XII SC (CHEMISTRY)-UNIT-10 :BIOMOLECULES(4 MARKS) by a.

k panda
page-1
I# DO AS DIRECTED
*1# What are reducing and non-reducing sugars ? Give two examples. .What is the structural feature characterizing reducing sugars ?
2# An optically active compound having molecular formula C 6H12O6 is found in two isomeric forms(A) and (B) .When (A) and (B)
are dissolved in water they show the following equilibrium:
(A)--------------🡪 equilibrium mixture 🡨--------------(B)
[α] = 1110 52.20 19.20
(i) What are such isomers called ? (ii) Can they be called enantiomers ? Justify your answer .
(iii) Draw the cyclic structure of isomer (A) .
3# An optically active amino acid (A) can exist in three forms depending on the P H of the medium.If the molecular formula of (A) is
C3H7NO2 then write (i) Structure of compound (A) in aqueous medium. What are such ions called ?
(ii) In which medium will the cationic form of compound (A) exist ?
(iii) In alkaline medium, towards which electrode will the compound (A) migrate in electric field ?
*4 # When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained unlike DNA ? What does this
fact suggest about the structure of RNA?
5 # What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?
6# What is the effect of denaturation on the structure of proteins ?
7# Write down the structures and names of the products obtained when D-Glucose is treated with
(a) HI (b) Br2 Water (c) HNO3 (d) NH2OH (e) Acetic Anhydride (f) HCN (g) ammoniacal AgNO3
8# What is the significance of D and (+) in α-D(+)-glucopyranose
9 # What are the expected products hydrolysis of (i) Lactose (ii) sucrose (iii) Maltose
10# Classify the following into monosaccharides and disaccharides. (Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose).
11# Name the type of bonding which stabilizes α-Helix in proteins ?
12# What are nucleic acids ?Name two important functions of nucleic acids in our bodies.
13# How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
14# What type of bonding helps in stabilising the α-helix structure of proteins?
15# What is the effect of denaturation on the structure of proteins?
16# Name two different types of structures of (i) secondary structures of proteins (ii) tertiary structures of proteins
17# What are the different types of RNA found in the cell?
18 # List four functions of carbohydrates in living organisms.

II # GIVE REASONS FOR THE FOLLOWING:

1) Despite having an aldehydic group glucose does not give 2,4-DNP and Schiff’s test,What does this indicate ?
**2) Amino acids have relatively higher melting point as compared to corresponding haloacids.
3) Except for vitamin B12 ,all other vitamins of group B , should be supplied regularly in diet.Why?
**4) The two strands in DNA are not identical but are complementary .
**5) Enumerate the reactions of D-glucose which cannot be explained by its open chain
structure.
**6) How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
** 7) Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring
compounds) are insoluble in water.
**8) Amino acids are amphoteric in behaviour.
9) On electrolysis in acidic solution amino acids migrate towards cathode while in alkaline solution these migrate towards anode .
**10) Why cannot vitamin –C be stored in our body ?
**11) Why are vitamin A and vitamin C essential to us? Give their important sources.
12) Why carbohydrates are generally optically active
13) Why are the amino acids , those constitute protein , called α- amino acids ?
14) Why is cellulose not digested in human body
15) Why sucrose is a non-reducing sugar?

III # DIFFERENTIATE BETWEEN THE FOLLOWING:

**1.Structural and functional differences between DNA and RNA **2.Globular and
Fibrous protein
**3. Essential and non-essential amino acids ** 4.Reducing and
non-reducing sugars
**5.Structural difference between starch and cellulose 6..α-Helix and β-Sheet
structure.
**7.Glycogen and starch 8. Amylose and
Amylopectin
9.Oligosaccharides and polysaccharides **10.Peptide linkage and
Glycosidic linkage
**11.Nucleoside and nucleotide 12. Native protein and
denatured protein
**13.α-glucose and β- glucose ** 14# Peptide linkage
and phosphodiester linkage . 15. Nucleotides of DNA and RNA
16# Primary and secondary structures of protein

BIOMOLECULES -- ANSWERS TO (I)

ANS-1- Reducing Sugars:- All those carbohydrates which contain free aldehydic or ketonic groups
which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. Eg. Glucose,
fructose, maltose, lactose.
Non- Reducing Sugars :- All those carbohydrates which do not contain free aldehydic or ketonic
groups cannot reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars .e.g
sucrose .
These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and
C2 of β-fructose. Since the
reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non
reducing sugar.
ANS-2-(i) Anomers (ii) No, they are not enantiomers . α- and β-anomers are not mirror images of each other . They are not having
non-super imposable mirror images ( which is the conditions of showing enantiomerism)

(iii)
ANS-3-(i) they are called Zwitter ions
(ii) In acidic medium ,the cationic form of compound will exist
(iii) In alkaline medium , it will behave as anion and migrates towards anode in electric field
ANS-4- This indicates that RNA does not exist as double strand but exist as single strand.
DNA exist as double strand . The sequence of bases on one strand of DNA is complimentary to the sequence of bases on the
other strand . Therefore , the quantities of four bases obtained on hydrolysis of DNA have relationships .
Since in case of RNA there is no relationship among the quantities of bases it indicates that RNA exists in the form of single strand
and not double strand
ANS-5- : β-D-2-deoxy ribose ( pentose sugar) + Thymine ( a nitrogeneous base) + A phosphate unit .
ANS-6- When a protein in its native form, is subjected to physical change like change in temperature
or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this,
globules unfold and helix get uncoiled and protein loses its biological
activity. This is called denaturation of protein. During denaturation 2° and 3° structures
are destroyed but 1º structure remains intact.
ANS-7- (a) n-hexane (b) Gluconic acid (c) Saccharic acid or glucaric acid (d) Glucose monooxime
(e) Glucose Pentaacetate (f) Glucose cyanohydrin (g) Gluconic acid
ANS-8- ‘D’ before the name of glucose represents the configuration whereas ‘(+)’ represents
dextrorotatory nature of the molecule.
[*** It may be remembered that ‘D’ and ‘L’ have no relation with the optical activity of the
compound.]
ANS-9- (i) β-D-Galactose and β-D-Glucose (C-1 and C-4 glycosidic linkage)
(ii) α- D(+)-Glucose and β-fructose (C1—C2 glycosidic linkage)
(iii) α-D Glucose and α-D- Glucose (C-1 and C-4 glycosidic linkage)
Ans-10- Disaccharides- maltose and lactose . Monosaccharides- Ribose, 2-deoxyribose,
galactose and fructose
ANS-11-All possible hydrogen bonds by twisting into a right handed screw (helix) with the –NH group of
each amino acid residue hydrogen bonded to the >C= O of an adjacent turn of the helix
ANS-12- nucleic acids are long chain polymers of nucleotides which are responsible for hereditary and protein synthesis present in the
nucleus of the cell.
Replication (A DNA molecule is capable of self duplication during cell division and identical DNA
strands are transferred
to daughter cells.) and protein synthesis in the cell.
Ans-13-Water soluble( B and C) and Fat soluble(ADEK). Vit-K.
Ans-14-Due all possible hydrogen bonds, by twisting into a right handed screw (helix) with the –NH group
of each amino acid residue hydrogen bonded to the >C= O of an adjacent turn of the helix .
Ans-15- During denaturation 2° and 3° structures are destroyed but 1º structure remains intact . Due to
this, globules unfold and helix get uncoiled and protein loses its biological activity
ANS-16- (i) α-Helix and β-Sheet structure. (ii) Globular and Fibrous protein
Ans-17- Messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA
ANS-18-(i) Carbohydrates are used as storage molecules as starch in plants and glycogen in animals.
(ii) Cell wall of bacteria and plants is made up of cellulose.
(iii) We build furniture, etc. from cellulose in the form of wood and clothe ourselves with cellulose in the
form of cotton fibre.
They provide raw materials for many important industries like textiles, paper, lacquers and
breweries.
(iv) Two aldopentoses viz. D-ribose ( present in RNA) and 2-deoxy-D-ribose ( present in DNA) are
present in nucleic acids.

ANSWERS TO II
1# Absence of open chain structures and confirmation of cyclic structures .
2# The presence of both acidic( carboxylic group) and basic(amino group) in the same molecule of amino acids
behave like salts rather than simple amines or carboxylic acids. These are water-soluble, high melting
solids .
Also form better hydrogen bonding ability of –NH 2 group as compared to halogen .
3# Vitmin-B12 can be stored but all other vitamins of group B are water soluble and be excreted . So it should be
supplied regularly in diet
4# Two nucleic acid chains are wound about each other and held together by hydrogen bonds
between pairs of bases.
The two strands are complementary to each other because the hydrogen bonds are
formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas
cytosine forms hydrogen bonds with guanine.
5# The following experimental facts could not be explained by its open chain structure
(i) Despite having the aldehyde group, glucose does not give 2,4-DNP test, Schiff’s test and it
does not form the hydrogensulphite addition product with NaHSO 3.
(ii). The pentaacetate of glucose does not react with hydroxylamine indicating the absence of
free —CHO group.
(iii) Glucose is found to exist in two different crystalline forms which are named as α and β-
form.

6# Pentaacetate of D-glucose does not react with hydroxylamine indicating the absence of
free –CHO group .
7# Glucose or sucrose ( contains –OH group) can form H-bonding with water but cyclohexane or benzene
cannot .
8# In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton,
giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and
negative charges.
In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids
and bases.
9# In acidic solution , amino acids behave as cation ( by accepting a proton or H +) and migrate
towards cathode .
In alkaline solution , amino acids behave as anion ( by releasing a H +) and migrate towards
anode
10# Vitamin-C is a water soluble in nature.
11# Deficiency of Vitamin-A will cause X e r o p h t h a l m i a (hardening of cornea of eye) Night
blindness.
Important source of Vit-A : Fish liver oil, carrots, butter and milk
Deficiency of Vitamin-C will cause Scurvy (bleeding gums).
Important sourceof Vit-C-Citrus fruits, amla and green leafy vegetables
12# Carbohydrates are polyhydroxy aldehydes and ketones . all the carbons are chiral in nature
because of attachment of four different group in each carbon .
13# Because the amino group is attached to the α- carbon atom of the carboxyl acid functional
group .
14# We do not have enzyme to break down β- glycosidic linkage .
15# These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose
and C2 of β-fructose.
Since the reducing groups of glucose and fructose are involved in glycosidic bond formation,
sucrose is a non reducing sugar. ( both the anomeric carbons engaged in glycosidic language , so absence of free aldehydic
and ketonic group .)
ANSWERS TO III
1# (i)In DNA sugar is deoxyribose while in RNA it is ribose .
(ii) Pyrimidine base thymine is present in DNA only while uracil is present in RNA only .
(iii) DNA exists as double strand helix whereas RNA exists as single strand .
Functional difference :- DNA has the property of replication .DNA controls the transmission of
hereditry characters.
RNA controls the synthesis of proteins.
2# (a) Fibrous proteins :- When the polypeptide chains run parallel and are held together by
hydrogen and disulphide bonds, then fibre– like structure is formed. Such proteins are generally
insoluble in water. E.g- keratin (present in hair, wool, silk) and myosin (present in muscles ) . They are
linear condensation products.

Globular proteins:- This structure results when the chains of polypeptides coil around to give a
spherical shape. These are usually soluble in water. Insulin and albumins are the common examples
of globular proteins. They are crossed linked condensation polymers of acidic and basic amino acids .
3# Nonessential amino acids:- amino acids, which can be synthesised in the body. E.g – Glycine ,
Alanine .
Essential amino acids are:- those which cannot be synthesised in the body and must be obtained
through diet.
E.g Valine, Leucine , Isoleucine
4# Reducing Sugars:- All those carbohydrates which contain free aldehydic or ketonic groups
which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. Eg. Glucose,
fructose, maltose, lactose.

Non- Reducing Sugars :- All those carbohydrates which do not contain free aldehydic or ketonic
groups cannot reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars .e.g
sucrose .
5# Starch is a polymer of α-glucose and consists of two components— Amylose(straight chain) and
Amylopectin(branched).
Cellulose is a straight chain polysaccharide composed only of β-D-glucose units which are joined by
glycosidic linkage between C1 of one glucose unit and C-4 of the next glucose unit. (non-reducing)
6# α −Helix is one of the most common ways in which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed screw (helix) with the –NH group of each amino acid
residue hydrogen bonded to the >C= O of an adjacent turn of the helix .
 FIGURE---(i) α-helix structures of proteins (ii) β- Pleated sheet structure of
proteins

In β-structure all peptide chains are stretched out to nearly maximum extension and then laid side
by side which are held together by intermolecular hydrogen bonds. The structure resembles the
pleated folds of drapery and therefore is known as β -pleated sheet.
7# Glycogen: The carbohydrates are stored in animal body as glycogen.
It is also known as animal starch because its structure is similar to amylopectin and is rather more
highly branched. It is present
in liver, muscles and brain
Starch is a polymer of α-glucose and consists of two components— Amylose(straight chain) and
Amylopectin(branched).
8# Amylose :- Amylose is water soluble component which constitutes about 15-20% of starch.
Chemically amylose is
a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1– C4 glycosidic linkage.
Amylopectin.:- Amylopectin is insoluble in water and constitutes about 80-85% of starch. It is a
branched chain polymer of α−D-glucose . units in which chain is formed by C1–C4 glycosidic linkage
whereas branching occurs by C1–C6 glycosidic linkage.
9# Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are
called oligosaccharides.
They are further classified as disaccharides, trisaccharides,tetrasaccharides, etc., depending upon
the number of monosaccharides, they provide on hydrolysis.
Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysis
are called polysaccharides.
starch, cellulose, glycogen, gums
10# Glycosidic linkage is linkage between the monosaccharide units.
proteins are the polymers of α-amino acids and they are connected to each other by peptide
bond or peptide linkage.
Chemically, peptide linkage is an amide formed between –COOH group and –NH 2 group.)
11# A unit formed by the attachment of a base to 1’ position of sugar is known as nucleoside.
When nucleoside is linked to phosphoric acid at 5’-position of sugar moiety, we get a
nucleotide
Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of the
pentose sugar.
 FIGURES:- (a) Nucleoside (b) Nucleotide

12# Native protein :- Protein found in a biological system with a unique three-dimensional structure
and biological activity
Denaturation of Protein :- When a protein in its native form, is subjected to physical change
like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed.
Due to this, globules unfold and helix get uncoiled and protein loses its biological activity.
During denaturation 2° and 3° structures are destroyed but 1º structure remains intact. The
coagulation of egg white on boiling is a common example of denaturation. Another example is
curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk
13# The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group
at C-1, called anomeric carbon .Such isomers, i.e., α-form and β-form, are called anomers. In α-form
the –OH group is below the plane of the ring
Anomeric carbon-- the aldehyde carbon before cyclisation

14# Peptide linkage is an amide formed between –COOH group and –NH 2 group between amino
acids molecules
i.e [ --CO—NH-- ] linkage
Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of
the pentose sugar.
15# (i)In the nucleotides of DNA sugar is deoxyribose while in RNA it is ribose .
(ii) Pyrimidine base( nitrogeneous base) only thymine is present in nucleotides of DNA
while uracil is present in
RNA only .
16# Primary structure of proteins: Proteins may have one or more polypeptide chains. Each
polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this
sequence of amino acids that is said to be the primary structure of that protein. Any change
in this primary structure i.e., the sequence of amino acids creates a different protein.
Secondary structure of proteins: These structures arise due to the regular folding of the
backbone of the polypeptide
chain due to hydrogen bonding between >C=O-- and –NH--- groups of the peptide bond.

IV# SOME IMPORTANT ADDITIONAL QUESTIONS:-

1# What structural difference is there between α-glucose and β- glucose
2 # What is zwitter ion .Describe this with suitable examples ?
3 # What do you understand by Invert Sugars
4# Name two component of starch .How do they differ from each other structurally ?
5# Name the purines present in DNA
6# Name the vitamin whose deficiency causes the disease “pernicious anaemia” /
7# Name the base that is found in nucleotide of RNA only ?
8# Name the sugars present in nucleic acids ?
9# Name the vitamins whose deficiency are responsible for (i) night blindness (ii) poor coagulation of blood .(iii) rickets (iv) scurvy
10# Name two (i) essential aminoacids. (ii) non-essential aminoacids .
11# Name the optically inactive aminoacid?
12# Name type of forces are responsible for stabilizing secondary and tertiary structures of proteins
13# What is the name given to the linkage which holds together two monomeric units in
(i) polysaccharides (ii) nucleotides (iii) proteins (iv) Two strands of DNA
14# Name the type of linkage is possible for primary structure of proteins?
15# Name the scientists who gave a double strand helix structure for DNA .
16# What changes occur in the nature of egg proteins on boiling ?
17# Give two examples each of (i) Fibrous proteins (ii) Globular proteins
18# How oligopeptides different from polypeptides ?
19 # If a fragment of one strand in DNA molecule has the base sequence CCATGCATG, what is the base sequence of complementary strand?
20#What is known as animal starch ? 21# State Two characteristic of enzyme catalyst .
22 # Give an examples of anomers 23# Give two common examples denaturation of proteins
24# Mention the one / two functions each of (a) Carbohydrates in plants and animals (b) Nucleic acids (c) Glycogen
(d) Vitamins (e) Enzymes
25# CONVERT THE FOLLOWING from glucose
(A) Glucose cyanohydrin (B) Glucose monooxime (C) Glucose Pentaacetate (D) Gluconic acid (E) Saccharic acid (F) n-Hexane
2 6 # Draw the open chain(Fischer projection) and ring structure(Howorth structure ) of α-D(+)-glucose.
and β-D(+)-glucose. [α-D(+)-glucopyranose]
27# Which enzyme dissolves blood clot formed in coronary artery which leads to heart trouble .
28 # What are the common types of secondary structure of proteins?

ANSWERS TO IV
ANS-1- The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl
group at C-1, called anomeric
carbon

.
ANS-2- It occurs in amino acids .In aqueous solution, the carboxyl group can lose a proton and amino
group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but
contains both positive and negative charges.

In zwitter ionic form, amino

acids show amphoteric behaviour as they react both with acids and bases
ANS-3- hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–)
and the product is named as invert sugar.
**** [Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory
fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°),
the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation,
from dextro (+) to laevo (–) and the product is named as invert sugar. ]
ANS-4- Amylose and amylopectin .
Amylose is a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1– C4 glycosidic
linkage.
Amylopectin is a branched chain polymer of α−D-glucose units in which chain is formed by C1–C4
glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage.
ANS-5-Adenine(A) and Guanine(G)
ANS-6- Vitamin-B-12
ANS-7- Uracil
ANS-8- Ribose sugar (a pentose sugar)
ANS-9-(i) Vit-A (ii) Vit-K (iii) Vit-D (iv) Vit-C
ANS-10-(i) Valine and Leucine (ii) glycine and alanine
ANS-11- Glycine
ANS-12- The main forces which stabilise the 2° and 3° structures of proteins are hydrogen bonds ,
disulphide linkages,
van der Waals and electrostatic forces of attraction.
ANS-13- (i) Glycosidic linkage (ii) Phosphodiester linkage (iii) Peptide linkage (iv) H-bonding
ANS-14- Peptide linkage
ANS-15- Watson and Francis Crick .
ANS-16- The coagulation of egg white on boiling is a common example of denaturation
ANS-17-(i) keratin (present in hair, wool, silk) and myosin (present in muscles) (ii) Insulin and
albumins
ANS-18- Oligopeptides contain upto 10 aminoacids. When the number of such amino acids is more than ten,
then the products are called polypeptides
Polypeptides with fewer amino acids are likely to be called proteins if they ordinarily have a well
defined conformation of a protein such as insulin which contains 51 amino acids.
ANS-19- GGTACGTAC
ANS-20-Glycogen
ANS-21- Highly efficient and specific. Enzymes are very specific for a particular reaction and for a particular
substrate
ANS-22 - α-glucose and β- glucose
ANS-23- (i)The coagulation of egg white on boiling is a common example of denaturation.
(ii)curdling of milk which is caused due to the formation of lactic acid by the bacteria present in
milk.
ANS--24 (e) Enzyme acts as catalyst . Enzymes are said to reduce the magnitude of activation energy.
For example, activation energy for acid hydrolysis of sucrose is 6.22 kJ mol –1, while the
activation energy is only
2.15 kJ mol–1 when hydrolysed by the enzyme, sucrase.
ANS-25- (A) Glucose + HCN (B) Glucose + NH2OH( Hydroxyl amine) (C) Glucose + Acetic anhydride (D) Glucose + Bromine
water / Tollen’s Reagent (E) Glucose + Conc.HNO3 (F) Glucose + HI (a redeucing agent)

ANS--26

(Howorth structure ) -D(+)-glucopyranose ( Fischer projection)

Ans-27:-Streptokinase
Ans-28- α-Helix and β-sheet structure

V # Judge the following statement as TRUE OR FALSE

1. Glucose give 2,4-DNP test and Schiff’s test .
2. In D(+)-Glucose , The ‘D’ represents the dextrorotatory nature of the molecule and (+) represents the configuration .
3. α-glucose and β- glucose are called anomers .
4.The linkage between two monosaccharide units through oxygen atom is called phosphodiester linkage.
5.Sucrose is a non-reducing sugar .
6. Sucrose is dextrorotatory and glucose is laevorotatory .
7. Maltose on hydrolysis produces two molecules of α-D-glucose.
8. Maltose and lactose are reducing sugar .
9.All monosaccharides are reducing sugar .
10. Cellulose is straight chain of glycosidic linkage between C1 of one α-D-glucose unit and C6 of next
glucose unit.
11. Amylose is water soluble and amylopectin is insoluble in water .
12. Starch is a polymer of α-D-glucose unit .
13. Glycine and alanine are non-essential amino acids .
14.Insulin and albumins are the common examples globular proteins.
15. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.
16. Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of the
pentose sugar .
17. When nucleoside is linked to phosphoric acid at 5’-position of sugar moiety, we get a nucleotide .
18.Adenine pairs with Thymine through three H-bond in DNA.
19.The spatial arrangement of subunits of two or more polypeptide chain with respect to each other is
known as quaternary structure.
20.Albinism is a disease caused due to deficiency of enzyme called tyrosinase.
ANSWERS TO V :-
1 2 3 4 5 6 7 8 9 10
F F T F T F T T T F
11 12 13 14 15 16 17 18 19 20
T T T T T T T F T T

CLASS- XII CHEMISTRY ---CO-ORDINATION COMPOUNDS (7M)

I # (A) Write the I U P A C NAME---
1.[Pt(NH3)2Cl2] 2.K3[Fe(C2O4)3] 3.[Pt(NH3)6]Cl4 4.[Pt Cl (NH2CH3) (NH3)2]Cl
5.Na3[Cr(OH)2F4] 6.[Cr(H2O)5Cl]Cl2 7.[Co(en)2(ONO)Cl]Cl 8.(NH4)3[Co(ONO)6] 9.[Ni(H2O)6](ClO4)2
10.Linkage isomer of [Co(ONO)(NH3)5]2+ 11.Ionisation isomer of [Co(NH3)5Br]SO4 14. LiAlH4
13. Ionisation isomer of [Pt(NH3)3(NO2)]Cl 14. Coordination isomer of [Co(NH3)6][Cr(CN)6] 15. [Co(en)3]3+
(B) Using IUPAC norms write the formula of the following:1.Hexaammine cobalt(III)sulphate 2. Pentaamminenitrito-o-cobalt(III)
(C) Write the formula and Draw the structures for
1.Mer-triamminetrichlorocobalt(III) 2 .Fac-triaquatrinitro-N-cobalt(III) 3. Potassium tri-oxalatochromate(III)
4. Hexaammineplatinum(IV)chloride 5.dichlorotetraamminecobalt(III) ion 6.cis-dichlorotetraamminechromate(III)
7. Pentaamminenitrito-N-cobalt(III) 8.Hexamethyldialuminium 9.Tetramethylsilane
10. Trimethylboron 11. Methyllithium 12.Trimethylarsane 13. Trimethylbismuth
(D) Give the IUPAC Name and Draw the structure of (A) Ni(CO)4 (B) Fe(CO)5 (C) [Co(CO)6]
(E) What is meant by the denticity of a ligand ? Describe the following with suitable examples (i) ambidentate ligand
(ii) multidentate ligands or chelating ligands (iii) unidentate ligand (iv) didentate ligand (v) coordination number
(vi) macrocyclic effect (vii) Chelate effect ( coordination entity, ligand,coordination number, coordination
polyhedron, homoleptic and heteroleptic.) , Synergic bond , Spectrochemical series, Overall stability constant
II # Using V B T predict the hybridization, geometry, shape and magnetic behaviour of the following :
1.(a) [Cr(NH3)6]3+ (b)[Ni(CO)4] (c) [Ni(CN)4]2‾ (d) Pentacarbonyliron(0) (e) [Co(NH3)6]3+
2+ 2‾ 3+ 3‾ 4‾ 3‾
(f) [Ni(NH3)6] (g)[CoCl4] (h) [Co(CO)6] (i) [Fe(CN)6] (j) [Fe(CN)6] (k) [FeF6] (l)[Co(C2O4)3]3‾ (M) [CoF6] 3‾
2. Compare the magnetic behaviour of the complex entities [Fe(CN)6]4‾ and [FeF6] 3‾
3.Explain on the basis of VBT ,the experimental findings that [Ni(CN)4]2‾ ion with a square-planar structure is diamagnetic and
[NiCl4]2‾ion with tetrahedral geometry is paramagnetic.
III # CFT / Bonding / ORGANOMETALLICS / MOT (back-bonding) / stability of complex etc.
1.(a) Which postulates did Werner use to explain the bonding in coordination cpds?What is the main weakness of Werner’s theory ?
(a)Discuss briefly the role of coordination compound (i) medicinal chemistry. (ii) biological processes
(iii)analytical chemistry (iv) extraction / metallurgy of metals (v) softening of water (vi) electroplating
(b)Define organometallic compound. How are organometallic compounds classified.Mention the application of organometallic
compound in the following areas (i) homogeneous catalysis (ii) heterogeneous catalysis (iii) organic synthesis
2.Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field. State for a d6 ion/ d 4 ion how the actual
configuration of the split d-orbitals in an octahedral crystal field is decided by the relative values of ∆ 0 and P .
3.Draw the figure to show splitting of degenerate d-orbitals in an octahedral crystal field.
(a) How does the magnitude of ∆0 decide the actual configuration of the d-orbitals in a complex entity?
(b) How does the magnitude of ∆0 affected by (i) nature of ligand and (ii) Oxidation states of metal ions
(c) Define crystal field orbital splitting energy (d) What the basis of the crystal field treatment of coordination compounds
 (e) For d4 ions what will be the pattern of electron distribution when (i) Δ0 > P (ii) Δ 0 < P
4.The values of dissociation constant of [Cu(NH3)4]2+ and [Co(NH3)6]3+ are 1.0 ×10 -12 and 6.2 ×10 -36 respectively.Which complex
would be more stable and why ?
5.Name two properties of the central metal ion which enable it to form stable complex entities .
6.A coordination compound has the formula CoCl3.4NH3 .It does not liberate NH3 but forms a precipitate of AgCl on treatement with
AgNO3 solution .Write the structure and IUPAC name of the complex
7.Name a ligand which is bidentate and give an example of the complex formed by this ligand .
8.Though CO is a weak lewis base yet it forms a number of stable metal carbonyls .Explain
9.Among the iron complex ,K3[Fe(CN)6] is weakely paramagnetic whereas K3[FeF6] is highly paramagnetic. Explain
10.Aqueous copper sulphate solution (blue in colour)gives:
(a) a green precipitate with aqueous potassium fluoride , and (b) a bright green solution with aq.potassium chloride . Explain these experimental results .
11.What is the coordination entity formed when excess of aq. KCN is added to an aq.solution of copper sulphate ?Why is it that no
precipitate of copper sulphide is obtained when H2S(g) is passed through the solution ?
12.Discuss the nature of bonding in [Ni(CO)4] (back-bonding / synergic bonding interactions in a carbonyl complex)
13.Mention two shortcomings of (a) VBT (b) CFT
14. What is understood by the generalization , “magnetic criteria of the bond type” ? Illustrate your answer with suitable examples .
IV# ISOMERISM
1.What type of isomerism is exhibited by (a) the following pair : [Co(NH3)5Br]SO4 and [ Co(NH3)5 SO4] Br (b) [Cr(H2O)5(NCS)]2+
3.Write the IUPAC Name and draw the structures of all the isomers with this formula of complex. (optical & geometrical)
(i) [Co(en)2Cl2]Cl (ii) [PtCl(NH3)5]Cl3 (iii)[CrCl2(en)(NH3)2]+ (draw geometrical & optical isomers)
3‾ +
(iv) [Co(OX)3] (v) [CoCl2(en)2] (vi) [CoCl2(en)(NH3)2]+ (vii) [CoCl (en)2(NH3)]+(viii)[CoCl3(NH3)3]+
4.Out of [Co(NH3)5Cl]SO4 and [ Co(NH3)5 SO4]Cl ,which compound will give white precipitate with BaCl2 solution and why
5.Write the correct formulae for the following coordination compounds
(a) CrCl3 .6H2O [violate ,with 3 chloride ions / unit formula] (b) CrCl3 .6H2O [light green colour, with 2 chloride ions / unit formula]
(c) CrCl3 .6H2O [dark green colour, with 1 chloride ions / unit formula]
6. Illustrate Hydrate , coordination , Linkage , Ionisation , geometrical , Optical isomerism with suitable examples
7. Draw all the isomers (geometrical and optical) of: (i) [CoCl 2(en)2]+ (ii) [Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+
8.Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
PREVIOUS QUESTION PAPERS - COORDINATION COMPOUNDS
CLASS XII - CHEMISTRY -2024
Q.1# D-2013 # Write the IUPAC Name of the following co-ordination compounds :
(a)[Cr(NH3)3Cl3] (ii) K3[Fe(CN)6] (iii)[CoBr2(en)2]+ (en = ethylenediamine) (3M)
Q.2## D-2013 # Write the isomerism exhibited by the following complexes :
(i)[Co(NH3)5Cl]SO4 (ii)[Co(en)3]3+ (iii) [Co(NH3)6][Cr(CN)6] (3M)
Q.3#AI-2013# For the complex[NiCl4]2- , write
(i) IUPAC Name , (ii) the hybridization type,
(iii) the shape of the complex (atomic number of Ni= 28) (3M)
Q.4# AI-2013# What is meant by crystal field splitting energy ? On the basis of crystal field
theory , Write the electronic configuration of d4 in terms of t2g and eg in an octahedral
field when (i) ∆0 > P (ii) ∆0 < P (3M)
Q.5# AI-12#SET-1# Name the following coordination entities and draw the structures of their stereo-isomers :
(i) [Co(en)2Cl2]+ (ii) [Cr(C2O4)3]3- (iii) [Co(NH3)3Cl3] (At. No. Cr =24 , Co = 27)
Q.6#AI-12#SET-2 # Name the following coordination entities and draw the structures.
(i) [Fe(CN)6]4- (ii) [Cr(NH3)4Cl2]+ (iii) [Ni(CN)4]2- (At.no. Fe =26 ,Cr =24 ,Ni = 28 )
Q.7#F-12#SET-1# Explain the following : (i) The π –complexes are known for transition elements only .
(ii) Nickel(II) does not form low spin octahedral complexes
(iii) [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions
Q.8#F-12#SET-2# Name the following coordination entities and draw the structures of one possible isomer of each : (i)
[Cr(C2O4)3]3- (ii) [ Pt(NH3)2Cl2] (iii) [Co(en)2Cl2]+
Q.9#D-12#SET-1# Give the formula of each of the following co-ordination entities:
(i) Co3+ ion is bound to one Cl- , one NH3 molecule and two bidentate ethylene diamine(en) molecules
(ii) Ni2+ ion is bound to two water molecules and two oxalate ions .
Write the name and magnetic behaviour of each of the above co-ordination entities . (At Nos of Co=27 , Ni=28)
Q.10#D-12#set-2 # State a reason for each of the following situations :
(i) Co2+ is easily oxidized to Co3+ in the presence of strong ligand .
(ii) CO is a stronger complexing reagent than NH3 .
(iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2-
Q.11#D-12#SET-3# Write the name , the structure and magnetic behaviour of each one of the following complexes : (i)
[Pt(NH3)2Cl(NO2)] (ii) [Co(NH3)4Cl2]Cl (iii) Ni(CO)4 (At No- Co=27 , Ni =28 , Pt =78)
Q.12#D-11#SET-1# Write the name , stereochemistry and magnetic behaviour of each one of the following complexes :
(At No- Mn=25 , Ni =28 , Co=27) (i) K4[Mn(CN)6] (ii) [Co(NH3)5Cl]Cl2 (iii) K2[Ni(CN)4]
Q.13#D-11#SET-2# For the complex [Fe(en)2Cl2]Cl , Identify (D-09)
(a) Oxidation Number of Iron -----------------------------------------------------------------------------
(b) The hybrid orbitals------------------ and the shape of the complex---------------------------
(c) the magnetic behavior of the complex-----------------------------------------------------------
(d) the number of geometrical isomers -----------------------------------------------------------
(e) whether there is an optical isomers also ---------------------------------------------------------
(f) Name the complex.(At.No. of Fe =26) -------------------------------------------------------------
Q.14#AI-11#SET-2# Write the structure and names of all the stereo-isomers of the following compounds ,
(i) [Co(en)3]Cl3 (ii) [Pt(NH3)2Cl2] (iii) [Fe(NH3)4Cl2]Cl
Q.15#AI-11#SET-3# Write the state of hybridization , the shape and magnetic behaviour of each one of the following
complexes : (At No- Mn=25 , Ni =28 , Co=27)
(i) Co(en)3]Cl3 (ii) [Co(NH3)4Cl2]Cl (iii) K2[Ni(CN)4]
Q.16#F-11# Set-1# Name the following co-ordination compounds and draw their structures :
(i) [Co(en)2Cl2]Cl (ii) [Pt(NH3)2Cl(NO2)]
Q.17#F-11# Set-2 # Name the following co-ordination compounds and draw their structures
(i) [Co(en)3]3+ (ii) [Co(NH3)4Cl2]+
Q.18#F-11#SET-3# Write the state of hybridization , the shape and magnetic behaviour of each one of the following
complexes : (At No- Mn=25 , Ni =28 , Co=27)
(i) K3[CrF6] (ii) [Co(NH3)5Cl]Cl2 (iii) K2[Ni(CN)4] (iv) [Co(NH3)2(en)2]3+ (F-10)
Q.19#AI-10#SET-1# Explain the following cases giving appropriate examples:
(i) Nickel does not form low spin octahedral complexes
(ii) The pi-complexes are known for the transition metals only .
(iii) Co2+ is easily oxidized to Co3+ in the presence of strong ligand
** Uses of Co-ordination compounds in various fields (i) Medicinal (ii) biological (iii) Analytical (iv) Extraction
Q.20#AI-10# Set-2# Write the name ,state of hybridization , the shape and magnetic behaviour of each one of
the following complexes : (i) [CoCl4]2- (ii) [Ni(CN)4]2- (iii) [Cr(H2O)2(C2O4)2]- (F-10)
Q.21#D-10#SET-1# Name the following co-ordination compounds according to the IUPAC system of
nomenclature (i) [Cr(en)2Cl2]Cl (ii) [Co(NH3)4(H2O)Cl]Cl2
Q.22#AI-10# Set-2# Write the name ,state of hybridization , the shape and magnetic behaviour of each one of
the following complexes : (i) [Co(NH3)6]3+ (ii) [Ni(CN)4]2-
Q.23# (a)1M# A co-ordination compound with the molecular formula CrCl3 .4H2O precipitates AgCl with
AgNO3 solution . Its molar conductivity is found to be equivalent to two ions . What is the structural formula
of the compound ?
(b) 2M# Draw all the isomers of [Co(NH3)2Cl2(en)]+ . Also write the IUPAC Names
Q.24## For the complex [Fe(en)2Cl2]Cl , Identify (D-09)
(a) Oxidation Number of Iron -----------------------------------------------------------------------------
(b) The hybrid orbitals------------------ and the shape of the complex---------------------------
(c) the magnetic behavior of the complex-----------------------------------------------------------
(d) the number of geometrical isomers -----------------------------------------------------------
(e) whether there is an optical isomers also ---------------------------------------------------------
(f) Name the complex.(At.No. of Fe =26) -------------------------------------------------------------
Q.25#Compare the following complex with respect to their shape , magnetic behavior and the hybrid orbitas involved :
(a) [CoF4]2ˉ : shape----------- magnetic behavior------------------ hybrid orbital ---------------------------------(D-09)
(b) [Cr(H2O)2(C2O4)2]ˉ : shape------- magnetic behavior---------------- hybrid orbital ---------------------------(D-09)
(c) [Ni(CO)4] : shape------------ magnetic behavior-------------- hybrid orbital --------------------------------(D-09)
(d) [Co(NH3)6]3+ : shape--------------- magnetic behavior----------- hybrid orbital -----------------------------(D-09)
(e) [Ni(CN)4]2-- : shape------------- magnetic behavior------------ hybrid orbital------------------------------ (AI-09)
(f) [NiCl4]2‾ : shape--------------- magnetic behavior---------- hybrid orbital ----------------------------------(AI-09)
(g) [CoF6]3ˉ : shape-------------- magnetic behavior----------- hybrid orbital -------------------------------(AI-09)
(h) [Fe(CN)6]4- : shape------------ magnetic behavior---------- hybrid orbital ---------------------------------------(F-09)
Q.26#Explain the following :
(i) Low spin octahedral complex are not known. (AI-09)
(ii) The π-complexes are known for transition metals only . (AI-09)
(iii) CO is the stronger ligand than NH3 for many metals . (AI-09)
Q.27# Describe and give suitable example for each of the following :
(a) Crystal field splitting in octahedral field (AI-09(AI-11) (b) Chealating ligand
(c) Ambidentate Ligand (AI-09)(F-10) (AI-11) (d) An outer orbital complex(F-09)
(e) Bidentate ligand(F-09) (f) Denticity of ligand (AI&F-11)
(g) Spectrochemical series(F-10) (h) Linkage isomerism with example (D-10) (AI-09)
(i) Co-ordination isomerism with example (D-10) set-2 (J) IONISATION ISOMERISM with example (AI-2014)
(K) SYNERGIC BONDING (AI-2014) AM I JOKING !!!!!!!!!!!!!! (l0) Octahedral splitting energy
Q.28# Three geometrical isomers are possible for [ Co(en)(H2O)2(NH3)2]3+ . Draw molecular structures of these
three isomers and indicate which one of them is chiral .(3M) (F-09)
Q.29# Describe the limitations of Valence Bond Theory (3M) (F-08)
Q.30#Draw the structure and write the hybridization state of Co in cis- [Co(NH3)4Cl2]+ (2) [ AI-2008]
Q.31#IUPAC Name (i) [Co(NH3)4Cl(ONO)]Cl (ii) [Co(NH3)6] [Cr(CN)6] (ii) [Cu(NH3)4] [PtCl4]
Q.32#Explain as to how the two complexes of Ni , [Ni(CN)4]2-- and [Ni(CO)4] have different structures but
do not differ in their magnetic behavior . (At.No. Ni=28) ( D-08)

CO-ORDINATION COMPOUNDS CARDS

CARD-1 : - Co-ordination compounds – FAQ – MLL CARD-2 : - Co-ordination compounds – FAQ – HOT
1# Give the IUPAC Name of it K3[Fe(C2O4)3] (1) 1# Give the IUPAC Name of it [Co(en)2(ONO)Cl]Cl (1)
2# Discuss briefly the role of following coordination compounds 2# . Indicate the types of isomerism exhibited by the following
(i) medicinal chemistry (ii) homogeneous catalyst (1) complexes and draw the structures for these isomers [Co(OX)3]3‾ (1)
3#. [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3– is 3# Discuss briefly the role of following coordination compounds (i)
weakly paramagnetic. Explain. (1) biological processes(systems) (ii) Estimation of hardness of water (1)
4#. Why is geometrical isomerism not possible in tetrahedral 4#.Calculate the overall complex dissociation equilibrium constant for
complexes having two different types of unidentate ligands the [Cu(NH3)4 ]2+ ion, given that β4 for this complex is 2.1 × 1013. (1)
coordinated with the central metal ion ? (1) 5#.Explain on the basis of VBT ,the experimental findings that
5#. FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio [Ni(CN)4]2‾ ion with a square-planar structure is diamagnetic and
gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous [NiCl4]2‾ion with tetrahedral geometry is paramagnetic (2)
ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. 6# . Draw the figure to show splitting of degenerate d-orbitals in an
Explain why? (1) tetrahedral crystal field.How Δ0 and Δt related (2)
6# Draw a sketch to show the splitting of d-orbitals in an octahedral 7# . Indicate the types of isomerism exhibited by the following
crystal field. State for a d6 ion how the actual configuration of the complexes and draw the structures for these isomers: [PtCl2(en)2]2+ (2)
split d-orbitals in an octahedral crystal field is decided by the
relative values of ∆0 and P . (2)
CARD-3 : - Co-ordination compounds – FAQ HOT CARD-4: - Co-ordination compounds – FAQ – MLL
1# Give the IUPAC Name of it [Co(H2NCH2CH2NH2)3]2(SO4)3 (1) 1# Give the IUPAC Name of it Ionisation isomer of [Co(NH3)5Br]SO4
2# Indicate the types of isomerism exhibited by the following (1)
complexes and draw the structures for these isomers: 2# . Indicate the types of isomerism exhibited by the following
[Co(NH3 )3(NO2 )3] (1) complexes and draw the structures for these isomers: [Fe(NH3)2(CN)4]–
3# Discuss briefly the role of following coordination compounds i) (1)
analytical chemistry (ii) electroplating (1) 3# Discuss briefly the role of following coordination compounds (i)
4#. [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though extraction / metallurgy of metals (ii) photography (1)
both are tetrahedral. Why? (1) 4#. [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute
5#.A coordination compound has the formula CoCl3.4NH3 .It does solutions. Why? (1)
not liberate NH3 but forms a precipitate of AgCl on treatement with 5#Though CO is a weak lewis base yet it forms a number of stable metal
AgNO3 solution .Write the structure and IUPAC name of the carbonyls .Explain (1)
complex . (1) 6#. Explain why [Co(NH3)6]3+ is an inner orbital complex whereas
6#.Among the iron complex , K3[Fe(CN)6] is weakely paramagnetic [Ni(NH3)6]2+ is an outer orbital complex. (1)
whereas K3[FeF6] is highly paramagnetic. Explain (2) 7#. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic.
7# Draw the figure to show splitting of degenerate d-orbitals in an Explain why? (1)
octahedral crystal field. (2) 8#. Draw the figure to show splitting of degenerate d-orbitals in an
(a) How does the magnitude of ∆0 decide the actual configuration octahedral crystal field. (2)
of the d-orbitals in a coordination entity? (a) Define crystal field orbital splitting energy (b) What the basis
(b) How does the magnitude of ∆0 affected by (i) nature of ligand of the crystal field treatment of coordination compounds
and (ii) Oxidation states of metal ions (c) For d4 ions what will be the pattern of electron distribution when
(i) Δ0 > P (ii) Δ 0 < P

MLL-1
1 mark questions
1. Explain coordination entity with example.

Ans: it constitute a central metal atom or ions bonded to a fixed number of molecules or ions ( ligands) .eg.
[Co(NH3)3Cl3].
2. What do you understand by coordination compounds?

Ans: coordination compounds are the compounds which contains complex ions. These compounds contain a
central metal atom or cation which is attached with a fixed number of anions or molecules called ligands
through coordinate bonds. eg. [Co(NH3)3Cl3]
3. What is coordination number?

Ans: the coordination number of a metal ion in a complex may be defined as the total number of ligand donor
atoms to which the metal ion is directly bonded. Eg. In the complex ion [Co(NH3)6]3+ has 6 coordination
number.
4. Name the different types of isomerisms in coordination compounds.

Ans: structural isomerism and stereoisomerism

5. What is spectrochemical series?

Ans: the series in which ligands are arranged in the order of increasing field strength is called spectrochemical
series. The order is :
I-<Br-<SCN-<Cl-<S2-<F-<OH-<C2O42-<H2O<NCS-<EDTA4-<NH3<en<CN-<CO
6. What do you understand by denticity of a ligand?

Ans: the number of coordinating groups present in ligand is called denticity of ligand. Eg.Bidentate ligand
ethane-1,2-diamine has 2 donor nitrogen atoms which can link to central metal atom.
7. Why is CO a stronger ligand than Cl-?

Ans: because CO has π bonds.

8. Why are low spin tetrahedral complexes not formed?

Ans : because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.
9. Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas
tetrahedral complexes do not. Why?

Ans: tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands
attached to the central metal atom are same with respect to each other.
10. What are crystal fields?

Ans: the ligands has around them negatively charged field because of which they are called crystal fields.
11. What is meant by chelate effect? Give an example .

Ans: when a didentate or polydentate ligand contains donor atoms positioned in such a way that when they
coordinate with the central metal atom, a 5 or 6 membered ring is formed , the effect is called chelate effect.
Eg. [PtCl2(en)]

12. What do you understand by ambidentate ligand?

Ans: a ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with
central metal atom or ion is called an ambidentate ligand. Eg.nitrito-N and nitrito-O.
13. What is the difference between homoleptic and heteroleptic complexes?

Ans: in homoleptic complexes the central metal atom is bound to only one kind of donor groups whereas in
heteroleptic complexes the central metal atom is bound to more than one type of donor atoms.
14. Give one limitation for crystal field theory.

Ans: i) as the ligands are considered as point charges, the anionic ligands should exert greater splitting effect.
However the anionic ligands are found at the low end of the spectrochemical series.
ii) it does not take into account the covalent character of metal ligand bond. ( any one )
15. How many ions are produced from the complex: [Co(NH3)6]Cl2

Ans: 3 ions
16. The oxidation number of cobalt in K[Co(CO)4]

Ans: -1
17. Which compound is used to estimate the hardness of water volumetrically?

Ans: EDTA
18. Magnetic moment of [MnCl4]2- is 5.92B.M explain with reason.

Ans: the magnetic moment of 5.9 B.M. corresponds to the presence of 5 unpaired electrons in the d- orbitals of
Mn2+ ion. As a result the hybridisation involved is sp3 rather than dsp2. Thus tetrahedral structure of [MnCl4]2-
complex will show 5.92 B.M magnetic moment value.
19. How many donor atoms are present in EDTA ligand?

Ans: 6
2 marks questions
1. Give the electronic configuration of the following complexes on the basis of crystal field
splitting theory.

i) [CoF6]3-

ii) [Fe(CN)6]4-

Ans: i) Co3+ (d6) t2g4eg2

iii) Fe2+ d6t2g6eg0

2. Explain the following with examples:

i) Linkage isomerism

ii) Outer orbital complex

Ans: i) this type of isomerism arises due to the presence of ambidentate ligand in a coordination compound. Eg.
[Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2
iii) When ns, np and nd orbitals are involved in hybridisation , outer orbital complex is formed.
Eg. [CoF6]2- in which cobalt is sp3d2 hybridised.

3. i)Low spin octahedral complexes of nickel are not found . Explain why?

ii)theπ complexes are known for transition elements only.explain.

Ans: i) nickel in its atomic or ionic state cannot afford 2 vacant 3d orbitals and hence d2sp3 hybridisation is not
possible.
ii) transition metals have vacant d orbitals in their atoms or ions into which the electron pairs can be donated by
ligands containing πelectrons.eg. benzene, ethylene etc. thus dπ-pπ bonding is possible.
4. How would you account for the following:

i) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.

ii) [Ni(CO)4] possess tetrahedral geometry whereas [Ni(CN)4]2- is square planar.

Ans: i) due to the presence of 1 electron in 3d subshell in [Ti(H2O)6]3+ complex d-d transition takes place by the
absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H2O)6]3+ does not
possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this
complex is not possible, therefore it is colourless.
ii) Ni in [Ni(CO)4] is sp3 hybridised. Hence it is tetrahedral. Whereas for [Ni(CN)4]2- is dsp2 hybridised hence
it has square planar geometry.
5. Give the stereochemistry and the magnetic behaviour of the following complexes:

i) [Co(NH3)5Cl]Cl2 ii) K2[Ni(CN)4]

Ans: i) d2sp3 hybridisation, structure and shape = octahedral

Magnetic behaviour- diamagnetic
ii) dsp2 hybridisation, structure and shape = square planar
magnetic behaviour- diamagnetic
6. Draw the structures of isomers if any and write the names of the following complexes:

i) [Cr(NH3)4Cl2]+ ii) [Co(en)3]3+

Ans: i) tetraamminedichloridochromium(III) ion ii) tris(ethane-1,2-diammine)obalt(III)ion.

7. Hydrated copper sulphate is blue in colour whereas anhydrous copper sulphate is colourless.
Why?

Ans: because water molecules act as ligands which splits the d orbital of the Cu2+ metal ion. This result in d-d
transition in which t2g6eg3 excited to t2g5eg4 and this impart blue colour to the crystal. Whereas when we talk
about anhydrous copper sulphate it does not contain any ligand which could split the d orbital to have CFSE
effect.
8. Calculate the magnetic moment of the metal ions present in the following complexes:

i) [Cu(NH3)4]SO4

ii) [Ni(CN)4]2-

Ans: i)electronicconfig. t2g6eg3, n=1, µs= √n(n+2) = 1.732 B.M

ii) electronicconfig. t2g6eg2 , n=2, µs= √n(n+2)=2.828 B.M
3 marks questions
1. (a) What is a ligand? Give an example of a bidentate ligand.

(b) explain as to how the 2 complexes of nickel,[Ni(CN)4]2- and Ni(CO)4 have different structures but donot
differ in their magnetic behaviour.( Ni=28)
Ans: (a) the ion , atom or molecule bound to the central atom or ion in the coordination entity is called ligand. A
ligand should have lone pair of electrons in their valence orbital which can be donated to central metal atom or
ion.
Eg.Bidentate ligand- ethylenediammine

(b)dsp2, square planar, diamagnetic (n=0)

 Sp3 hybridisation , tetrahedral geometry, diamagnetic (n=0)
2. Nomenclate the following complexes:
i) [Co(NH3)5(CO3)]Cl
ii)[COCl2(en)2]Cl
iii) Fe4[Fe(CN) 6]
Ans: i) pentaamminecarbonatocobalt(III)chloride
ii) dichloridobis(ethane-1,2-diamine)cobalt(III)chloride
iii)iron(III)hexacyanidoferrate(II)
3. (a)why do compounds with similar geometry have different magnetic moment?
(b)what is the relationship between the observed colour and wavelength of light absorbed by the
complex?
Ans: (a) it is due to the presence of weak and strong ligands in complexes, if CFSE is high the complex will
show low value of magnetic moment and if it is low the value of magnetic moment is high. Eg. [CoF 6]3- and
[Co(NH3)6]3+ , the former is paramagnetic and the latter is diamagnetic.
(b) higher the CFS lower will be the wavelength of absorbed light. Colour of the complex is obtained from
the wavelength of the leftover light.
4. Explain the following terms giving a suitable example.
(a) ambident ligand
(b) denticity of a ligand
(c) crystal field splitting in an octahedral field
Ans (a) Aligand which contains two donor atoms but only one of them forms a coordinate bond at a time with
central metal atom or ion is called an ambidentate ligand. Eg.nitrito-N and nitrito-O.
(b)The number of coordinating groups present in ligand is called denticity of ligand. Eg.Bidentateligand
ethane-1,2-diamine has 2 donor nitrogen atoms which can link to central metal atom.
(c) the splitting of the degenerated d orbital into 3 orbitals of lower energy t2g and 2 orbitals of higher energy eg
due to presence of a ligand in a octahedral crystal field is known as crystal field splitting in an octahedral
complex.

5. draw structures of geometrical isomers of the following complexes:

(a) [Fe(NH3)2(CN)4]- (b)[CrCl2(ox)2]3- (c)[Co(en)3]Cl3
6. write the state of hybridisation, the shape and the magnetic behaviour of the following complexes:
(i) [Co(en)3]Cl3 (II) K2[Ni(CN)4](III)[Fe(CN)6]3-
7 how would you account for the following:
(i) [Ti(H2O)6]3+is coloured while [Sc(H2O)6]3+ is colourless .
(II) [ Fe(CN)6]3- is weakly paramagnetic while [ Fe(CN)6]4- is diamagnetic.
(III) Ni(CO)4 possess tetrahedral geometry while [Ni (CN)4]2- is square planar.
Ansi) due to the presence of 1 electron in 3d subshell in [Ti(H2O)6]3+ complex d-d transition takes place by the
absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H2O)6]3+ does not
possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this
complex is not possible, therefore it is colourless.
(ii) paramagnetism is attributed to the presence o f unpaired electrons. Greater the number of unpaired
electron greater is the paramagnetism. Due to the presence of one electron in the 3d subshell in [ Fe(CN) 6]3- it is
weakly paramagnetic. On the other hand [ Fe(CN)6]4- is diamagnetic because all electrons are paired.
iii) Ni in [Ni(CO)4] is sp3 hybridised. Hence it is tetrahedral. Whereas for [Ni(CN)4]2- is dsp2 hybridised
hence it has square planar geometry.
9. Explain the following ::

(i) low spin octahedral complexes of Ni are not known.

(ii) The pi – complexes are known for the transition elements only.

(iii) CO is a stronger ligand than NH3 for many metals

Ans. i) nickel in its atomic or ionic state cannot afford 2 vacant 3d orbitals and hence d 2sp3 hybridisation is not
possible.
ii) transition metals have vacant d orbitals in their atoms or ions into which the electron pairs can be donated by
ligands containing πelectrons.eg. benzene, ethylene etc. thus dπ-pπ bonding is possible.
(iii) because in case of CO back bonding takes place in which the central atom uses its filled d orbitals with
empty anti bonding π*molecular orbital of CO.
10. What is meant by stability of a coordination compounds in solutions? State the factors which
govern the stability of complexes.

Ans : the stability of a complex in solution refers to the degree of association between the two species involved
in the state of equilibrium. The magnitude of the equilibrium constant for the association expresses the stability
M +4L ML4K = [ML4]/[M][L]4
Factors on which stability of complex depends (i) charge on central metal ion (ii) nature of the metal ion (iii)
basic nature of the ligand (iv) presence of the chelate ring (v) effect of multidentate cyclic ligand .
5 marks questions
1. Draw the structures of the following molecules:

(a) [Fe(NH3)2(CN)4]- (b)[CrCl2(ox)2]3- (c)[Co(en)3]Cl3(d) [Co(en)3]Cl3(e)[Fe(CN)6]3-

2. Write the state of hybridisation the shape and the magnetic behaviour of the following complex
entities:

i) [Cr(NH3)4Cl2]Cl ii) [Co(en)3]Cl3 iii) K2[NiCl4] iv)[Fe(H2O)6]2+ v)[NiCl4]2-

3. Using valence bond theory explain the following questions in relation to [Co(NH3)6]3+.

(i) Nomenclature

(ii) Type of hybridisation

(iii) Inner or outer orbital complex

(iv) Magnetic behaviour

(v) Spin only magnetic moment

4. Compare the following complexes with respect to structural shape of units, magnetic behaviour
and hybrid orbitals involved in units:

[Co(NH3)6]3+, [Cr(NH3)6]3+,[ Ni(CO)4]

MLL-2
ONE MARK QUESTIONS:
Q.1Write the formula forTetraamineaquachloridocobalt(III) chloride
Q.2 Write the IUPAC name of [Co (NH3)4 Br2]2 [Zn Cl4]
Q.3 Which of these cannot act as ligand and why: NH3, H2O, CO, CH4. Give reason?
Q.4 How many EDTA (lethylendiamine tetra acetic acid) molecules are required to make an octahedral
complex with a Ca2+ ion.
Q.5 Why tetrahedral complexes do not exhibit geometrical isomerism ?
Q.6 What is the hybridisation of central metal ion and shape of Wilkinson’s catalyst ?
Q.7 Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+ and (ii) [CoBr2(en)2]+
Q.8 Draw the structure of optical isomers of [Co(en)3]3+.
Q.9 Name the types of isomerism exhibited by [Co(NH3)5(NO2)](NO3)2
Q.10 Write the formula of Amminebromidochloridonitrito-N-platinate(II) ion
Q.11 which one is more stable complex among(i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+(iii) [Fe(C2O4)3]3−
Q.12 How many ions are produced from the complex Co(NH3)6Cl2 in solution?
Q.13 What do you meant by degenerate d-orbitals?
Q.14 Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl2 (ox)2]3- (b)
trans-[CrCl2 (ox)2]3-
Q.15 The spin only magnetic moment of [MnBr4]2- is 5.9 BM. Predict the geometry of the complex ion?
Q.16 What is the oxidation state of Ni in [Ni(CO)4].
Q.17 What is the magnetic behavior of [Ni(CN)4]2- .
Q.18 Name the metal ion present in vitamin B12.
Q.19 Name the isomerism shown by complex K[Cr(H2O)2(C2O4)2]
Q.20 Name the compound used for inhibiting the growth of tumours. (cancer treatment)
TWO MARK QUESTIONS:
Q.1Write the IUPAC names of the following coordination compounds:
(i) [Pt(NH3)2Cl(NO2 )] (ii) K3 [Cr(C2O4)3 ]
Q.2 A cationic complex has two isomers A & B. Each has one Co3+, five NH3, one Br and one SO42-. A gives a
white precipitate with BaCl2 solution while B gives a yellow precipitate with AgNO3 solution.
(a) What are the possible structures of the complexes A and B?
(a) Write the name of structural isomerism shown by A and B.
Q.3 FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4
solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why ?
Q.4 What is meant by ambidentate ligands? Give two examples.
Q.5 [Co(NH3)6]3+ is diamagnetic whereas [Co(F6)]3- is paramagnetic. Give reasons.
Q.6 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
Q.7 Draw figure to show the splitting of d orbitals in an octahedral crystal field. How is the magnitude of ∆0
affected by
(i) Nature of ligand. (ii) Oxidation State of metal ion.
Q.8 Metal carbonyl are much more stable than normal complexes, why?
Q.9 Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers.
Q.10 Write the formulas for the following coordination compounds and name the isomerism shown by them:
i. potassium tetracyanonickelate(II)

ii. tris(ethane−1,2−diamine) chromium(III) chloride

THREE MARK QUESTIONS:

Q.1 How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]
Q.2 (a) Write the IUPAC name of [Ti(H2O)6]+3.
(b) [Ti(H2O)6]+3 is coloured why?
(c) A Complex having scandium in +3 oxidation-state was found colorless why?
Q.3 (a) Write IUPAC name of [Co(en)3]3+
(b) [NiCl4]2- is paramagnetic while [Ni(CO)]4 is diamagnetic though both are tetrahedral. Why?
(c) Explain why K3[Fe(CN)6] is more stable than K4[Fe(CN)6].
Q.4 (a) What is the hybridization state of nickel in [Ni(CN)4]2−
(b) Draw the structure of [Ni(CN)4]2−
(c) A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.
Q.5 Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Q.6 For the complexes (i) [Ni(CN)4]2− (ii) [Ni(Cl)4]2− (iii) [Ni(CO)4] Identify:
(a) The oxidation No. of nickel (b) The hybrid orbitals and the shape of the complexes
Q.7 Specify the (i) oxidation numbers (ii) coordination numbers and (iii) IUPAC name of the following
coordination entities:
(a) [Co(H2O)(CN)(en)2]2+(b) [PtCl4]2−
Q.8 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(a) [Fe(CN)6]4−(b) [FeF6]3−
Q.9 Dimethyl glyoxime is added to alcoholic solution of NiCl2. When ammonium hydroxide is slowly added to
it, a rosy red precipitate of a complex appears.
(a) Give the str. of the complex showing hydrogen bond.
(b) Give oxidation state and hybridization of central metal ion.
(c) Identify whether it is paramagnetic or diamagnetic.
Q.10 Draw all the isomers (geometrical and optical) of:
(a) [CoCl2(en)2]+(b) [Co(NH3)Cl(en)2]2+
FIVE MARK QUESTIONS:
Q.1 For the complex [Fe(en)2Cl2]Cl identify :
(a) the oxidation No. of Iron.
(b) the hybrid orbitals and the shape of the complex.
(c) the magnetic behavior of the complex.
(d) No. of geometrical isomers.
(e) whether there is an optical isomer also
(f) name of the complex.
Q.2 State a reason for each of the following situations
(a) Co2+ ion is easily oxidized to Co3+ in the presence of strong ligand.
(b) CO is a stronger complexing reagent than NH3 .
(c) The molecular shape of [Ni(CO)4 ] is not the same as that of [Ni(CN)4]2−.
(d) Ni does not form low spin octahedral complexes.
(e) The π complexes are known for the transition metals only.
Q.3 Write down the IUPAC name for each of the following complexes and indicate the oxidation state,
electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the
complex:
(i)K[Cr(H2O)2(C2O4)2].3H2O (ii) [Co(NH3)5Cl]Cl2(iii)CrCl3(py)3
Q.4 (i) Write the IUPAC name and type of isomerism shown by following complex compounds
a) [Co(NH3)5(NO2)](NO3)2

b) [Pt(NH3)(H2O)Cl2]

(ii) Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical
isomers?
Q.5 (i) If to an aqueous solution of CuSO4 in two tubes, we add ammonia solution in one tube and HCl(aq) to the
other tube, how the colour of the solutions will change? Explain with the help of reaction.
(ii) Write the IUPAC name of complex compounds formed during the process
MLL-2
Answers(one mark questions)
Ans.1 [Co(NH3)4(H2O)Cl]Cl2
Ans.2 Tetraamminedibromocobalt (III) tetrachlorozincate (II)
Ans.3 CH4 can’t act as a ligand due to absence of lone pair of electron.
Ans.4 EDTA is a hexadentate ligand therefore only one EDTA molecule is required.
Ans.5 Because relative position of ligands attached to central atom are same with respect to one another
Ans.6 Wilkinson’s catalyst is (PH3P)3RhCl. In this Rh has dsp² hybridisation and square planar shape.
Ans.7 (i) +3 (ii) +3
Ans.8
Ans 9. It can show linkage isomerism.
[Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(ONO)](NO3)2
It can also show ionization isomerism.
[Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(NO3)](NO3)(NO2)
Ans 10.[Pt(NH3)Br Cl (NO2)]-
Ans 11. [Fe(C2O4)3]3− (due to chelation)
Ans 12. The given complex can be written as [Co(NH3)6]Cl2.
Thus, [Co(NH3)6]+ along with two Cl− ions are produced.
Ans.13 d-orbitals having same energy are called as degenerate d- orbitas.
Ans. 14 cis-[CrCl2 (ox)2]3-
Ans. 15 Tetrahedral (sp3)
Ans 16 Zero (0)
Ans 17. Diamagnetic
Ans 18. Cobalt (Co3+)
Ans 19. Both geometrical (cis-, trans-) isomers for K[Cr(H2O)2(C2O4)2]can exist. Also, optical isomers for cis-
isomer exist.
Ans 20.compounds of platinum (for example, cis-platin)
Answers(two mark questions)
Ans. 1
(i) [Pt(NH3)2Cl(NO2 )] Diaminechloridonitro(o)platinum(II)
(ii) K3 [Cr(C2 O4 )3 ] Potassium trioxalatochromate(III)

Ans. 2
(a) [CO (NH3)5Br]SO4and [CO (NH3)5SO4]Br
(b) Ionisation Isomerism
Ans. 3
When FeSO4 and (NH4)2SO4 solution are mixed in 1 : 1 molar ratio, a double salt is formed. It has the formula
FeSO4 (NH4)2SO4 .6 H2O. In aqueous solution, the salt dissociates in to its constituent ions.
When CuSO4 and NH3 are mixed in the molar ratio of 1 : 4 in solution, a complex [Cu (NH3)4] SO4 is
formed, which does not dissociates into constituent ions.
Ans. 4
Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate
ligands.
For example:

(a) (The donor atom is N) (The donor atom is

oxygen)
(b) (The donor atom is S) (The donor atom is N)
Ans 5.
In [Co(NH3)6]3+, cobalt ion is in + 3 oxidation state and has the electronic configuration 3d6. It undergoes d²sp³
hybridisation. Each hybrid orbital receives 1 pair of electrons from ammonia. Since all electrons are paired it is
diamagnetic.
In [CoF6]3+, cobalt ion is in + 3 oxidation state and has the electronic configuration 3d6. It undergoes sp³
d² hybridisation. Each hybrid orbital receives a pair of electrons from F-. THe 3d electrons of Co remain
unpaired making it paramagnetic.
Ans 6.
Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH3 is a weak field ligand that does not cause the
pairing of the electrons in the 3d orbital.
Cr3+

Therefore, it undergoes d2sp3 hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is
paramagnetic in nature.
In [Ni(CN)4]2−, Ni exists in the +2 oxidation state i.e., d8 configuration.
Ni2+:

CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+ undergoes dsp2
hybridization.

As there are no unpaired electrons, it is diamagnetic.

Ans 7
The splitting of the d orbitals in an octahedral field takes palce in such a way that , experience a rise
in energy and form the eglevel, while dxy, dyzanddzx experience a fall in energy and form the t2g level.

● (a) Stronger the ligand, more is the splitting.

● (b) Higher the oxidation state, greater is the magnitude of 0.

Ans 8.
The metal-carbon bonds in metal carbonyls have both σ and π characters. A σ bond is formed when the
carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A π bond is formed by the
donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π* orbital (also known
as back bonding of the carbonyl group). The σ bond strengthens the π bond and vice-versa. Thus, a synergic
effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the
metal.
Ans 9.
When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react
differently with different reagents to give different products.
[Co(NH3)5Cl]SO4 + Ba2+ BaSO4 white precipitate
+
[Co(NH3)5Cl]SO4 + Ag No Reaction
[Co(NH3)5SO4]Cl + Ag+ AgCl white precipitate
[Co(NH3)5SO4]Cl + Ba2+ No Reaction
Ans 10
i. K2[Ni(CN)4] No Isomerism shown by it

ii. [Cr(en)3]Cl3 Optical isomerism

Answers(three mark questions)

Ans 1. (a) For [Cr (C2O4)3]3−, no geometric isomer is possible as it is a bidentate ligand.

(b) [Co(NH3)3Cl3] Two geometrical isomers are possible.

Ans 2. (a) hexaaquotitanium(III) ion

(b) Due to presence of single electron in d orbitals so that d-d transition is possible.
(c) Due to absence of unpaired electron in d orbital
Ans. 3 (a) Tris-(1,2-ethanediamine) cobalt (III) ion
(b) Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is
due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of
unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic.

In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s
electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are
present in this case, [Ni(CO)4] is diamagnetic
(c) It is because the stability of complex depends upon the charge density (i. e. charge/radius ratio) on central
ion. More is the charge density greater is the stability.
Ans 4. (a) dsp2hybridisation
(b) Square planer structure
(c) In [Ni(H2O)6]2+, H2O is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex,
the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d
transition is present. Hence, Ni(H2O)6]2+ is coloured.
In [Ni(CN)4]2−, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not
possible in [Ni(CN)4]2−. Hence, it is colourless
Ans. 5
[Co(NH3)6]3+ [Ni(NH3)6]2+

Oxidation state of cobalt = +3 Oxidation state of Ni = +2

Electronic configuration of cobalt = d6 Electronic configuration of nickel = d8

If NH3 causes the pairing, then only one 3d

NH3 being a strong field ligand causes the pairing. orbital is empty. Thus, it cannot undergo
Therefore, Ni can undergo d2sp3 hybridization. d2sp3 hybridization. Therefore, it undergoes
sp3d2 hybridization.

Hence, it is an inner orbital complex.

Hence, it forms an outer orbital complex

Ans. 6
complex Oxidation no. of Ni hybrid orbitals shape
(i) [Ni(CN)4]2− +2 dsp2 Square planer
(ii) [Ni(Cl)4]2− +2 sp3 Tetrahedral
(iii) [Ni(CO)4] Zero (0) sp3 Tetrahedral

Ans. 7
(a) [Co(H2O)(CN)(en)2]2+
oxidation numbers = +3;
coordination number = 6
IUPAC name = aquacynidobis-(ethane 1,2diamine) cobalt (III) ion
2−
(b) [PtCl4]
oxidation numbers = +2;
coordination number = 4
IUPAC name = tetachloridoplatinate(II) ion
Ans. 8
(a) [Fe(CN)6]4−
In the above coordination complex, iron exists in the +2 oxidation state.
Fe2+ : Electronic configuration is 3d6

Orbitals of Fe2+ ion:

As CN− is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since there are six ligands around the central metal ion, the most feasible hybridization is d2sp3.
d2sp3hybridized orbitals of Fe2+ are:
6 electron pairs from CN− ions occupy the six hybrid d2sp3orbitals.

Then,
Hence, the geometry of the complex is octahedral and the complex is diamagnetic
(as there are no unpaired electrons).
(b) [FeF6]3−
In this complex, the oxidation state of Fe is +3.
Orbitals of Fe+3ion:

There are 6 F− ions. Thus, it will undergo d2sp3 or sp3d2 hybridization. As F− is a weak field ligand, it does not
cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp3d2.
sp3d2 hybridized orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.