons} = 8, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(8-4) = 2 $

$\mathrm{O}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $

$ \text{B.O.} = \frac{1}{2}(10-6) = 2 $

$\mathrm{He}_2^{2+}$:

$ \text{Number of bonding electrons} = 2, \text{Number of antibonding electrons} = 0 $

$ \text{B.O.} = \frac{1}{2}(2-0) = 1 $

Total Bond Order for Set (d):

$ 2 + 2 + 1 = 5 $

Conclusion

Set (c) has the maximum total bond order sum of $7$.

Diborane - $\mathrm{B}_2 \mathrm{H}_6$

Each boron atom is $s p^3$ hybridised.

$ \theta_1=97^{\circ} \text { and } \theta_2=120^{\circ} $

$\mathrm{H}_t$ is terminal hydrogen atom and two boron atoms lie in the same plane.

The correct order of covalent bond character is:

$ \mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4 $

According to Fajan's rules, the covalent character of a bond increases with higher charge on the cation. In this sequence, as the charge on the cation increases from LiCl to $\mathrm{CCl}_4$, the covalent character of the bonds also increases.

Dipole moment, mathematically given by, $\mu=\sqrt{n(n+2)}$ BM only occur when there is unpaired electron present in the ion/molecules.

For $\mathrm{CO}_2$ and $\mathrm{BCl}_3$ there is no unpaired electrons.

In $\mathrm{BCl}_3$ and $\mathrm{NF}_3$ pair only $\mathrm{NF}_3$ has dipole' moment due to present of unpaired electron. ( 0.24 D )

In case of $\mathrm{SO}_2$ and $\mathrm{NF}_3$ both have unpaired electrons and hence, both have dipole moments, 1.60 D and 0.24 D respectively.

Statement given in (i), (iii) and (iv) are correct, while (ii) is incorrect. It's correct from is, Covalent character is the measure of the tendency of an atom to attract the shared pair of electron towards itself. LiF has more covalent character than KF because LiF has more electronegativity difference. Electronegativity difference is directly proportional to covalent character.

The explanation involves analyzing the order of various properties among different molecules, and identifying which sequences are correct.

For sets (ii), (iii), and (iv), the sequences provided are correct:

(ii) Dipole Moment: The order $ \mathrm{H}_2\mathrm{O}>\mathrm{NH}_3>\mathrm{H}_2\mathrm{S} $ is correct due to the polarity and shapes of the molecules, with water having the highest dipole moment.

(iii) Bond Enthalpy: The sequence $ \mathrm{N}_2>\mathrm{O}_2>\mathrm{H}_2 $ is accurate because nitrogen gas has a triple bond, making it much stronger compared to the double bond in oxygen and the single bond in hydrogen.

(iv) Bond Order: The order $ \mathrm{NO}^{+}>\mathrm{O}_2>\mathrm{O}_2^{2-} $ correctly reflects the number of bonds between the atoms, with $ \mathrm{NO}^{+} $ having the highest bond order.

However, for set (i), the provided order for bond angles is incorrect. The correct order should be:

(i) Bond Angle: $ \mathrm{H}_2\mathrm{O}<\mathrm{NH}_3<\mathrm{SO}_2 $, with numerical values $ 104.5^\circ<107^\circ<119^\circ $. This is based on the interplay of lone pair repulsion and the geometric configurations of the molecules.

In the reaction $ \mathrm{XeF}_6 + \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeOF}_4 + 2 \mathrm{HF} $, the hybridization of the xenon (Xe) atom changes.

For $\mathrm{XeF}_6$:

Calculating hybridization using the formula:

$ \text{Hybridization} = \frac{1}{2}[V + M - C + A] $

where:

$ V $ is the number of valence electrons of the central atom (Xe), which is 8.

$ M $ is the number of monovalent atoms attached, here 6 fluorine atoms.

$ C $ is the charge on the cation.

$ A $ is the charge on the anion.

$ = \frac{1}{2}[8 + 6 - 0 + 0] = 7 $

The hybridization for $\mathrm{XeF}_6$ is $ sp^3d^3 $.

For $\mathrm{XeOF}_4$:

The hybridization number is calculated to be 6, resulting in a hybridization of $ sp^3d^2 $.

Thus, the xenon atom goes through a change in hybridization from $ sp^3d^3 $ in $\mathrm{XeF}_6$ to $ sp^3d^2 $ in $\mathrm{XeOF}_4$.

Isostructural species have the same shape or structure. Here's how to determine the hybridization and shape of each species:

For $\mathrm{XeO}_3$:

Hybridization formula: $\frac{1}{2}[V + M - C + A]$

$V =$ Valence electrons of the central atom (Xe), which is 8

$M =$ Number of monovalent atoms attached, which is 0

$C =$ Positive charge, which is 0

$A =$ Negative charge, which is 0

$ = \frac{1}{2}[8 + 0 - 0 + 0] = 4 = \text{sp}^3 $

Xe has one lone pair, resulting in a pyramidal shape.

For $\mathrm{H}_3\mathrm{O}^+$:

Apply the same hybridization formula:

$ = \frac{1}{2}[6 + 3 - 1 + 0] = 4 = \text{sp}^3 $

The structure is pyramidal.

For $\mathrm{SO}_3$:

$ = \frac{1}{2}[6 + 0 - 0 + 0] = 3 = \text{sp}^2 $

Results in a trigonal planar shape.

For $\mathrm{CO}_3^{2-}$:

$ = \frac{1}{2}[4 + 0 - 0 + 2] = 3 = \text{sp}^2 $

Also results in a trigonal planar shape.

Conclusion:

$\mathrm{XeO}_3$ and $\mathrm{H}_3\mathrm{O}^+$ both have a pyramidal shape.

$\mathrm{SO}_3$ and $\mathrm{CO}_3^{2-}$ both have a trigonal planar shape.

These alignments help identify which species are isostructural pairs based on their shapes.

$ \begin{aligned} &\text { The correct order of bond angle is }\\ &\mathrm{HgCl}_2>\mathrm{SO}_3>\mathrm{SiCl}_4>\mathrm{NH}_3 \end{aligned} $

The formula of formal charge is

$ \mathrm{FC}=V-N-\frac{B}{2} $

$V=$ Number of valence electrons of neutral atom.

$N=$ Number of non-bonding electrons on this atom.

$B=$ Number of electrons shared in bond with other atom.

For $1 \mathrm{st}, \mathrm{FC}=6-4-\frac{1}{2} \times 4=0$

For 2 nd, FC $=5-2-\frac{1}{2} \times 6=0$

For 3 rd, FC $=6-6-\frac{1}{2} \times 2=-1$

Hence, the formal charges are $0,0,-1$.

$\mathrm{Al}_2 \mathrm{Cl}_6$ (Dimer) is a $s p^3$ hybridised and tetrahedral at aluminum.

$ b_1=118^{\circ}, b_2=79^{\circ}, b_3=101^{\circ} $

Hence, $\mathrm{XeF}_6$ and $\mathrm{IF}_7$ have same hybridisation of their central atoms.

The correct order is $\mathrm{SO}_3<\mathrm{SnCl}_2<\mathrm{ClF}_3<\mathrm{XeF}_2$.

Bond length/distance is defined as average distance between niaclei of two bonded atom in a molecules.

Bond length $=r_1+r_2$

Element $(x)=143$ pm means it is close to bond length of fluorine $\left(\mathrm{F}_2\right)$.

Atomic number $=9$

Element $(Y)=110 \mathrm{pm}$ is equal to bond length of nitrogen $\left(N_2\right)$.

Atomic number $=7$

Element $(Z)=121$ pm is equal to bond length of axygen $\left(\mathrm{O}_2\right)$.

Atomic number $=8$

A formal charge is the charge assigned to an atom in a molecule in the covalent view of bonding assuming that electrons in all chemical bonds are shared equally between atoms regardless of relative electronegativity.

It is the difference between an atoms number of valence electrons in its neutral state and the number allocated to that atom in a Lewis structure.

$ Q_f=v-(V+B) $

where, $V$ is sumber of valence electron in free atom.

$U=$ Number of unshared electron in atom.

$B=$ Number of bond around the atom.

The correct match is

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

Bond enthalpy also known as bond dissociation energy describes the amount of energy stored in a bond between atoms in a molecules.

Specifically it is the energy required to be added for cleavage of bond in a gas phase.

$ \begin{aligned} & \mathrm{Si}-\mathrm{Si} \longrightarrow 297 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}-\mathrm{C} \longrightarrow 348 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{Sn}-\mathrm{Sn} \longrightarrow 248 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{Ge}-\mathrm{Ge} \longrightarrow 260 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Down the group, bond enthalpy decreases.

To find the bond dissociation energy of $ X_2 $, let's assume the bond dissociation energy of $ X_2 $ is $ a \, \text{kJ/mol} $.

Therefore, $ \text{BE}(X_2) = a \, \text{kJ/mol} $.

Given that the bond dissociation energy ratios are $ 1:0.5:1 $, then:

$ \text{BE}(Y_2) = 0.5a \, \text{kJ/mol} $

$ \text{BE}(XY) = a \, \text{kJ/mol} $.

We know that the formation reaction of $ XY $ is:

$ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightarrow XY, \quad \Delta H = -200 \, \text{kJ/mol} $

Using the enthalpy change equation:

$ \Delta_r H = \text{BE}(\text{Reactants}) - \text{BE}(\text{Products}) $

Substituting the bond dissociation energies into the equation gives:

$ \Delta_r H = \frac{1}{2} \text{BE}(X_2) + \frac{1}{2} \text{BE}(Y_2) - \text{BE}(XY) $

Plugging in the known values:

$ -200 = \frac{a}{2} + \frac{0.5a}{2} - a $

Simplifying:

$ -200 = \frac{a}{2} + \frac{0.25a}{2} - a = \frac{a}{2} + \frac{a}{4} - a $

Combine the terms:

$ -200 = \frac{a}{2} + \frac{0.25a}{2} - \frac{2a}{2} $

$ -200 = \frac{0.5a + 0.25a - 2a}{2} = \frac{-0.75a}{2} $

Solving for $ a $:

$ -200 = \frac{-0.75a}{2} $

$ -200 \times 2 = -0.75a $

$ -400 = -0.75a $

Dividing both sides by $-0.75$:

$ a = \frac{400}{0.75} = 800 \, \text{kJ/mol} $

Thus, the bond dissociation energy of $ X_2 $ is $ 800 \, \text{kJ/mol} $.

$ I_{3}^{-} $is linear with central iodine atom undergoing $ s p^{3} d $-hybridisation and three lone pairs occupying equatorial position.

Thus, this structure include $ d $-orbital contribution to the hybridisation scheme.

The structure is,

$ \mathrm{PCl}_{3} $ structure is $ s p^{3} $-hybridised with one lone pair. $ \mathrm{NO}_{3}^{-} $is $ s p^{2} $-hybridised and $ \mathrm{H}_{2} \mathrm{Se} $ is $ s p^{3} $ - hybridised with two lone pairs of Se atom.

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

A : No. of lone pairs on oxygen = 2

B :

($\theta$ < Bond angle in H$_2$O (104.5$^\circ$)

D : molecule is bent "v" shaped

(A) IF$_7$ $-$ 0 lone pairs

(B) ICI$_4^ - $ $-$ 2 lone pairs

(C) XeF$_6$ $-$ 1 lone pair

(D) XeF$_2$ $-$ 3 lone pairs