(i) \[n=4,l=1,\] |
(ii) \[n=4,l=0\] |
(iii) \[n=3,l=2\] |
and |
(iv) \[n=3,l=1\] |
A) (i) < (ii) < (iii) < (iv)
B) (iv) < (iii) < (ii) < (i)
C) (iv) < (ii) < (iii) < (i)
D) (iv) < (i) < (ii) < (iii)
Correct Answer: C
Solution :
According to (n + 0 rule, the lower the value of (n + 0 for an orbital, the lower is its energy. If two orbitals have the same value of\[(n+l),\] the orbital with lower value of n will have the lower energy. Hence, the order of increasing energy of electron is asIV < II < III < I | |||
\[\begin{array}{*{35}{l}} n=3 \\ \end{array}\] | \[n=4\] | \[\begin{array}{*{35}{l}} n=3 \\ \end{array}\] | \[n=4\] |
\[l=1\] | \[l=0\] | \[l=2\] | \[l=1\] |
\[n+l=4\] | \[n+l=4\] | \[n+l=5\] | \[n+l=5\] |
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