Question

Consider the following statements and arrange in the order of true/false as given in the codes.
Here $L$ is the lone pair
$S_1:$ In $A{B}_2{L}_2$ type, the BAB bond angle will be always greater than normal tetrahedral.
$S_2:$ Maleic acid has higher $K_{a1}$ 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

The correct option is C FTT

$S_1:O{F}_2<{109.5}^o$ and $OB{r}_2>{109.5}^o$
$O{F}_2$ $(appx.$ ${102}^o$): Fluorine being more electronegative attracts the bond pairs of electrons towards itself and thus there is more lp-lp repulsion decreasing the bond angle.
$OB{r}_2(appx.{113}^o):$ 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 increase the bond angle.
$S_2:$ Anion of maleic acid (cis-form) gets stabilised 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 $s{p}^3$ hybridisation and one lone pair of electrons.