Question

Consider the following statements and arrange in the order of true/false as given in the codes:
$S_1$: In $A{B}_2{L}_2$ type, the $BAB$ bond angle will always be greater than normal tetrahedral ($L$ is a lone pair).

$S_2$: Maleic acid has higher $K_{a_1}$ value than fumaric acid due to intramolecular hydrogen bonding.
$S_3$: In ${Cl{O}_3}^{-},N{H}_3$ and $Xe{O}_3$, the hybridisation and the number of lone pair(s) on central atoms are same.

• A. $TTT$
• B. $FTF$
• C. $FTT$
• D. $FFT$

Answer 1

Answer: (C)

$FTT$

$S_1:$ The bond angle of $O{F}_2<{109.5}^o$ and $OB{r}_2>{109.5}^o$.
$O{F}_2$ (approx. ${102}^0$): Fluorine being more electronegative attracts the bond pairs of electrons towards itself and thus, there is more lp-lp repulsion decreasing the bond angle.
$OB{r}_2$ (approx. ${113}^0$): The bond pairs are close to more electronegative oxygen atom and thus, there is repulsion between bp-bp increasing the bond angle. The bulkier Br atoms also create steric repulsion which also increases the bond angle.
$S_2:$ Anion of maleic acid (cis-form) gets stabilized through intramolecular hydrogen bonding after giving one proton as given in following structure. In fumaric acid (trans-form), no such intramolecular H-bonding is possible.
$S_3:$ All have sp${}^3$ hybridization and one lone pair of electrons.