Q 1 :    

Number of complexes from the following with even number of unpaired ''d'' electrons is __________ .

 

[V(H2O)6]3+,[Cr(H2O)6]2+,[Fe(H2O)6]3+,[Ni(H2O)6]3+,[Cu(H2O)6]2+

 

[Given atomic numbers: V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29]                 [2024]

  • 2

     

  • 4

     

  • 1

     

  • 5

     

(A)

Complex Electronic
configuration of
central metal ion
without crystal
field splitting
Electronic
configuration of
central metal ion
after crystal field
splitting
Number of
unpaired d
electrons in
complex
[V(H2O)6]3+ V3+=[Ar]183d2 t2g1,1,0,eg0,0            2
[Cr(H2O)6]2+ Cr2+=[Ar]183d4 t2g1,1,1,eg1,0            4
[Fe(H2O)6]3+ Fe3+=[Ar]183d5 t2g1,1,1,eg1,1            5
[Ni(H2O)6]3+ Ni3+=[Ar]183d7 t2g2,2,1,eg1,1            3
[Cu(H2O)6]2+ Cu2+=[Ar]183d9 t2g2,2,2,eg2,1            1

 



Q 2 :    

The correct sequence of ligands in the order of decreasing field strength is :                      [2024]

  • S2->OH->EDTA4->CO

     

  • OH->F->NH3>CN-

     

  • CO>H2O>F->S2-

     

  • NCS->EDTA4->CN->CO

     

(C)

Ligands can be arranged in a series in the order of increasing field strength as given below:

I-<Br-<SCN-<Cl-<S2-<F-<OH-<C2O42-<H2O<NCS-<edta4-<NH3<en<CN-<CO

 



Q 3 :    

If an iron (III) complex with the formula [Fe(NH3)x(CN)y]- has no electron in its eg orbital, then the value of x + y is                  [2024]

  • 4

     

  • 3

     

  • 5

     

  • 6

     

(D)

    Iron in +3 oxidation state forms complexes with C.N. = 6.

    So x+y=6.

    [Fe(NH3)x(CN)y]-

     Ammonia is a neutral ligand and cyanide is a unit negatively charged ligand.

     Since sum of oxidation number = charge

     O.N of Fe+x×0+y×(-1)=-1

     3+(-y)=-1

      y=4

      Thus x=2

      The complex is [Fe(NH3)2(CN)4]-

 



Q 4 :    

The number of unpaired d-electrons in [Co(H2O)6]3+ is ________ .                     [2024]

  • 1

     

  • 0

     

  • 4

     

  • 2

     

(B)

      [Co(H2O)6]3+

      Oxidation number of Co = +3

       Co3+=[Ar]183d6

       With Co3+, ligands like H2O acts as strong field ligands. With strong field ligands, d6 complexes split as t2g2,2,2,eg0,0

        So it has no unpaired electrons.

 



Q 5 :    

The correct order of ligands arranged in increasing field strength.                        [2024]

  • H2O<OH-<CN-<NH3

     

  • Cl-<OH-<Br-<CN-

     

  • Br-<F-<H2O<NH3

     

  • F-<Br-<I-<NH3

     

(C)

    In general, ligands can be arranged in a series in the order of increasing field strength as given below:

    I-<Br-<SCN-<Cl-<S2-<F-<OH-<C2O42-<H2O<NCS-<edta4-<NH3<en<CN-<CO

 



Q 6 :    

Which one of the following complexes will exhibit the least paramagnetic behaviour ?

 

[Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27]                       [2024]

  • [Co(H2O)6]2+

     

  • [Mn(H2O)6]2+

     

  • [Fe(H2O)6]2+

     

  • [Cr(H2O)6]2+

     

(A)

Complex Electronic configuration
of central metal ion
without splitting
Electronic configuration
after splitting
Number of unpaired
electrons
[Co(H2O)6]2+ Co2+=[Ar]183d7 t2g2,2,1eg1,1              3
[Mn(H2O)6]2+ Mn2+=[Ar]183d5 t2g1,1,1eg1,1              5
[Fe(H2O)6]2+ Fe2+=[Ar]183d6 t2g2,1,1eg1,1              4
[Cr(H2O)6]2+ Cr2+=[Ar]183d4 t2g1,1,1eg1,0              4

 

Paramagnetic nature  number of unpaired electrons.



Q 7 :    

The number of complexes from the following with no electrons in the t2 orbital is _______ .

 

TiCl4,[MnO4]-,[FeO4]2-,[FeCl4]-,[CoCl4]2-                      [2024]

  • 1

     

  • 4

     

  • 2

     

  • 3

     

(D)

Compound Electronic configuration
of central atom without
tetrahedral splitting
Electronic configuration
of central atom after
tetrahedral splitting
TiCl4 Tiü+=[Ar]183d4s e0t20
MnO4- Mn7+=[Ar]183d04s0 e0t20
FeO42- Fe6+=[Ar]183d24s0 e2t20
FeCl4- Fe2+=[Ar]183d64s0 e3t23
CoCl42- Co2+=[Ar]183d74s0 e4t23

 

Three complexes donot have electron in t2 orbitals



Q 8 :    

Match List I with List II.

                     List I                      List II
    Tetrahedral Complex         Electronic Configuration
A.                   TiCl4 I.                        e2,t20
B.                  [FeO4]2- II.                        e4,t23
C.                  [FeCl4]- III.                        e0,t20
D.                  [CoCl4]2- IV.                        e2,t23

 

Choose the correct answer from the options given below:                           [2024]

  • (A) - (III), (B) - (I), (C) - (IV), (D) - (II)

     

  • (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

     

  • (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

     

  • (A) - (I), (B) - (III), (C) - (IV), (D) - (II)

     

(A)

Complex Electronic
configuration of central
metal without tetrahedral
splitting

Electronic
configuration of central
metal with tetrahedral splitting

TiCl4 Ti4+=[Ar]183d04s0 e0,0,t20,0,0
[FeO4]2- Fe6+=[Ar]183d24s0 e1,1,t20,0,0
[FeCl4-] Fe3+=[Ar]183d54s0 e1,1, t21,1,1
[CoCl4]2- Co2+=[Ar]183d74s0 e2,2,t21,1,1

 



Q 9 :    

Number of complexes with even number of electrons in t2g orbitals is - [Fe(H2O)6]2+, [Co(H2O)6]2+, [Co(H2O)6]3+, [Cu(H2O)6]2+, [Cr(H2O)6]2+  [2024]

  • 3

     

  • 5

     

  • 2

     

  • 1

     

(A)

Complex Electronic
configuration
of central metal
ion without
octahedral
splitting
Electronic
configuration of
central metal ion
after octahedral
splitting
Number of
electrons
in t2g
[Fe(H2O)6]2+ Fe2+=[Ar]183d6         t2g2,1,1eg1,1          4
[Co(H2O)6]2+ Co2+=[Ar]183d7         t2g2,2,1eg1,1          5
[Co(H2O)6]3+ Co3+=[Ar]183d6         t2g2,2,2eg0,0          6
[Cu(H2O)6]2+ Cu2+=[Ar]183d9         t2g2,2,2eg2,1          6
[Cr(H2O)6]2+ Cr2+=[Ar]183d4          t2g1,1,1eg1,0          3

Note: In complexes with C.N = 6, Co3+ undergoes d2sp3 hybridization (except in complexes with C.N = 6, where Co3+ undergoes sp3d2 hybridization).



Q 10 :    

The Spin only magnetic moment value of square planar complex [Pt(NH3)2Cl(NH2CH3)]Cl is ________ B.M. (Nearest integer) (Given atomic number for Pt = 78)              [2024]



(0)

   In [Pt(NH3)2Cl(NH2CH3)]ClPt is in +2 oxidation state.

    Pt2+=[Xe]544f145d8

    Metal ions with 4d8 or 5d8 configuration with coordination number 4 are always dsp2 hybridized and diamagnetic, irrespective of ligand. This is because Zeff for these 4d8 or 5d8 is high on account of poor shielding of d and/or f electrons, hence all ligands are sufficiently attracted towards the metal ion resulting large crystal field splitting which causes in pairing up of d8 electrons.