Correct increasing energy of the orbitals 3d, 4p, 4f, 5p orbitals
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a
4p, 3d, 5p, 4f
b
3d, 4p, 4f, 5p
c
4p, 3d, 4f, 5p
d
3d, 4p, 5p, 4f
answer is D.
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Detailed Solution
Orbitals (n+l) 3d (3+2)= 5 4p (4+1) = 5 4f (4+3) = 7 5p (5+1) = 6In 3d and 4p, (n+l) value is same, in such cases higher the ‘n’ value, higher will be the energy.∴ Energy of 3d < 4p < 5p<4f