MCQ
Effective atomic number of $Co(CO)_4$ is $35$, hence it is less stable. It attains stability by
- AOxidation of $Co$
- BReduction of $Co$
- CDimerization
- ✓Both $(B)$ and $(C)$
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$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2 $ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is
(Rounded off to the nearest integer)
$\left[\right.$ Given $\left.: \frac{2.303 RT }{ F }=0.059\right]$
$STATEMENT-2$: The $\mathrm{Fe}$ in $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}_3 \mathrm{SO}_4\right.$ has three unpaired electrons.