$_{84}^{210} Po ⟶_{82}^{206} Pb +_{2}^{4} He$
In this reaction, predict the position of group of $P o$ in periodic table when lead is in the $I V A$ group:

I I A

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I V B

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V I A

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V I B

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Solution

The correct option is C $V I A$
When a radioactive element looses one alpha particle, it loooses two proton and shfts to two position left to parent element, so if $P o$ is producing $P b$ and lead is in $I V$ group so Po should belongs to $V I A$ group.

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Similar questions

_{84}^{210} P o ⟶

_{82}^{206} P b +_{2}^{4} H e

P o

I V A

_{84}^{210} P o \to_{82}^{206} P b +_{2}^{4} H e

_{84} P o^{210} \to_{82} P b^{206} +_{2} H e^{4}

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