Question

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

• A. $4f$
• B. $6p$
• C. $6s$
• D. $5d$

Answer 1

The Correct Option is D

To determine which orbital the 71^st electron of an element with an atomic number of 71 enters, we must understand the electron configuration of an element. The element with atomic number 71 is Lutetium (Lu). The electron configurations fill in order of increasing energy levels following the Aufbau principle.

The general order of filling is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, and so on. As we proceed through the periodic table, we fill the orbitals according to the increasing value of n+l (where n is the principal quantum number and l is the azimuthal quantum number), and for orbitals having the same n+l value, electrons fill orbitals in order of increasing n value.

Now, let's look at the filling process relevant for Lutetium (Lu):

• The first 54 electrons will fill up to the 5p orbital (Xe configuration).
• The next 2 electrons fill the 6s orbital.
• The subsequent 14 electrons fill the 4f orbital (This happens concurrently with filling of 5d, but 4f fills prior).
• The 71^st electron will enter the 5d orbital.

Thus, the correct electron configuration of Lu ends in the 5d orbital:

[Xe] 4f^{14} 5d^1 6s^2

Therefore, the 71^{st} electron of element X (Lutetium) enters into the 5d orbital.

The correct answer is 5d.

Let's rule out other options:

• 4f is already filled before the 71^{st} electron.
• 6p is filled after it.
• 6s fills before 4f and 5d, so it is already filled.