Question

The correct option(s) regarding the complex $\left[Co\right(en\left)\right(N{H}_3{)}_3\left(H_{2}O\right){]}^{3+}(en=H_{2}NC{H}_{2}C{H}_{2}N{H}_2)$ is (are)

• A. It has two geometrical isomers
• B. It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands
• C. It is paramagnetic
• D. It absorbs light at longer wavelength as compared to $\left[co\right(en\left)\right(N{H}_3{)}_4{]}^{3+}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A It has two geometrical isomers
B It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands
D It absorbs light at longer wavelength as compared to $\left[co\right(en\left)\right(N{H}_3{)}_4{]}^{3+}$

a) $\left[Co\right(en\left)\right(N{H}_3{)}_3\left(H_{2}O\right){]}^{3+}$ has two geometrical isomers. The two geometrical isomers are



(b) It will have three geometrical isomers, if bidentate 'en' is replaced by two cyanide ligands. These are:


c) Since both (en) and $N{H}_3$ are strong field ligands, so it will pair up the unpaired electrons of $Co$, thus it should be a diamagnetic complex.
d) $\left(N{H}_3\right)$ is a stronger ligand than
$H_{2}O$. So, the splitting will be more in $\left[Co\right(en\left)\right(N{H}_3{)}_4{]}^{3+}$ and hence it should absorb light at shorter wavelength (higer energy) as compared yo $\left[Co\right(en\left)\right(N{H}_3{)}_3\left(H_{2}O\right){]}^{3+}$.