A) \[n=4,\text{ }l=0,\text{ }m=0\text{ }and\text{ s}=+\frac{1}{2}\]
B) \[n=3,\text{ }l=0,\text{ }m=0\text{ }and\text{ }s=+\frac{1}{2}\]
C) \[n=3,\text{ }l=1,\text{ }m=+1\text{ }and\text{ }s=+\frac{1}{2}\]
D) \[n=3,\text{ }l=2,\text{ }m=+1\text{ }and\text{ }s=+\frac{1}{2}\]
Correct Answer: D
Solution :
\[n=4\]and\[l=0\]stands for 4s orbital \[n=3\]and\[l=0\]stands for 3s orbital \[n=3\] and\[l=1\]stands for 3p orbital \[n=3\]and\[l=2\]stands for 3d orbital The sequence of energy of above orbitals is \[3d>4s>3p>3s\]
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