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The quantum numbers of four electrons are given below :
I. $n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$
II. $n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$
III. $n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$
IV. $n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$
The correct decreasing order of energy of these electrons is
Options:
I. $n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$
II. $n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$
III. $n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$
IV. $n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$
The correct decreasing order of energy of these electrons is
Solution:
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Verified Answer
The correct answer is:
I $\gt$ III $\gt$ II $\gt$ IV
(I) $n=4, I=2, m_I=-2, s=-\frac{1}{2}$; represents $4 d(n+I=6)$
(II) $n=3, I=2, m_l=1, s=+\frac{1}{2}$; represents $3 d(n+I=5)$
(III) $n=4, I=1, m_I=0, s=+\frac{1}{2}$; represents $4 p(n+I=5)$
(IV) $n=3, I=1, m_l=-1, s=+\frac{1}{2}$; represents $3 p(n+I=4)$
Order of energy depends on the $(n+I)$, greater is the $(n+I)$ value greater is the energy, if $(n+I)$ is same, then it depends on $n$; if ' $n$ ' is more, energy is more.
Step-1: According to $(\mathrm{n}+\mathrm{I}$ )
Energy $=(\mathrm{I})\gt(\mathrm{II})=($ III $)\gt(\mathrm{IV})$
Step-2 : If $\mathrm{n} \uparrow$, then energy increases
Energy $=(\mathrm{I})\gt($ III $)\gt($ II $)\gt($ IV)
(II) $n=3, I=2, m_l=1, s=+\frac{1}{2}$; represents $3 d(n+I=5)$
(III) $n=4, I=1, m_I=0, s=+\frac{1}{2}$; represents $4 p(n+I=5)$
(IV) $n=3, I=1, m_l=-1, s=+\frac{1}{2}$; represents $3 p(n+I=4)$
Order of energy depends on the $(n+I)$, greater is the $(n+I)$ value greater is the energy, if $(n+I)$ is same, then it depends on $n$; if ' $n$ ' is more, energy is more.
Step-1: According to $(\mathrm{n}+\mathrm{I}$ )
Energy $=(\mathrm{I})\gt(\mathrm{II})=($ III $)\gt(\mathrm{IV})$
Step-2 : If $\mathrm{n} \uparrow$, then energy increases
Energy $=(\mathrm{I})\gt($ III $)\gt($ II $)\gt($ IV)
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