MCQ
The lanthanide contraction is responsible for the fact that
- A$Zr$ and $Y$ have about the same radius
- B$Zr$ and $Nb$ have similar oxidation state
- ✓$Zr$ and $Hf$ have about the same radius
- D$Zr$ and $Zn$ have the same oxidation state
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$\begin{array}{*{20}{c}}
{C{H_3} - C = CH - {C_6}{H_5}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{HBr\,/\,Peroxide}}'A'$
$\begin{array}{*{20}{c}}
{C{H_3} - C = CH - {C_6}{H_5}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow[{HBr}]{}'B'$
Reason: $He$ is soluble in blood.
