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QUESTION ON ATOMIC STRUCTURE
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hi the e configuration is 1s^2,2s^2,2p^6,3e^2,3p^6,4s^2,3d^10,4p^5 ii)e with l=1 are e in 2p,3p and 4p orbitals i.e 6+6+5=17 iii)e with n+l=5 are e in 3d and 4p orbitals i.e l for d is 2 and for p is 1. i m doubtful about the i) and iv). !!!!!!!!
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Varsha be cool, tomorrow is what we make it today, so if u can dream it , u can make it . life is an ice cream ,eat it before it melts away!!!!!!!!!     
 
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