(i) \[n=4,l=1\] |
(ii) \[n=4,l=0\] |
(iii)\[n=3,l=2\] |
(iv) \[n=3,l=1\] |
A) (iii)<(i)<(iv)<(ii)
B) (iv)<(ii)<(iii)<(i)
C) (i)<(iii)<(ii)<(iv)
D) (ii)<(iv)<(i)<(iii)
Correct Answer: B
Solution :
(i) 4 p (ii) 4s (iii) 3 d (iv) 3p According to Bohr Bury's \[(n+\ell )\] rule, increasing order of energy will be (iv)<(ii)<(iii)<(i). Note : If the two orbitals have same value of (n + i) then the orbital with lower value of n will be filled first.You need to login to perform this action.
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