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Coloured complexes - class-XI

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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.

Why are the compounds of transition metal generally coloured ?

  1. due to s being filled before d (Aufbau principle)

  2. Due to presence of unpaired electron

  3. unfilled d orbital

  4. d-d transition


Correct Option: B,D
Explanation:

coloured compound of transition elements is assosiated with partially filled (n-1)d orbitals. the transition metal ions containing unpaired d-electrons undergoes electronic transition from one d-orbital to another. during this d-d transition process the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. the color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in visible region for all transition elements.

Statement 1: Transition metal compounds are often colored.
Statement 2: They frequently possess partially filled d-orbitals.

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

  2. Both the Statement 1 and Statement 2 are correct, but Statement 2 is NOT the correct explanation of Statement 1.

  3. Statement 1 is correct, but Statement 2 is not correct.

  4. Statement 1 is not correct, but Statement 2 is correct.


Correct Option: A
Explanation:

Transition metals are often coloured.

reason - they frequently posses partially filled of orbitals.

Salt of which of the elements are mostly blue in colour ?

  1. Cu

  2. Zn

  3. Fe

  4. Cr


Correct Option: A
Explanation:

$Cu\longrightarrow $ Blue colour

$Zn\longrightarrow $ White colour
$Fe\longrightarrow $ Green colour
$Cr\longrightarrow $ Orange colour

Colour of $Fe^{+2}$ ions is:

  1. blue

  2. light green

  3. very dark green

  4. yellow


Correct Option: B
Explanation:

Colour of $Fe^{2+}$ is light green.

Due to this many salts having $Fe^{2+}$ ion are also green in colour.
For ex: $FeSO _4$

The silver salts like $AgBr$ and $AgI$ are coloured because of:

  1. d-d transition of electrons

  2. charge transfer

  3. polirisation of halid${Ag}^{+}$ by

  4. both (b) and (c)


Correct Option: A
Explanation:

The silver salts like$AgBr$ and $AgI$ are colored, the color of transition metal ions is due  to the presence of unpaired electron . The electrons which absorbs radiations of color that from the visible or UV spectrum and undergo transition from ground state to excited state within d-sub shells,so it is d-d transitions of electrons.

Which of the following compounds is not coloured ?

  1. $ Na _{2}[CuCl _{4}] $

  2. $ Na _{2}[CdCl _{2}] $

  3. $ [Cr(H _{2}O) _{6}]Cl _{3} $

  4. all of the above


Correct Option: B
Explanation:

The outer shell electronic configuration of $Cu^{2+}$ has $3d^{9}$ configuration and contain 1  unpaired electron, thus it is coloured.

$Na _2[CdCl _2]$, compound is not coloured because of central atom $Cd$ which has $4d^{10} 5s^2$ electronic configuration so transition do not occur.
Hence option B 

The color of $KMn{ O } _{ 4 }$ is due to:

  1. L-> M charge transfer transition

  2. $\sigma -{ \sigma }^{ * }$ transition

  3. M-> L charge transfer transition

  4. d-d transition


Correct Option: A
Explanation:

The color in $KMnO _4$ arises from an electronic transition.

In the permanganate ion $MnO _4^{-}$, manganese is in the +7 oxidation state, so it has no d electrons. Photons promote an electron from the highest energy molecular orbital to an empty d orbital on the manganese. When a photon of light is absorbed, this charge transfer takes place from L to M.

Choose the correct statement.

  1. Only few transition metal complexes are coloured

  2. d-orbital are degenerated hence, they form complexes

  3. Transition metal complexes reflect the complimentary colour of absorbed colour

  4. Energy difference between ${ t } _{ 2(g) }$ and ${ t } _{ g }$ level is very large


Correct Option: C
Explanation:

During this d-d transition process, the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. The color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in the visible region for all transition elements.

Which of the following  is expected to form colourless complex?

  1. ${ Ni }^{ 2+ }$

  2. ${ Cu }^{ + }$

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

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


Correct Option: B
Explanation:
$Cu^+$ is colourless 
D block elements of periodic table are called as Transition elements. Transition elements have partially filled d orbitals. The colour for the elements of D block is due to transition of electrons which is called as  d -Transition for which presence of partially filled d electrons is must.
copper  Cu , has 29 electrons. 
so  electronic configuration will be $1s^22s^22p^63s^23p^64s^13d^{10}$
but in case of $Cu^+$ ion, 28 electrons so:
electronic configuration will be $1s^22s^22p^63s^23p^63d^{10}$
$Cu^+$ ion will loose its $4s^1$ electron,  and as it has filled $3d^{10}$ orbital,therefore no transition  and hence $Cu^+$ ion will not have any colour.


Out of $TiF _{6}^{2-},\ CoF _{6}^{3-},\ Cu _{2}Cl _{2}$ and $NiCl _{4}^{2-}$, the colourless sphere are..... 

  1. $Cu _{2}Cl _{2},\ NiCl _{4}^{2-}$

  2. $TiF _{6}^{2-},\ Cu _{2}Cl _{2}$

  3. $CoF _{6}^{3-},\ NiCl _{4}^{2-}$

  4. $TiF _{6}^{2-},\ CiF _{6}^{3-}$


Correct Option: B

Which of the following compound is expected to be coloured?

  1. $\,{\text{AgN}}{{\text{O}} _3}$

  2. $\,{\text{CuS}}{{\text{O}} _4}$

  3. ${\text{CuCl}}$

  4. ${\text{ZnC}}{{\text{l}} _{\text{2}}}$


Correct Option: A

When $ZnO$ is heated, it appears of _______colour due to metal excess defect.

  1. pink

  2. yellow

  3. violet

  4. blue


Correct Option: A

The color of $KMnO _{4}$  is due to :

  1. $d-d$ transition

  2. $L\rightarrow M$ charge transfer transition

  3. $\sigma-\sigma^{*}$ transition

  4. $M\rightarrow L$ charge transfer transition


Correct Option: A

Pair of ions which give blue colour in aqueous state: $V^{+4},\ Ni^{+2},\ Ti^{+3},\ Co^{+2},\ Fe^{+3}\ and\ Cu^{+2}$.

  1. $V^{+4}, Cu^{+2}$

  2. $Co^{+2}, Ni^{+2}$

  3. $Fe^{+3}, Ti^{+3}$

  4. $Co^{+2}, Fe^{+3}$


Correct Option: A

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 (1) and (2)


Correct Option: D

Which of the following is not correct?

  1. ${\text{TiC}}{{\text{l}} _{\text{4}}}$ is a colourless compound.

  2. $\left[ {{\text{Cr}}{{\left( {{\text{N}}{{\text{H}} _{\text{3}}}} \right)} _{\text{6}}}} \right]{\text{C}}{{\text{L}} _{\text{3}}}$ is a coloured compound

  3. ${{\text{K}} _{\text{3}}}\left[ {{\text{V}}{{\text{F}} _{\text{6}}}} \right]$ is a colourless compound

  4. $\;\left[ {{\text{Cu}}{{\left( {{\text{NCC}}{{\text{H}} _{\text{3}}}} \right)} _{\text{4}}}} \right]\left[ {{\text{B}}{{\text{F}} _{\text{4}}}} \right]$ is a colourless compound


Correct Option: A

The colour imparted by $Co(II)$ compounds to glass is 

  1. green

  2. deep blue

  3. yellow

  4. red


Correct Option: B
Explanation:
 Vanadium yellow-green 
 Chrome emerald green 
 Iron Coke bottle green 
 Manganese amethyst 
 Cobalt violet blue 
 Copper greenish blue to blue 
 Nickel grayish brown 
 Selenium salmon pink 
 Cerium- Titanium yellow 
 Neodymium dichroic violet-pink 
 Uranium yellow (fluorescent 

Which of the ions will give colorless aqueous solution?

  1. ${Ni}^{2+}$

  2. ${Fe}^{2+}$

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

  4. ${Cu}^{+}$


Correct Option: D
Explanation:

${Ni}^{2+}$ is light green, ${Fe}^{2+}$ is pale green, 


${Cu}^{2+}$ is blue and ${Cu}^{1+}$ is colorless in aqueous solution as its outermost configuration is $3d^{10}$ so thei is no unpaired electron which causes colour.

Hence ,option D is correct.

The color of the transition metal ions is due to:

  1. $d-d$ transition

  2. charge transfer

  3. change in the geometry

  4. none


Correct Option: A,B
Explanation:

In case of transition elements, where the d-level is not completely filled, it is possible to promote an electron from a lower energy d-orbital to a higher energy d-orbital by absorption of a photon of electromagnetic radiation having an appropriate energy. Electromagnetic radiations in the visible region of the spectrum often possess the appropriate energy for such transitions. 

Which of the following statements are correct with reference to the ferrous and ferric ions?

  1. ${Fe}^{3+}$ gives brown color with potassium ferricyanide

  2. ${Fe}^{2+}$ gives blue precipitate with potassium ferricyanide

  3. ${Fe}^{3+}$ gives red color with potassium thiocyanate

  4. ${Fe}^{2+}$ gives brown color with ammonium thiocyanate


Correct Option: B,C

Which one of the following ionic species will not impart color to an aqueous solution?

  1. ${Ti}^{4+}$

  2. ${Cu}^{+}$

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

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


Correct Option: A,B,C
Explanation:
The transition metal ions which have completely filled d-orbitals are colorless.

The transition metal ions which have completely empty d-orbitals are also colorless.

$ Cr^{3+}$ has a blue-green color.  because contain unpair electrons.

Rest all are colorless. 

The colorless species is:

  1. $V{Cl} _{3}$

  2. $VO{SO} _{4}$

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

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


Correct Option: C
Explanation:

$ VCl _3$ is green.

 
$ VOSO _4$ is blue. 

$ Na _3VO _4$ is colorless vanadium is present in +5 oxidation state .it does not have any unpaired electron.
 
$\left[ V{ \left( { H } _{ 2 }O \right)  } _{ 6 }{ SO } _{ 4 } \right]$ .${ H } _{ 2 }O$ is purple.

The aqueous solution of the salt will colored in the case of:

  1. $Zn{({NO} _{3})} _{2}$

  2. $Li{NO} _{3}$

  3. $Co{({NO} _{3})} _{2}$

  4. $Cr{Cl} _{3}$


Correct Option: C,D
Explanation:

$ Co^{2+}$ = $3d^7$  


$Cr^{3+}$ = $3d^3$
 
$ Co^{2+}$ and $Cr^{3+}$   will have unpaired electrons and hence are expected to have aqueous solution of the salt colored.

What is/are true statement?

  1. Ions of $d$-block elements are colored due to $d-d$ transition

  2. Ions of $f$-block elements are colored due to $f-f$ transtition

  3. ${ \left[ Sc{ \left( { H } _{ 2 }O \right) } _{ 6 } \right] }^{ 3+ },{ \left[ Ti{ \left( { H } _{ 2 }O \right) } _{ 6 } \right] }^{ 4+ }$ are colored complexes.

  4. ${Cu}^{+}$ is colorless ion.


Correct Option: A,B,D
Explanation:
$Cu^+$ is colourless because it doesn't contain any electrons which can absorb light. .

In ${ \left[ Sc{ \left( { H } _{ 2 }O \right)  } _{ 6 } \right]  }^{ 3+ }$, $ Sc^{3+}$  is $ d^0$ system and hence colorless.

Hence,option C is incorrect.

The aqueous solutions of the following salts will be colored in the case of:

  1. $Zn{({NO} _{3})} _{2}$

  2. $Li{NO} _{3}$

  3. $Co{({NO} _{3})} _{2}$

  4. $Cr{Cl} _{3}$

  5. Potash alum


Correct Option: C,D
Explanation:
${Co}^{2+}$
$\uparrow \downarrow $ $\uparrow \downarrow $ $\uparrow $ $\uparrow $ $\uparrow $
$z=27$
$Co=[Ar]{3d}^{7}{4s}^{2}$
${Co}^{2+}=[Ar]{3d}^{7}$
Two unpaired electrons present.

${Cr}^{3+}$:
$z=24$
$Cr=[Ar]{3d}^{5}{4s}^{1}$
${Cr}^{3+}=[Ar]{3d}^{3}$
$\uparrow $ $\uparrow $ $\uparrow $
$Co{({NO} _{3})} _{2}$ and $Cr{Cl} _{2}$ have unpaired electron; hence, they are colored.

The compounds have $ Zn^{2+}, Li^+, Co^{2+} and Cr^{3+}$ with 0, 0, 2 and 3 unpaired electrons. Hence option C and D are correct. Potash alum is colorless. 

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 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. 

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