question_answer

The electronic configuration of a tripositively charged ion\[({{X}^{+3}})\]is 2, 8. Then (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. (c) Name the element whose dinegatively charged ion is isoelectronic with\[{{X}^{+3}}\]. (d) What is the valency of 'X' ? Name the group in periodic table to which it belongs. (e) Find the number of 'M' electrons present in its neutral atom.

Answer:

The electronic configuration of a tripositively charged ion\[({{X}^{+3}})\]is 2, 8. Therefore the number of electrons in its neutral atom \[10+3=13\] (a) Aluminium\[{{(}_{13}}A{{l}^{27}}\text{ })\]\[\Rightarrow \]The number of neutrons in it\[=27-13=14\]. (b)\[AlC{{l}_{3}}\]and\[A{{l}_{2}}{{O}_{3}}\] (c) Oxygen (d) 3 (e) EC of X - 2, 8, 3. Therefore the number of ?M? electrons present in it= 3.