The 71^{st} electron of an element X with an | vedclass

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(C) The atomic number of element $X$ is $71$. The electronic configuration of this element (Lutetium) is $[Xe]4f^{14} 5d^1 6s^2$. According to the Auf

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$5d$

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(C) The atomic number of element $X$ is $71$. The electronic configuration of this element (Lutetium) is $[Xe]4f^{14} 5d^1 6s^2$. According to the Aufbau principle,the electrons fill orbitals in increasing order of energy. For $Z = 71$,the configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^1$. The last electron ($71^{st}$ electron) enters the $5d$ orbital.

The $71^{st}$ electron of an element $X$ with an atomic number of $71$ enters into the orbital:

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