MCQ
Phenol reacts with dilute $HN{O_3}$ at normal temperature to form
- A

- ✓

- C

- D







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$C_2H_4(g) + H_2(g) \to C_2H_6(g)$
.......$kJ$
| Bond | Bond energy $(kJ)$ |
| $C-H$ |
$413$ |
| $C-C$ | $348$ |
| $C=C$ | $610$ |
| $H-H$ | $436$ |
Reason : Low spin complexes have lesser number of unpaired electrons.