Question

Identify the pair in which the specified bond length of first is greater than second.

$PCl_3F_2, PF_3Cl_2 : BL_{P-Cleq} X$

$SO_2Cl_2, SO_2F_2 : BL_{S=O} \checkmark$

$BF_3, BCl_3 : BL_{B-X} X = F/Cl \checkmark$

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

$O_3, O_2 : BL_{O-O} \checkmark$

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

Answer 1

The pairs where the specified bond length of the first is greater than the second are:

• $SO_2Cl_2, SO_2F_2 : BL_{S=O}$
• $NO_3^-, NO_2 : BL_{N-O}$
• $O_3, O_2 : BL_{O-O}$

Answer 2

Question 9 Explanation:

1. $PCl_3F_2$ vs $PF_3Cl_2$ ($BL_{P-Cleq}$): In $PCl_3F_2$, F atoms are axial, Cl atoms are equatorial. In $PF_3Cl_2$, Cl atoms are equatorial, 1 F atom is equatorial, 2 F atoms are axial. The P-Cl equatorial bond in $PF_3Cl_2$ is slightly longer than in $PCl_3F_2$ due to the presence of a highly electronegative F atom in the equatorial plane competing for electron density. So, $BL_{P-Cleq}$ ($PCl_3F_2$) < $BL_{P-Cleq}$ ($PF_3Cl_2$).

2. $SO_2Cl_2$ vs $SO_2F_2$ ($BL_{S=O}$): Fluorine is more electronegative than chlorine. In $SO_2F_2$, F atoms pull more electron density from S, making S more positive, thus strengthening and shortening the S=O bonds compared to $SO_2Cl_2$. So, $BL_{S=O}$ ($SO_2Cl_2$) > $BL_{S=O}$ ($SO_2F_2$).

3. $BF_3$ vs $BCl_3$ ($BL_{B-X}$): Comparing B-F and B-Cl bond lengths. Chlorine is larger than fluorine, so B-Cl bond is inherently longer than B-F bond. Back-bonding in $BF_3$ makes B-F even shorter. So, $BL_{B-F}$ ($BF_3$) < $BL_{B-Cl}$ ($BCl_3$).

4. $NO_3^-$ vs $NO_2$ ($BL_{N-O}$): Bond length is inversely proportional to bond order. Bond order in $NO_3^-$ is 1.33. Bond order in $NO_2$ is 1.5. Higher bond order means shorter bond. So, $BL_{N-O}$ ($NO_3^-$) > $BL_{N-O}$ ($NO_2$).

5. $O_3$ vs $O_2$ ($BL_{O-O}$): Bond order in $O_3$ is 1.5. Bond order in $O_2$ is 2. So, $BL_{O-O}$ ($O_3$) > $BL_{O-O}$ ($O_2$).

6. $CO$ vs $CO_2$ ($BL_{C-O}$): Bond order in $CO$ is 3. Bond order in $CO_2$ is 2. So, $BL_{C-O}$ ($CO$) < $BL_{C-O}$ ($CO_2$).