NEET Test Series from KOTA - 10 Papers In MS WORD
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CHEMISTRY(KCET)
285487
The complex hexaamineplatinum(IV) chloride will give\(\qquad\) number of ions on ionisation.
1 3
2 2
3 5
4 4
Explanation:
(c)\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}\)
The given complex provide 5 ions on ionisation.
Karnataka CET 2022
CHEMISTRY(KCET)
285488
The correct IUPAC name of cis-platin is
1 diamminedichloridoplatinum(0)
2 dichloridodiammineplatinum(IV)
3 diamminedichloridoplatinum(II)
4 diamminedichloridoplatinum(IV).
Explanation:
(c) Cis-platin is\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\).
The IUPAC name is diammine dichloridoplatinum(II)
Karnataka CET 2022
CHEMISTRY(KCET)
285489
Crystal Field Splitting Energy (CFSE) for\(\left[\mathrm{CoCl}_6\right]^{-4}\) is \(18000 \mathrm{~cm}^{-1}\). The Crystal Field Splitting Energy (CFSE) for \(\left[\mathrm{CoCl}_4\right]^{2-}\) will be
285487
The complex hexaamineplatinum(IV) chloride will give\(\qquad\) number of ions on ionisation.
1 3
2 2
3 5
4 4
Explanation:
(c)\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}\)
The given complex provide 5 ions on ionisation.
Karnataka CET 2022
CHEMISTRY(KCET)
285488
The correct IUPAC name of cis-platin is
1 diamminedichloridoplatinum(0)
2 dichloridodiammineplatinum(IV)
3 diamminedichloridoplatinum(II)
4 diamminedichloridoplatinum(IV).
Explanation:
(c) Cis-platin is\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\).
The IUPAC name is diammine dichloridoplatinum(II)
Karnataka CET 2022
CHEMISTRY(KCET)
285489
Crystal Field Splitting Energy (CFSE) for\(\left[\mathrm{CoCl}_6\right]^{-4}\) is \(18000 \mathrm{~cm}^{-1}\). The Crystal Field Splitting Energy (CFSE) for \(\left[\mathrm{CoCl}_4\right]^{2-}\) will be
285487
The complex hexaamineplatinum(IV) chloride will give\(\qquad\) number of ions on ionisation.
1 3
2 2
3 5
4 4
Explanation:
(c)\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}\)
The given complex provide 5 ions on ionisation.
Karnataka CET 2022
CHEMISTRY(KCET)
285488
The correct IUPAC name of cis-platin is
1 diamminedichloridoplatinum(0)
2 dichloridodiammineplatinum(IV)
3 diamminedichloridoplatinum(II)
4 diamminedichloridoplatinum(IV).
Explanation:
(c) Cis-platin is\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\).
The IUPAC name is diammine dichloridoplatinum(II)
Karnataka CET 2022
CHEMISTRY(KCET)
285489
Crystal Field Splitting Energy (CFSE) for\(\left[\mathrm{CoCl}_6\right]^{-4}\) is \(18000 \mathrm{~cm}^{-1}\). The Crystal Field Splitting Energy (CFSE) for \(\left[\mathrm{CoCl}_4\right]^{2-}\) will be
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHEMISTRY(KCET)
285487
The complex hexaamineplatinum(IV) chloride will give\(\qquad\) number of ions on ionisation.
1 3
2 2
3 5
4 4
Explanation:
(c)\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}\)
The given complex provide 5 ions on ionisation.
Karnataka CET 2022
CHEMISTRY(KCET)
285488
The correct IUPAC name of cis-platin is
1 diamminedichloridoplatinum(0)
2 dichloridodiammineplatinum(IV)
3 diamminedichloridoplatinum(II)
4 diamminedichloridoplatinum(IV).
Explanation:
(c) Cis-platin is\(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\).
The IUPAC name is diammine dichloridoplatinum(II)
Karnataka CET 2022
CHEMISTRY(KCET)
285489
Crystal Field Splitting Energy (CFSE) for\(\left[\mathrm{CoCl}_6\right]^{-4}\) is \(18000 \mathrm{~cm}^{-1}\). The Crystal Field Splitting Energy (CFSE) for \(\left[\mathrm{CoCl}_4\right]^{2-}\) will be