Question

${}_{92}^{238}U$ (IIIB) undergoes following emission:

${}_{92}^{238}U\stackrel{-\alpha }{⟶}A\stackrel{-\alpha }{⟶}B\stackrel{-\beta }{⟶}C$

Which is/are correct statement(s)?

• A. A will be of IB group
• B. A will be of IIIB group
• C. B will be of IIA (alkaline earth metal) group
• D. C will be of IIIA (boron family group)

Answer 1

Answer: (B), (C)

A will be of IIIB group
B will be of IIA (alkaline earth metal) group

${}_{92}^{238}U\stackrel{-\alpha }{\to }{}_{90}^{234}Th\stackrel{-\alpha }{\to }{}_{88}^{230}Ra\stackrel{-\beta }{\to }{}_{89}^{230}Ac$

(B) A (Th) belongs to the group 3rd or IIIB group. It is an actinide.

(C) B (Ra) is an alkaline earth metal and belongs to group 2nd or IIA.

(D) C (Ac) is an actinide. It belongs to group IIIB.

Thus, options B and C are correct.

Options A, D is incorrect.