KCET · Chemistry · Coordination Compounds
The complex hexamineplatinum(IV)chloride will give number of ions on ionisation.
- A 4
- B 3
- C 2
- D 5
Answer & Solution
Correct Answer
(D) 5
Step-by-step Solution
Detailed explanation
The formula of hexamineplatinum (IV) chloride is \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4\).
\[
\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \longrightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}
\]
Thus, five ions are produced.
\[
\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \longrightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}
\]
Thus, five ions are produced.
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