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Chemistry Test - 11

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Chemistry Test - 11
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  • Question 1
    5 / -1

    Which one of the following d orbitals has a different shape but degenerate with the rest of the d orbitals?

    Solution

    Concept:

    • An atom consists of many orbitals that are distinguished from each other based on their shape, size, and orientation in space.
    • There are four quantum numbers n, l, m, s which are used to identify an electron.

    • The azimuthal quantum number determines the three-dimensional shape of the orbital quantum number and is responsible for the shapes of 5 d orbitals

    Explanation:

    • Orbitals having the same energy levels are called degenerate orbitals.
    • Orbitals having the same principal quantum number ' n ' and azimuthal quantum number ' l ' are generally degenerate.
    • The d-orbitals have ' l ' value = 2, so when they have the same value of ' n ', they will be degenerate.
    • The shape is called baby-soother type shape whereas the others are clover-shaped.

    Hence, the d orbitals that have a different shape but degenerate with the rest of the d orbitals are dz2

    Additional Information

    • The shapes are obtained through the wave functions of electron distribution which is different for different values of magnetic quantum number ' m '.
  • Question 2
    5 / -1

    What is the shape of [PtCl4]2- ?

    Solution

     

    Concept:

    Crystal field theory:

    • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
    • An electric field is produced at the metal ion by the surrounding ligands.
    • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
    • CFT theory considers the ligands as point charges.
    • There is no overlap between the ligand orbitals and metal ion orbitals.
    • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

    The geometry of the complex formed.

    • Placement of the donor atoms at different stereochemical positions will perturb the metal ions to a different extent.
    • This will result in the formation of different stereochemistries or structures.
    • The stereochemistry also depends on the coordination number and the type of orbitals involved.
    H number Hybridization Geometry
    2 sp linear
    3 sp2 trigonal planar
    4 sp3 tetrahedral
    5 dsp2 square planar
    6 sp3d trigonal bipyramidal
    7 sp3d2 octahedral
    8 sp3d3 pentagonal
     

    Explanation:

    [PtCl4]2- :
    • Platinum (Pt) in the complex  [PtCl4]2- is in a +2 oxidation state. Hence the electronic configuration is [Xe]4f145d8.
    • Out of 8 5d electrons, 6 are paired and two are unpaired in the ground state.
    • The CFSE of a square planar complex is:

    Crystal Field Theory

    • In presence of CFSE, the degeneracy of the dx2-y2 and dz2 breaks further.
    • Thus, all the electrons gets paired leaving one of the d orbitals vacant.
    • Hence, inner 5d, 6s and 6p orbitals are involved in hybridisation.
    • The resulting hybridisation is dsp2 and the geometry is square planar.

    Solved: Draw a plausible structure to represent:(a) [PtCl4]2−(b... |  Chegg.com

    Hence, the shape of [PtCl4]2- is square planar.

  • Question 3
    5 / -1
    The number of water molecule(s) not coordinated to copper ion directly in CuSO4.5H2O is:
    Solution

    Concept:

    According to Werner’s theory,

    • one of the five molecules of H2O molecules will attach to the SO4 molecules in the form of hydrogen bonding.
    • It will act as both a primary and secondary molecules so the coordination number will be 4.

    Explanation:

    • Copper sulphate pentahydrate contains copper (II) in a geometry best described as distorted octahedral.
    • The copper (II) is bound to four water molecules in a square-planar geometry and two oxygen atoms from two sulphate ions (one H2O is H-Bonded here).

    This salt dissolves in water to produce the pale-blue [Cu(H2O)6]2+ ion, in which two of the water molecules are less tightly held and have longer bond distances.

    • In CuSO4.5H2O, four H2O molecules are directly coordinated to the central metal ion (Cu+2) while one H2O molecule is hydrogen bonded with SO42-.

  • Question 4
    5 / -1
    Two complexes [Cr(H2O)6]Cl3 (a) and  [Cr(NH3)6]Cl3 (b) are violet and yellow coloured, respectively. The incorrect statement regarding them is
    Solution

    Explanation:

    Δ0 order will be compared by Spectro chemical series, not by energies of violet and yellow light

    Δ0 order is [Cr(H2O)6]Cl3 < [Cr(NH3)6]Cl3 

    Concept:

    When ligands attach to a transition metal to form a coordination complex, electrons in the d orbital split into high energy and low energy orbitals.

    The difference in energy of the two levels is denoted as ∆, and it is a characteristic property both of the metal and the ligands, the "o" subscript on the ∆ indicates that the complex has octahedral geometry.

    • ‘A’ [Cr(H2O)6]Cl3 Hexaaquachromium (III) chloride absorbs yellow light of less energy and emits the violet light of high energy with complementary colour because  H2O a is weak field ligand.
    • But when we consider a case ‘B’ [Cr(NH3)6]Cl3 Hexaamminechromium (III) chloride, due to presence of strong field ligand NH3, it absorbs high energy violet light and emits low energy complementary colour.


    The energy difference, Δo, determines the color of the coordination complex. According to the spectrochemical series, the high spin ligands are considered “weak field”, and absorb longer wavelengths of light (weak Δo), while complexes with low spin ligands absorb light of greater frequency (high Δo). ∆₀ order will be compared by spectro chemical series not by energies of violet & yellow light so Δo order is

    [Cr(H2O)6] Cl3<[Cr(NH3)6 ] Cl3

  • Question 5
    5 / -1
    Which among the following statements is CORRECT about [Cu (NH3)4]SO?
    Solution

    Explanation:

    • The complex given here is [Cu (NH3)4]SO4.
    • Hybridization is dsp2, square planar with one unpaired electron (Paramagnetic)
    • The coordination sphere is non-ionizable, when the complex breaks, it forms:

     [Cu (NH3)4]SO4 ⇔  [Cu (NH3)4]+2 + SO42-

    • The bond between Cu and ammonia is not broken due to the presence of a strong Co-ordinate bond between them.

    Additional Information Crystal field theory:

    • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
    • An electric field is produced at the metal ion by the surrounding ligands.
    • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
    • CFT theory considers the ligands as point charges.
    • There is no overlap between the ligand orbitals and metal ion orbitals.
    • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.
    • The bond formed by the ligands and the metal ion forms a coordination sphere.
    • The coordination sphere is non-ionizable in solution because of the co-ordinate bond between them.
    • The oxidation number or primary valency of the metal ion is satisfied by the side ions which form the ionizable sphere.
  • Question 6
    5 / -1
    The number of bridging CO ligand(s) and Co-Co bond Co2(CO)8, respectively are
    Solution

    Concept:

    The structure of Co2(CO)8 (a polynuclear metal carbonyl) can be written as:

    The total number of bridging CO ligands = 2 and the Co-Co bond =1.
  • Question 7
    5 / -1
    The number of ions furnished per molecule of the complex [Ni(NH3)4]Cl2 is
    Solution

    Concept:

    Co-ordination complexes and CFT:

    • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
    • An electric field is produced at the metal ion by the surrounding ligands.
    • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
    • CFT theory considers the ligands as point charges.
    • There is no overlap between the ligand orbitals and metal ion orbitals.
    • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

    Explanation:

    • The metal ion has two valencies:
      • The oxidation state or the primary valency.
      • The coordination number or the secondary valency.
    • The secondary valencies are satisfied by coordinating with primary ligands.
    • The bond formed by the ligands and the metal ion forms a coordination sphere.
    • The coordination sphere is non-ionizable in solution because of the co-ordinate bond between them.
    • The oxidation number or primary valency of the metal ion is satisfied by the side ions which form the ionizable sphere.
    • For example:

    Werner's co-ordination theory-primary valency-secondary valency. | by  KAKALI GHOSH , Teacher,blogger. M.Sc chemistry. | Medium

    • In the complex [Ni(NH3)4]Cl2, Nickel is the central metal atom. Its coordination number is four.
    • The Coordination sites are occupied by four ammonia ligands, which are neutral in charge.
    • The overall charge on Ni ion is +2, which is satisfied by the two chloride ions.
    • In solution, the ammine molecules are non-ionizable but the chloride ions are hence, the number of ions furnished will be two.
    • The dissociation takes place as:

    [Ni(NH3)4]Cl→ [Ni(NH3)4]2+ + 2Cl-

    On Ionisation, we get 2 Cl- ions & 1 [Ni(NH3)4]2+ ion

    Hence, the number of ions furnished per molecule of the complex [Ni(NH3)4]Cl2 is Three.

  • Question 8
    5 / -1
    Mn2(CO)10 is an organometallic compound due to the presence of?
    Solution

    Concept:

    Mn2(CO)10 is an organometallic compound due to the presence of Mn–C bond. The metal-carbon bond in organometallic compounds possess both σ and –π character. The M–C, σ bond is formed by the donation of lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal. The M–C, and π–bonds are formed by the donation of pair of electrons from a filled d-orbital of metal into vacant antibonding π* orbital of CO. The M–L bonding creates a synergic effect which strengthens the bond between CO and the metal.

    The structure of Mn2(CO)10 is shown below.

  • Question 9
    5 / -1

    The complex ion [Co(NH3)6]3+ is formed by sp3 d2 hybridisation. Hence the ion should possess

    Solution

    Concept-

    Crystal field theory:

    • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
    • An electric field is produced at the metal ion by the surrounding ligands.
    • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
    • CFT theory considers the ligands as point charges.
    • There is no overlap between the ligand orbitals and metal ion orbitals.
    • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

    Geometry of the complex formed.

    • Placement of the donor atoms at different stereochemical positions will perturb the metal ions to a different extent.
    • This will result in the formation of different stereochemistries or structure.
    • The stereochemistry depends on the co-ordination number.


    Explanation:

    • The complex given here is [Co(NH3)6]3+  
    • The complex has sp3d2 hybridization which means 4d, 4s, and 4p orbitals of Cobalt are involved and the cation has +3 charge.
    • The ligand ammonia is a strong field ligand and forms low spin complexes.
    • The cation undergoes hybridisation in the following way-

    • Six ligands are attached and it corresponds to C.N number six which gives octahedral geometry.

    Additional Information

    • Strong field ligands produce a greater amount of perturbation
    • Weak field ligands produce a lower amount of perturbation.
    • Ammonia is was strong field ligand and it produces spin pairing in the following way

    • Experimental data also suggest that it has d2sp3 hybridisation.
  • Question 10
    5 / -1
    Pick the correct name of [Co(NH3) 5 Cl]Cl 2
    Solution

    Concept:

    IUPAC stands for International Union of Pure and Applied Chemistry. In this, the naming of the compound is followed by a certain rule to avoid confusion all over the world.

    As per IUPAC rules, additive nomenclature was founded in order to describe the structures of coordination entities or complexes, but this method is readily extended to other molecular entities as well. 

    Mononuclear complexes are considered to consist of a central atom, often a metal ion, which is bonded to surrounding small molecules or ions, which are referred to as ligands

    Explanation:

    The oxidation state of Co in [Co(NH35 Cl]Cl 2 is x : x + 5 × 0 - 1 = + 2 ⇒ x = +3. 

    From the above explanation we know that in mononuclear complexes we have a central atom which is metal in our case it is Cobalt which is surrounded by ammonia and Clmolecules then its IUPAC name will be Pentaamminenitritocobalt (III) ion 

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