B has a smaller first ionization enthalpy tha | vedclass

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Question

(A) $Be$ has the electronic configuration $1s^{2} 2s^{2}$.$B$ has the electronic configuration $1s^{2} 2s^{2} 2p^{1}$.Statement $(I)$ is correct: It i

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$(I), (II)$ and $(III)$

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(A) $Be$ has the electronic configuration $1s^{2} 2s^{2}$.$B$ has the electronic configuration $1s^{2} 2s^{2} 2p^{1}$.Statement $(I)$ is correct: It is easier to remove a $2p$ electron than a $2s$ electron because the $2p$ orbital is higher in energy and less stable.Statement $(II)$ is correct: The $2p$ electron in $B$ experiences greater shielding from the $1s^{2} 2s^{2}$ core compared to the $2s$ electrons in $Be$,which are less shielded.Statement $(III)$ is correct: The $2s$ orbital is closer to the nucleus and has higher penetration power than the $2p$ orbital.Statement $(IV)$ is incorrect: $B$ has a smaller atomic radius than $Be$ due to the increase in effective nuclear charge across the period.Therefore,statements $(I), (II),$ and $(III)$ are correct.

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