Question

One isotope of a non-metallic element has mass number $127$ and $74$ neutrons in the nucleus. The anion derived from the isotope has $54$ electrons. Hence, symbol for the anion is

• A. ${ }_{54}^{127} X^{-}$
• B. ${ }_{53}^{127} X^{-}$
• C. ${ }_{53}^{74} X^{-}$
• D. ${ }_{54}^{74} X^{-}$

Answer 1

Answer: (B)

${}_{Z} {}^{ A } X ={ }_{ A - N }^{ A} X={ }_{127-74}^{127} X ={ }_{53}^{127} X$
There are $54$ electrons. Hence, ionic species is ${ }_{53}^{127} X ^{-}$