Coordination Chemistry- 1 - Chemistry Inorganic Free MCQ Test with solutions

Because DMG is a polydentate ligand that binds the central atom nickel forming a ring-like structure.

Metal carbonyls are coordination complexes of transition metals with carbon monoxide ligands. Metal carbonyls are useful in organic synthesis and as catalysts or catalyst precursors in homogeneous catalysis,

Fe(CO)₅, Ni(CO)₄ etc. 

• NH₃ is a neutral ligand.
• NO⁺ is a cationic ligand.
• Cl^– is an anionic ligand.
• However, BF₃ is not a ligand as it does not have lone pairs to donate and can not share a pair of electrons.

Cl^- have extra e and also empty orbital so act as sigma as well as pi donor..NH3 is only sigma donor bcoz of unavailbility of vacant orbital.

• In 1^st compound, CN is a unidentate ligand and hence it will surround the Fe atom from 6 different sides.
• On the other hand, the 2^nd compound contains bidentate ligand so each (en) will have two different sites of donating electrons and hence it required 3 (en) atoms for surrounding Co atom from 6 different sides.
  Hence A is correct.

• Oxidation number:
  x + (-2 x 2) + 0 x 2 = -1
  x = +3
• The coordination number is 6 because C₂O₄^2-  is a bidentate ligand and NH₃ is monodentate. 
• Magnetic moment = √n(n+2)
  where n is no. of unpaired electrons.
  For Cr, n = 3 so the Magnetic moment will be √15 BM.

Ti⁴⁺ is the most stable ion because it has a completely filled argon core with an empty 3d orbital, which contributes to its stability. The other ions have the following characteristics:

• V²⁺: Has a partially filled d orbital, making it less stable.
• Mn³⁺: Also has a partially filled d orbital, which contributes to lower stability.
• Cr³⁺: Similar to the others, it possesses a partially filled d orbital, resulting in reduced stability.

In summary, the presence of a completely filled d orbital in Ti⁴⁺ enhances its stability compared to the other ions.

The correct option is C

• Six empty orbitals (two d, one s and three p) are available, so whether the ligand is weak (H₂O) or strong (CN⁻)
• The six orbitals hybridize to give six equivalent d²sp³ hybrid orbitals.
• Hence, in both [Cr(H₂O)₆]³⁺ 
  and [Cr(CN)₆]³⁻ ions chromium is in a state of d²sp³ hybridization.

• In [Co(NH₃)₆]³⁺, the oxidation state of cobalt is +3. Ammonia is a strong field ligand so it pairs up 4 unpaired electrons and frees up 2,3−d orbitals.
• These 3−d orbitals are involved in hybridisation with one 4s and three 4p orbitals forming an inner orbital complex, so hybridisation of [Co(NH₃)₆]³⁺ is d²sp³.

Correct option is A)

From the formula CrCl_2​Br.6H₂​O following complexes can be derived
- [Cr(H₂​O)₆​]BrCl_2​[CrBr(H₂​O)₅​]Cl₂​.H_2​O
- [CrCl(H₂​O)₅​]BrCl.H_2​O[CrCl₂​(H_2​O)₄​]Br.2H₂​O
- [CrBrCl(H_2​O)_4​]Cl.2H₂​O.

In [Co(NH₃)₄Cl₂]⁺ 

• (O.S. of Co + charge on ligands) = +1
• ∴ (O.S. of Co + 4 × charge of NH₃ molecule + 2 × charge of Cl^– ion) = +1
• ∴ O.S. of Co + 4 × 0 + 2 × –1 = +1
• Oxidation state of Co = +3

In [Co(NH₃)₄ (H₂O)Cl]²⁺

• (O.S. of Co + charge on ligands) = +2
• (O.S. of Co + 4 × charge of NH₃ molecule + charge of H₂O + charge of Cl^– ion) = +2
• ∴ O.S. of Co + 4 × 0 + 0 –1 = +2
• Oxidation state of Co = +3

Hence, Oxidation State of Cobalt does not change.

The correct option is B. Δt<P

• Tetrahedral complexes always show high spin complexes i.e. no pairing of electrons, because in tetrahedral complexes the coordination number of metal is four only, thus crystal field splitting is less in tetrahedral complexes.
• Hence, pairing energy is more compared to the crystal field splitting energy.

Oxidation state of Ni in [NiF₆]^2- :
x − 6 = −2
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand.
⇒ low spin d⁶ Ni(IV) complex.
⇒ unpaired electrons = 0. 
⇒ zero magnetic moment.

• The electron configuration is [Ar]3d⁵4s¹.
• We have to accommodate the 6 Ligands and the fact that CO is a strong ligand, this results in d²sp³ hybridization.
• Therefore, there are no unpaired electrons in Cr(CO)6. So n = 0. Thus, the spin only magnetic moment is also 0.

Be is the only group 2 element that does not form a stable complex with [EDTA]⁴⁻⋅Mg²⁺ and Ca²⁺ have the greatest tendency to form complexes with [EDTA]⁴⁻.

• When EDTA solution is added to the Mg²⁺ ion solution, all six-coordinate sites of Mg2+ are occupied, and a colourless [Mg−EDTA]²⁻ chelate is formed. This reaction also results in the lowering of the pH of the solution.
• So, the correct answer is 'Four coordinates sites of Mg2+ are occupied by EDTA and the remaining two sites are occupied by water molecules.'

The total number of geometrical isomers for the complex [RhCl(CO)(PPh₃​)(NH₃)] is 3 since it is a square planar. 

• [Cr(H₂O)₆]²⁺ is distorted because of more number of electrons i.e. 4 electrons (electron-electron repulsion).
• But in the case of Mn & Fe both of them has 3d⁵ configuration which is more stable & decreases the chances of distortion.
• In the case of Cr³⁺ less number of electrons are present, so, less electron-electron repulsion hence lesser will be the distortion.
• When there is unsymmetrical filling of electron in Eg orbital occur, the shape will distorted e.g. in the case of d⁴. This is known as Jahn-Teller distortion.
  Hence A is correct.

The correct option is B

• The complexes in which the metal ion have a unsymmetrical eg or t_2​g orbital shows Jahn-taller distortions.
• In the complex [Cu(H2​O)_6​]²⁺ the configuration is  t_2​g⁶eg³ i.e eg orbital is unsymmetrical and hence will show Jahn-Taller distortions.