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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| $List-I$ (Compound) | $List-II$ (Shape) |
| $(A)$ $BrF _{5}$ | $(I)$ bent |
| $(B)$ $\left[ CrF _{6}\right]^{3-}$ | $(II)$ square pyramidal |
| $(C)$ $O _{3}$ | $(III)$ trigonal bipyramidal |
| $(D)$ $PCl _{5}$ | $(IV)$ octahedral |
Choose the correct answer from the options given below.
