JEE Main20209 Jan 2020Morning ShiftChemistryClassification of Elements and Periodicity in PropertiesActual
B has a smaller first ionization enthalpy than B e . Consider the following statement: (I) it is easier to remove 2 p electron than 2 s electron (II) 2 p electron of B is more shielded from the nucleus by the inner core of electrons than the 2 s electrons of B e (III) 2 s electron has more penetration power than 2 p electron (IV) atomic radius of B is more than B e (atomic number B : 5 , B e = 4 ) The correct stateme
Options
- A(I), (II) and (IV)
- B(II), (III) and (IV)
- C(I), (II) and (III)
- D(I), (III) and (IV)
Correct answer
C. (I), (II) and (III)
Step-by-step solution
Since, Penetration of s orbital is the maximum because of the closeness to the nucleus than the p, d and f orbitals it is easier to remove 2 p electron than 2 s electron Z eff ∝ 1 shielding effect Z eff B > Z eff Be So, shielding effect is more on 2p electron of B than 2s electron of Be So, Be I .E 1 > B I .E 1