Arrange the electrons represented by the foll | vedclass

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $(i): n = 4, l = 0 \implies n + l = 4 + 0 = 4 \ (4s 	ext{ orbital})$.For $(ii):

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$(iii) > (i) > (ii) > (iv)$

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $(i): n = 4, l = 0 \implies n + l = 4 + 0 = 4 \ (4s 	ext{ orbital})$.For $(ii): n = 3, l = 1 \implies n + l = 3 + 1 = 4 \ (3p 	ext{ orbital})$.For $(iii): n = 3, l = 2 \implies n + l = 3 + 2 = 5 \ (3d 	ext{ orbital})$.For $(iv): n = 3, l = 0 \implies n + l = 3 + 0 = 3 \ (3s 	ext{ orbital})$.According to the $(n + l)$ rule,if $(n + l)$ values are different,the orbital with the higher $(n + l)$ value has higher energy.If $(n + l)$ values are the same,the orbital with the higher $n$ value has higher energy.Comparing the values: $(iii) \ (5) > (i) \ (4, n=4) > (ii) \ (4, n=3) > (iv) \ (3)$.Thus,the decreasing order of energy is $(iii) > (i) > (ii) > (iv)$.

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