hots cmp nutshell group G d f block and Co Ordination Compounds

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					                                    High Order Thinking Skill Question

                                           d & f block elements:

                                                 Group - G

One mark Questions

Q 1: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Ans: As Oxygen & Fluorine have small size & high electronegetivity.

Q 2 : Why do Zr & Hf exhibit simimlar Properties?

Ans: Due to Lanthanide contraction .

Q3: Which of the 3d series element exihibits the largest number of oxidation state &why?

Ans: Mn as it has 5 unpaired electrons.

Q4 : Out of Co & Zn salts which one is attarcted by a magnet . why?

Ans: Co salts Asthey have unpaired electrons.

Q5: Why Cu prous ion is not stable in aquous solutions ?

Ans: As hydration energy is much more than 2nd ionisation enthalpy which compansate it to give Cupric
ion.

Two Marks Questions

Q 6 :Use Hunds rule to drive the electronic configuration Ce+++ & calculate its magnetic moment on the
basis of “spin - only “ formula .

Ans: Ce (Z=58)[Xe]4f15d16s

Ce+++ = 4f1 4f1

µ=           =               = 1.73 B

Q7: Chemisry of all lanthanoids isnso identical Explain ?

Ans : All lanthanoids have similar outer electronic configuration & show +3 oxidation state in their
compound s & t hey show lanthanoid contraction so they have similar chemical properties, although the
lanthanoids differ in the no of 4f electrons which are burried deep in the atoms & hence do not
influence the properties.

Q 8 : Write the chemical equation for
   a : Oxidation of Fe ++ by Cr2O7 – – in acidic medium.

  B : Oxidation of S2O3 – by MnO4 – in neutral medium.

Ans: a:        6 Fe ++ + Cr2O7 – – + 14 H+  2 Cr+++ + 6 Fe+++ + 7H2O

        B:     8 MnO4 - + 2 S2 O3 + H2 O      8MnO2 + 6 SO4 – - + 2 OH –

Q 9 : Write down the equations for the preparation of potassium di cromate from chromite ore.

Ans :         4 FeCr2 O4 + 16 NaOH + 7 O2  8 Na2 CrO4 + 2 Fe2 O3 + 8 H2O

             2 Na2 CrO4 + 2 H +          Na2 Cr2O7 + 2 Na+ + H2O

             Na2 Cr2O7 + 2 K Cl             K2 Cr2O7 + 2 NaCl
                            High order thinking Skill Questions
                               “Coordination Compounds”
                                        Group - G
One Marks Question:

   1. Which of the following is expected to be more stable?
      [Co (en)3]3+ or [Co(en)3]3+

      Ans. [Co (en)3]3+ as it is a Chelate Complex.

   2 Give an example of metal carbonyl having Metal- Metal bond

      Ans. Mn2(CO)10

   3. Give an example of a compound showing coordination isomerism.

      Ans. [Co (NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]

   4. Which kind of Isomerism is shown in the following example?

  Ans. [Co (NH3)5 ONO]Cl2 & [Co (NH3)5 NO2]Cl2

  5. What is the coordination number of Co in [Co (en)2 H2O Br]Cl ?

   Ans: 6

  Two Marks Questions:

   1. The species [CuCl4]2- exists while [CuI4]2- does not exists. Why?

   Ans: Due to larger size of Iodide ion bonding between Cu2+ and I- does not exist.

   2. How do you account for the following [Ti (H2O)6]3+ is coloured while [Sc(H2O)6]3+ is
      colour less?
      Ans:
   3. [Ti(H2O)6]3+ is coloured as Ti3+ having one unpaired electron while Sc3+ has no unpaired
      electron.
   4. Which of the complex is more stable?
      K3[Fe(CN)6] & K4[Fe(CN)6]
      Ans: K3[Fe(CN)6] is more stable as Fe3+ ion is more stable
   5. Why oxalic acid is is used to remove rust stain?
      Ans: Rust is hydrated Iron (III) oxide, Fe2O3.xH2O
      It dissolves in oxalic acid due to the formation of soluble complex of Iron (III)
      Fe2O3      + 6 H2C2O4 => 2[Fe(C2O4)3]3- + 3H2O +6H+
   6. [Cr(NH3)6]3+ is paramagnetic while[ Ni(CN)4]2- is diamagnetic . Explain why?
      Ans: Cr is in the +3 Oxidation state i.e d3 configuration. Also, NH3 is a weak ligand.


                                       Three Marks Questions:
       1. Aqueous copper sulphate (blue in colour) gives:-
          (a) A green ppt. with aqueous potassium fluoride and
          (b) A bright green solution with aqueous Potassium chloride. Explain these
              experimental results

           Ans: Copper (II) ions reacts with fluoride ions to form the complex [CuF4]2- & with
           chloride ions bright green coloured [CuCl4]2- complex ion.

           CuSO4 (aq) + 4KF  K2[CuF4] + K2SO4

           CuSO4 (aq) + 4KCl  K2[CuCl4] + K2SO4

       2. A compound of CrCl3.4H2O precipitates AgCl when treated with AgNO3 . The molar
          conductance of the solution correspondence 2 ions .Write the structural formula of the
          compound.

           Ans:- {Cr(H2O)4Cl2]Cl (tetraaquadichloridochromium(III) chloride

       3. Square planner complexes with coordination No. four exhibit geometrical isomerism
          whereas tetra hedral complexes do not why.
          Ans: In the square planner complexes, the four corners of a square are not equivalent
          with respect to central metal ion and there fore they exhibit geometrical isomerism. In
          tetra hedral complexes the four corners are4 equivalent with respect to central metal
          ion.

       4 the hexaaquamanganese(II) ion contains five unpaired electrons while the
hexacynomanganes(II) ion contain one unpaired electron.Explain

         Ans: In hexaaquamanganese(II) ion the ligand H2O is a weak field ligand and so can not
cause pairing of electrons in 3d orbital while in hexacynomanganes(II) ion the ligand CN- is a
strong field ligand so it causes the pairing of electron in the 3d orbital

       5.FeSO4 solution mixed with ammonium sulphate solution in 1: 1 molar ratio give the test
of ferrous ion but copper sulphate solution mixed with aqueous ammonium in 1:4 molar ratio
they do not give the test of copper ion. Explain Why

      Ans: FeSO4 solution mixed with ammonium sulphate solution does not form any complex
and give the test of ferrous ion but copper sulphate solution mixed with aqueous ammonium
solution gives complex[Cu(NH3)4]2+
                                                CMP

                                       d & f block elements

1 mark questions

Q-1) Write the general electronic configuration of d –block elements

Ans.) (n-1)d1-10 ns1-2

Q-2) On what ground can you say that scandium(Z=21) is a transition element but           zinc is not
?

Ans.) Scandium is a transition element as it has partially failed d-orbital in its elemental state
where as zinc does not contain the partially filled d-orbital either in elemental state or in its most
common oxidation state.

Q-3) Which is stronger reducing agent Cr2+ or Fe2+ and why?

Ans.) Cr2+ is a stronger reducing agent than Fe 2+ as its configuration changes from d4 to d3and d3
configuration is more stable due to half-filled t2g configuration.

Q-4) Explain why Cu+ ion is not stable in aqueous solution?

Ans.) Cu2+is more stable than Cu+ as the hydration enthalpy of Cu2+ is much more -ve &
compensates for the second ionization enthalpy of Cu.

Q-5) Name a transition element which is not exhibit variable oxidation state in it compounds.

Ans.) Scandium.

                                       2 Marks question

Q-6) Zn, Cd and Hg are not considered as transition metals. Why?

Ans.)They are not considered as transition metals as their d-orbital is completely filled in ground
state & in their most common oxidation state.

Q-7) Why La(OH)3 is more stronger base than Lu(OH) 3?

Ans.) Because Lu3+ is smaller in size than La3+ due to lanthanoid contraction. Lu-O bond is
stronger than La-O bond.

Q-8)What is meant by disproportionation of an oxidation state? Give an example.
  Ans) When a particular oxidation state becomes less stable relative to other oxidation states
,one lower, one higher it is said to undergo disproportionation.

  3MnO42- +4H+---2 MnO-4 + MnO2+2H2O

Q-9) Why Cr+2 reducing and Mn+2 oxidising when both have d4 configuration?

Ans.) Cr+2 is reducing as its configuration changes from d4 to d3, the latter has a half filled t2g
configuration. Mn+2 to . Mn+3 change result in the half filled d5configuration which has extra
stability.

Q-10) Draw the structure of the following ions.

   (1) CrO4-2          (IICr2O7-2

Ans.




                                       5 Marks Questions

Q-11) Give reason for the following

a) The enthalpies of atomization of transition metals are high.

b) The transition metals generally form colored compounds.

c) Transition metals & their many compounds act as good catalyst.

d) Transition metals and many of their compound show paramagnetic behavior.
e) Transition metals for interstitial compounds.

Ans.) a) Due to the presence of large no. of unpaired electrons in their atoms & strong bond
between them.

b) Due to the presence of unpaired electrons and d-d transition.

c) Due to variable oxidation state and suitable surface

d) Due to unpaired electrons & partially filled d-orbital.

e) Because they have a tendency to trap the small size atom like H,B,C,N etc.

Q-12) i) How would you prepare potassium dichromate from chromite ore?

ii) Write down the oxidizing action of potassium dichromate to oxidised Ferrous to Ferric ion

iii) Write down the conversion of iodide ion to iodine by potassium dichromate in acidic
medium.

Ans:
1.




II.



Over all equation :




III.



Over all Reaction:

Cr2O72- + I- + 14H+- 2Cr3+ + 3I2 + 7 H2O
Q-13) i) Give the preparation of potassium permagnate from pyrolusite.

ii) Draw the structure of MnO42-and MnO4-

2 iii) a) Write down the reaction of ferrous ion by potassium pemagnate to ferric ion(ionic
reaction)

      a) What happens when oxalic acid react with KMnO4 in acidic medium?


Ans.) I. Preparation of KMnO4




      II.




III. a.5Fe2+ + MnO4- +8H+     -- Mn2+      + 4H2O     +5Fe3+

   b. 5C2O42- + 2MnO4- +16H+        -- 2Mn2+     + 8H2O     +10CO2

Q-14) What do you mean by lanthanoid contraction ? Also write its cause & consequences

Ans.) Lanthanoid contraction: A steady decrease in the size of the lanthanoids with increase in
atomic number is known as lanthanoid contraction.

Cause: Due to imperfect shielding effect.

Consequences: i) Similar properties among lanthanoids

                 ii) Chemical separation of lanthanoids become difficult.

                 iii) Zr & Hf have same properties

Q-15) Give reasons

   i) Why transition elements form complexes?
   ii)    Transition element form alloy.
   iii)   Transition elements show variable oxidation state.
   iv)    What is the effect of change of pH of K2Cr2O7
   v)     Write the common oxidation state of lanthanoids.

Ans.) i) Due to small size & high nuclear charge also they have vacant d-orbitals.

          ii) Due to similar atomic size

          iii) Due to participation of ns & (n-1)d electrons.

          iv) Chromate ion can be changed into dichromate & vice-versa.

          V)     +3
                                 Common Minimum Programme

                                   Co-Ordination Compounds

One Mark Questions:

Q 1: Give IUPAC name of Ionisation Isomers of [ Pt(NH3)3 (NO2)]Cl .

Ans: Ionisation Isomers of [ Pt(NH3)3 (NO2)]Cl is [ Pt(NH3)3Cl] (NO2)].

Name: Triamminechlorido platinum(II) nitrite.

Q2: What is the coordination number of central metal ion in [Fe(C2O4)3]3- ?

Ans: 6

Q3: Write IUPAC name of linkage isomer of [ Co(NH3)5 (NO2)]Cl2 .

Ans: Linkage isomer of [ Co(NH3)5 (NO2)]Cl2 . is [ Co(NH3)5 (ONO)]Cl2 .

Pentaamminenitrito- N- cobalt (III)chloride.

Q4 :Write the formula of the following coordination compound:

Potassium tetracyano nickelate(II).

Ans: K2 [Ni(CN)4]

Q5:What is the oxidation state of Ni in [Ni(CO)4] ?

Ans: 0

Two marks questions:

Q6: Write IUPAC name of coordination isomer of [ Co(NH3)6 ] [ Cr(CN)6 ] .

Ans: Coordination isomer of [ Co(NH3)6 ] [ Cr(CN)6 ] is of [ Cr(NH3)6 ] [ Co(CN)6 ]

Name : hexaammine chromium (III)hexa cyno cobaltate(III).

Q7: How is double salt different from complex salt?

Ans: Double salt retain its identity only in solid state where as complex salt retain its identity in
solid as well as solution state.

Q8:What do you meant by chelate ligands & ambidentate ligands ?
Ans: When di or polydentate ligands uses its two or more donor atoms to bind a single metal
ionis called chelate ligand.eg ethylene diamine.

Ambidentate ligands :ligands which can ligate through two different donor sites.eg NO2-

Q 9: The Spin magnetic moment of [MnBr4]2- is 5.9 BM .Predict the geometry of the complex
ion?

Ans: Since coordination number of Mn2+ ion is 4 it will either sp3 with tetrahedral shape or dsp2
with square planar shape but it is tetrahedral as d orbital will occupy by the presence of 5
unpaired electrons.

Q10: Define spectro chemical series.

Ans: It is a series in which ligands can be arranged in the order of increasing field strength.




Three marks questions

Q11:Discuss the role of coordination compounds in

     i) Biological systems
     ii)     Analytical chemistry
     iii)    Medicinal chemistry

Ans: i) Haemglobin and myoglobin are complexes of Fe.

    ii)EDTA is used in determining hardness of water .

    iii)Cis platin is used in the treatment of cancer.

Q12)[NiCl4]2-is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral.why?

    Ans: In [NiCl4] 2- Ni is in +2 oxidation state with the configuration 3d8.Cl- is a weak ligand
so pairing of electrons does not take place .It is paramagnetic due to 2 unpaired electrons.In
[Ni(CO)4] Ni is in zero oxidation state with the configuration 3d8 4s2.CO is a strong ligand which
results in pairing of electrons.as it contains no unpaired electrons it is diamagnetic.

Q13) Discuss the nature of bonding [Fe(CN)6] 4-using VBT

 Ans:Oxidation state of Fe in[Fe(CN)6] 4-is +2

Orbitals of Fe2+ ions
                                                          Donated by 6 CN- ions



Q14) Discuss the nature of bonding in metal carbonyls.

 Ans:The metal –carbon bond in metal carbonyls possess both s and p character. The metal –
carbon sigma bond is formed by donation of lone pair of electrons on the carbonyl carbon into a
vacant d orbital of the metal.The metal- carbon pi bond is formed by the donation of a pair of
electrons from a filled d orbital of metal into the vacant antibonding pi molecular orbital .The
metal to ligand bonding creates a synergic effect which strengthens the bond between CO and
the metal.




                                            s
                                   ‘d’ and ‘f’ block elements

                                          NUT SHELL

General electronic Configuration of d- block elements-(n-1)d1-10 ns1-2

Properties of d- block elements:

Atomic size: Decreases with increasing atomic number but increases for Cu and Zn.

MP and BP: Due to strong metallic bonds these are high.

Oxidation state: Most common oxidation state is (+2) and highest (+7) of Mn.

Magnetic Properties: Due to partially filled‘d’ orbital

Colour property: Due to d-d transition state of partially filled‘d’ orbital.

Alloy formation: Due to approximate equal size

Complex Compounds: 1. Due to Partially filled ‘d; orbital

   2. High nuclear Charge
   3. Small size

 Interstitial Compounds: Due to presence of approximate size of interstitial spaces in their
lattice

Catalytic Property: Due to available of Free Valances at their surface.

   A. Preparation of KMnO4




   Structures of Ions:
   B. Preparation of K2Cr2O7




Structures of ions:




General electronic Configuration of f- block elements- (n-2)f1-14 (n-1)0-1 ns2

Atomic size (Lanthanoid Contraction):The size of lanthanoids decreases continuously with
the increasing atomic number.

Reason: Due to poor shielding effect of ‘d’ and ‘f’ orbitals

Consequences:

   1. Difficult to separate them.
   2. Same physical and chemical properties

Oxidation state: Most common oxidation state is (+3)

Comparison of properties of‘d’ and ‘f’ block elements

Actinoids Contraction: Similar to lanthanoids contration
                   NUT SHELL : CO-ORDINATION COMPOUNDS

 Coordination compounds:- compounds in which transition metal atom or ion is
surrounded with ligans.

Ligands- an atom or group of atoms or ion which donate lone pair.

Double salts. Addition compounds stable in solid state and break up in water.

Isomerism’s:

   A. Structural Isomerism- Due to difference in structures
      1. Linkage Isomerism- Due to presence of ambident ligand
         [Co(NH3)5ONO]Cl2 and [Co(NH3)5NO2]Cl2
      2. Coordination Isomerism’s : Having two complex ions
         [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
      3. Ionization Isomerism’s: [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4
      4. Solvated Isomerism’s: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2H2O
   B. Stereo Isomerism’s :
      a. Geometrical Isomerism’s: Cis and Trans isomers , facial and meridional
      b. Optical isomerism’s: Leavo and dextro.

Bonding in Complex Complex Compounds:-

Werner theory

VBT

MOT

Splitting of‘d’ orbital in octahedral Complex compounds:

				
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