Complexes MCQ Quiz in मल्याळम - Objective Question with Answer for Complexes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is fac-mer isomer

Concept:-

• Enantiomers: Enantiomers are non-superimposable mirror images, much like your left hand and your right hand. They are chiral and will rotate plane-polarized light. In an octahedral complex for it to be chiral and exhibit enantiomerism, it typically needs to satisfy certain characteristics. Some of these are having two or more bidentate ligands, or three different monodentate ligands, or unsymmetrical bidentate ligands. "Bidentate" refers to ligands that have two atoms available for bonding with the central metal atom/ion.
• Linkage Isomers: Linkage isomerism occurs when the ligand can coordinate to the same metal through different donor atoms. For example, a nitro group NO₂ can attach to the metal either through the nitrogen or one of the oxygens. In this case, the ligands we are considering are a simple chloride ion Cl- (which can only bind through Cl) and PPh₃ (which can only bind through P), so there's no possibility for linkage isomerism.
• Concept of Chirality: Chirality in chemistry is a property where a molecule differs from its mirror image, and it's impossible to superimpose the molecule and its mirror image on top of each other. This property gives rise to what we call enantiomers. An object that is not identical to its mirror image is called chiral. In the case of molecular chemistry, we say a molecule is chiral if it cannot be superimposed onto its mirror image.

Explanation:-

• Oxidative addition of Cl₂ to the square planar complex [IrCl(PPh₃)₃] can lead to the creation of an octahedral complex with the formula [IrCl₃(PPh₃)₃]. When Cl₂ is added, the oxidation state of the Ir increases by two and two new ligands (Cl) are added to the complex. In an octahedral geometry, both of these could bind to the central metal atom from either the same face (forming a fac-isomer) or from opposite faces (forming a mer-isomer). Therefore, the possible isomers here are:
• fac-[IrCl₃(PPh3)₃]: the three equivalent ligands (either Cl or PPh₃) are all in one face.
• mer-[IrCl3(PPh₃)₃]: the three equivalent ligands are in a meridional position, implying they occupy a 'T' shape within the coordination sphere of the metal.
• The oxidation addition of Cl₂ will not generate cis-trans isomer as they generally apply to octahedral complexes containing two different types of bidentate ligands or square planar complexes. They do not apply to this case since we have just monocidentate ligands.
• Enantiomers would not be produced as the complex is not chiral. Chirality arises in octahedral complexes when they have bidentate ligands, three different monodentate ligands or unsymmetrical bidentate ligands. In this case, we just have Cl and PPh3 as monodentate ligands.
• Linkage isomers occur when a ligand is capable of binding to a central atom in multiple places. Here, both Cl and PPh3 can only bind in one specific way, so linkage isomers are not possible.
• Hence, the correct answer to the given problem would be:
• fac-mer isomer. The oxidative addition leads to the creation of fac-[IrCl₃(PPh₃)₃] and mer-[IrCl₃(PPh₃)₃] isomers.

The correct answer is Kurnakov test

Concept:-

• Coordination compounds and Isomerism: Coordination compounds can show different types of isomerism, depending on the arrangement of the ligands around the central metal atom. The [PtA₂X₂] compound is a square planar complex where 'cis' and 'trans' isomerism could occur, given A and X are different ligands.
• Cis and Trans Isomers: In cis isomers, the same type of ligands are adjacent to each other, while in trans isomers, the same type of ligands are opposite to each other.
• Chemical Tests: Different tests for distinguishing isomers are based on the unique chemical properties and behaviors of the compounds. These tests exploit the different reactivities or physical properties of the isomers.

• The Chromyl Chloride test, Carbylamine test, and Ring test are also chemical tests, but they are not used to differentiate cis and trans isomers of Pt(II) complex. Instead, they're used to identify particular types of compounds or ions.

  a. Chromyl Chloride Test: This test is used to detect chloride ions in a solution. It's not applicable in differentiating cis and trans isomers.

  b. Carbylamine Test: This test is used to distinguish primary amines from secondary and tertiary amines. It doesn't apply to our problem of cis-trans isomer determination.

  c. Ring Test: This test is also known as the confirmation test for nitrates, where a brown ring indicates the presence of nitrate ions. This too is not relevant to distinguishing cis and trans platinum complexes.

Explanation:-

Cis and trans isomers in coordination complexes refer to the spatial arrangement of the ligands attached to the central metal atom. In simple terms, "cis" configuration means the like ligands are on the same side of the metal complex, while "trans" configuration means the like ligands are opposite to each other.

The [PtA₂X₂] represents a square-planar complex of platinum where it can form both cis and trans isomers, depending upon whether similar ligands (A and A or X and X) are adjacent (cis) or opposite (trans).

The cis and trans isomers of these types of complexes can be distinguished using the Kurnakov test (also known as Kurnakov's reaction).

The Kurnakov test can distinguish between cis-platin ([Pt(NH₃)₂Cl₂]) and trans-platin ([Pt(NH₃)₂Cl₂]), where cis-platin reacts with thiourea to form a colorless solution, while trans-platin does not react.

So, the correct answer for distinguishing cis and trans complexes of the type [PtA₂X₂] is option 3) Kurnakov test.

Concept:

Coordination number -

• Coordination number is the number of coordinate bonds with which the ligands are bound to central ion.
• Ligand can be unidentate, bidentate and polydentate and ambidentate. 

Oxidation state - 

• Oxidation state also known as the oxidation number of an element is the number of positive or negative charges assigned to that element in a compound.
• It is the number of electrons lost or gained by an atom of that element.

Explanation:

Coordination number of Cr in  K3 [Cr(C2O4)3]

• The oxalate (C₂O₄)^2- is a bidentate ligand. 
• Each oxalate ligand coordinate through two oxygen atoms.
• A total of three oxalate ligands are present in the coordination sphere with the Cr metal ion.
• Therefore, the coordination number of Cr is 6.

Oxidation state of Cr in  K3 [Cr(C2O4)3]

• The overall charge on the complex is zero.
• The Sum of oxidation states of all elements in the complex is equal to 0.
• The charge on each oxalate ion is -2, so, the oxidation state of each oxalate ion is -2.
• Let the oxidation state of Cr is 'x'.

Calculation for the oxidation state of Cr - 

K3 [Cr(C2O4)3]

+3 + x + 3(-2) = 0

x - 3 = 0

x = +3

Therefore, the oxidation state of Cr is +3.

Conclusion:

Therefore, The coordination number and oxidation state of Cr in K3 [Cr(C2O4)3] are respectively 6 and 3

Concept:

Coordination number -

• Coordination number is the number of coordinate bonds with which the ligands are bound to central ion.
• Ligand can be unidentate, bidentate and polydentate and ambidentate. 

Oxidation state - 

• Oxidation state also known as the oxidation number of an element is the number of positive or negative charges assigned to that element in a compound.
• It is the number of electrons lost or gained by an atom of that element.

Explanation:

Coordination number of Cr in  K3 [Cr(C2O4)3]

• The oxalate (C₂O₄)^2- is a bidentate ligand. 
• Each oxalate ligand coordinate through two oxygen atoms.
• A total of three oxalate ligands are present in the coordination sphere with the Cr metal ion.
• Therefore, the coordination number of Cr is 6.

Oxidation state of Cr in  K3 [Cr(C2O4)3]

• The overall charge on the complex is zero.
• The Sum of oxidation states of all elements in the complex is equal to 0.
• The charge on each oxalate ion is -2, so, the oxidation state of each oxalate ion is -2.
• Let the oxidation state of Cr is 'x'.

Calculation for the oxidation state of Cr - 

K3 [Cr(C2O4)3]

+3 + x + 3(-2) = 0

x - 3 = 0

x = +3

Therefore, the oxidation state of Cr is +3.

Conclusion:

Therefore, The coordination number and oxidation state of Cr in K3 [Cr(C2O4)3] are respectively 6 and 3

Concept:

Cis-Trans Isomers are isomers that differ in the arrangement of two ligands in square planar and octahedral geometry.

Cis isomers are isomers where the two ligands are 90 degrees apart from one another in relation to the central molecule. This is because Cis isomers have a bond angle of 90^o, between two same atoms.

Trans isomers are isomers where the two ligands are on opposite sides in a molecule because trans isomers have a bond angle of 180^o, between the two same atoms.

When naming cis or trans isomers, the name begins either with cis or trans, whichever applies, followed by a hyphen and then the name of a molecule.

Cis-trans Isomerism is possible with [Pt(en)₂Cl₂]²⁺

Only square planar and octahedral geometries can have cis or trans isomers.

(Dichloro(ethelyenediamine)platinum (II)) shows only optical isomerism. The other complexes do not show stereoisomerism.

Stereoisomerism, is also called as spatial isomerism, is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space. This contrasts with structural isomers which shares the same molecular formula, but the bond connections or their order differs.

Concept:

Crystal Field Stabilization Energy -

• It is also abbreviated as CFSE.
• It is the amount of stabilization provided by the splitting of d-orbitals in two levels.
• It is denoted by Δo in the octahedral field and Δt in the tetrahedral field.
• If  Δo > Pairing energy, the complex will be a low spin complex.
• If  Δo  < Pairing energy, the complex will be a high spin complex.

Explanation:

Given that the d4 complex has an electronic configuration as  \(\rm t_{2g}^3, e_g^1\), indicates complex is a high spin.

For high spin complexes,  Δo < Pairing energy.

For a high spin d4 complex splitting is - 

CFSE = [-0.4× 3 + 0.6 × 1] Δo .

= - 0.6 Δo 

Conclusion:

Thus, for a d4 complex has an electronic configuration as  \(\rm t_{2g}^3, e_g^1\),  Δo is always < Pairing energy.

Hence, the correct answer is option 1.

Concept:

Effective Atomic Number (EAN) rule -

• The total no. of electrons present with the central metal atom in a coordination complex is called as the EAN.
• It is also known as the 18e^- rule.

It is calculated by using the formula - ​T

Total electron on central metal = (Z-x) + 2nl, 

where

• Z = atomic no. of metal
• x = oxidation state of metal
• n = number of ligands attached
• L = no. of coordinate bonds formed

Explanation:

The total number of electrons accommodated by Fe in [Fe(CO)₅] is given by the Effective Atomic Number(EAN) rule.

According to the EAN rule, 

Total electron on Fe = (Z-x) + 2nl

where

• Z = atomic no. of Fe
• x = oxidation state of Fe
• n = number of ligands attached
• L = no. of coordinate bonds formed

Oxidation no. of Fe in [Fe(CO)₅] is 0.

Thus, EAN for Fe = (26-0) + 2 × 1 × 5 = 26+10 = 36

Hence, the total number of electrons accommodated by Fe = 36

Conclusion:

The total number of electrons accommodated by Fe in  [Fe(CO)₅] is 36.

Correct answer: 4)

Concept:

• According to Crystal Field Theory, as a ligand approaches the metal ion, the electrons in the d-orbitals and those in the ligand repel each other due to repulsion between like charges.
• Thus the d-electrons closer to the ligands will have a higher energy than those further away which results in the d-orbitals splitting in energy. 
• The d_xy, d_xz and d_yz orbitals all lie between the bond axes.
• These three orbitals will be changed in energy only a little.
• These orbitals are more like non-bonding orbitals.
• These orbitals are sometimes called the "t_2g" set of orbitals.

Explanation:

• d-subshell has five orbitals, i.e. d_x²−y²​,d_z²​,d_xy,​d_xz​ and d_yz​.
• There are two d orbitals that will interact very strongly with these ligands: the dx2−y2, which lies directly on the x and y axes, and the dz2, which lies directly on the z-axis.
• Together, these two metal orbitals and the ligand orbitals that interact with them will form new bonding and antibonding molecular orbitals.
• The dx2−y2 and the dz2 orbitals lie along the bond axes.
• These two orbitals will be raised relatively high in energy.
• These orbitals are like antibonding levels.
• These orbitals are sometimes called the "e_g" set of orbitals. The term "e_g" comes from the mathematics of symmetry.

Conclusion:

Thus, dx2 - y2 and dz2 sets of the d- orbitals are directly oriented towards the ligands in octahedral coordination compounds.

CONCEPT:

Isoelectronic Species

• Two species are said to be isoelectronic if they have the same number of electrons.
• This includes atoms or ions, regardless of their chemical nature.
• To find the total number of electrons:

  Total Electrons = Atomic Number ± Charge

EXPLANATION:

• Sn₉⁴⁻:
  • Atomic number of Sn (in carbon Family)
  • Charge = 4− → Add 4 electrons
  • Total electrons = 9x4 + 4 = 40
• Bi₉⁵⁺:
  • Atomic number of Bi (Nitrogen family)
  • Charge = 95+ → Remove 5 electrons
  • Total electrons = 9x5 - 5 = 40

Therefore, the isoelectronic species are  Sn94-, Bi95+

CONCEPT:

Metal Clusters

• Metal clusters are chemical species in which two or more metal atoms are directly bonded together.
• These are commonly formed by transition metals that:
  • Have partially filled d orbitals
  • Exhibit multiple oxidation states
  • Show a high tendency for metal–metal bonding
• Clusters can be stabilized by ligands like CO, halides, etc.

EXPLANATION:

• Niobium (Nb), Molybdenum (Mo), and Technetium (Tc) belong to the middle of the d-block (groups 5–7).
• These elements have a high tendency to form stable metal–metal bonds, especially in their carbonyl and halide complexes.
• Examples include:
  • [Nb₆Cl₁₂]²⁻: a known niobium cluster
  • [Mo₆Cl₁₄]²⁻: molybdenum cluster with octahedral Mo–Mo bonding
  • Tc carbonyl clusters: like [Tc₂(CO)₁₀], [Tc₆(CO)₁₅]

Therefore, the metals with the greatest tendency to form metal clusters are:
Nb, Mo, Tc