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Find out the number of angular nodes in the orbital to which the last electron of Cr enter.

$(a)\;4\qquad(b)\;3\qquad(c)\;2\qquad(d)\;1$

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$Cr=[Ar]3d^5,4s^1$
The last electron of Cr enters the 3d-subshell.
$\therefore$ No of angular nodes $l=2$
Hence (c) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
 

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