MCQ
Number of $\pi $ electrons present in naphthalene is
- A$4$
- B$6$
- ✓$10$
- D$14$
$\pi $ bonds $=5$
hence electrons are double
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Use $K_{ sp }( ZnS )=1.25 \times 10^{-22}$ and
Overall dissociation constant of $H _2 S , K_{ NET }=K_1 K_2=1 \times 10^{-21}$
