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Coordination Compounds · Question 4

JEE Advanced 2025 Paper 2NumericalEasy📈 Topic in 33% papers+4
1

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br)6]3- and [Mn(CN)6]3- is .

Answer

7.5 to 7.8

Solution

Both contain Mn3+ (3d4); = n(n+2) B.M.

Br- (weak field): high spin t2g3eg1, n = 4 ⇒ = 24 = 4.90 B.M.

CN- (strong field): low spin t2g4, n = 2 ⇒ = 8 = 2.83 B.M.

Sum = 4.90 + 2.83 = 7.73 B.M.

Answer: 7.73

JEE Advanced 2025 Paper 2

Keep practising — Coordination Compounds

2

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 = H2NCH2CH2NH2]

  1. Set-I: [Ni(CO)4] and [PdCl2(PPh3)2] Set-II: [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl
  2. Set-I: [Co(en)(NH3)2Cl2] and [PdCl2(PPh3)2] Set-II: [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
  3. Set-I: [Co(NH3)3(NO2)3] and [Co(en)2Cl2] Set-II: [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl
  4. Set-I: [Cr(NH3)5Cl]Cl2 and [Co(en)(NH3)2Cl2] Set-II: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 H2O
JEE Advanced 2024 Paper 1
3

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 . [Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]

JEE Advanced 2024 Paper 1
4

Among the following complexes, the total number of diamagnetic species is . [Mn(NH3)6]3+, [MnCl6]3-, [FeF6]3-, [CoF6]3-, [Fe(NH3)6]3+, and [Co(en)3]3+ [Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = H2NCH2CH2NH2]

JEE Advanced 2024 Paper 1
5

Among [Co(CN)4]4-, [Co(CO)3(NO)], XeF4, [PCl4]+, [PdCl4]2-, [ICl4]-, [Cu(CN)4]3- and P4 the total number of species with tetrahedral geometry is .

JEE Advanced 2024 Paper 2

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