Correct statement for 3d- series is | vedclass

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(D) $1$. $Sc$ $(Z=21)$ has the configuration $[Ar] 3d^1 4s^2$. By losing three electrons,it attains the stable noble gas configuration $[Ar]$,making $

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All are correct

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(D) $1$. $Sc$ $(Z=21)$ has the configuration $[Ar] 3d^1 4s^2$. By losing three electrons,it attains the stable noble gas configuration $[Ar]$,making $+3$ the most stable oxidation state.$2$. $Zn$ $(Z=30)$ has a completely filled $3d^{10} 4s^2$ configuration. Removing an electron from this stable configuration requires a very high amount of energy,hence it has a very high first ionization energy compared to other $3d$ elements.$3$. In the $3d$ series,the metallic radius generally decreases from $Sc$ to $Cr$. However,$Mn$ has a slightly larger radius than expected due to half-filled $d^5$ configuration leading to weak inter-atomic metallic bonding. Thus,the metallic radius of $Mn$ is indeed greater than that of $Fe$.$4$. Since all statements are correct,the correct option is $D$.

Correct statement for $3d-$ series is

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