Chemistry

Coordination Chemistry

319 Questions

Coordination chemistry focuses on coordination compounds, their structures, and their magnetic properties. The questions cover primary and secondary valencies, ligand types, and geometries of complexes. This topic is heavily tested in chemistry competitive exams and requires a good grasp of molecular structures.

primary and secondary valencycomplex geometriesligand field theorymagnetic properties of complexesisomerism in complexes

Coordination Chemistry Questions

Multiple choice geography in the land of kerala introduction to metallurgy occurrence of metals study of kerala

A tin chloride $Q$ undergoes the following reaction (not balanced)
$Q + Cl^{-} \rightarrow X$
$Q + Me _{3}N \rightarrow Y$
$Q + CuCl _{2}\rightarrow Z + CuCl$
$X$ is monoanion having pyramidal geometry. Both $Y$ and $Z$ are neutral compounds. Choose the correct options(s)

  1. The central atom in $X$ is $sp^{3}$ hybridized
  2. There is a coordinate bond in $Y$
  3. The oxidation state of the central atom in $Z$ is $+2$
  4. The central atom in $Z$ has one lone pair of electrons
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$\underset {(P)}{SnCl _{2}} + Cl^{-} \rightarrow \underset {(X)}{SnCl _{3}^{-}}$
$\underset {(P)}{SnCl _{2}} + Me _{3}N \rightarrow \underset {(Y)}{SnCl _{2} [N(CH _{3}) _{3}]}$
$\underset {(P)}{SnCl _{2}} + 2CuCl _{2}\rightarrow \underset {(Z)}{SnCl _{4}} + CuCl$.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Arrange the following complexes in order of increasing electrical conductivity.
(I) $[CoCl _3(NH _3) _3]$     (II) $[CoCl(NH _3) _5]CI _2$
(III) $[Co(NH _3) _6]Cl _3$   (IV) $[CoCl _2(NH _3) _]1Cl$ 

  1. III > II > IV > I

  2. II > III > IV > I

  3. II > III > I > IV

  4. III > IV > II > I
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electrical conductivity $\times $ no. / moles of ions present after dissociation.

$I$ $\therefore \quad \left[ Co{ Cl } _{ 3 }{ \left( { NH } _{ 3 } \right)  } _{ 3 } \right] $
$II$  $\left[ CoCl{ \left( { NH } _{ 3 } \right)  } _{ 5 } \right] { Cl } _{ 2 }\rightarrow { \left[ CoCl{ \left( { NH } _{ 3 } \right)  } _{ 5 } \right]  }^{ 2+ }+2{ Cl }^{ \left( - \right)  }$
$IV$  $\left[ Co{ Cl } _{ 2 }{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right] Cl\rightarrow { \left[ Co{ Cl } _{ 2 }{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ + }+{ Cl }^{ \left( - \right)  }$
$III$  ${ \left[ Co{ \left( { NH } _{ 3 } \right)  } _{ 6 } \right] Cl } _{ 3 }\rightarrow { \left[ Co{ \left( { NH } _{ 3 } \right)  } _{ 6 } \right]  }^{ 3+ }+3{ Cl }^{ \left( - \right)  }$
Hence the correct answer is,
$\boxed { III>II>IV>I } $

Multiple choice chemistry principles of metallurgy concentration of ore concentration of ores general principles of metallurgy

During concentration of Tin stone separation of wolframates of $Fe^({II})$ and $Mn^({II})$ is based on the fact that:

  1. $SnO _2$ is paramagnetic
  2. $SnO _2$ is diamagnetic
  3. $FeWO _4$ and $MnWO _4$ are paramagnetic

  4. Both B and C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

During concentration of Tin stone separation of wolframates of $Fe({II})$ and $Mn({II})$ is based on the fact $SnO _2$ is diamagnetic and both $FeWO _4$ & $MnWO _4$ are paramagnetic so magnetic separation method is used for concentration of ore.

Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

Which of the following has maximum complex forming ability with a given metal ion?

  1. $ \mathrm{P}\mathrm{H} _{3}$
  2. $ \mathrm{B}\mathrm{i}\mathrm{H} _{3}$
  3. $ \mathrm{N}\mathrm{H} _{3}$
  4. $ \mathrm{S}\mathrm{b}\mathrm{H} _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Due to higher tendency of nitrogen to donate its lone pair of electron to form coordinate bond which is not common in other hydrides of VA group elements, it has maximum complex forming ability.

Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

The ion or group detected by $K _2[HgI _4]$ is:

  1. NO
  2. $Cl^-$
  3. $NH^{-2}$
  4. $NH _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ K } _{ 2 }[Hg{ I } _{ 4 }]$ ( potassium mercuric iodide) is known as nesseler's reagent.

It is used for the detection of ammonia.
The reaction taking place is:
$2{ K } _{ 2 }[Hg{ I } _{ 4 }]+3KOH+{ NH } _{ 3 }\rightarrow [{ OHg } _{ 2 }.{ NH } _{ 2 }]I+7KI+2{ H } _{ 2 }O$
The colour of nesseler's reagent changes to brown in basic medium.

Multiple choice chemistry coordination chemistry werner's theory werner's theory of coordination compounds coordination compounds

Among the following complexes (K-P) :
$K _3[Fe(CN _6]-K  $ and $  [Co(NH _3) _6]Cl _3-L$ ; 
$Na _3[Co(oxalate) _3]-M $ and    $  [Ni(H _2O) _6]Cl _2-N$ ;
$K _2[Pt(CN) _4]-O$ and $[Zn(H _2O) _6](NO _3) _2 - P$;
the diamagnetic complexes are:

  1. K,L,M,N

  2. K,M,O,P

  3. L,M,O,P

  4. all

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K _3[Fe(CN _6]-K  $ = $ 3d^5$ it is paramagnetic 


 $  [Co(NH _3) _6]Cl _3-L$ = $3d^6$ is diamagnetic

$Na _3[Co(oxalate) _3]-M $ = $3d^6$ is diamagnetic 

 $  [Ni(H _2O) _6]Cl _2-N$ = $3d^8$ is paramagnetic

$K _2[Pt(CN) _4]-O$ = $d^3 $ it is diamagnetic

 and $[Zn(H _2O) _6](NO _3) _2 - P$;= $ d^6$ it is diamagnetic

Multiple choice chemistry coordination chemistry werner's theory werner's theory of coordination compounds coordination compounds

Which of the following sets of examples and geometry of the compounds is not correct?

  1. Octahedral - $[Co(NH _3) _6]^{3+}, [Fe(CN) _6]^{3-}$
  2. Square planar - $[Ni(CN) _4]^{2-}, [Cu(NH _3) _4]^{2+}$
  3. Tetrahedral- $[Ni(CO) _4], [ZnCl _4]^{2-}$
  4. Trigonal bipyramidal - $[Fe(NH _3) _6]^{2+}, [CuCl _4]^{2-}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$[Fe(NH _3) _6]^{2+}=Fe^{2+}=[Ar]3d^6=$ Octahedral.

Hybridisation is $=sp^3d^2$ .

$[CuCl _4]^{2+}=Cu^{2+}=[Cu]3d^{9}=$ Tetrahedral.
Hybridisation $=$ $sp^3$

These two do not have trigonal bipyramidal geometry.