321838
Correct relationship between pairing energy (p) and C.F.S.E \(\left(\Delta_{\circ}\right)\) in complex ion \(\left[\operatorname{Ir}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is:
1 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ < P}}\)
2 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ > P}}\)
3 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ = P}}\)
4 cannot comment
Explanation:
Ir belongs to \(5 \mathrm{~d}\)-series. In \(5 \mathrm{~d}\)-series central metal, pairing always occurs hence, \(\Delta {\rm{. > P}}\)
CHXII09:COORDINATION COMPOUNDS
321839
Low spin complex of \({{\text{d}}^{\text{6}}}\) -cation in an octahedral field will have the following energy: \(\left(\Delta_{0}=\right.\) Crystal field splitting energy in an octahedral field, \(\mathrm{P}=\) Electron pairing energy)
\({{\text{d}}^{\text{6}}}\) -cation with low spin has electronic configuration \({\text{t}}_{{\text{2g}}}^{\text{6}}{\text{e}}_{\text{g}}^{\text{0}}\) . Total energy \(=\left(-0.4 \Delta_{0}\right.\) per \(\left.e^{-} \times 6\right)+\left(e^{-}\right.\)pairing energy of 3 pairs) \( = - 2.4{\Delta _0} + {\text{3P}} = - \frac{{12}}{5}{\Delta _0} + {\text{3P}}\)
CHXII09:COORDINATION COMPOUNDS
321840
Which of the following system is an octahedral complex and has maximum unpaired electrons?
1 \(\mathrm{d}^{4}\) (low spin)
2 \(\mathrm{d}^{7}\) (high spin)
3 \(\mathrm{d}^{9}\) (high spin)
4 \(\mathrm{d}^{6}\) (low spin)
Explanation:
KCET - 2023
CHXII09:COORDINATION COMPOUNDS
321841
Assertion : In high spin situation, configuration of \(\mathrm{d}^{5}\) ions will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{2}\). Reason : In high spin situation, pairing energy is less than crystal field energy.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a high spin situation, the crystal field splitting energy \(\left(\Delta_{\mathrm{o}}\right)\) is smaller compared to the pairing energy \((\mathrm{P})\). This is particularly relevant in the \(d^{5}\) electron configuration, where the\(4^{\text {th }}\) and \(5^{\text {th }}\) electrons are added to the degenerate ' \(\mathrm{e}_{\mathrm{g}}\) ' orbitals rather than the ' \(\mathrm{t}_{2 \mathrm{~g}}\) ' orbitals. As a result, the electron configuration of the \(\mathrm{d}^{5}\) ion will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}{ }^{2}\). So the option (3) is correct.
321838
Correct relationship between pairing energy (p) and C.F.S.E \(\left(\Delta_{\circ}\right)\) in complex ion \(\left[\operatorname{Ir}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is:
1 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ < P}}\)
2 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ > P}}\)
3 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ = P}}\)
4 cannot comment
Explanation:
Ir belongs to \(5 \mathrm{~d}\)-series. In \(5 \mathrm{~d}\)-series central metal, pairing always occurs hence, \(\Delta {\rm{. > P}}\)
CHXII09:COORDINATION COMPOUNDS
321839
Low spin complex of \({{\text{d}}^{\text{6}}}\) -cation in an octahedral field will have the following energy: \(\left(\Delta_{0}=\right.\) Crystal field splitting energy in an octahedral field, \(\mathrm{P}=\) Electron pairing energy)
\({{\text{d}}^{\text{6}}}\) -cation with low spin has electronic configuration \({\text{t}}_{{\text{2g}}}^{\text{6}}{\text{e}}_{\text{g}}^{\text{0}}\) . Total energy \(=\left(-0.4 \Delta_{0}\right.\) per \(\left.e^{-} \times 6\right)+\left(e^{-}\right.\)pairing energy of 3 pairs) \( = - 2.4{\Delta _0} + {\text{3P}} = - \frac{{12}}{5}{\Delta _0} + {\text{3P}}\)
CHXII09:COORDINATION COMPOUNDS
321840
Which of the following system is an octahedral complex and has maximum unpaired electrons?
1 \(\mathrm{d}^{4}\) (low spin)
2 \(\mathrm{d}^{7}\) (high spin)
3 \(\mathrm{d}^{9}\) (high spin)
4 \(\mathrm{d}^{6}\) (low spin)
Explanation:
KCET - 2023
CHXII09:COORDINATION COMPOUNDS
321841
Assertion : In high spin situation, configuration of \(\mathrm{d}^{5}\) ions will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{2}\). Reason : In high spin situation, pairing energy is less than crystal field energy.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a high spin situation, the crystal field splitting energy \(\left(\Delta_{\mathrm{o}}\right)\) is smaller compared to the pairing energy \((\mathrm{P})\). This is particularly relevant in the \(d^{5}\) electron configuration, where the\(4^{\text {th }}\) and \(5^{\text {th }}\) electrons are added to the degenerate ' \(\mathrm{e}_{\mathrm{g}}\) ' orbitals rather than the ' \(\mathrm{t}_{2 \mathrm{~g}}\) ' orbitals. As a result, the electron configuration of the \(\mathrm{d}^{5}\) ion will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}{ }^{2}\). So the option (3) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII09:COORDINATION COMPOUNDS
321838
Correct relationship between pairing energy (p) and C.F.S.E \(\left(\Delta_{\circ}\right)\) in complex ion \(\left[\operatorname{Ir}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is:
1 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ < P}}\)
2 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ > P}}\)
3 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ = P}}\)
4 cannot comment
Explanation:
Ir belongs to \(5 \mathrm{~d}\)-series. In \(5 \mathrm{~d}\)-series central metal, pairing always occurs hence, \(\Delta {\rm{. > P}}\)
CHXII09:COORDINATION COMPOUNDS
321839
Low spin complex of \({{\text{d}}^{\text{6}}}\) -cation in an octahedral field will have the following energy: \(\left(\Delta_{0}=\right.\) Crystal field splitting energy in an octahedral field, \(\mathrm{P}=\) Electron pairing energy)
\({{\text{d}}^{\text{6}}}\) -cation with low spin has electronic configuration \({\text{t}}_{{\text{2g}}}^{\text{6}}{\text{e}}_{\text{g}}^{\text{0}}\) . Total energy \(=\left(-0.4 \Delta_{0}\right.\) per \(\left.e^{-} \times 6\right)+\left(e^{-}\right.\)pairing energy of 3 pairs) \( = - 2.4{\Delta _0} + {\text{3P}} = - \frac{{12}}{5}{\Delta _0} + {\text{3P}}\)
CHXII09:COORDINATION COMPOUNDS
321840
Which of the following system is an octahedral complex and has maximum unpaired electrons?
1 \(\mathrm{d}^{4}\) (low spin)
2 \(\mathrm{d}^{7}\) (high spin)
3 \(\mathrm{d}^{9}\) (high spin)
4 \(\mathrm{d}^{6}\) (low spin)
Explanation:
KCET - 2023
CHXII09:COORDINATION COMPOUNDS
321841
Assertion : In high spin situation, configuration of \(\mathrm{d}^{5}\) ions will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{2}\). Reason : In high spin situation, pairing energy is less than crystal field energy.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a high spin situation, the crystal field splitting energy \(\left(\Delta_{\mathrm{o}}\right)\) is smaller compared to the pairing energy \((\mathrm{P})\). This is particularly relevant in the \(d^{5}\) electron configuration, where the\(4^{\text {th }}\) and \(5^{\text {th }}\) electrons are added to the degenerate ' \(\mathrm{e}_{\mathrm{g}}\) ' orbitals rather than the ' \(\mathrm{t}_{2 \mathrm{~g}}\) ' orbitals. As a result, the electron configuration of the \(\mathrm{d}^{5}\) ion will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}{ }^{2}\). So the option (3) is correct.
321838
Correct relationship between pairing energy (p) and C.F.S.E \(\left(\Delta_{\circ}\right)\) in complex ion \(\left[\operatorname{Ir}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is:
1 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ < P}}\)
2 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ > P}}\)
3 \({{\rm{\Delta }}_{\rm{^\circ }}}{\rm{ = P}}\)
4 cannot comment
Explanation:
Ir belongs to \(5 \mathrm{~d}\)-series. In \(5 \mathrm{~d}\)-series central metal, pairing always occurs hence, \(\Delta {\rm{. > P}}\)
CHXII09:COORDINATION COMPOUNDS
321839
Low spin complex of \({{\text{d}}^{\text{6}}}\) -cation in an octahedral field will have the following energy: \(\left(\Delta_{0}=\right.\) Crystal field splitting energy in an octahedral field, \(\mathrm{P}=\) Electron pairing energy)
\({{\text{d}}^{\text{6}}}\) -cation with low spin has electronic configuration \({\text{t}}_{{\text{2g}}}^{\text{6}}{\text{e}}_{\text{g}}^{\text{0}}\) . Total energy \(=\left(-0.4 \Delta_{0}\right.\) per \(\left.e^{-} \times 6\right)+\left(e^{-}\right.\)pairing energy of 3 pairs) \( = - 2.4{\Delta _0} + {\text{3P}} = - \frac{{12}}{5}{\Delta _0} + {\text{3P}}\)
CHXII09:COORDINATION COMPOUNDS
321840
Which of the following system is an octahedral complex and has maximum unpaired electrons?
1 \(\mathrm{d}^{4}\) (low spin)
2 \(\mathrm{d}^{7}\) (high spin)
3 \(\mathrm{d}^{9}\) (high spin)
4 \(\mathrm{d}^{6}\) (low spin)
Explanation:
KCET - 2023
CHXII09:COORDINATION COMPOUNDS
321841
Assertion : In high spin situation, configuration of \(\mathrm{d}^{5}\) ions will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{2}\). Reason : In high spin situation, pairing energy is less than crystal field energy.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a high spin situation, the crystal field splitting energy \(\left(\Delta_{\mathrm{o}}\right)\) is smaller compared to the pairing energy \((\mathrm{P})\). This is particularly relevant in the \(d^{5}\) electron configuration, where the\(4^{\text {th }}\) and \(5^{\text {th }}\) electrons are added to the degenerate ' \(\mathrm{e}_{\mathrm{g}}\) ' orbitals rather than the ' \(\mathrm{t}_{2 \mathrm{~g}}\) ' orbitals. As a result, the electron configuration of the \(\mathrm{d}^{5}\) ion will be \(\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}{ }^{2}\). So the option (3) is correct.
