321868
The CFSE for tetrahedral \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is 20,000 \(\mathrm{cm}^{-1}\). The CFSE for octahedral \(\left[\mathrm{PtCl}_{6}\right]^{2-}\) will be
1 \(8,000 \mathrm{~cm}^{-1}\)
2 \(20,000 \mathrm{~cm}^{-1}\)
3 \(45,000 \mathrm{~cm}^{-1}\)
4 \(16,000 \mathrm{~cm}^{-1}\)
Explanation:
According to question, \(\Delta_{o}=20,000 \mathrm{~cm}^{-1}\) \(\begin{aligned}\therefore \quad \Delta_{o}=\dfrac{9}{4} \Delta_{t} & =\dfrac{9}{4} \times 20,000 \mathrm{~cm}^{-1} \\& =45,000 \mathrm{~cm}^{-1} .\end{aligned}\)
NCERT Exemplar
CHXII09:COORDINATION COMPOUNDS
321869
Give the number of weak field ligand(s) from the following \(\mathrm{S}^{2-}, \stackrel{\ominus}{\mathrm{O}} \mathrm{H}, \mathrm{Cl}^{\ominus}, \mathrm{H}_{2} \mathrm{O}, \mathrm{Py}, \mathrm{NO}_{2}^{\ominus}, \mathrm{NO}_{3}^{\ominus}\)
321870
Which among the following cations will form lowest stability complex if the ligand remains the same?
1 \(\mathrm{Cu}^{2+}\)
2 \(\mathrm{Fe}^{2+}\)
3 \(\mathrm{Cd}^{2+}\)
4 \(\mathrm{Ni}^{2+}\)
Explanation:
Stability order for given divalent metal ions is : \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Fe}^{2+}>\mathrm{Cd}^{2+}\) The charge to size ratio of \(\mathrm{Cd}^{2+}\) is lower and hence it forms lowest stability complex if the ligand remains same.
MHTCET - 2021
CHXII09:COORDINATION COMPOUNDS
321871
The magnitude of crystal field stabilisation energy in tetrahedral complexes \({\rm{(CFSE}}\,\,{\rm{of}}\,\,{{\rm{\Delta }}_{\rm{t}}}{\rm{)}}\) is considerably less than that in the octahedral field because
1 there are only four ligands instead of six so the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\)
2 the direction of the orbitals does not coincide with the direction of the ligands. This reduce the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\)
3 Both points (1) and (2) are correct
4 Both points (1) and (2) are incorrect
Explanation:
Both (a) and (b) statements are correct. (a) There are only four ligands instead of six. So, the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\). (b) The direction of the orbital does not coincide with the direction of the ligands. This reduces the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\).
321868
The CFSE for tetrahedral \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is 20,000 \(\mathrm{cm}^{-1}\). The CFSE for octahedral \(\left[\mathrm{PtCl}_{6}\right]^{2-}\) will be
1 \(8,000 \mathrm{~cm}^{-1}\)
2 \(20,000 \mathrm{~cm}^{-1}\)
3 \(45,000 \mathrm{~cm}^{-1}\)
4 \(16,000 \mathrm{~cm}^{-1}\)
Explanation:
According to question, \(\Delta_{o}=20,000 \mathrm{~cm}^{-1}\) \(\begin{aligned}\therefore \quad \Delta_{o}=\dfrac{9}{4} \Delta_{t} & =\dfrac{9}{4} \times 20,000 \mathrm{~cm}^{-1} \\& =45,000 \mathrm{~cm}^{-1} .\end{aligned}\)
NCERT Exemplar
CHXII09:COORDINATION COMPOUNDS
321869
Give the number of weak field ligand(s) from the following \(\mathrm{S}^{2-}, \stackrel{\ominus}{\mathrm{O}} \mathrm{H}, \mathrm{Cl}^{\ominus}, \mathrm{H}_{2} \mathrm{O}, \mathrm{Py}, \mathrm{NO}_{2}^{\ominus}, \mathrm{NO}_{3}^{\ominus}\)
321870
Which among the following cations will form lowest stability complex if the ligand remains the same?
1 \(\mathrm{Cu}^{2+}\)
2 \(\mathrm{Fe}^{2+}\)
3 \(\mathrm{Cd}^{2+}\)
4 \(\mathrm{Ni}^{2+}\)
Explanation:
Stability order for given divalent metal ions is : \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Fe}^{2+}>\mathrm{Cd}^{2+}\) The charge to size ratio of \(\mathrm{Cd}^{2+}\) is lower and hence it forms lowest stability complex if the ligand remains same.
MHTCET - 2021
CHXII09:COORDINATION COMPOUNDS
321871
The magnitude of crystal field stabilisation energy in tetrahedral complexes \({\rm{(CFSE}}\,\,{\rm{of}}\,\,{{\rm{\Delta }}_{\rm{t}}}{\rm{)}}\) is considerably less than that in the octahedral field because
1 there are only four ligands instead of six so the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\)
2 the direction of the orbitals does not coincide with the direction of the ligands. This reduce the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\)
3 Both points (1) and (2) are correct
4 Both points (1) and (2) are incorrect
Explanation:
Both (a) and (b) statements are correct. (a) There are only four ligands instead of six. So, the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\). (b) The direction of the orbital does not coincide with the direction of the ligands. This reduces the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXII09:COORDINATION COMPOUNDS
321868
The CFSE for tetrahedral \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is 20,000 \(\mathrm{cm}^{-1}\). The CFSE for octahedral \(\left[\mathrm{PtCl}_{6}\right]^{2-}\) will be
1 \(8,000 \mathrm{~cm}^{-1}\)
2 \(20,000 \mathrm{~cm}^{-1}\)
3 \(45,000 \mathrm{~cm}^{-1}\)
4 \(16,000 \mathrm{~cm}^{-1}\)
Explanation:
According to question, \(\Delta_{o}=20,000 \mathrm{~cm}^{-1}\) \(\begin{aligned}\therefore \quad \Delta_{o}=\dfrac{9}{4} \Delta_{t} & =\dfrac{9}{4} \times 20,000 \mathrm{~cm}^{-1} \\& =45,000 \mathrm{~cm}^{-1} .\end{aligned}\)
NCERT Exemplar
CHXII09:COORDINATION COMPOUNDS
321869
Give the number of weak field ligand(s) from the following \(\mathrm{S}^{2-}, \stackrel{\ominus}{\mathrm{O}} \mathrm{H}, \mathrm{Cl}^{\ominus}, \mathrm{H}_{2} \mathrm{O}, \mathrm{Py}, \mathrm{NO}_{2}^{\ominus}, \mathrm{NO}_{3}^{\ominus}\)
321870
Which among the following cations will form lowest stability complex if the ligand remains the same?
1 \(\mathrm{Cu}^{2+}\)
2 \(\mathrm{Fe}^{2+}\)
3 \(\mathrm{Cd}^{2+}\)
4 \(\mathrm{Ni}^{2+}\)
Explanation:
Stability order for given divalent metal ions is : \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Fe}^{2+}>\mathrm{Cd}^{2+}\) The charge to size ratio of \(\mathrm{Cd}^{2+}\) is lower and hence it forms lowest stability complex if the ligand remains same.
MHTCET - 2021
CHXII09:COORDINATION COMPOUNDS
321871
The magnitude of crystal field stabilisation energy in tetrahedral complexes \({\rm{(CFSE}}\,\,{\rm{of}}\,\,{{\rm{\Delta }}_{\rm{t}}}{\rm{)}}\) is considerably less than that in the octahedral field because
1 there are only four ligands instead of six so the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\)
2 the direction of the orbitals does not coincide with the direction of the ligands. This reduce the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\)
3 Both points (1) and (2) are correct
4 Both points (1) and (2) are incorrect
Explanation:
Both (a) and (b) statements are correct. (a) There are only four ligands instead of six. So, the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\). (b) The direction of the orbital does not coincide with the direction of the ligands. This reduces the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\).
321868
The CFSE for tetrahedral \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is 20,000 \(\mathrm{cm}^{-1}\). The CFSE for octahedral \(\left[\mathrm{PtCl}_{6}\right]^{2-}\) will be
1 \(8,000 \mathrm{~cm}^{-1}\)
2 \(20,000 \mathrm{~cm}^{-1}\)
3 \(45,000 \mathrm{~cm}^{-1}\)
4 \(16,000 \mathrm{~cm}^{-1}\)
Explanation:
According to question, \(\Delta_{o}=20,000 \mathrm{~cm}^{-1}\) \(\begin{aligned}\therefore \quad \Delta_{o}=\dfrac{9}{4} \Delta_{t} & =\dfrac{9}{4} \times 20,000 \mathrm{~cm}^{-1} \\& =45,000 \mathrm{~cm}^{-1} .\end{aligned}\)
NCERT Exemplar
CHXII09:COORDINATION COMPOUNDS
321869
Give the number of weak field ligand(s) from the following \(\mathrm{S}^{2-}, \stackrel{\ominus}{\mathrm{O}} \mathrm{H}, \mathrm{Cl}^{\ominus}, \mathrm{H}_{2} \mathrm{O}, \mathrm{Py}, \mathrm{NO}_{2}^{\ominus}, \mathrm{NO}_{3}^{\ominus}\)
321870
Which among the following cations will form lowest stability complex if the ligand remains the same?
1 \(\mathrm{Cu}^{2+}\)
2 \(\mathrm{Fe}^{2+}\)
3 \(\mathrm{Cd}^{2+}\)
4 \(\mathrm{Ni}^{2+}\)
Explanation:
Stability order for given divalent metal ions is : \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Fe}^{2+}>\mathrm{Cd}^{2+}\) The charge to size ratio of \(\mathrm{Cd}^{2+}\) is lower and hence it forms lowest stability complex if the ligand remains same.
MHTCET - 2021
CHXII09:COORDINATION COMPOUNDS
321871
The magnitude of crystal field stabilisation energy in tetrahedral complexes \({\rm{(CFSE}}\,\,{\rm{of}}\,\,{{\rm{\Delta }}_{\rm{t}}}{\rm{)}}\) is considerably less than that in the octahedral field because
1 there are only four ligands instead of six so the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\)
2 the direction of the orbitals does not coincide with the direction of the ligands. This reduce the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\)
3 Both points (1) and (2) are correct
4 Both points (1) and (2) are incorrect
Explanation:
Both (a) and (b) statements are correct. (a) There are only four ligands instead of six. So, the ligand field is only \(2 / 3\) the size, hence the \(\Delta_{t}\) is \(2 / 3\). (b) The direction of the orbital does not coincide with the direction of the ligands. This reduces the crystal field stabilisation energy \((\Delta)\) by further \(2 / 3\).