Question:
The IUPAC name of the following complex is
\[
[Co(NH_2CH_2CH_2NH_2)_2Br_2]Br
\]
The IUPAC name of the following complex is
\[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]Br \]
\[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]Br \]
Show Hint
In coordination compounds, ligands inside the coordination sphere are named first, followed by the metal with oxidation state in Roman numerals, and then the counter ion. For repeated complex ligands like ethylene diamine, use prefixes such as \(bis\), \(tris\), etc.
Updated On: Jun 26, 2026
- Bis(ethane-1,2-diamine) dibromido cobalt (III) bromide
- Di(ethane-1,2-diamine) dibromido cobalt (III) bromide
- Tribromido bis(ethane-1,2-diamine) cobalt (III)
- Dibromido bis(ethylene diamine) cobalt (III) bromide
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Identify the coordination entity.
The given complex is \[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]Br \] The coordination sphere is \[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]^+ \] and the counter ion outside the coordination sphere is \[ Br^- \]
Step 2: Identify the ligands.
The ligand \[ NH_2CH_2CH_2NH_2 \] is ethylene diamine, commonly written as \[ en \] It is a neutral bidentate ligand.
There are two ethylene diamine ligands, so the prefix used is \[ bis \] because ethylene diamine is a polydentate ligand.
There are also two bromide ligands inside the coordination sphere. These are named as \[ dibromido \]
Step 3: Find the oxidation state of cobalt.
Let the oxidation state of cobalt be \[ x \] Ethylene diamine is neutral, so its charge is \[ 0 \] Each bromido ligand has charge \[ -1 \] There are two bromido ligands, so total charge due to bromido ligands is \[ -2 \] The complex ion has charge \[ +1 \] Therefore, \[ x+0-2=+1 \] \[ x=+3 \] Thus, cobalt is in the \[ +3 \] oxidation state.
Step 4: Write the IUPAC name.
Ligands are named before the metal.
The anionic ligand bromido is written as \[ dibromido \] The neutral ligand ethylene diamine is written as \[ bis(ethylene diamine) \] The metal is cobalt with oxidation state \[ (III) \] The counter ion is bromide.
Therefore, the name is Dibromido bis(ethylene diamine) cobalt (III) bromide
Step 5: Final conclusion.
Hence, the IUPAC name of the complex is \[ \boxed{\text{Dibromido bis(ethylene diamine) cobalt (III) bromide}} \] Therefore, the correct option is \[ \boxed{(4)} \]
The given complex is \[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]Br \] The coordination sphere is \[ [Co(NH_2CH_2CH_2NH_2)_2Br_2]^+ \] and the counter ion outside the coordination sphere is \[ Br^- \]
Step 2: Identify the ligands.
The ligand \[ NH_2CH_2CH_2NH_2 \] is ethylene diamine, commonly written as \[ en \] It is a neutral bidentate ligand.
There are two ethylene diamine ligands, so the prefix used is \[ bis \] because ethylene diamine is a polydentate ligand.
There are also two bromide ligands inside the coordination sphere. These are named as \[ dibromido \]
Step 3: Find the oxidation state of cobalt.
Let the oxidation state of cobalt be \[ x \] Ethylene diamine is neutral, so its charge is \[ 0 \] Each bromido ligand has charge \[ -1 \] There are two bromido ligands, so total charge due to bromido ligands is \[ -2 \] The complex ion has charge \[ +1 \] Therefore, \[ x+0-2=+1 \] \[ x=+3 \] Thus, cobalt is in the \[ +3 \] oxidation state.
Step 4: Write the IUPAC name.
Ligands are named before the metal.
The anionic ligand bromido is written as \[ dibromido \] The neutral ligand ethylene diamine is written as \[ bis(ethylene diamine) \] The metal is cobalt with oxidation state \[ (III) \] The counter ion is bromide.
Therefore, the name is Dibromido bis(ethylene diamine) cobalt (III) bromide
Step 5: Final conclusion.
Hence, the IUPAC name of the complex is \[ \boxed{\text{Dibromido bis(ethylene diamine) cobalt (III) bromide}} \] Therefore, the correct option is \[ \boxed{(4)} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Coordination chemistry Questions
- Cobalt (III) chloride forms a green-colored complex ‘X’ with NH\(_3\). Number of moles of AgCl formed when excess AgNO\(_3\) solution is added to 100 mL of 1M solution of ‘X’ is:
- The paramagnetic complex ion which has no unpaired electrons in \( t_{2g} \) orbitals is
Match the following.
- Observe the following complexes \[ [\text{Mn(CN)}_6]^{3-},\quad [\text{Fe(CN)}_6]^{3-},\quad [\text{Co(C}_2\text{O}_4)_3]^{3-},\quad [\text{MnCl}_6]^{3-},\quad [\text{Fe(CN)}_6]^{4-},\quad [\text{CoF}_6]^{3-} \] From the above, the number of inner orbital complexes with paramagnetic nature is:
- Which of the following statements are correct? i. \( [\text{Mn(CN)}_6]^{3-},\ [\text{MnCl}_6]^{3-} \) both are paramagnetic
ii. Both \( [\text{NiCl}_4]^{2-},\ [\text{Ni(CO)}_4] \) are tetrahedral and diamagnetic
iii. \( [\text{Co(NH}_3)_6]^{3+} \) is inner orbital complex, but \( [\text{Ni(NH}_3)_6]^{2+} \) is outer orbital complex
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions