$_{84}^{210} P o ⟶$ $_{82}^{206} P b +_{2}^{4} H e$
In the above reaction, predict the position of $P o$ in the periodic table when lead belongs to $I V A$ group.

I I A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V I A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I V B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $V I A$
$_{84}^{210} P o ⟶$ $_{82}^{206} P b +_{2}^{4} H e$

An element which is formed by losing one α−particle occupies position left to the parent element, Pb is in IVA, thus Po should be in VIA.

Hence, option B is correct.

Suggest Corrections

Similar questions

_{84}^{210} Po ⟶_{82}^{206} Pb +_{2}^{4} He

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}

P o - α - - \to P b - β - \to B i

Join BYJU'S Learning Program