A) C>B>Be>Li
B) C > Be > B > Li
C) B>C>Be>Li
D) Be > Li > B > C
Correct Answer: B
Solution :
First ionisation energy: The energy required to remove the most loosely bound electron from the isolated gaseous atom is called its first ionisation energy or \[\Delta \operatorname{iH}\] \[\operatorname{C}=1{{s}^{2}}{{s}^{2}}2{{p}^{2}}, Be=1{{s}^{2}}2{{s}^{2}}B=1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\] \[\operatorname{Li}=1{{s}^{2}}2{{s}^{2}}\] Removal of 2 electrons from neutral atom gives doubly positively charged ion which will held its remaining electrons even more tightly. From Li to C,\[\Delta \operatorname{iH}\]keeps on increasing due to increasing nuclear charge and decreasing atomic radii.
You need to login to perform this action.
You will be redirected in
3 sec
