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Write down all the four quantum numbers for (i) 19th electron of `._(24)Cr` (ii) 21st electron for `._(21)Sc`.

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(i) 19th electron of `._(24)C`.
Cr (Z=24) ` 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6) 4s^(1)3d^(5)`
The 19th electron is present in the 4s orbital. For the 4s electron
`n=4, l=0, m=0, s=+(1)/(2) or -(1)/(2)`.
(ii) 21st electron for `._(21)Sc`.
Sc (Z=21)` 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5)`
The 21st electron in present in the 3d orbital. For 3d electron,
`n=3, l=2, m=-2, -1,0,+1, +2, s+(1)/(2) or -(1)/(2)`

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