Question

An ion with mass number 37 contains one unit of negative charge. If the ion contains 11.1% more neutrons than electrons, find the symbol of ion

Answer 1

Answer:

Let the number of electrons in the ion carrying a negative charge be x.

Then,

Number of neutrons present

= x + 11.1% of x

= x + 0.111 x

= 1.111 x

Number of electrons in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

∴ Number of protons in the neutral atom = x – 1

Therefore 37 = 1.111x + x -1

Or 2.111 x = 38

Or x = 18

Therefore no of protons = atomic no = x -1 = 18 – 1 = 17

∴The symbol of the ion is: Cl

Explanation:

May this helps uh☺

Answer 2

$\fbox\textcolor{orange}{\texttt{Answer = Cl(37,17)}}$

$\textcolor{indigo}{\texttt{Explanation :-}}$

Suppose no. of electrons = x

no. of neutrons = $x + \frac{11.1}{100}\times{x}\: = \:1.111x$

No. of electrons in the neutral atom = $x - 1$

No. of protons in the neutral atom =$x - 1$

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

ie, $37\:=\:1.111x+x-1$

$2.111x\:=\:37+1\:=\:38$

$x = \frac{38}{2.111}\:=\texttt{18}$

No. of protons = Atomic no. =

$x-1\:=\:18-1\:=\texttt{17}$

Hence the symbol of the ion is: $</h3><h3>\fbox{^{37}_{17}Cl}$