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Coordination Compounds · Question 97

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

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)

AIPMT 2001

Keep practising — Coordination Compounds

2

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]
AIPMT 2001
3

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
AIPMT 2000
4

Shape of Fe(CO)5 is

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

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
AIPMT 2000

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