Question

Identify the pair in which the specified bond length of first is greater than second. $PCl_3F_2$, $PF_3Cl_2$: $BL_{P-Cleq}$ $SO_2Cl_2$, $SO_2F_2$: $BL_{S=O}$ $BF_3$, $BCl_3$: $BL_{B-X}$ X = F/Cl $NO_3^-$, $NO_2$: $BL_{N-O}$ $O_3$, $O_2$: $BL_{O-O}$ CO, $CO_2$: $BL_{C-O}$

10. How many of the following species have no d-orbitals involved in their hybridisation? $XeF_2$, $SnF_2$, $PbCl_2$, $SF_4$, $CCl_4$, $BF_3$, $SO_2Cl_2$, $XeO_2F_2$, $XeOF_4$, $POCl_3$

Answer 1

Question 9: The pairs in which the specified bond length of the first is greater than the second are:

1. $PCl_3F_2$, $PF_3Cl_2$: $BL_{P-Cleq}$
2. $SO_2Cl_2$, $SO_2F_2$: $BL_{S=O}$
3. $NO_3^-$, $NO_2$: $BL_{N-O}$
4. $O_3$, $O_2$: $BL_{O-O}$

Question 10: 6

Answer 2

Solution for Question 9:

The question asks to identify the pairs in which the specified bond length of the first species is greater than that of the second species. We will analyze each pair:

1. $PCl_3F_2$, $PF_3Cl_2$: $BL_{P-Cleq}$

   • Both molecules adopt a trigonal bipyramidal (TBP) geometry. According to Bent's Rule, more electronegative substituents prefer positions with less s-character (axial positions in TBP), and less electronegative substituents prefer positions with more s-character (equatorial positions).
   • In $PCl_3F_2$: The two more electronegative F atoms occupy axial positions. The three Cl atoms occupy equatorial positions. All three equatorial bonds are P-Cl. The s-character of the equatorial orbitals is equally distributed among these three P-Cl bonds.
   • In $PF_3Cl_2$: The two F atoms occupy axial positions. The remaining F atom and the two Cl atoms occupy equatorial positions. So, the equatorial plane contains one P-F bond and two P-Cl bonds. Since F is more electronegative than Cl, the P-F bond will draw less s-character (more p-character) to itself compared to a P-Cl bond. Consequently, the two P-Cl equatorial bonds will have a larger share of the remaining s-character.
   • Higher s-character in a bond leads to a shorter and stronger bond. Therefore, the P-Cl equatorial bonds in $PF_3Cl_2$ will be shorter than those in $PCl_3F_2$.
   • Thus, $BL_{P-Cleq}$ in $PCl_3F_2$ > $BL_{P-Cleq}$ in $PF_3Cl_2$. This statement is TRUE.

2. $SO_2Cl_2$, $SO_2F_2$: $BL_{S=O}$

   • Both molecules have a tetrahedral geometry around the sulfur atom.
   • Fluorine is significantly more electronegative than chlorine. The more electronegative F atoms in $SO_2F_2$ withdraw electron density from the central sulfur atom more effectively than the Cl atoms in $SO_2Cl_2$.
   • This electron withdrawal makes the sulfur atom more electron-deficient. To compensate, sulfur strengthens its double bond character with oxygen (e.g., by increasing the p-character in the S-X bonds, leaving more s-character for the S=O bonds, or by enhancing back-donation from oxygen p-orbitals to sulfur d-orbitals, if d-orbitals are involved).
   • Alternatively, according to Bent's Rule, the more electronegative F atoms will prefer hybrid orbitals with more p-character, leaving more s-character for the S=O bonds. More s-character leads to shorter bonds.
   • Therefore, the $S=O$ bond length in $SO_2F_2$ is shorter than in $SO_2Cl_2$.
   • Thus, $BL_{S=O}$ in $SO_2Cl_2$ > $BL_{S=O}$ in $SO_2F_2$. This statement is TRUE.

3. $BF_3$, $BCl_3$: $BL_{B-X}$ (X = F/Cl)

   • Both molecules are trigonal planar. We are comparing the B-F bond length in $BF_3$ with the B-Cl bond length in $BCl_3$.
   • Fluorine is smaller and more electronegative than chlorine.
   • In $BF_3$, there is significant pπ-pπ backbonding from the lone pairs on F (2p orbital) to the empty p-orbital on B (2p orbital). This backbonding gives the B-F bond partial double bond character, leading to a shorter bond length.
   • In $BCl_3$, the pπ-pπ backbonding from Cl (3p orbital) to B (2p orbital) is much less effective due to the larger size of the Cl atom and poorer orbital overlap.
   • Consequently, the B-F bond in $BF_3$ is significantly shorter than the B-Cl bond in $BCl_3$. (Experimental: B-F ~130 pm, B-Cl ~173 pm).
   • Thus, $BL_{B-X}$ of $BF_3$ < $BL_{B-X}$ of $BCl_3$. The statement "first > second" is FALSE.

4. $NO_3^-$, $NO_2$: $BL_{N-O}$

   • $NO_3^-$ (Nitrate ion): It has three equivalent N-O bonds due to resonance. The resonance structures show one N=O double bond and two N-O single bonds. The average bond order is (2+1+1)/3 = 4/3 ≈ 1.33.
   • $NO_2$ (Nitrogen dioxide): It has two equivalent N-O bonds due to resonance. The resonance structures show one N=O double bond and one N-O single bond, with an odd electron. The average bond order is (2+1)/2 = 1.5.
   • A higher bond order corresponds to a shorter bond length. Since the bond order of N-O in $NO_2$ (1.5) is higher than in $NO_3^-$ (1.33), the N-O bond in $NO_2$ is shorter.
   • Thus, $BL_{N-O}$ in $NO_3^-$ > $BL_{N-O}$ in $NO_2$. This statement is TRUE.

5. $O_3$, $O_2$: $BL_{O-O}$

   • $O_3$ (Ozone): It has two equivalent O-O bonds due to resonance. The resonance structures show one O=O double bond and one O-O single bond. The average bond order is (2+1)/2 = 1.5.
   • $O_2$ (Oxygen molecule): It has a double bond (O=O). The bond order is 2.
   • Since $O_2$ has a higher bond order (2) than $O_3$ (1.5), the O-O bond in $O_2$ is shorter.
   • Thus, $BL_{O-O}$ in $O_3$ > $BL_{O-O}$ in $O_2$. This statement is TRUE.

6. CO, $CO_2$: $BL_{C-O}$

   • CO (Carbon Monoxide): It has a triple bond (C≡O). The bond order is 3.
   • $CO_2$ (Carbon Dioxide): It has two C=O double bonds. The bond order is 2.
   • Since CO has a higher bond order (3) than $CO_2$ (2), the C-O bond in CO is shorter.
   • Thus, $BL_{C-O}$ in CO < $BL_{C-O}$ in $CO_2$. The statement "first > second" is FALSE.

Pairs where the specified bond length of the first is greater than the second:

1. $PCl_3F_2$, $PF_3Cl_2$: $BL_{P-Cleq}$
2. $SO_2Cl_2$, $SO_2F_2$: $BL_{S=O}$
3. $NO_3^-$, $NO_2$: $BL_{N-O}$
4. $O_3$, $O_2$: $BL_{O-O}$

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Solution for Question 10:

The question asks to identify how many of the given species have no d-orbitals involved in their hybridization. This means we need to find species where the central atom uses only s and p orbitals for hybridization (i.e., sp, sp2, or sp3 hybridization). D-orbital involvement implies sp3d, sp3d2, sp3d3 hybridization.

To determine hybridization, we can use the steric number (SN = number of lone pairs + number of sigma bonds).

1. $XeF_2$:

   • Central atom: Xe. Valence electrons = 8.
   • 2 F atoms form 2 sigma bonds.
   • Remaining electrons = 8 - (2 * 1) = 6 electrons = 3 lone pairs.
   • SN = 2 (sigma bonds) + 3 (lone pairs) = 5.
   • Hybridization: $sp^3d$. D-orbitals involved.

2. $SnF_2$:

   • Central atom: Sn. Group 14, Valence electrons = 4.
   • 2 F atoms form 2 sigma bonds.
   • Remaining electrons = 4 - (2 * 1) = 2 electrons = 1 lone pair.
   • SN = 2 (sigma bonds) + 1 (lone pair) = 3.
   • Hybridization: $sp^2$. No d-orbitals involved.

3. $PbCl_2$:

   • Central atom: Pb. Group 14, Valence electrons = 4.
   • 2 Cl atoms form 2 sigma bonds.
   • Remaining electrons = 4 - (2 * 1) = 2 electrons = 1 lone pair.
   • SN = 2 (sigma bonds) + 1 (lone pair) = 3.
   • Hybridization: $sp^2$. No d-orbitals involved.

4. $SF_4$:

   • Central atom: S. Valence electrons = 6.
   • 4 F atoms form 4 sigma bonds.
   • Remaining electrons = 6 - (4 * 1) = 2 electrons = 1 lone pair.
   • SN = 4 (sigma bonds) + 1 (lone pair) = 5.
   • Hybridization: $sp^3d$. D-orbitals involved.

5. $CCl_4$:

   • Central atom: C. Valence electrons = 4.
   • 4 Cl atoms form 4 sigma bonds.
   • Remaining electrons = 4 - (4 * 1) = 0 electrons = 0 lone pairs.
   • SN = 4 (sigma bonds) + 0 (lone pairs) = 4.
   • Hybridization: $sp^3$. No d-orbitals involved.

6. $BF_3$:

   • Central atom: B. Valence electrons = 3.
   • 3 F atoms form 3 sigma bonds.
   • Remaining electrons = 3 - (3 * 1) = 0 electrons = 0 lone pairs.
   • SN = 3 (sigma bonds) + 0 (lone pairs) = 3.
   • Hybridization: $sp^2$. No d-orbitals involved.

7. $SO_2Cl_2$:

   • Central atom: S. Valence electrons = 6.
   • 2 O atoms form 2 double bonds (count as 2 sigma bonds for SN).
   • 2 Cl atoms form 2 sigma bonds.
   • Total sigma bonds = 2 (from O) + 2 (from Cl) = 4.
   • Remaining electrons = 6 - (2 * 2) - (2 * 1) = 0 electrons = 0 lone pairs.
   • SN = 4 (sigma bonds) + 0 (lone pairs) = 4.
   • Hybridization: $sp^3$. No d-orbitals involved.

8. $XeO_2F_2$:

   • Central atom: Xe. Valence electrons = 8.
   • 2 O atoms form 2 double bonds (count as 2 sigma bonds).
   • 2 F atoms form 2 sigma bonds.
   • Total sigma bonds = 2 (from O) + 2 (from F) = 4.
   • Remaining electrons = 8 - (2 * 2) - (2 * 1) = 2 electrons = 1 lone pair.
   • SN = 4 (sigma bonds) + 1 (lone pair) = 5.
   • Hybridization: $sp^3d$. D-orbitals involved.

9. $XeOF_4$:

   • Central atom: Xe. Valence electrons = 8.
   • 1 O atom forms 1 double bond (count as 1 sigma bond).
   • 4 F atoms form 4 sigma bonds.
   • Total sigma bonds = 1 (from O) + 4 (from F) = 5.
   • Remaining electrons = 8 - (1 * 2) - (4 * 1) = 2 electrons = 1 lone pair.
   • SN = 5 (sigma bonds) + 1 (lone pair) = 6.
   • Hybridization: $sp^3d^2$. D-orbitals involved.

10. $POCl_3$:

    • Central atom: P. Valence electrons = 5.
    • 1 O atom forms 1 double bond (count as 1 sigma bond).
    • 3 Cl atoms form 3 sigma bonds.
    • Total sigma bonds = 1 (from O) + 3 (from Cl) = 4.
    • Remaining electrons = 5 - (1 * 2) - (3 * 1) = 0 electrons = 0 lone pairs.
    • SN = 4 (sigma bonds) + 0 (lone pairs) = 4.
    • Hybridization: $sp^3$. No d-orbitals involved.

Species with no d-orbitals involved in hybridization:

• $SnF_2$ ($sp^2$)
• $PbCl_2$ ($sp^2$)
• $CCl_4$ ($sp^3$)
• $BF_3$ ($sp^2$)
• $SO_2Cl_2$ ($sp^3$)
• $POCl_3$ ($sp^3$)

There are 6 such species.

The final answer is $\boxed{6}$

Explanation of the solution:

Question 9: To compare bond lengths, consider factors like bond order, atomic size, electronegativity, hybridization, and resonance.

1. $PCl_3F_2$ vs $PF_3Cl_2$ ($BL_{P-Cleq}$): In TBP geometry, equatorial bonds have more s-character. More electronegative atoms prefer less s-character. In $PF_3Cl_2$, the equatorial P-F bond (F is more electronegative than Cl) takes less s-character, leaving more s-character for the equatorial P-Cl bonds, making them shorter than P-Cl bonds in $PCl_3F_2$ where all equatorial positions are occupied by Cl. Thus, $BL_{P-Cleq}$ of $PCl_3F_2$ > $BL_{P-Cleq}$ of $PF_3Cl_2$.
2. $SO_2Cl_2$ vs $SO_2F_2$ ($BL_{S=O}$): More electronegative substituents (F) on the central atom (S) withdraw electron density, making the central atom more electron-deficient. This strengthens the pi-bonding with oxygen, leading to shorter S=O bonds in $SO_2F_2$ compared to $SO_2Cl_2$. Thus, $BL_{S=O}$ of $SO_2Cl_2$ > $BL_{S=O}$ of $SO_2F_2$.
3. $BF_3$ vs $BCl_3$ ($BL_{B-X}$): Significant pπ-pπ backbonding occurs in $BF_3$ (from F 2p to B 2p), giving partial double bond character and shortening the B-F bond. This backbonding is much weaker in $BCl_3$ (Cl 3p to B 2p) due to poor orbital overlap. Thus, $BL_{B-F}$ in $BF_3$ < $BL_{B-Cl}$ in $BCl_3$.
4. $NO_3^-$ vs $NO_2$ ($BL_{N-O}$): Bond length is inversely proportional to bond order. $NO_2$ has an average N-O bond order of 1.5 (due to resonance), while $NO_3^-$ has an average N-O bond order of 1.33. Higher bond order means shorter bond. Thus, $BL_{N-O}$ in $NO_3^-$ > $BL_{N-O}$ in $NO_2$.
5. $O_3$ vs $O_2$ ($BL_{O-O}$): $O_2$ has a double bond (bond order 2). $O_3$ has an average O-O bond order of 1.5 (due to resonance). Thus, $BL_{O-O}$ in $O_3$ > $BL_{O-O}$ in $O_2$.
6. CO vs $CO_2$ ($BL_{C-O}$): CO has a triple bond (bond order 3). $CO_2$ has double bonds (bond order 2). Thus, $BL_{C-O}$ in CO < $BL_{C-O}$ in $CO_2$.

Question 10: Determine hybridization using steric number (SN = number of sigma bonds + number of lone pairs). If SN is 2, 3, or 4, hybridization is sp, sp2, or sp3, respectively, with no d-orbital involvement. If SN is 5, 6, or 7, hybridization is sp3d, sp3d2, or sp3d3, respectively, involving d-orbitals.

• $XeF_2$: SN=5 ($sp^3d$).
• $SnF_2$: SN=3 ($sp^2$).
• $PbCl_2$: SN=3 ($sp^2$).
• $SF_4$: SN=5 ($sp^3d$).
• $CCl_4$: SN=4 ($sp^3$).
• $BF_3$: SN=3 ($sp^2$).
• $SO_2Cl_2$: SN=4 ($sp^3$).
• $XeO_2F_2$: SN=5 ($sp^3d$).
• $XeOF_4$: SN=6 ($sp^3d^2$).
• $POCl_3$: SN=4 ($sp^3$). Species with no d-orbital involvement are $SnF_2$, $PbCl_2$, $CCl_4$, $BF_3$, $SO_2Cl_2$, $POCl_3$. Total 6 species.