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The electronic configuration of an element is $1s^2,2s^22p^6,3s^23p^63d^5,4s^1$.This represents its

$(a)\;\text{Excited state and ground state}\qquad(b)\;\text{Ground state}\qquad(c)\;\text{Cationic form}\qquad(d)\;\text{Anionic form}$

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The given electronic configuration is for ground state of $_{24}Cr$ and of $Mn^+$
Hence (a) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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