STD 12 - 8.2 Carboxylic acids and their derivative — JEE STD 11 Science — Question

$(A)\, n=4,l=1$                        $(B)\, n=4,l=0$

$(C) \,n=3,l=2$                        $(D)\, n=3,l=1$

can be placed in order of increasing energy as :

$1.$  Boron is approximately $sp^3$ hybridized

$2.$  $B-H-B$ angle is $180^o$

$3.$  There are two terminal $B-H$ bonds for each boron atom

$4.$  There are only $12$ bonding electrons available

Of these statements

$C{H_3} - CH = O + {C_6}{H_5} - CH =O \xrightarrow[\Delta ]{{\mathop O\limits^\Theta  H\,\left( {aq} \right)}}\mathop X\limits_{{\text{(}}{{\text{C}}_4}{{\text{H}}_6}{\text{O)}}}  + \mathop Y\limits_{{\text{(}}{{\text{C}}_9}{{\text{H}}_8}{\text{O)}}} $