- ✓$CrO_2Cl_2$ and $MnO_4^-$
- B$MnO_2$
- C$[Fe(CN)_6)]^{3-}$ and $[Co(CN)_6]^{3-}$
- D$MnO$
$Ni ( CO )_4 \Rightarrow 28+4 \times 2=36$
$(i)$ $V ( CO )_6 \Rightarrow 23+2 \times 6=35$
$(ii)$ $Cr ( CO )_5 \Rightarrow 24+2 \times 5=34$
$(iii)$ $Cu ( CO )_3 \Rightarrow 29+2 \times 3=35$
$(iv)$ $Mn ( CO )_5 \Rightarrow 25+2 \times 5=35$
$(v)$ $Fe ( CO )_5 \Rightarrow 26+2 \times 5=36$
$(vi)$ $\left[ Co ( CO )_3\right]^{3-} \Rightarrow 27+3+2 \times 3=36$
$(vii)$ $\left[ Cr ( CO )_4\right]^{4-} \Rightarrow 24+4+2 \times 4=36$
$(viii)$ $\left[\operatorname{Ir}( CO )_3\right] \Rightarrow 77+2 \times 3=83$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

The above reaction is called
$Pt(s)|\mathop {{H_2}(g)}\limits_{1\ atm} |HA\,\,\,\,\,\mathop {1\,M}\limits_{({K_a} = {{10}^{ - 7}})} \,\,||\,HB\,\,\,\,\,\mathop {1\,M}\limits_{({K_1} = {{10}^{ - 5}})}$
$\,|\mathop{{H_2}(g)}\limits_{1\ atm} |Pi(s)$
(Given atomic number: $\mathrm{C}: 6, \mathrm{Na}: 11, \mathrm{O}: 8$, $\mathrm{Fe}: 26, \mathrm{Cr}: 24 \mathrm{~J}$