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Coordination Compounds

AIPMT 2001MCQEasy📈 Topic in 38% papers+4−1
Ques. 1AIPMT 2001

Which statement is incorrect?

  1. Ni(CO)4 – tetrahedral, paramagnetic
  2. [Ni(CN)4]2- – square planar, diamagnetic
  3. Ni(CO)4 – tetrahedral, diamagnetic
  4. [NiCl4]2- – tetrahedral, paramagnetic.

Answer

(1) Ni(CO)4 – tetrahedral, paramagnetic

Solution

Ni(CO)4: Ni(0), 3d84s2; CO is a strong field ligand, so the 4s electrons pair into 3d 3d10, sp3, tetrahedral, no unpaired electron diamagnetic.

[Ni(CN)4]2-: Ni2+, 3d8; CN- pairs the electrons dsp2, square planar, diamagnetic.

[NiCl4]2-: Ni2+, 3d8; Cl- is a weak field ligand sp3, tetrahedral, 2 unpaired electrons paramagnetic.

"Ni(CO)4 – tetrahedral, paramagnetic" is incorrect.

Answer: (1)

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Ques. 2AIPMT 2001

Which of the following will exhibit maximum ionic conductivity?

  1. K4[Fe(CN)6]
  2. [Co(NH3)6]Cl3
  3. [Cu(NH3)4]Cl2
  4. [Ni(CO)4]
Ques. 3AIPMT 2000

In qualitative analysis of second group in laboratory, H2S gas is passed in acidic medium for precipitation. When Cu2+ and Cd2+ react with KCN, then for product, true statement is

  1. K2[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable
  2. K2[Cu(CN)4] – less stable; K2[Cd(CN)4] – more stable
  3. K3[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable
  4. K3[Cu(CN)4] – less stable; K3[Cd(CN)4] – more stable
Ques. 4AIPMT 2000

Shape of Fe(CO)5 is

  1. octahedral
  2. square planar
  3. trigonal bipyramidal
  4. square pyramidal
Ques. 5AIPMT 2000

Which complex compound will give four isomers?

  1. [Fe(en)3]Cl3
  2. [Co(en)(NH3)2Cl2]Cl
  3. [Fe(PPh3)3NH3ClBr]Cl
  4. [Co(PPh3)3Cl]Cl3

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