Tag: coloured complexes

Questions Related to coloured complexes

Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state? 

  1. $Ag _2SO _4$

  2. $CuF _2$

  3. $ZnF _2$

  4. $Cu _2Cl _2$


Correct Option: B
Explanation:
The respective transition elements exists in
$Ag^{+}-{4d}^{10}5s^0$, (have completely filled d orbital)
$Cu^{+2}-3d^{9}4s^0$, (have incompletely filled d orbital)
$Zn^{+2}-3d^{10}4s^0$ , (have completely filled d orbital)
$Cu^{+1}-3d^{10}4s^0$ (have completely filled d orbital),
Since, $Cu^{+2}$ has unpaired electron it will show electron transitions. 
Hence will be coloured.

In which of the following ions, the colour is not due to $d-d$ transition?

  1. $[Ti(H _2O) _6]^{3+}$

  2. $[Cu(NH _3) _4]^{2+}$

  3. $[CoF _6]^{3-}$

  4. $CrO _4^{2-}$


Correct Option: D
Explanation:
Colour is due to d-d transtition ,then such complexes have colour due to d-d transition are surely octahedral complexes.
Among the given options,                                                   
                                         ${ \left[ Ti{ \left( { H } _{ 2 }O \right)  } _{ 6 } \right]  }^{ 3+ }$
                                        ${ \left[ Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ 2+ }$
                                           ${ \left[ { COF } _{ 6 } \right]  }^{ 3- }$
This are octahedral complexes.
$CrO _4^{2-}$ is a salt,it colour is not due to d-d transition.

Match the column I with column II and mark the appropriate choice.

Column I Column II
(A) $FeSO _4.7H _2O$ (i) Green
(B) $NiCl _2.4H _2O$ (ii) Light Pink
(C) $MNCl _2.4H _2O$ (iii) Pale green
(D) $CoCl _2.6H _20$ (iv) Pink
(E) $Cu _2Cl _2$ (v) Colourless
  1. (A) - (iii), (B) - (iv), (C) - (i), (D) - (ii), (E) - (v)

  2. (A) - (ii), (B) - (iii), (C) - (iv), (D) - (i), (E) - (v)

  3. (A) - (v), (B) - (ii), (C)-(iii), (D) - (iv), (E) - (i)

  4. (A) - (iii), (B)-(i), (C)-(ii), (D) - (iv), (E).- (v)


Correct Option: D
Explanation:

The colour of the following compounds are:

(i) $FeSO _4$. & $H _2O$ - Pale green.
(II) $NiCl _2$.$4H _2O$ -Green
(iii) $MnCl _2$.$4H _2O$ - Light pink
(iv) $CoCl _2$.$6H _2O$ - Pink
(v) $CuCl _2$ - Colourless.
The colour of the transition metal compounds is due to partially filled d orbitals. Due to that,  d-d transition its complementary colour is seen. 
Hydrated complexes are coloured because in that come H$ _2$O molecules act as a ligand and gets arranged around the central metal atom so as to bring d-d transition.

For $Zn^{2+}$, $Ni^{2+}$, $Cu$, and $Cr^{2+}$ which of the following statements is correct?

  1. Only $Zn^{2+}$ is colourless and $Ni^{2+}$, $Cu^{2+}$ and $Cr$ are coloured

  2. All the ions are coloured

  3. All the ions are colourless

  4. $Zn^{2+}$and $Cu^{2+}$ are colourless while $Ni^{2+}$ and $Cr^{2+}$ are coloured


Correct Option: A
Explanation:

Since $Zn^{+2}$ have fully filled $d$ orbital having no unpaired electrons,it does not undergo electronic transitions,hence colourless.However,rest of the elements given as $Ni^{+2}$ , $Cu^{+2}$ and $Cr^{+2}$ has unpaired electrons which can undergo electronic transitions and hence show colours.

Which of the following group contains coloured ions?
1. $Cu^+$
2. $Ti^{4+} $
3. $Co^{2+}$
4.  
$Fe^{2+}$

  1. $1,4$

  2. $3,4$

  3. $2,3$

  4. $1,2$


Correct Option: B
Explanation:

$Cu^{+}$ and $Ti^{+4}$ have fully filled and empty d orbital respectively. Hence due to absence of unpaired electrons,they will be colourless. But $Fe^{+2}$ and $Co^{+2}$chave unpaired electrons. Hence they will be coloured.

Choose the correct answer from the alternative given.
Amongst $TiF _6^{2-}$, $CoF$,   $Cu _2$ $Cl _2$ and $NiCl$, which are the colourless species? (atomic number of Ti = 22, Co = 27, Cu = 29, Ni = 28)

  1. $CoF _6^{3-}$and $NiCl _4^{2-}$

  2. $TiF _6^{2-}$and $Cu _2Cl _2$

  3. $Cu _2Cl _2$ and $NiCl _4^{2-}$

  4. $TiF _6^{2-}$ and $CoF _6^{ 3-}$


Correct Option: B
Explanation:

TiF$ _6^{2-}$ -

The oxidation state of Ti is +4.
Electronic configuration of Ti$^{4+}$ is [Ar] 3d$^0 4s^0$.
So, unpaired e$^-$ s is present in d-orbitals so it is colourless.
Cu$ _2Cl _2$-
The oxidation state of Cu is +1.
Electronic configuration of Cu$^+$ is [Ar] 3d$^{10} 4s^0$.
So, no unpaired d e$^-$ s are present . Hence, the given compound is colourless.

Colour of transition metal ions are due to absorption of some wavelength. This results in:

  1. $d-f$ transition

  2. $s-s$ transition

  3. $s-d$ transition

  4. $d-d$ transition


Correct Option: D
Explanation:

Colour of transition metal ions are due to absorption of the same wavelength. This results in $d-d$ transition.

The coloured species is ?

  1. ${ VCl } _{ 3 }$

  2. ${ VOSO } _{ 4 }$

  3. ${ Na } _{ 3 }{ VO } _{ 4 }$

  4. $[{ V(H } _{ 2 }O) _{ 6 }]{ SO } _{ 4 }{ H } _{ 2 }O$


Correct Option: C

Which of the following compounds are coloured due to charge transfer spectra?

  1. ${K} _{2}{Cr} _{2}{O} _{7}$

  2. $KMn{O} _{4}$

  3. $Cu{SO} _{4}.5{H} _{2}O$

  4. Both a and b


Correct Option: C

Which of the following ion will give colourless aqueous solution ?  

  1. $Ni^{2+}$

  2. $Sc^{3+}$

  3. $Cu^{2+}$

  4. $Mn^{3+}$


Correct Option: B
Explanation:

Solution:- (B) ${Sc}^{+3}$

Only the ions that have electrons in $d$-orbital and in which $d-d$ transition is possible will be coloured.
Electronic configuration of ${Sc}^{+3} \left( \text{At. no. = 21} \right) - 1 {s}^{2} 2 {s}^{2} 2{p}^{6} 3{s}^{2} 3 {p}^{6}$
Since it do not contains electrons in $d$-orbital and hence will not undergo $d-d$ transition and do not show colour.