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Coordination Compounds · Question 52

JEE Main 2025 (Online) 2nd April Evening ShiftMCQEasy📈 Topic in 42% papers+4−1
1

The type of hybridization and the magnetic property of [MnCl6]3- are,

  1. sp3d2, paramagnetic with four unpaired electrons.
  2. d2sp3, paramagnetic with four unpaired electrons.
  3. sp3d2, paramagnetic with two unpaired electrons.
  4. d2sp3, paramagnetic with two unpaired electrons.

Answer

(1) sp3d2, paramagnetic with four unpaired electrons.

Solution

Oxidation state of Mn: x + 6(-1) = -3 x = +3

Mn3+: [Ar] 3d4

Cl- is a weak-field ligand, so the 3d electrons do not pair: 4 unpaired electrons remain in 3d.

The six ligand pairs go into 4s, 4p and 4d orbitals sp3d2 (outer orbital, high spin, octahedral).

= 4(4+2) = 4.9 BM paramagnetic with four unpaired electrons.

Answer: (1)

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2

The d-orbital electronic configuration of the complex among [Co(en)3]3+, [CoF6]3-, [Mn(H2O)6]2+ and [Zn(H2O)6]2+ that has the highest CFSE is:

  1. t2g6eg0
  2. t2g3eg2
  3. t2g4eg2
  4. t2g6eg4
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3

The correct increasing order of stability of the complexes based on 0 value is : I. [Mn(CN)6]3- II. [Co(CN)6]4- III. [Fe(CN)6]4- IV. [Fe(CN)6]3-

  1. I < II < IV < III
  2. IV < III < II < I
  3. III < II < IV < I
  4. II < III < I < IV
JEE Main 2025 (Online) 29th January Morning Shift
4

Match List - I with List - II. Choose the correct answer from the options given below :

  1. (A)-(III), (B)-(I), (C)-(I), (D)-(IV)
  2. (A)-(III), (B)-(II), (C)-(III), (D)-(IV)
  3. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
JEE Main 2025 (Online) 29th January Morning Shift
5

Match List - I with List - II. Choose the correct answer from the options given below:

  1. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  4. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
JEE Main 2025 (Online) 28th January Evening Shift

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