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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Classification of Elements and Periodicity in Properties

JEE Main 2020 · 9 January · Morning Shift · Chemistry · Classification of Elements and Periodicity in Properties

B has a smaller first ionization enthalpy than Be . Consider the...

BB has a smaller first ionization enthalpy than BeBe. Consider the following statement:
(I) it is easier to remove 2p2p electron than2s2s electron
(II) 2p2p electron of BB is more shielded from the nucleus by the inner core of electrons than the 2s2s electrons of BeBe
(III) 2s2s electron has more penetration power than 2p2p electron
(IV) atomic radius of BB is more than BeBe
(atomic number B:5,Be=4B: 5, Be = 4 )
The correct statements are.

Options

  1. A(I), (II) and (IV)
  2. B(II), (III) and (IV)
  3. C(I), (II) and (III)
  4. 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 2p2p electron than 2s2s electron
Zeff∝1shielding effect{\text{Z}}_{\text{eff}} ∝ \dfrac{1}{\text{shielding effect}}

(Z eff ){\left({\text{Z}}_{\text{\text{ eff }}}\right)}_{\text{B}} > {\left({\text{Z}}_{\text{eff}}\right)}_{\text{Be}} So, shielding effect is more on 2p electron of B than 2s electron of Be

So, (Be)I.E1>(B)I.E1{\left(\text{Be}\right)}_{\text{I}{\text{.E}}_{1}} > {\left(\text{B}\right)}_{\text{I}{\text{.E}}_{1}}
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