MCQ
The lowest value of heat of neutralization is obtained for
- A$HCl + NaOH$
- ✓$C{H_3}COOH + N{H_4}OH$
- C$N{H_4}OH + HCl$
- D$NaOH + C{H_3}COOH$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$1.$ 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