JEE Main 2020 — Chemistry Inorganic Chemistry

2020mcqeasy

$B$ has a smaller first ionization enthalpy than $Be$ . Consider the following statement:

(I) it is easier to remove $2p$ electron than $2s$ electron

(II) $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of $Be$

(III) $2s$ electron has more penetration power than $2p$ electron

(IV) atomic radius of $B$ is more than $Be$

(atomic number $B:5,Be=4$ )

The correct statements are.

Official previous-year question

Held on 9 Jan 2020 · Verified 6 Jul 2026.

Options

A
(I), (II) and (IV)
B
(II), (III) and (IV)
C
(I), (II) and (III)
D
(I), (III) and (IV)

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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 $2p$ electron than $2s$ electron

${\text{Z}}_{ \text{eff}} \propto \frac{1}{ \text{shielding effect}}$

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

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

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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 $2p$ electron than $2s$ electron

${\text{Z}}_{ \text{eff}} \propto \frac{1}{ \text{shielding effect}}$

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

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

JEE Main 2020 — Chemistry Inorganic Chemistry

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Solution

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JEE Main Inorganic Chemistry in other years

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JEE Main 2020 — Chemistry Inorganic Chemistry

Options

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Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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