Ncert Exercises & Solutions

The electronic configuration of gadolinium (atomic number of 64) is:

1. $[Xe] 4 f^{3} 5 d^{5} 6 s^{2}$ $[Xe] 4 f^{3} 5 d^{5} 6 s^{2}$

2. $[Xe] 4 f^{7} 5 d^{2} 6 s^{1}$ $[Xe] 4 f^{7} 5 d^{2} 6 s^{1}$

3. $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$ $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$

4. $[Xe] 4 f^{8} 5 d^{6} 6 s^{2}$ $[Xe] 4 f^{8} 5 d^{6} 6 s^{2}$

HINT: Gadolium is a f block element .

Explanation:

The electronic configuration of La (Z=57) is $[Xe] 5 d^{1} 6 s^{2}$ $[Xe] 5 d^{1} 6 s^{2}$ . Therefore, further addition of electrons occurs in the lower energy 4f-orbital till it is exactly half-filled at Eu (Z=64). Thus, the electronic configuration of Eu is $[Xe] 4 f^{7} 6 s^{2}$ $[Xe] 4 f^{7} 6 s^{2}$ .

Thereafter, addition of next electron does not occur in the more stable exactly half-filled $4 f^{7}$ $4 f^{7}$ shell but occurs in the little higher energy 5d-orbital. Thus, the electronic configuration of Gd (Z=64) is $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$ $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$ .

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