Chemistry

Coordination Chemistry

325 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 chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

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

Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

Which of the following  is expected to form colourless complex?

  1. ${ Ni }^{ 2+ }$
  2. ${ Cu }^{ + }$
  3. ${ Ti }^{ 3+ }$
  4. ${ Fe }^{ 3+ }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.


Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

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. $TiF _{6}^{2-},\ CiF _{6}^{3-}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

TiF6^2- is colorless (Ti^4+ is d^0, no d-d transitions possible). Cu2Cl2 is also colorless/slightly yellowish as it's copper(I) with full d^10 configuration. The other complexes have partially filled d-orbitals giving color: CoF6^3- (green, Co^3+ d^6) and NiCl4^2- (blue, Ni^2+ d^8). Note: Option D has a typo 'CiF' should be 'CoF'.

Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

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. $d _{z^2},dxy$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Multiple choice chemistry p- block elements-ii anomalous properties of nitrogen nitrogen - 15 group group 15 elements

Which of the following statements is wrong?

  1. Single N-N bond is stronger than the single P-P bond.
  2. $PH _{3}$ can act as a ligand in the formation of coordination compound with transition elements.
  3. $NO _{2}$ is paramagnetic in nature.
  4. Covalency of nitrogen in $N _{2}O _{5}$ is four.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

N-N sigma bond (single bod)  is weaker than P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogen  repel each other making the single bon between $N-N$  weaker than P-P sigma bond.In $P-P$  bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.

So statement A is wrong.

Hence option A is correct.

Multiple choice chemistry chemical bond and chemical equation coordinate bond types of bonds and properties chemical bonding and structure

The bonds present in $K _4[Fe(CN) _6]$ are:

  1. all ionic

  2. all covalent

  3. ionic and covalent

  4. ionic, covalent and coordinate covalent

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

The bonds present in  $\displaystyle K _4 [Fe(CN) _6]$ are ionic, covalent and coordinate covalent bonds. Ionic bonds are present between $\displaystyle K^+$ ion and $\displaystyle [Fe(CN) _6]^{4-}$ ion.
Covalent bonds are present in $\displaystyle C \equiv N$ ligand.
Co-ordinate covalent bonds are present between $\displaystyle Fe$ and $\displaystyle C$ and also in between $\displaystyle C$ and $\displaystyle N$.

Multiple choice chemistry chemical bonding and structure coordinate bond types of bonds and properties chemical bond and chemical equation

Which of the following species contain coordinate covalent bond :

  1. $AlCl _3$
  2. $CO$
  3. $[Fe(CN) _6]^{4-}$
  4. $N _3^-$
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation
$:C\equiv O:$ $\rightarrow $ In Carbonyl two covalent bonds are present and one coordinate bond is present from O$\rightarrow $C.
in $[Fe(N{ ) } _{ 6 }{ ] }^{ 4- }\rightarrow $ All bonds are coordinate bond between ${ CN }^{ - }$ and ${ Fe }^{ +2 }$ metal ion.
in ${ N } _{ 3 }^{ - }$$\rightarrow [\overset { \cdot \cdot  }{ \underset { \cdot \cdot  }{ N }  } =N=\overset { \cdot \cdot  }{ \underset { \cdot \cdot  }{ N }  } { ] }^{ - }\longleftrightarrow [\ddot { N } \equiv N-\overset { \cdot \cdot  }{ \underset { \cdot \cdot  }{ N }  } :{ ] }^{ - }$
In ${ N } _{ 3 }^{ - }$ covalent and coordinate both type of bonds are present.
in $Al{ Cl } _{ 3 }$$\rightarrow $ Icovalent bond is present only between $Al $ and $Cl$.
Multiple choice chemistry chemical bond and chemical equation coordinate bond types of bonds and properties chemical bonding and structure

Consider following compounds:


$I: K _4[Fe(CN) _6], II:NH _4Cl, III: H _2SO _4$ 

Ionic, covalent and coordinate bonds are present in :

  1. I, II and III

  2. I and III

  3. II and III

  4. I and II
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$H _{2}SO _{4}$  has $3$ types of bonds- ionic, covalent and coordinate. Sulphur is the central atom,surrounded by $4$ oxygen atoms and $2$ hydrogen atoms places on left of any $2$ oxygen atoms. Now, oxygen shares one electron each with the sulphur and hydrogen atom. But by this only $2$ oxygen atoms will be able to complete octet. The remaining $2$ will complete by coordinate bonding. Both the oxygen atoms will take electrons from sulphur but sulphur won't take theirs ie sharing from only one end. This is called coordinate bond. Hydronium ions are attached with sulphate ions using ionic bonding
${ K } _{ 4 }[{ Fe(CN) } _{ 6 }]\ $: has ionic bond between ${ K }^{ + }and  { [{ Fe(CN) } _{ 6 }] }^{ 4- }$, coordinate bonding in $CN$ and covalent bond in ${ [{ Fe(CN) } _{ 6 }] }^{ 4- }$
$\ { NH } _{ 4 }Cl\ $: has ionic bond between $\ { { NH } _{ 4 } }^{ + }and  { Cl }^{ -}$, coordinate bonding in $\ { { NH } _{ 4 } }^{ + }$ and covalent bond in $\ { { NH } _{ 3 } .}^{  }\quad \$

Multiple choice photosystems photosynthesis in plants metabolism, cell respiration, and photosynthesis photosynthesis in higher plants biology

Complex IV refers to cytochrome c oxidase complex containing cytochromes.

  1. b and $c _1$ and one copper centre
  2. a and $a _3$ and four copper centres
  3. $c _1$ and c and three copper centres
  4. a and $a _3$ and two copper centres
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The enzyme cytochrome c oxidase or Complex IV is a large transmembrane protein complex found in bacteria, archaea, and in eukaryotes in their mitochondria. It is the last enzyme in the respiratory electron transport chain of cells located in the membrane. It receives an electron from each of four cytochrome c molecules, and transfers them to one dioxygen molecule, converting the molecular oxygen to two molecules of water. In this process it binds four protons from the inner aqueous phase to make two water molecules, and translocates another four protons across the membrane, increasing the transmembrane difference of proton electrochemical potential which the ATP synthase then uses to synthesize ATP. The complex contains two hemes, a cytochrome a and cytochrome a3, and two copper centers, the CuA and CuB centers. The cytochrome a3 and CuB form a binuclear center that is the site of oxygen reduction.


So the correct option is 'a and a$ _3$ and two copper centres'.

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

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