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(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If $(n+l)$ values are the same,the orbital wit

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$(i) < (ii) < (iv) < (iii)$

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(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If $(n+l)$ values are the same,the orbital with the lower $n$ value has lower energy.$(i)$ $n=2, l=0 \implies n+l = 2+0 = 2$ ($2s$ orbital)$(ii)$ $n=2, l=1 \implies n+l = 2+1 = 3$ ($2p$ orbital)$(iii)$ $n=3, l=2 \implies n+l = 3+2 = 5$ ($3d$ orbital)$(iv)$ $n=3, l=1 \implies n+l = 3+1 = 4$ ($3p$ orbital)Comparing the $(n+l)$ values: $2 Therefore,the increasing order of energy is: $(i) < (ii) < (iv) < (iii)$.

Arrange the following in the increasing order of their energy:
$(i)$ $n = 2, l = 0, m_l = 0, m_s = +\frac{1}{2}$
$(ii)$ $n = 2, l = 1, m_l = +1, m_s = -\frac{1}{2}$
$(iii)$ $n = 3, l = 2, m_l = +1, m_s = -\frac{1}{2}$
$(iv)$ $n = 3, l = 1, m_l = -1, m_s = +\frac{1}{2}$

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Arrange the following in the increasing order of their energy:$(i)$ $n = 2, l = 0, m_l = 0, m_s = +\frac{1}{2}$$(ii)$ $n = 2, l = 1, m_l = +1, m_s = -\frac{1}{2}$$(iii)$ $n = 3, l = 2, m_l = +1, m_s = -\frac{1}{2}$$(iv)$ $n = 3, l = 1, m_l = -1, m_s = +\frac{1}{2}$