Question

A binary ionic compound is known to contain a cation with 51 protons and 48 electrons. The anion contains one-third the number of protons as the cation. The number of electrons in the anion is equal to the number of protons plus 1. What is the formula of this compound? What is the name of this compound?

Answer 1

Answer:

$SbCl_3$

Explanation:

The cation contains 51 protons and 48 electrons. The electrons are only lost not protons, so the element which corresponds to atomic number 51 is the cation which is Antimony (Sb).

Also, the charge on the cation is 51 - 48 = 3

This is because neutral atom contains the same number of electrons as protons.

Given the anion has one-third protons as cation which means thta:

Number of protons in anion = $\frac {51}{3}=17$

This corresponds to chlorine atom. The electrons are 1 greater than protons means that it has gain 1 electron.

So,

Sb     Cl

3         1

Cross multiply the valency.

So, formula is $SbCl_3$