Question

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. The ion is

• A. ${}_{24}^{56}C{r}^{3+}$
• B. ${}_{27}^{56}C{o}^{3+}$
• C. ${}_{26}^{56}F{e}^{3+}$
• D. ${}_{28}^{56}N{i}^{3+}$

Answer 1

The correct option is C

${}_{26}^{56}F{e}^{3+}$

Let the no. of electrons in the ion $M^{3+}=x$

$\therefore$ No.of neutrons $=x+\frac{30.4x}{100}=1.304x$

No.of electrons in the neutral atom = x + 3

$\therefore$ No.of protons = x + 3

Mass no = No. of protons + No. if neutrons

$56=x+3+1.304x$

or, $2.304x=53$

or, $x=23$

No.of protons = x + 3 = 23 + 3 = 26

Hence, the ion is ${}_{26}^{56}F{e}^{3+}$