- 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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$A.\,\,F -CH_2\,\,CH_2\,\,COOH$
$\begin{array}{*{20}{c}}
{B.\,\,Cl - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$C.\,\,F -CH_2 -COOH$
$D.\,\,Br -CH_2-CH_2 -COOH$
Correct answer is