Question
Consider the following ionization enthalpies of two elements $$'A'$$ and $$'B'.$$
|
Element |
Ionization |
enthalpy $$\left( {kJ/mol} \right)$$ |
|
1st |
2nd |
3rd |
| A |
899 |
1757 |
14847 |
| B |
737 |
1450 |
7731 |
Which of the following statements is correct ?
A.
Both $$'A'$$ and $$'B'$$ belong to group - 1 where $$'B'$$ comes below $$'A'.$$
B.
Both $$'A'$$ and $$'B'$$ belong to group - 1 where $$'A'$$ comes below $$'B'.$$
C.
Both $$'A'$$ and $$'B'$$ belong to group - 2 where $$'B'$$ comes below $$'A'.$$
D.
Both $$'A'$$ and $$'B'$$ belong to group - 2 where $$'A'$$ comes below $$'B'.$$
Answer :
Both $$'A'$$ and $$'B'$$ belong to group - 2 where $$'B'$$ comes below $$'A'.$$
Solution :
Generally, the ionization enthalpies or energy increases from left to right in a period and decreases from top to bottom in a group. Several factor such as atomic radius, nuclear charge, shielding effect are responsible for change of ionization enthalpies.
Here, Ist ionization enthalpy of $$A$$ and $$B$$ is greater than group $$I\left( {Li = 520\,kJmo{l^{ - 1}}\,{\text{to}}\,Cs = 374\,kJ\,mo{l^{ - 1}}} \right),$$ which means element $$A$$ and $$B$$ belong to group –2 and all three given ionization enthalpy values are less for element $$B$$ means $$B$$ will come below $$A.$$