MCQ
Visible light will be absorbed by
- A$[Sc(H_2O)_3(NH_3)_{3}]^{3+}$
- B$[Ti(en)_2(NH_3)_2]^{4+}$
- ✓$[Cr(NH_3)_6]^{3+}$
- D$[Zn(NH_3)_6]^{2+}$
Also, when an electron can go from lower energy orbital to higher energy orbital, then they can absorb visible light.
In $\left[ Cr \left( NH _3\right)_6\right]^{3+}, Cr$ is present as $Cr ^{+3}$
Electronic configuration of $Cr { }^3=[A r] 3 d^3 4 s^0$
Now, since the complex has 3 unpaired electrons, excitation of electrons is possible \& therefore, it will absorb visible light.
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$(a)$ Ionisation energy $(Na < Li < B < Be)$
$(b)$ Electron affinity $(N < P < O < S)$
$(c)$ Electron negativity $(S < O < Cl < F)$
$(d)$ Atomic radii $(F < N < O < Na)$
