Chemical Bonding & Molecular Structure

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.

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

The bond angle of $\mathrm{SO}_2$ is more than $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ because $\mathrm{SO}_2$ is $s p^2$-hybridised while $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ are $s p^3$-hybridised. Hence, in $\mathrm{SO}_2$ bond angle will be $120^{\circ}$ while in $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ it will be approx. $109.5^{\circ}$.

Out of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3, \mathrm{NH}_3$ has more bond angle as in $\mathrm{NH}_3$ there is one lone pair and in $\mathrm{H}_2 \mathrm{O}$ there are two lone pair of electrons. Hence, repulsion is less in $\mathrm{NH}_3$ and bond angle is more.

Thus, the order of bond angles will be

$ \mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{SO}_2 $

Bond order $=\frac{1}{2}$ [No. of bonding electrons - No. of anti-bonding electrons]

$ =\frac{1}{2}\left[N_b-N_a\right] $

Molecular orbital configuration

$ \begin{aligned} & C_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2 \\ & \text { Bond order }=\frac{1}{2}[8-4]=2 \end{aligned} $

$ \begin{aligned} &\begin{array}{rl} \mathrm{N}_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma * & 2 s^2, \pi 2 p_x^2 \\ & \approx \pi 2 p_y^2, \sigma 2 p_z^2 \end{array}\\ &\begin{aligned} & \text { Bond order }=\frac{1}{2}[10-4]=3 \\ & \begin{array}{l} \mathrm{He}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2=\frac{1}{2}[2-2]=0 \\ \mathrm{~B}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1, \pi 2 p_y^1 \\ \quad=\frac{1}{2}[6-4]=1 \end{array} \end{aligned}\\ &\text { Thus, order of bond order will be }\\ &\mathrm{He}_2<\mathrm{B}_2<\mathrm{C}_2<\mathrm{N}_2 \end{aligned} $

Molecular orbital configuration of $\mathrm{C}_2$ is

$ \sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2, \pi^* 2 p_y^2 $

So, $\mathrm{C}_2$ molecule contains only $\pi$-bonds according to MOT.

$\mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}$

$\mathrm{NH}_3$ is $s p^3$-hybridised and it has 3 -bond pairs and 1-lone pair.

$\mathrm{NH}_4^{+}$is also $s p^3$-hybridised, it has 4 -bond pair and no lone pair.

$ \begin{aligned} \text { Lone pair } & =6 \\ \text { Bond pair } & =3 \\ & 6: 3 \text { or }(2: 1) \end{aligned} $