Question

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. What would be the symbol of this ion?

• A. ${}_{26}^{56}F{e}^{3+}$
• B. ${}_{26}^{56}F{e}^{2+}$
• C. ${}_{27}^{56}C{o}^{2+}$
• D. ${}_{27}^{58}C{o}^{3+}$

Answer 1

The correct option is A ${}_{26}^{56}F{e}^{3+}$

Let the $X$ and is represented as: ${}_Z^{56}{X}^{3+}$

Let the atomic number $\left(Z\right)$/ number of potons $\left(p\right)$ be ${}^{\prime }{x}^{\prime }$

Given: 3 units of positive charge

thus, number of electrons $=x-3$

Number of neutrons,n=30.4% more neutrons than electrons

$n=(x-3)+\frac{30.4}{100}\times (x-3)$

$n=(\frac{30.4}{100}+1)\times (x-3)$

$n=1.304(x-3)$

Mass number of X=56

Mass number of an atom=number of protons + number of neutrons

$\therefore 56=x+1.304(x-3)$

or $56=2.304x-3.912$

$59.912=2.304x$

$x=26$

Thus, number of protons (p) $=x=26$, element is Iron (Fe)

number of electrons (e) $=x-3=26-3=23$.

Therefore, symbol of the ${}_{26}^{56}F{e}^{3+}$