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  • Given below are the quantum numbers for  electrons.  A.  <img alt="\mathrm{\mathrm{n}=3,1=2

Given below are the quantum numbers for 4 electrons. 

A.  \mathrm{\mathrm{n}=3,1=2, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2}
B.  \mathrm{n}=4,1=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2
C.  \mathrm{n}=4,1=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2
D.  \mathrm{n}=3,1=1, \mathrm{~m}_{1}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2

The correct order of increasing energy is 

Option: 1

\mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C}


Option: 2

\mathrm{D}<\mathrm{A}<\mathrm{B}<\mathrm{C}


Option: 3

\mathrm{B}<\mathrm{D}<\mathrm{A}<\mathrm{C}


Option: 4

\mathrm{B}<\mathrm{D}<\mathrm{C}<\mathrm{A}


Answers (1)

best_answer

According to \mathrm{\left ( m+l \right )} role , lower the value of  n,
lower is energy of orbital.

If \mathrm{\left ( m+l \right )} value are same , then lower is value of n
lower is energy of orbital.

\mathrm{A \quad n= 3 \quad l= 2 \quad \quad n+l= 5}
\mathrm{B \quad n= 4 \quad l= 1 \quad \quad n+l= 5}
\mathrm{C \quad n= 4 \quad l= 2 \quad \quad n+l= 6}
\mathrm{D \quad n= 3 \quad l= 1 \quad \quad n+l= 4}

Energy of electrons is 
\mathrm{D< A< B< C}

Correct option is 2.

Posted by

Kuldeep Maurya

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