Tag: coloured complexes

Questions Related to coloured complexes

Which ion is colourless?

  1. $C{ r }^{ +4 }$

  2. $Sc^{ +3 }$

  3. $Ti^{ +3 }$

  4. ${ V }^{ +3 }$


Correct Option: B
Explanation:

$(A)$ $Cr\quad [Ar]3d^54s^1$

$Cr^{+4}\quad [Ar]3d^2$
$2$ unpaired $e^-s\longrightarrow$ paramagnetic.

$(B)$ $Sc\quad [Ar]3d^14s^2$
$Sc^{+3}\quad [Ar]3d^2$
$0$ unpaired $e^-s\longrightarrow$ dimagnetic.

$(C)$ $Ti\quad [Ar]3d^24s^2$
$Ti^{+3}\quad [Ar]3d^1$
$1$ unpaired $e^-s\longrightarrow$ paramagnetic.

$(D)$ $V\quad [Ar]3d^34s^2$
$V^{+3}\quad [Ar]3d^2$
$2$ unpaired $e^-s\longrightarrow$ paramagnetic.

Ions which do not have unpaired electrons do not show colour in aqueous solution. These are known as diamagnetic ions.

When $MnO _{2}$ is fused with KOH in the presence of air, a coloured compound is formed, the product and its colour is :

  1. $K _{2}MnO _{4}$, dark green

  2. $KMnO _{4}$, purple

  3. $Mn _{2}O _{3}$, brown

  4. $Mn _{3}O _{4}$, black.


Correct Option: A
Explanation:

The reaction between MnO$ _2$ and KOH in presence of air is given as:

$2MNO _2 +4KOH +O _2$ $\rightarrow$ $2K _2 MnO _4 +2H _2O$
                                         (potassium manganate)
The $K _2Mn O _4$ formed is dark green in colour.

Which one of the following elements forms compounds that are all coloured?

  1. Magnesium

  2. Aluminium

  3. Iron

  4. Chromium


Correct Option: D
Explanation:
Most of the chromium compounds are coloured due to excitation of an electron from a lower energy d-orbital to higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency lies in a visible region. The colour observed corresponds to the complementary colour of light absorbed.

Which of the following compounds is not colored?

  1. ${Na} _{2}Cu{Cl} _{4}$

  2. ${Na} _{2}Cd{Cl} _{4}$

  3. ${Na} _{3}[Fe{(CN)} _{6}]$

  4. ${K} _{3}[Fe{(CN)} _{6}]$


Correct Option: B
Explanation:
Electronic configuration: $[Kr] 4d^{10} 5s^{2}$

Cadmium is not considered as transition element which has completely filled d - configuration and is not having unpaired electrons to exhibit color property.
So $Na _2CdCl _4$ do not exhibit color
Hence option B is correct.

Which of the following ion will not form coloured aqueous solution?

  1. $Ni^{2+}$

  2. $Fe^{3+}$

  3. $Ti^{4+}$

  4. $Cu^{2+}$


Correct Option: C
Explanation:
$ \displaystyle Ni^{2+}$ ion with $ \displaystyle 3d^8$ outer electronic configuration has two unpaired electrons. It will form a green coloured aqueous solution.

$ \displaystyle Fe^{3+}$ ion with $ \displaystyle 3d^5$ outer electronic configuration has five unpaired electrons. It will form yellow coloured aqueous solution.

$ \displaystyle Ti^{4+}$ ion will not form coloured aqueous solutionaas it does not contain unpaired electrons.
$ \displaystyle Ti^{4+}$ ion with $ \displaystyle 3d^0$ outer electronic configuration has zero unpaired electrons.


$ \displaystyle Cu^{2+}$ ion with $ \displaystyle 3d^9$ outer electronic configuration has one unpaired electron. It will form blue coloured aqueous solution.

Hence, option $C$ is correct.

Which of the following orbitals are degenerate for $[Cr(H _2O) _6]^{3+}$?

  1. $d _{x^2-y^2},d _{xy}$

  2. $d _{xy},d _{yz}$

  3. $d _{x^2-y^2},d _{yz}$

  4. $d _{z^2},dxy$


Correct Option: B
Explanation:

Solution:- (B) ${d} _{xy}, {d} _{yz}$

$[Cr(H _2O) _6]^{3+} $ has $d^2sp^3$ hybridization .$dxy,dyz,dzx$,orbitals are degenerate.

Ammonium dichromate is used in fire works. The green colored powder blow in the air is:

  1. $Cr{O} _{3}$

  2. ${Cr} _{2}{O} _{3}$

  3. $Cr$

  4. $CrO ({O} _{2})$


Correct Option: B
Explanation:
$(\mathrm{NH} _4) _2\mathrm{Cr} _2\mathrm{O} _7\xrightarrow{\Delta} \mathrm{Cr} _2\mathrm{O} _3+\mathrm{N} _2+4\mathrm{H} _2\mathrm{O}$

The green-coloured $ Cr _2O _3$ powder is blown in the air by the large volume of $ N _2$ and water vapour produced and settles like dust from the volcano. 

Hence, the correct option is $\text{B}$

Which one of the following forms a colorless solution in aqueous medium?

  1. ${Cr}^{3+}$

  2. ${Ti}^{3+}$

  3. ${Sc}^{3+}$

  4. ${V}^{3+}$


Correct Option: C
Explanation:

$Cr^{3+}$ = $3d^3$ = coloured

$Ti^{3+}$ =  $3d^1$ = coloured

$V^{3+}$ =  $3d^2$ = coloured

$Sc^{3+}$ has electron configuration of [Ar] $4s^0  3d^0$. Therefore without any unpaired electrons it forms a colorless solution in aqueous medium.