MCQ
Oxidation state of $+ 1$ for phosphorus is found in
- A${H_3}P{O_3}$
- B${H_3}P{O_4}$
- ✓${H_3}P{O_2}$
- D${H_4}{P_2}{O_7}$
$3+x+(-2 \times 2)=0$
$x=+1$
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$X + Y\mathop {\xrightarrow{{NaOH}}}\limits_{5\,^oC} \begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - CH - CHO} \\
{\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(X)$ and $(Y)$ will respectively be :
$(i)\,\, Cl^-$ can give up an electron more easily than $F^-$
$(ii) \,\,Cl^-$ is a better reducing agent than $F^-$
$(iii)\,\, Cl^-$ is smaller in size than $F^-$
$(iv)\,\, F^-$ can be oxidized more readily than $Cl^-$