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Engineering
130 Views   |  

The correct combination is :

  • Option 1)

    \left [ Ni\left ( CN \right )_{4} \right ]^{2-}-tetrahedral ;

    \left [ Ni\left ( CO \right )_{4} \right ]-paramagnetic

     

     

     

     

     

     

     

     

     

  • Option 2)

    \left [ NiCl_{4} \right ]^{2-}-paramagnetic;

    \left [ Ni\left ( CO \right )_{4} \right ]-tetrahedral

  • Option 3)

    \left [ NiCl_{4} \right ]^{2-}-square-planar;

    \left [ Ni\left ( CN \right )_{4} \right ]^{2-}-paramagnetic

  • Option 4)

    \left [ NiCl_{4} \right ]^{2-}-diamagnetic;

    \left [ Ni\left ( CO \right )_{4} \right ]-square-planar

 

As we learned 

 

Weak field Ligand -

These are the ligands which do not pair the electron in inner d-orbital.

- wherein

For example: N_{2}O, F^{-}, Cl^{-} etc    

 

 

Strong Field Ligand -

These are the ligands which pair the electron in inner d-orbital and makes required d-orbitals empty/available. 

- wherein

For example: CO, CN etc

 

 

Ni^{+2}\rightarrow \left [ Ar \right ]3d^{8},4S^{0},4p^{0}                         Ni\left ( o \right )\left [ Ar \right ]3d^{8},4S^{2},4p^{0}

Cl- is weak field ligand                                             CO is strong field ligand

(no pairing of electron)                                    \left [ Ar \right ]3d^{10} 4S^{0}4p^{0}

                                                                                              Sp^{3}

                                                                                    tetrahedral 

 

 


Option 1)

\left [ Ni\left ( CN \right )_{4} \right ]^{2-}-tetrahedral ;

\left [ Ni\left ( CO \right )_{4} \right ]-paramagnetic

 

 

 

 

 

 

 

 

 

Option 2)

\left [ NiCl_{4} \right ]^{2-}-paramagnetic;

\left [ Ni\left ( CO \right )_{4} \right ]-tetrahedral

Option 3)

\left [ NiCl_{4} \right ]^{2-}-square-planar;

\left [ Ni\left ( CN \right )_{4} \right ]^{2-}-paramagnetic

Option 4)

\left [ NiCl_{4} \right ]^{2-}-diamagnetic;

\left [ Ni\left ( CO \right )_{4} \right ]-square-planar

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Engineering
145 Views   |  

The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :

  • Option 1)

    +3, 0, and +4

  • Option 2)

    +3, +4, and +6

  • Option 3)

    +3, +2, and +4

  • Option 4)

    +3, 0, and +6

 
As we learned Oxidation number of central metal atom - The charge the central atom would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. -     Option 1) +3, 0, and +4 Option 2) +3, +4, and +6 Option 3) +3, +2, and +4 Option 4) +3, 0, and +6
Engineering
94 Views   |  

Consider the following reaction and statements :

[Co(NH_{3})_{4}Br_{2}]^{+}+Br^{-}[Co(NH_{3})_{3}Br_{3}]+NH_{3}

(I) Two isomers are produced if the
     reactant complex ion is a cis-isomer.

(II) Two isomers are produced if the
     reactant complex ion is a transisomer.

(III) Only one isomer is produced if the
       reactant complex ion is a transisomer

(IV) Only one isomer is produced if the
        reactant complex ion is a cis-isomer.

The correct statements are :

  • Option 1)

    (II) and (IV)

  • Option 2)

    (I) and (II)

  • Option 3)

    (I) and (III)

  • Option 4)

    (III) and (IV)

 

As we learned

Optical Isomerism -

Optical isomers are mirror images that cannot be suprimposed on one another

-

 

 \left [ Ma_{4}b_{2} \right ]+b\rightarrow \left [ Ma_{3}b_{3} \right ]+a

 

Two possibilities for \left [ Ma_{4}b_{2} \right ]

 


Option 1)

(II) and (IV)

Option 2)

(I) and (II)

Option 3)

(I) and (III)

Option 4)

(III) and (IV)

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Engineering
165 Views   |  

Which of the following complexes will show geometrical isomerism ?

  • Option 1)

    aquachlorobis(ethylenediamine) cobalt(II) chloride

  • Option 2)

    pentaaquachlorochromium(III) chloride

  • Option 3)

    potassium amminetrichloroplatinate( II)

  • Option 4)

    potassium tris(oxalato)chromate(III)

 

As we have learned

Geometrical Isomers -

This is due to the difference in relative position of the ligands attached to the central metal atom with respect to each other

-

Aqua chlorobis (ethylencdiamine) Cobalt (II) Chloride.Will show geometrical isomerism

[Co(H2O)Cl(en)2]Cl

 


Option 1)

aquachlorobis(ethylenediamine) cobalt(II) chloride

This is correct

Option 2)

pentaaquachlorochromium(III) chloride

This is incorrect

Option 3)

potassium amminetrichloroplatinate( II)

This is incorrect

Option 4)

potassium tris(oxalato)chromate(III)

This is incorrect

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Engineering
340 Views   |  

In Wilkinson’s catalyst, the hybridization of central metal ion and its shape are
respectively :

  • Option 1)

    sp^{3}d , trigonal bipyramidal

  • Option 2)

    sp^{3},tetrahedral

  • Option 3)

    dsp^{2},square planar

  • Option 4)

    d^{2}sp^{3},octahedral

 

As we have learned

Hybridisation -

sp3d2 - square bipyramidal or octahedral 

d2sp3 - octahedral 

sp3 - tetradedral 

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - [Ni(Cl)_{4}]^{2-}

dsp2 - [Pt(CN)_{4}]^{2-}

 

The wilikinson catalyst is [Rh Cl(pph_{3})_{3}]  and the hybridisation and its shape are dsp^{2}

and square planer respectively 

 

 

 

 

 


Option 1)

sp^{3}d , trigonal bipyramidal

This is incorrect

Option 2)

sp^{3},tetrahedral

This is incorrect

Option 3)

dsp^{2},square planar

This is correct

Option 4)

d^{2}sp^{3},octahedral

This is incorrect

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