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Coordination Compounds — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryCoordination Compounds4 Marks
Question
Read the passage given below and answer the following questions:
To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., $(n - 1)d$ or outer d-orbitals i.e., nd. For example, $Co^{3+}$ forms both inner orbital and outer orbital complexes, with ammonia it forms $[Co(NH_3)_6]^{3+}$ and with fluorine it forms $[CoF_6]^{3-}$ complex ion.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following is not true for $[CoF_6]^{3-}?$
  1. It is paramagnetic.
  2. It has coordination number of $6.$
  3. It is outer orbital complex.
  4. It involves $d^2sp^3$ hybridisation.
  1. $[Cr(H_2O)_6]Cl_3 ($at. no. of $Cr = 24)$ has a magnetic moment of $3.83B.M.$ The correct distribution of $3d-$electrons in the central metal of the complex is:
  1. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_\text{yz}$
  2. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$
  3. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{zy}},3\text{d}^1_{\text{z}^2}$
  4. $3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_{\text{z}^2},3\text{d}^1_\text{xz}$
  1. Which of the following is true for $[Co(NH_3)_6]^{3+}?$
  1. It is an octahedral, di magnetic and outer orbital complex.
  2. It is an octahedral, paramagnetic and outer orbital complex.
  3. It is an octahedral, paramagnetic and inner orbital complex.
  4. It is an octahedral, di magnetic and inner orbital complex.
  1. The paramagnetism of $[CoF_6]^{3-}$ is due to.
  1. $3$ electrons.
  2. $4$ electrons.
  3. $2$ electrons.
  4. $1$ electron.
  1. Which of the following is an inner orbital or low spin complex?
  1. $[Ni(H_2O)_6]^{3+}$
  2. $[FeF_6]^{3-}$
  3. $[Co(CN)_6]^{3-}$
  4. $[NiCl_4]^{2-}$
✓

Answer

  1. (d) It involves $d^2sp^3$ hybridisation.
Explanation:

lt involves $sp^3d^2$ hybridisation and not $d^2sp^3.$​​​​​​​
  1. (b) $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$
Explanation:

Magnetic moment of $3.83B.M.$ suggests that it has $3$ unpaired electrons,

$\therefore n = 3$ i.e., $Cr^{3+}: 3d^3$​​​​​​​

It involves $d^2sp^3$ hybridisation so correct distribution of electrons is $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}.$
  1. (d) It is an octahedral, di magnetic and inner orbital complex.
Explanation:

$[Co(NH_3)_6]^{3+}$ is $d^2sp^3$ hybridised with all electrons paired hence, it is diamagnetic and inner orbital complex.
  1. (b) $4$ electrons.
Explanation:

  1. (c) $[Co(CN)_6]^{3-}$​​​​​​​
Explanation:

Inner orbital complexes are formed with strong ligands as they force electrons to pair up and hence the complex will be either diamagnetic or will have less number of unpaired electrons.

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