MCQ
$[Mn(CO)_4 NO]$ is diamagnetic because
- A$Mn$ metal is diamagnetic in free state
- B$Mn$ is in $+1$ oxidation state in this complex
- C$NO$ is present as positive ligand
- ✓All of the above
$[\overset{-1}{\mathop{Mn}}\,{{(CO)}_{4}}(N{{O}^{+}})]:$ No unpaired $e^-$ either on ligands or on $\overset{-1}{\mathop{Mn}}$ hence it is diamagnetic
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| Column $-I$ | Column $-II$ |
| $(a) \,Xe{F_6}$ | $(i)$ Distorted Octahedral |
| $(b)\, XeO_3$ | $(ii)$ Square planar |
| $(c)\, XeOF_4$ | $(iii)$ pyramidal |
| $(d)\, XeF_4$ | $(iv)$ Square pyramidal |
$\left[\right.$ Use $: \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}: \Delta_{\mathrm{\gamma}} \mathrm{H}=-57.1\, \mathrm{k} \mathrm{J} \,\mathrm{mol}^{-1}$
Specific heat of $\mathrm{H}_{2} \mathrm{O}=4.18 \mathrm{Jk}^{-} \mathrm{g}^{-}$
density of $\mathrm{H}_{2} \mathrm{O}=1.0\, \mathrm{~g} \mathrm{~cm}^{-3}$
Assume no change in volume of solution on mixing.]
$C_6H_5-CHO + C_6H_5CHO$ $\xrightarrow{{N{a_2}[Fe{{(CO)}_4}]}}$ $\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{C_6}{H_5} - C - O - C{H_2}{C_6}{H_5}}
\end{array}$
$C{H_2} = C{H_2}\xrightarrow{{HBr}}X\xrightarrow{{{\text{Hydrolysis}}}}Y\mathop {\xrightarrow{{N{a_2}C{O_3}}}}\limits_{{I_2}{\text{ excess}}} Z$