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(B) The energy of an orbital is determined by the $(n + l)$ rule.$A$ $\Rightarrow 3d$ $\Rightarrow n + l = 3 + 2 = 5$$B$ $\Rightarrow 4p$ $\Rightarrow

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$D < A < B < C$

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(B) The energy of an orbital is determined by the $(n + l)$ rule.$A$ $\Rightarrow 3d$ $\Rightarrow n + l = 3 + 2 = 5$$B$ $\Rightarrow 4p$ $\Rightarrow n + l = 4 + 1 = 5$$C$ $\Rightarrow 4d$ $\Rightarrow n + l = 4 + 2 = 6$$D$ $\Rightarrow 3p$ $\Rightarrow n + l = 3 + 1 = 4$According to the $(n + l)$ rule,the energy increases as the $(n + l)$ value increases.If the $(n + l)$ values are the same,the orbital with the lower $n$ value has lower energy.Comparing the values:$D (n+l = 4) Therefore,the correct order of increasing energy is $D < A < B < C$.

Given below are the quantum numbers for $4$ electrons.
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
The correct order of increasing energy is

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Given below are the quantum numbers for $4$ electrons.$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$The correct order of increasing energy is