question_answer

According to Bohr-Bury rules The electronic configuration of a tripositively charged ion \[({{X}^{+3}})\]is 2, 8. (a) Identify the element and find the number of neutrons in it. (b) Write the formula of the compound that it forms with chlorine and oxygen respectively.

Answer:

(a) A tripositively charged ion is formed by the loss of three electrons. Therefore, the number of electrons in its neutral atom = No. of electrons in the ion\[+3\] \[=10+3=13\] \[\Rightarrow \]The atomic number of the element\[(Z)=13\] \[\Rightarrow \]The element is aluminium. Mass number of aluminium \[(A)=27\] \[\therefore \]No. of neutrons \[(n)=AZ=27-13=14\] (b) To find out the formula, we need to know the valency of aluminium. Electronic configuration = 2, 8, 3 Valency = No. of electrons in outermost orbit = 3 Therefore, it forms\[XC{{l}_{3}},\text{ }{{X}_{2}}{{O}_{3}}\]with chlorine and oxygen respectively.