JEE MainChemistryClassification of Elements and Periodicity in Properties
A plot of first ionization enthalpy versus atomic number for the elements of the second period ( Li to Ne ) shows points sequentially labeled A through H. The graph exhibits a general upward trend but features two distinct dips: the ionization enthalpy of C is lower than that of B, and the ionization enthalpy of F is lower than that of E. Identify the incorrect statement regarding this trend.
Options
- AThe element A has the lowest first ionization enthalpy in the period because its valence electron experiences
- BThe decrease in ionization enthalpy from B to C occurs because the 2p electron in C is easier to remove than t
- CThe element H has the highest first ionization enthalpy in the period due to its highly stable ns^2np^6 electr
- DThe higher ionization enthalpy of E compared to F is primarily due to the completely filled 2s subshell in E.
Correct answer
D. The higher ionization enthalpy of E compared to F is primarily due to the completely filled 2s subshell in E.
Step-by-step solution
The elements of the second period sequentially from A to H are Li , Be , B , C , N , O , F , and Ne . Element A ( Li ) has the lowest effective nuclear charge in the period, resulting in the lowest first ionization enthalpy. Element B is Be ( 1s^2 2s^2 ) and element C is B ( 1s^2 2s^2 2p^1 ). The dip from B to C is due to the removal of a 2p electron in Boron, which is at a higher energy level and easier to remove than the 2s electron in Beryllium. Element H ( Ne ) has a stable octet configuration ( 1s^2 2s^2 2p^6