MCQ
$CO$ has same electrons as or the ion that is isoelectronic with $CO$ is
- A$N_2^ + $
- ✓$C{N^ - }$
- C$O_2^ + $
- D$O_2^ - $
$CO = 6 + 8 = 14$ and $C{N^ - } = 6 + 7 + 1 = 14$.
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$4C{a_5}{(P{O_4})_3}F + 18Si{O_2} + 30C \to 3{P_4} + 2Ca{F_2} + 18CaSi{O_3} + 30CO$
(Given : The solubility product of $\mathrm{Zn}(\mathrm{OH})_{2}$ is $\left.2 \times 10^{-20}\right)$
