A) \[Al,Ga\]
B) \[Be,Mg\]
C) \[Mg,Al\]
D) \[B,Be\]
Correct Answer: A
Solution :
Electronic configuration of \[\therefore \] and \[\frac{1}{\lambda }=109678\times {{(1)}^{2}}\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]c{{m}^{-1}}\] are written as \[=\frac{109678\times 3}{4}c{{m}^{-1}}\] \[\lambda =\frac{4\times {{10}^{7}}\times {{10}^{-7}}}{3\times 109678}cm\] Electronic configuration of Ga exhibits that it contains 10 electrons in d-orbitals which do not screen the nucleus effectively because of the large size of d-orbitals. Consequently, the electrons are attracted by a greater force towards nucleus. Hence, the atomic radii of Ga decreases or Ga has almost same radii as that of \[=121.65nm.\].
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