Question

An ion with mass number $37$ possesses one unit of negative charge. If ion contains $11.1\%$ more neutrons than electrons, the symbol of the ion, $X$, is

• A. $_{17}^{35}X$
• B. $_{17}^{35}X^{\ominus}$
• C. $_{17}^{37}X$
• D. $_{17}^{37}X^{\ominus}$

Answer 1

Answer: (D)

$_{17}^{37}X^{\ominus}$

Answer 2

Suppose number of electrons in an ion $=X$ Number of neutrons $=x+\frac{11.1}{100} x=1.11 x$
$=1.111 \, x$
Number of electrons in the neutral atom $=x-1$
$\therefore$ Number of protons $=x-1$
Mass number $=$ number of neutrons $+$ number of protons
$\therefore 37=1.111 \,x + x - 1$
or $2.111 \,x=38$
or $x=18$
Number of protons = atomic number
$=x-1=18-1=17$
Hence, the symbol of the ion will be ${ }_{17}^{37} X^{-1}$