274282
Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$ the correct increasing order of their first ionization enthalpy is :
1 $\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<$ P
2 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<$ P $<$ S
3 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<\mathrm{S}<$ P
4 Al $<$ Mg $<$ S $<$ Si $<$ P
Explanation:
(A) : The first ionization energy of an element is the energy needed to remove the outermost or highest energy electron from a neutral atom in the gas phase. As we know, the first ionization energy in the periodic table is decreases down the group and increases in the left to right in period. Al has less ionization energy than other because the removal of one electron from $3 p$ orbital it goes to stable configuration. The same reason occure between $\mathrm{S}$ and $\mathrm{P}$ i.e. sulphur has less ionization energy than phosphorus. Hence, the correct order will be $-\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<\mathrm{P}$.
JEE Main 24.02.2021
COORDINATION COMPOUNDS
274292
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(A) : The spectrochemical series is given as- $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<$Urea $<\mathrm{OH}^{-}<$ etOH $<$ ox $<\mathrm{O}^{2-}<\mathrm{H}_{2} \mathrm{O}<$ EDTA $<\mathrm{NCS}^{-}<\mathrm{Py},<\mathrm{NH}_{3}$ $<$ en $<$ bpy, phen $<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CH}_{3}^{-}<\mathrm{C}_{6} \mathrm{H}_{5}^{-}<\mathrm{CN}^{-}$ $\mathrm{CO}$. Thus, option (a) has the correct order
NEET 2020)
COORDINATION COMPOUNDS
274293
Crystal field splitting in octahedral coordination entities, ' $\mathbf{g}_{\mathrm{g}}$ ' set of orbital's are
(C) : Crystal field splitting in octahedral coordination entities, 'eg 'set of orbital's are $\left[\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}, \mathrm{~d}_{\mathrm{z}^{2}}\right]$
JHARKHAND - 2019
COORDINATION COMPOUNDS
274298
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)
1
2
3
4
Explanation:
(A) : Octahedral complex contains the two axial ligands, if we remove the both axial ligand then it converted into square planar field then following energy diagram take place-
JEE Main 2019
COORDINATION COMPOUNDS
274300
Crystal field splitting energy for octahedral $\left(\Delta_{0}\right)$ and tetrahedral $\left(\Delta_{t}\right)$ complexes is related as
(A) : The relation between crystal field stabilisation energy for octahedral $\left(\Delta_{\mathrm{o}}\right)$ and tetrahedral $\left(\Delta_{\mathrm{t}}\right)$ complexes are - $\Delta_{\mathrm{t}} \approx \frac{4}{9} \Delta_{\mathrm{o}}$
274282
Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$ the correct increasing order of their first ionization enthalpy is :
1 $\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<$ P
2 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<$ P $<$ S
3 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<\mathrm{S}<$ P
4 Al $<$ Mg $<$ S $<$ Si $<$ P
Explanation:
(A) : The first ionization energy of an element is the energy needed to remove the outermost or highest energy electron from a neutral atom in the gas phase. As we know, the first ionization energy in the periodic table is decreases down the group and increases in the left to right in period. Al has less ionization energy than other because the removal of one electron from $3 p$ orbital it goes to stable configuration. The same reason occure between $\mathrm{S}$ and $\mathrm{P}$ i.e. sulphur has less ionization energy than phosphorus. Hence, the correct order will be $-\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<\mathrm{P}$.
JEE Main 24.02.2021
COORDINATION COMPOUNDS
274292
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(A) : The spectrochemical series is given as- $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<$Urea $<\mathrm{OH}^{-}<$ etOH $<$ ox $<\mathrm{O}^{2-}<\mathrm{H}_{2} \mathrm{O}<$ EDTA $<\mathrm{NCS}^{-}<\mathrm{Py},<\mathrm{NH}_{3}$ $<$ en $<$ bpy, phen $<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CH}_{3}^{-}<\mathrm{C}_{6} \mathrm{H}_{5}^{-}<\mathrm{CN}^{-}$ $\mathrm{CO}$. Thus, option (a) has the correct order
NEET 2020)
COORDINATION COMPOUNDS
274293
Crystal field splitting in octahedral coordination entities, ' $\mathbf{g}_{\mathrm{g}}$ ' set of orbital's are
(C) : Crystal field splitting in octahedral coordination entities, 'eg 'set of orbital's are $\left[\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}, \mathrm{~d}_{\mathrm{z}^{2}}\right]$
JHARKHAND - 2019
COORDINATION COMPOUNDS
274298
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)
1
2
3
4
Explanation:
(A) : Octahedral complex contains the two axial ligands, if we remove the both axial ligand then it converted into square planar field then following energy diagram take place-
JEE Main 2019
COORDINATION COMPOUNDS
274300
Crystal field splitting energy for octahedral $\left(\Delta_{0}\right)$ and tetrahedral $\left(\Delta_{t}\right)$ complexes is related as
(A) : The relation between crystal field stabilisation energy for octahedral $\left(\Delta_{\mathrm{o}}\right)$ and tetrahedral $\left(\Delta_{\mathrm{t}}\right)$ complexes are - $\Delta_{\mathrm{t}} \approx \frac{4}{9} \Delta_{\mathrm{o}}$
274282
Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$ the correct increasing order of their first ionization enthalpy is :
1 $\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<$ P
2 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<$ P $<$ S
3 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<\mathrm{S}<$ P
4 Al $<$ Mg $<$ S $<$ Si $<$ P
Explanation:
(A) : The first ionization energy of an element is the energy needed to remove the outermost or highest energy electron from a neutral atom in the gas phase. As we know, the first ionization energy in the periodic table is decreases down the group and increases in the left to right in period. Al has less ionization energy than other because the removal of one electron from $3 p$ orbital it goes to stable configuration. The same reason occure between $\mathrm{S}$ and $\mathrm{P}$ i.e. sulphur has less ionization energy than phosphorus. Hence, the correct order will be $-\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<\mathrm{P}$.
JEE Main 24.02.2021
COORDINATION COMPOUNDS
274292
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(A) : The spectrochemical series is given as- $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<$Urea $<\mathrm{OH}^{-}<$ etOH $<$ ox $<\mathrm{O}^{2-}<\mathrm{H}_{2} \mathrm{O}<$ EDTA $<\mathrm{NCS}^{-}<\mathrm{Py},<\mathrm{NH}_{3}$ $<$ en $<$ bpy, phen $<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CH}_{3}^{-}<\mathrm{C}_{6} \mathrm{H}_{5}^{-}<\mathrm{CN}^{-}$ $\mathrm{CO}$. Thus, option (a) has the correct order
NEET 2020)
COORDINATION COMPOUNDS
274293
Crystal field splitting in octahedral coordination entities, ' $\mathbf{g}_{\mathrm{g}}$ ' set of orbital's are
(C) : Crystal field splitting in octahedral coordination entities, 'eg 'set of orbital's are $\left[\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}, \mathrm{~d}_{\mathrm{z}^{2}}\right]$
JHARKHAND - 2019
COORDINATION COMPOUNDS
274298
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)
1
2
3
4
Explanation:
(A) : Octahedral complex contains the two axial ligands, if we remove the both axial ligand then it converted into square planar field then following energy diagram take place-
JEE Main 2019
COORDINATION COMPOUNDS
274300
Crystal field splitting energy for octahedral $\left(\Delta_{0}\right)$ and tetrahedral $\left(\Delta_{t}\right)$ complexes is related as
(A) : The relation between crystal field stabilisation energy for octahedral $\left(\Delta_{\mathrm{o}}\right)$ and tetrahedral $\left(\Delta_{\mathrm{t}}\right)$ complexes are - $\Delta_{\mathrm{t}} \approx \frac{4}{9} \Delta_{\mathrm{o}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274282
Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$ the correct increasing order of their first ionization enthalpy is :
1 $\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<$ P
2 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<$ P $<$ S
3 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<\mathrm{S}<$ P
4 Al $<$ Mg $<$ S $<$ Si $<$ P
Explanation:
(A) : The first ionization energy of an element is the energy needed to remove the outermost or highest energy electron from a neutral atom in the gas phase. As we know, the first ionization energy in the periodic table is decreases down the group and increases in the left to right in period. Al has less ionization energy than other because the removal of one electron from $3 p$ orbital it goes to stable configuration. The same reason occure between $\mathrm{S}$ and $\mathrm{P}$ i.e. sulphur has less ionization energy than phosphorus. Hence, the correct order will be $-\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<\mathrm{P}$.
JEE Main 24.02.2021
COORDINATION COMPOUNDS
274292
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(A) : The spectrochemical series is given as- $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<$Urea $<\mathrm{OH}^{-}<$ etOH $<$ ox $<\mathrm{O}^{2-}<\mathrm{H}_{2} \mathrm{O}<$ EDTA $<\mathrm{NCS}^{-}<\mathrm{Py},<\mathrm{NH}_{3}$ $<$ en $<$ bpy, phen $<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CH}_{3}^{-}<\mathrm{C}_{6} \mathrm{H}_{5}^{-}<\mathrm{CN}^{-}$ $\mathrm{CO}$. Thus, option (a) has the correct order
NEET 2020)
COORDINATION COMPOUNDS
274293
Crystal field splitting in octahedral coordination entities, ' $\mathbf{g}_{\mathrm{g}}$ ' set of orbital's are
(C) : Crystal field splitting in octahedral coordination entities, 'eg 'set of orbital's are $\left[\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}, \mathrm{~d}_{\mathrm{z}^{2}}\right]$
JHARKHAND - 2019
COORDINATION COMPOUNDS
274298
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)
1
2
3
4
Explanation:
(A) : Octahedral complex contains the two axial ligands, if we remove the both axial ligand then it converted into square planar field then following energy diagram take place-
JEE Main 2019
COORDINATION COMPOUNDS
274300
Crystal field splitting energy for octahedral $\left(\Delta_{0}\right)$ and tetrahedral $\left(\Delta_{t}\right)$ complexes is related as
(A) : The relation between crystal field stabilisation energy for octahedral $\left(\Delta_{\mathrm{o}}\right)$ and tetrahedral $\left(\Delta_{\mathrm{t}}\right)$ complexes are - $\Delta_{\mathrm{t}} \approx \frac{4}{9} \Delta_{\mathrm{o}}$
274282
Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$ the correct increasing order of their first ionization enthalpy is :
1 $\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<$ P
2 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<$ P $<$ S
3 $\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}<\mathrm{S}<$ P
4 Al $<$ Mg $<$ S $<$ Si $<$ P
Explanation:
(A) : The first ionization energy of an element is the energy needed to remove the outermost or highest energy electron from a neutral atom in the gas phase. As we know, the first ionization energy in the periodic table is decreases down the group and increases in the left to right in period. Al has less ionization energy than other because the removal of one electron from $3 p$ orbital it goes to stable configuration. The same reason occure between $\mathrm{S}$ and $\mathrm{P}$ i.e. sulphur has less ionization energy than phosphorus. Hence, the correct order will be $-\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}<\mathrm{S}<\mathrm{P}$.
JEE Main 24.02.2021
COORDINATION COMPOUNDS
274292
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(A) : The spectrochemical series is given as- $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<$Urea $<\mathrm{OH}^{-}<$ etOH $<$ ox $<\mathrm{O}^{2-}<\mathrm{H}_{2} \mathrm{O}<$ EDTA $<\mathrm{NCS}^{-}<\mathrm{Py},<\mathrm{NH}_{3}$ $<$ en $<$ bpy, phen $<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CH}_{3}^{-}<\mathrm{C}_{6} \mathrm{H}_{5}^{-}<\mathrm{CN}^{-}$ $\mathrm{CO}$. Thus, option (a) has the correct order
NEET 2020)
COORDINATION COMPOUNDS
274293
Crystal field splitting in octahedral coordination entities, ' $\mathbf{g}_{\mathrm{g}}$ ' set of orbital's are
(C) : Crystal field splitting in octahedral coordination entities, 'eg 'set of orbital's are $\left[\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}, \mathrm{~d}_{\mathrm{z}^{2}}\right]$
JHARKHAND - 2019
COORDINATION COMPOUNDS
274298
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)
1
2
3
4
Explanation:
(A) : Octahedral complex contains the two axial ligands, if we remove the both axial ligand then it converted into square planar field then following energy diagram take place-
JEE Main 2019
COORDINATION COMPOUNDS
274300
Crystal field splitting energy for octahedral $\left(\Delta_{0}\right)$ and tetrahedral $\left(\Delta_{t}\right)$ complexes is related as
(A) : The relation between crystal field stabilisation energy for octahedral $\left(\Delta_{\mathrm{o}}\right)$ and tetrahedral $\left(\Delta_{\mathrm{t}}\right)$ complexes are - $\Delta_{\mathrm{t}} \approx \frac{4}{9} \Delta_{\mathrm{o}}$
