NCERT MOST IMPORTANT QUESTIONS CLASS – 12 | CHEMISTRY IMPORTANT QUESTIONS | CHAPTER – 9 | COORDINATION COMPOUNDS | EDUGROWN |

In This Post we are  providing Chapter-9 COORDINATION COMPOUNDS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON COORDINATION COMPOUNDS

Question 1.
When a coordination compound NiCl₂.6H₂0 is mixed with AgNO₃, 2 moles of AgCI are precipitated per mole of the compound. Write
(i) Structural formula of the complex
(ii) IUPAC name of the complex
Answer:
For one mole of the compound, 2 moles of AgCI are precipitated with AgNO₃, which indicates 2 ionisable Cl” ions in the complex.

1. Structural formula: [Ni (H₂0)₆]Cl₂
2. IUPAC name: Hexaaquanickel (II) chloride

Question 2.
Write IUPAC name of the complex [Cr(NH₃)₄Cl₂]⁺. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Pentaamminenitrito-O-cobalt(III) chloride
(ii) Potassium tetracyanidonickelate(II) (CBSE Delhi 2019)
Answer:
IUPAC name: Tetramminedichloridochromi um(III) ion.
Geometrical isomers:

OR
(i) [CO(NH₃)₅(ONO)]Cl₂
(ii) K₂[Ni(CN)₄]

Question 3.
Write the hybridisation and magnetic character of the following complexes:
(i) [Fe(H₂O)₆]²⁺
(ii) [Fe(CO)₅] (Atomic no. of Fe = 26)
Answer:
[Fe(H₂0)6]²⁺
Hybridisation: sp³d²
Magnetic character: Paramagnetic due to 4 unpaired electrons.
Fe(C0)₅
Hybridisation: dsp³

Question 4.
Draw one of the geometrical isomers of the complex [Pt(en)₂Cl₂]²⁺ which is optically inactive. Also, write the name of this entity according to the IUPAC nomenclature.
Or
Discuss the bonding in the coordination entity [Co(NH₃)₆ ]³⁺ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co= 27) (CBSE Sample Paper 2019)
Answer:

IUPAC Name of the entity:
Dichloridobis(ethane – 1,2-diamine) platinum(IV) ion
Or

Question 5.
(a) Draw the structures of geometrical isomers of [Fe(NH₃)₂ (CN)₄]^–
(b) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? [Atomic number of Ni = 28]
OR
Define the following:
(a) Ambidentate ligands
(b) Spectra chemical series
(C) Heteroleptic complexes
Answer:

(b) In the presence of strong field ligand CO, the unpaired d-electrons of Ni pair up so [Ni(CO)₄] is diamagnetic but Cl^– being a weak ligand is unable to pair up the unpaired electrons, so [Ni(Cl₄)]^2- is paramagnetic.

and

OR
Answer:
(a) Ambidentate ligands: The monodentate ligands which can coordinate with the central atom through more than one site are called ambidentate ligands.
These ligands contain more than one coordinating atoms in their molecules.
For example, NO₂ can coordinate to the metal atom through N or 0 as

Similarly, CN^– can coordinate through C or N as

Thiocyanato (SCN) can coordinate through S or N

(b) The arrangement of ligands in the increasing order of crystal field splitting is called spectrochemical series. This is shown below:
I^– < Br^– < SCN^– < Cl^– < F^– < OH^– < Ox^2- < O^2- < H₂0 < NCS^– < py = NH₃ <en <NO₂^– < CN^– < CO.

Weak field ligands are those which cause less crystal field splitting. These form high spin complexes. For example, Cl^–, F^–, etc.

Strong field ligands are those which cause greater crystal field splitting. These form low spin complexes. For example, CN^–, NO₂^–, CO.

(c) The complexes in which the metal is bound to more than one kind of donor groups (ligands) are called heteroleptic complexes. Some common examples of heteroleptic complexes are [NiCl₂(H₂O)₄], [CoCl₂(NH₃)₄]⁺, etc.

Question 6.
Write the structures and names of all the stereoisomers of the following compounds:
(i) [Co(en)₃]Cl₃
Answer:

(ii) [Pt(NH₃)₂Cl₂]
Answer:

(iii) [Fe(NH₃)4Cl₂]Cl (CBSE 2011)
Answer:

Question 7.
Write the name, stereochemistry and magnetic behaviour of the following:
(i) K₄[Mn(CN)₆]
(ii) [Co(NH₃)₂ C1]Cl₂
(iii) K₂[Ni(CN)₄] (CBSE Delhi 2011)
Answer:

Complex	Name	Stereochemistry	Magnetic behaviour
(i) K4[Mn(CN)6]	Potassium hexacyanomanganate (I)	Octahedral	Paramagnetic
(ii) [Co(NH3)5Cl]Cl2	Pentaamminechtorido cobalt (III) chloride	Octahedral	Diamagnetic
(iii) K2[Ni(CN)4)	Potassium tetracyanonicketate (II)	Square planar	Diamagnetic

Question 8.
For the complex [Fe(en)₂Cl₂]Cl identify the following:
(i) Oxidation number of iron
Answer:
III.

(ii) Hybrid orbitals and shape of the complex
Answer:
d²sp³ hybridisation, octahedral

(iii) Magnetic behaviour of the complex
Answer:
Paramagnetic due to one unpaired electron

(iv) Number of its geometric isomers
Answer:
Two

(v) Whether there may be optical isomer also
Answer:
One optical, an isomer of cls-geometrical isomer.

(vi) Name the complex. (CBSE 2011)
Answer:
Dichloridobis(ethylenediamine) iron (III) chloride.

Question 9.
Give the formula of each of the following coordination entities:
(i) CO³⁺ ion is bound to one Cl^–, one NH₃ molecule and two bidentate ethylene diamine (en) molecules.
(ii) Ni²⁺ ¡on is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entitles. (At. nos. CO = 27, Ni = 28) (CBSE 2012)
Answer:
(i) [Co(NH₃) (Cl) (en)₂]²⁺ Amminechtoriðo bls-(ethane -1, 2-diamine) cobalt (III) ion Co(27):

Since there are no unpaired electrons,
complex is diamagnetic.

(ii) [Ni(H₂O)₂ (C₂O₄)₂]^2- Diaquadioxatatonickelate (II) ion Ni(28):

The complex has two unpaired electrons, therefore, it will be paramagnetic.

Question 10.
State a reason for each of the following situations:
(i) CO²⁺ is easily oxidised to CO³⁺ in presence of a strong ligand.
Answer:
The configuration of CO²⁺ is t_2g⁶ e_g¹ and for CO³⁺, it is t_2g⁶. The crystal field stabilisation energy is more than compensated for the third ionisation enthalpy. Therefore, CO²⁺ is easily oxidised to CO³⁺ in the presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH₃.
Answer:
CO is a stronger complexing ligand than NH₃ because it contains both σ and π character and can form a back bond (M → CO) also. Therefore, CO forms a stronger bond with the metal. It is also called a strong field ligand.

(iii) The molecular shape of Ni(CO)₄ is not the same as that of [Ni(CN)₄]^2-. (CBSE Delhi 2012)
Answer:
The molecular shape of [Ni(C0)₄] is tetrahedral because this complex nickel involves sp³ hybridisation. In [Ni(CN)₄]^2-, nickel involves dsp² and its shape is square planer.