JEE Advanced Previous Year Questions (2021 - 2024): Coordination Compounds | Chemistry for JEE Main & Advanced PDF Download

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Q1: Among [Co(CN) 4]4 ?, [Co(CO) 3(NO)], XeF 4, [PCl 4]?, [PdCl 4]² ?, [ICl 4] ?, [Cu(CN) 4]³ ?, and P 4, the total number
of species with tetrahedral geometry is ______.     [JEE Advanced 2024 Paper 2]
Ans: 4 or 5
[Co(CN) 4] 4? ? Co ° ? 3d 7 4s²
Due to SFL, CN ? pairing and transference of electrons take place, and hybridisation is dsp².
Geometry ? Square planar
OR
? Hybridisation [Co(CN) 4]4 ? ? sp³
[Co(CO) 3(NO)] ? Co ?¹ ? 3d 8 4s² ? 3d¹ ° (In presence of S.F.L)
Here NO is present in +1 state.
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Q1: Among [Co(CN) 4]4 ?, [Co(CO) 3(NO)], XeF 4, [PCl 4]?, [PdCl 4]² ?, [ICl 4] ?, [Cu(CN) 4]³ ?, and P 4, the total number
of species with tetrahedral geometry is ______.     [JEE Advanced 2024 Paper 2]
Ans: 4 or 5
[Co(CN) 4] 4? ? Co ° ? 3d 7 4s²
Due to SFL, CN ? pairing and transference of electrons take place, and hybridisation is dsp².
Geometry ? Square planar
OR
? Hybridisation [Co(CN) 4]4 ? ? sp³
[Co(CO) 3(NO)] ? Co ?¹ ? 3d 8 4s² ? 3d¹ ° (In presence of S.F.L)
Here NO is present in +1 state.
? Hybridisation ? [Co(CO) 3(NO)] ? sp³
Geometry = T etrahedral
XeF 4 ? 4bp + 2lp ? sp³d²
Geometry = Square planar
PCl 4? ? 4lb + 0lp
sp³ ? tetrahedral
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Q1: Among [Co(CN) 4]4 ?, [Co(CO) 3(NO)], XeF 4, [PCl 4]?, [PdCl 4]² ?, [ICl 4] ?, [Cu(CN) 4]³ ?, and P 4, the total number
of species with tetrahedral geometry is ______.     [JEE Advanced 2024 Paper 2]
Ans: 4 or 5
[Co(CN) 4] 4? ? Co ° ? 3d 7 4s²
Due to SFL, CN ? pairing and transference of electrons take place, and hybridisation is dsp².
Geometry ? Square planar
OR
? Hybridisation [Co(CN) 4]4 ? ? sp³
[Co(CO) 3(NO)] ? Co ?¹ ? 3d 8 4s² ? 3d¹ ° (In presence of S.F.L)
Here NO is present in +1 state.
? Hybridisation ? [Co(CO) 3(NO)] ? sp³
Geometry = T etrahedral
XeF 4 ? 4bp + 2lp ? sp³d²
Geometry = Square planar
PCl 4? ? 4lb + 0lp
sp³ ? tetrahedral
[PdCl 4]² ? ? Pd² ?, Cl ? behaves as SFL
Pd² ? ? 4d 8 ? dsp² ? square planar
ICl 4? ? 4bp + 2lp
Hybridisation = sp³d²
Geometry = square planer
[Cu(CN) 4]³ ? ? Cu ?¹ ? 3d¹ ° ? sp³
Geometry = T etrahedral
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Q1: Among [Co(CN) 4]4 ?, [Co(CO) 3(NO)], XeF 4, [PCl 4]?, [PdCl 4]² ?, [ICl 4] ?, [Cu(CN) 4]³ ?, and P 4, the total number
of species with tetrahedral geometry is ______.     [JEE Advanced 2024 Paper 2]
Ans: 4 or 5
[Co(CN) 4] 4? ? Co ° ? 3d 7 4s²
Due to SFL, CN ? pairing and transference of electrons take place, and hybridisation is dsp².
Geometry ? Square planar
OR
? Hybridisation [Co(CN) 4]4 ? ? sp³
[Co(CO) 3(NO)] ? Co ?¹ ? 3d 8 4s² ? 3d¹ ° (In presence of S.F.L)
Here NO is present in +1 state.
? Hybridisation ? [Co(CO) 3(NO)] ? sp³
Geometry = T etrahedral
XeF 4 ? 4bp + 2lp ? sp³d²
Geometry = Square planar
PCl 4? ? 4lb + 0lp
sp³ ? tetrahedral
[PdCl 4]² ? ? Pd² ?, Cl ? behaves as SFL
Pd² ? ? 4d 8 ? dsp² ? square planar
ICl 4? ? 4bp + 2lp
Hybridisation = sp³d²
Geometry = square planer
[Cu(CN) 4]³ ? ? Cu ?¹ ? 3d¹ ° ? sp³
Geometry = T etrahedral
P   is tetrahedral.
Q2: Among V(CO) 6, Cr(CO) 5, Cu(CO) 3, Mn(CO) 5, Fe(CO) 5, [Co(CO) 3]³ ?, [Cr(CO) 4] 4 ?, and Ir(CO) 3, the total
number of species isoelectronic with Ni(CO) 4 is ______.     [JEE Advanced 2024 Paper 1]
[Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]
Ans: 1
T otal number of electrons in Ni(CO) 4 = 84
4
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Q1: Among [Co(CN) 4]4 ?, [Co(CO) 3(NO)], XeF 4, [PCl 4]?, [PdCl 4]² ?, [ICl 4] ?, [Cu(CN) 4]³ ?, and P 4, the total number
of species with tetrahedral geometry is ______.     [JEE Advanced 2024 Paper 2]
Ans: 4 or 5
[Co(CN) 4] 4? ? Co ° ? 3d 7 4s²
Due to SFL, CN ? pairing and transference of electrons take place, and hybridisation is dsp².
Geometry ? Square planar
OR
? Hybridisation [Co(CN) 4]4 ? ? sp³
[Co(CO) 3(NO)] ? Co ?¹ ? 3d 8 4s² ? 3d¹ ° (In presence of S.F.L)
Here NO is present in +1 state.
? Hybridisation ? [Co(CO) 3(NO)] ? sp³
Geometry = T etrahedral
XeF 4 ? 4bp + 2lp ? sp³d²
Geometry = Square planar
PCl 4? ? 4lb + 0lp
sp³ ? tetrahedral
[PdCl 4]² ? ? Pd² ?, Cl ? behaves as SFL
Pd² ? ? 4d 8 ? dsp² ? square planar
ICl 4? ? 4bp + 2lp
Hybridisation = sp³d²
Geometry = square planer
[Cu(CN) 4]³ ? ? Cu ?¹ ? 3d¹ ° ? sp³
Geometry = T etrahedral
P   is tetrahedral.
Q2: Among V(CO) 6, Cr(CO) 5, Cu(CO) 3, Mn(CO) 5, Fe(CO) 5, [Co(CO) 3]³ ?, [Cr(CO) 4] 4 ?, and Ir(CO) 3, the total
number of species isoelectronic with Ni(CO) 4 is ______.     [JEE Advanced 2024 Paper 1]
[Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]
Ans: 1
T otal number of electrons in Ni(CO) 4 = 84
4
Q3: Among the following complexes, the total number of diamagnetic species is _______.
[Mn(NH 3)6]³ ?, [MnCl 6]³ ?, [FeF 6]³ ?, [CoF 6]³ ?, [Fe(NH 3)6]³ ?, and [Co(en) 3]³ ?
[Given, atomic number: Mn = 25, Fe = 26, Co = 27]
en = H 2NCH 2CH 2NH 2     [JEE Advanced 2024 Paper 1]
Ans: 1
Mn³ ? ? [Ar]3d 4
d 4 confi guration in t 2g and e 9 orbitals will always have unpaired electrons irrespective of SFL and WFL.
Fe³ ? ? [Ar]3d 5
d 5 confi guration will also have unpaired electrons irrespective of SFL and WFL.
Co³ ? ? [Ar]3d 6
d 6 ? it can be both paramagnetic or diamagnetic based on the fi eld of ligands.
In case of F ? ? weak fi eld ligand, confi guration will be t e hence it is paramagnetic but in case of en ?
strong fi eld ligand, confi guration will be t e hence it will be diamagnetic.
Q4: Among the following options, select the option in which each complex in Set-I shows geometrical
isomerism and the two complexes in Set-II are ionization isomers of each other.
[en = H 2NCH 2CH 2NH 2]      [JEE Advanced 2024 Paper 1]
(a) Set-I: [Ni(CO) 4] and [PdCl 2(PPH 3) 2]
Set-II: [Co(NH 3)5Cl]SO 4 and [Co(NH 3) 5(SO 4)]Cl
(b) Set-I: [Co(en)(NH 3)2Cl 2] and [PdCl 2(PPH 3)2]
Set-II: [Co(NH 3)6][Cr(CN) 6] and [Cr(NH 3)6][Co(CN) 6]
4
2g
2
g
6
2g
0
g