question_answer

The electronic configuration of a tripositively charged ion\[({{X}^{+3}})\]is 2, 8. Then (a) (i) Identify the element. (ii) 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 valency of 'X'? (e) Find the number of electrons present in 'M' shell of its neutral atom.

	(a)	(b)	(c)	(d)	(e)
		\[i\]	\[ii\]
(a)	Boron	10	\[BaC{{l}_{3}}\And {{B}_{2}}{{O}_{3}}\]	Oxygen	2	2
(b)	Aluminium	14	\[AlC{{l}_{3}}\And A{{l}_{2}}{{O}_{3}}\]	Oxygen	3	3
(c)	Aluminium	13	\[AlC{{l}_{3}}\And A{{l}_{2}}{{O}_{3}}\]	Fluorine	2	3
(d)	Boron	9	\[BC{{l}_{3}}\And {{B}_{2}}{{O}_{3}}\]	Nitrogen	3	2

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\[{{(}_{13}}A{{l}^{27}})\] \[\Rightarrow \]The number of neutrons in it \[=27-13=14\]. (b) \[AlC{{l}_{3}}\] and \[A{{l}_{2}}{{O}_{3}}\] (c) Oxygen (\[{{O}^{2}}\]has 10 electrons) (d) 3 (e) Electronic configuration of X is 2, 8, 3. Therefore, the number of ?M? electrons present in it is 3.