MCQ
The oxidation state of nitrogen is highest in
- ✓${N_3}H$
- B$N{H_2}OH$
- C${N_2}{H_4}$
- D$N{H_3}$
$3x + 1 = 0$
$3x = - 1,\, \Rightarrow x = - \frac{1}{3}\,\,{\rm{in}}\,\,{N_3}H$
$x + 2\,( + 1) + 1\,( - 2) + 1(1) = 0$
$x = - 1\,\,{\rm{in}}\,\,N{H_2}OH$
$x \times 2 + 4\,(1) = 0$$x = - \frac{4}{2} = - 2\,{\rm{in}}\,\,{N_2}{H_4}$
$x + 3\,(1) = 0$$x = - 3\,\,in\,\,N{H_3}$
Hence, highest in ${N_3}H$.
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$(A)$ Emit or absorb energy in the form of electromagnetic radiation
$(B)$ Frequency distribution of the emitted radiation depends on temperature
$(C)$ At a given temperature, intensity vs frequency curve passes through a maximum value
$(D)$ The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature
Element $-$ $IP$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}\xrightarrow[{(ii)\,NaB{H_4}}]{{(i)\,Hg{{(OAc)}_2};{H_2}O}}$