Tag: d and f block elements

Questions Related to d and f block elements

The ion that gives colourless compound:

  1. $Sc^{+3}$

  2. $V^{+3}$

  3. $Cr^{+3}$

  4. All of these


Correct Option: A

Which pair of ions is colourless.

  1. $Mn^{+3}, CO^{+3}$

  2. $Fe^{3+}, Cr^{+3}$

  3. $Zn^{2+}, Sc^{3+}$

  4. $Ti^{2+}, Cu^{2+}$


Correct Option: C
Explanation:

As both $Zn^{2+}  = d^{10}, Sc^{3+} = d^0$ have zero unpaired d-electron so they both are colourless.

Out of the ions $Ag^+, Co^{2+}, Ti^{4+}$, which one will be coloured in aqueous solutions.


[Atomic no : Ag = 47, Co = 27, Ti = 22]

  1. $Ag^+$

  2. $Co^{2+}$

  3. $Ti^{4+}$

  4. None of these


Correct Option: B
Explanation:

compound with unpaired d-electron will show color.


$Ag^+ = 5d^{10} \,\,\,\,\,\,,n = 0$

$Ti^{+4} = 3s^2 3p^6 \,\,\,\,\,\,, n = 0$

$Co^{+2} = [Ar]3d^7\,\,\,\,\,\,\,,n = 3$ It will show color due to unpaired d electrons.

Most copper $(I)$ compounds are found to be colourless. This is due to :

  1. presence of low oxidation state of copper

  2. completely filled d-level in $Cu(I)$

  3. diamagnetic nature of the compound

  4. high polarizability of $Cu(I)$ ion


Correct Option: B
Explanation:

In copper (I) ion there are no vacant d orbitals as it is diamagnetic. Copper(I) ion being less charged has small ligand field effect and the transition is in the infrared region in which no color is perceived by human eye.

The ion which exhibits green colour is:

  1. $Cu^{2+}$

  2. $Mn^{2+}$

  3. $Co^{2+}$

  4. $Ni^{2+}$


Correct Option: D
Explanation:

Transition metal ion                     Colour

            ${ Cu }^{ 2+ }$                                blue-green
            ${ Fe }^{ 2+ }$                                 olive green
            ${ Ni }^{ 2+ }$                                  bright green
            ${ Fe }^{ 3+ }$                                 brown to yellow
${ Ni }^{ 2+ }$ only exhibits green colour, other exhibits partial green colours / mixed green colours.

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 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.