The quantum numbers of four electrons are giv | vedclass

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $I: n = 4, l = 2, (n + l) = 4 + 2 = 6$ ($4d$ orbital).For $II: n = 3, l = 2, (n +

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$IV < II < III < I$

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(C) The energy of an orbital is determined by the $(n + l)$ rule.For $I: n = 4, l = 2, (n + l) = 4 + 2 = 6$ ($4d$ orbital).For $II: n = 3, l = 2, (n + l) = 3 + 2 = 5$ ($3d$ orbital).For $III: n = 4, l = 1, (n + l) = 4 + 1 = 5$ ($4p$ orbital).For $IV: n = 3, l = 1, (n + l) = 3 + 1 = 4$ ($3p$ orbital).According to the $(n + l)$ rule,lower $(n + l)$ value means lower energy. If $(n + l)$ values are equal,the orbital with lower $n$ has lower energy.Comparing the values: $IV (4) Thus,the increasing order of energy is $IV < II < III < I$.

The quantum numbers of four electrons are given below:
$I. \ n = 4, l = 2, m_l = -2, m_s = -1/2$
$II. \ n = 3, l = 2, m_l = 1, m_s = +1/2$
$III. \ n = 4, l = 1, m_l = 0, m_s = +1/2$
$IV. \ n = 3, l = 1, m_l = 1, m_s = -1/2$
The correct order of their increasing energies will be:

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The quantum numbers of four electrons are given below:$I. \ n = 4, l = 2, m_l = -2, m_s = -1/2$$II. \ n = 3, l = 2, m_l = 1, m_s = +1/2$$III. \ n = 4, l = 1, m_l = 0, m_s = +1/2$$IV. \ n = 3, l = 1, m_l = 1, m_s = -1/2$The correct order of their increasing energies will be: