Question

An element of group $III$ with at. no. 90 and mass number 238 undergoes decay of one $\alpha$-particle. The newly formed element belongs to:

• A. $Igp$
• B. $IIgp$
• C. $IIIgp$
• D. $IVgp$

Answer 1

The correct option is B $IIgp$

The balanced nuclear reaction is as shown.
$\underset{IIIgroup}{{}_{90}^{238}\text{Th}}⟶\underset{IIgroup}{{}_{88}^{234}\text{Ra}}{+}_2^4\text{He}$;
Since one alpha particle is lost, the atomic number decreases from 90 to 88. The element with atomic number 88 is Ra. It is alkaline earth metal and it belongs to II group.