Question

If spin quantum number are $-\frac{1}{2},0,+\frac{1}{2}$ based on the above assumption electronic configuration of the elements with atomic number 30 is:

• A. $1s^{2}2s^{2}2p^{9}3s^{2}3p^{9}4s^3$
• B. $1s^{3}1p^{15}2s^{3}1d^9$
• C. $1s^{9}1p^{15}2s^6$
• D. $1s^3{19}^{p}2s^{3}2p^{9}1d^6$

Answer 1

The correct option is C $1s^{9}1p^{15}2s^6$

Since, there are $3$ values of spin quantum numbers, therefore $3$ electron must be present in an orbital. Since $\ell ϵ(-n,n)$ and azimuthual quantum number $-\varphi +1$ to $\varphi +1$.

Therefore configuration is ${1s}^9{1p}^{15}{2s}^3{1d}^9$