9.

COORDINATION
COMPOUNDS

Chemistry Smart Booklet

Theory + NCERT MCQs + Topic Wise Practice
MCQs + NEET PYQs

1
 PC
M C
O S
B- C
74 IE
14 N
93 CE
66 A
06 CA
D
EM
Y

2
 COORDINATION COMPOUNDS
COORDINATION COMPOUNDS
• Differences between coordination compound and double bond

Coordination compound Double salt

A coordination compound contains a central When two salts in stoichiometric ratio
metal atom or ion surrounded by several are crystallised together from their
oppositely charged ions or neutral saturated solution, they are called
molecules. These ions or double salts.
molecules re-bonded to the metalatom or
ion by a coordinate bond.

Y
Example: K4[Fe(CN)6] Example: FeSO4.(NH4)2SO4.6H2O

EM
(Mohr’s salt)
They do not dissociate into simpleions They dissociate into simple ions when
when dissolved in water. dissolved in water.

D
06 CA
• Coordination entity: A coordination entity constitutes a central metal atom or ion
bonded to a fixed number of ions or molecules. Example: In K4[Fe(CN)6], [Fe(CN)6]4−
represents a coordination entity.
66 A
• Central atom or ion: In a coordination entity, the atom/ion to which a fixed number
93 CE

of ions/groups are bound in a definite geometrical arrangement is called the central atom
or ion. Example: In K4[Fe(CN)6],Fe²+ is the central metal ion.
• Ligands: A molecule, ion or group which is bonded to the metal atom or ion in a complex
14 N

or coordination compound by a coordinate bond is called a ligand. It may be neutral,

74 IE

positively ornegatively charged. Examples: H2O, CN−, NO+ etc.

• Donor atom: An atom of the ligand attached directly to the metal is called the donor
B- C

atom. Example:In the complex K4[Fe(CN)6], carbon is a donor atom.

O S

• Coordination number: The coordination number (CN) of a metal ion in a complex can
be defined asthe number of ligand donor atoms to which the metal is directly bonded.
M C

Example: In the complex K4[Fe(CN)6], the coordination number of Fe is 6.

PC

• Coordination sphere: The central atom/ion and the ligands attached to it are enclosed
in square bracket and is collectively termed the coordination sphere. Example: In the
complex K4[Fe(CN)6],[Fe(CN)6]4− is the coordination sphere.
• Counter ions: The ions present outside the coordination sphere are called counter ions.
Example: Inthe complex K4[Fe(CN)6], K+ is the counter ion.
• Coordination polyhedron: The spatial arrangement of the ligand atoms which are
directly attached tothe central atom/ion defines a coordination polyhedron about the
central atom. The most common coordination polyhedra are octahedral, square planar
and tetrahedral. Examples: [PtCl4]2− is square planar, Ni(CO)4 is tetrahedral and
[Cu(NH3)6]3+ is octahedral.
• Charge on the complex ion: The charge on the complex ion is equal to the algebraic

3
 sum of thecharges on all the ligands coordinated to the central metal ion.
• Denticity: The number of ligating (linking) atoms present in a ligand is called denticity.
• Unidentate ligands: The ligands whose only donor atom is bonded to a metal atom
are calledunidentate ligands. Examples: H2O, NH3, CO, CN−
• Didentate ligands: The ligands which contain two donor atoms or ions
through which they are bonded to the metal ion. Example: Ethylene
diamine (H2NCH2CH2NH2) has two
nitrogen atoms, and oxalate ion has two oxygen atoms which can bind with the metal
atom.
• Polydentate ligand: When several donor atoms are present in a single ligand, the
ligand is called a polydentate ligand. Example: In N(CH2CH2NH2)3, the ligand is said to

Y
be polydentate. Ethylenediaminetetraacetate ion (EDTA4–) is an important hexadentate
ligand. It can bind through twonitrogen and four oxygen atoms to a central metal ion.

EM
• Chelate: An inorganic metal complex in which there is a close ring of atoms caused by
attachment ofa ligand to a metal atom at two points. An example is the complex ion

D
formed between ethylene diamine and cupric ion [Cu(NH2CH2NH2)2]2+.
• Ambidentate ligands: Ligands which can ligate (link) through two different atoms

06 CA
present in it are called ambidentate ligands. Examples: NO − and SCN−. NO − can link
through N as well as O, while SCN− can link through S as well as N.
66 A
• Werner’s coordination theory: Werner was able to explain the nature of bonding
in complexes. The postulates of Werner’s theory are
93 CE

o Metal shows two kinds of valencies—primary valence and secondary valence.

14 N

Primary valence Secondary valence

This valence is normally This valence is non-ionisable.
74 IE

ionisable.
B- C

It is equal to the positive chargeon the The secondary valency equals to the
central metal atom. number of ligand atoms coordinated to the
O S

metal. It is also called the coordination

M C

number of the metal.

These valencies are satisfied by It is commonly satisfied by neutral and
PC

negatively charged ions. negatively charged, sometimes by

positively charged ligands.
Example: In CrCl3, the primaryvalency
is three. It is equal to the
oxidation state of the centralmetal ion.

o The ions/groups bound by secondary linkages to the metal have characteristic spatial
arrangements corresponding to different coordination numbers.
o The most common geometrical shapes in coordination compounds are octahedral,
square planarand tetrahedral.
• Oxidation number of the central atom: The oxidation number of the central atom

4
 in a complex is defined as the charge it would carry if all the ligands are removed along
with the electron pairs whichare shared with the central atom.
• Homoleptic complexes: Those complexes in which metal or ion is coordinately
bonded to only onekind of donor atom. Example: [Co(NH3)6]3+
• Heteroleptic complexes: Those complexes in which metal or ion is coordinately
bonded to more thanone kind of donor atom. Example: [CoCl2(NH3)4]+, [Co(NH3)5Br]2+
• Isomers: Two or more compounds which have the same chemical formula but different
arrangementof atoms are called isomers.
• Types of isomerism
o Structural isomerism
▪ Linkage isomerism

Y
▪ Solvate isomerism or hydrate isomerism
▪ Ionisation isomerism

EM
▪ Coordination isomerism
o Stereoisomerism

D
▪ Geometrical isomerism
▪ Optical isomerism

06 CA
• Structural isomerism: This type of isomerism arises due to the difference in
structures of coordination compounds. Structural isomerism, or constitutional
66 A
isomerism, is a form of isomerism inwhich molecules with the same molecular formula
have atoms bonded together in different orders.
93 CE

o Ionisation isomerism: This form of isomerism arises when the counter ion in a
complex salt isitself a potential ligand and can displace a ligand which can then become
14 N

the counter ion.

Examples: [Co(NH3)5Br] SO4 and [Co(NH3)5 SO4] Br
74 IE

o Solvate isomerism: It is isomerism in which the solvent is involved as the ligand. If

B- C

the solvent is water, then it is called hydrate isomerism. Example: [Cr(H2O)6]Cl3 and
[CrCl2(H2O)4] Cl2.2H2O
O S

o Linkage isomerism: Linkage isomerism arises in a coordination compound

M C

containing an ambidentate ligand. In the isomerism, a ligand can form linkage with
metal through different atoms.
PC

Examples: [Co(NH3)5ONO]Cl2 and [Co(NH3)5NO2]Cl2

o Coordination isomerism: This type of isomerism arises from the interchange of
ligands betweencationic and anionic entities of different metal ions present in a complex.
Examples: [Co(NH3)6][Cr(C2O4)3] and [Cr(NH3)6][Co(C2O4)3]
• Stereoisomerism: This type of isomerism arises because of different spatial
arrangement.
o Geometrical isomerism: It arises in heteroleptic complexes due to different possible
geometricalarrangements of ligands.
o Optical isomerism: Optical isomers are those isomers which are non-superimposable
mirrorimages.

5
• Valence bond theory:
According to this theory, the metal atom or ion under the influence of ligands can use
its (n − 1)d, ns, np or ns, np or nd orbitals for hybridization to yield a set of equivalent
orbitals of definite geometry such as octahedral, tetrahedral and square planar.These
hybridised orbitals are allowed to overlap with ligand orbitals which can donate electron
pairs for bonding.

Distribution of
Coordinatio
Type of hybridisation hybridorbitals in
nnumber
space
4 sp3 Tetrahedral
4 dsp2 Square planar

Y
5 sp3d Trigonal bipyramidal

EM
sp3d2 (nd orbitals are involved; outer
6 orbital complex or high-spin or spin- Octahedral

D
free complex)

6 06 CA d2sp3 [(n − 1)d orbitals are involved;

inner orbital complex or low-spin or Octahedral
66 A
spin-paired complex]
93 CE

• Magnetic properties of coordination compounds: A coordination

14 N

compound is paramagnetic in nature if it has unpaired electrons and diamagnetic

74 IE

if all the electrons in the coordination compoundare paired.

Magnetic moment 𝜇 = √𝑛(𝑛 + 2) where n is the number of unpaired electrons.
B- C

• Crystal Field Theory: It assumes the ligands to be point charges and there is an
O S

electrostatic force of attraction between ligands and the metal atom or ion. It is a
theoretical assumption.
M C

• Crystal field splitting in octahedral coordination complexes

PC

6
• Crystal field splitting in tetrahedral coordination complexes

• For the same metal, the same ligands and metal–ligand distances, the difference in

Y
energy between egand t2g level is

EM
4
Δ t = Δ0
9
• Metal carbonyls: Metal carbonyls are homoleptic complexes in which carbon

D
monoxide (CO) acts asthe ligand. Example: Ni(CO)4

06 CA
The metal–carbon bond in metal carbonyls possesses both σ and 𝜋 characters. The metal–
carbon bond in metal carbonyls possess both s and p characters. The M–C σ bond is formed
66 A
by the donation of a lone pair of electrons from the carbonyl carbon into a vacant orbital of
the metal. The M–C 𝜋 bond is formed by the donation of a pair of electrons from a filled
93 CE

d orbital of metal into the vacant anti-bonding 𝜋* orbital of carbon monoxide. The metal to
ligand bonding creates a synergic effect which strengthensthe bond between CO and the
metal.
14 N
74 IE
B- C
O S
M C
PC

7
 NCERT LINE BY LINE QUESTIONS
(1.) Which of the following coordination compound is also known as heteroleptic complex? [Page: 240]
+
Co ( NH3 )4 Cl2 
3+
(a.) Co ( NH3 )6  (b.)

 PtCl4 
2− 3−
(c.) (d.)Cu ( CN )4 
(2.) The crystal field stabilisation energy (CFSE) for  COCl0 is 15000cm−1 . The CFSE for  CoCl4 
4− 2−

will be [Odisha NEET‐2019, Page: 252]

(a.) 6000cm −1 (b.) 16000cm−1
(c.) 18000cm−1 (d.) 8000cm−1

Y
(3.) Match the Column I with Column II and select the correct codes given below.

EM
Column I Column II
Complex IUPAC name
(P) [Co(H 2 NCH 2 CH 2 NH 2 )3 ]2 SO 4 (i) Pentaaminine carbonatocobalt (III) chloride

D
(Q) Co ( NH 3 )5 ( CO3 )  Cl (ii) Tris(ethane-1,2 -diammine) cobalt (III) sulphate

06 CA
(R) CoCl2 ( en )2  Cl (iii) Dichloridobis (ethane-1,2-diammine) cobalt (III) chloride

+ (iv) Dichloridobis (ethane-1,2‐diammione) cobalt (III)

(S) CoCl2 ( en )2 
66 A
(a.) P-(i),Q-(ii),R-(iii),S-(iv) (b.) P-(iv),Q-(iii),R-(ii),S-(i)
93 CE

(c.) P-(ii),Q-(i),R-(iii),S-(iv) (d.) P-(iii),Q-(ii),R-(i),S-(iv) [Page: 24l]

(4.) Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium
14 N

complexes of the type  Pd ( C6H 5 )2 ( SCN )2  and  Pd ( C6H 5 )2 ( NCS )2  are

74 IE

[QR code, NCERT Exemplar, Page: 245, HOTS]

(a.) linkage isomers (b.) coordination isomers
B- C

(c.) ionisation isomers (d.) geometrical isomers

O S

(5.) Which of the following formula belongs to Mohr’s salt?

M C

(a.) KCl  MgCl2  6H 2O (b.) FeSO4  ( NH 4 )2 SO4  6H 2O

PC

(c.) KAI (SO4 )2 12H 2O (d.) None of these

(6.) Match the column I with column II and select the correct code given below. [Page: 247]
Column I Column II

Hybridisation Geometry
(P) sp
3 (i) Tetrahedral

(Q) dsp2 (ii) Square planar

(R) sp d
3 (iii) Trigonal bipyramidal
3 2
(S) sp d (iv) Octahedral

(a.) P-(i),Q-(ii),R-(iii),S-(iv) (b.) P-(i),Q-(iii),R-(iv),S-(ii)

(c.) P-(iii),Q-(iv),R-(ii),S-(i) (d.) P-(ii),▁O-(i),R-(iii),S-(iv)

8
(7.) [Page: 241]
The given compounds is
(a.) hexadenate ligand. (b.) pentadentate ligand.
(c.) tetradentate ligand. (d.) didenate ligand.

(8.) What is the oxidation state of the compound tetracarbonyl nickel? [Page: 242]
(a.) Zero (b.) One
(c.) Two (d.) Four

(9.) Which of the metal carbonyl have metal ( M ) − metal (M) bond?

Y
(a.)  Fe ( CO )5  (b.) Cr ( CO )6 

EM
(c.)  Ni ( CO )4  (d.)  Mn 2 ( CO )10 

D
(10.) Some statements are given below regarding coordination complexes. Select the correct statement. [Page:
249]

06 CA
(I)  MnCl6  has outer orbital complex.
3−

(II)  FeF6 
3−
involve sp3d 2 hybridisation.
66 A
(III)  CoF6 
3−
has paramagnetic in nature.
93 CE

(a.) I, II and III (b.) II and III

(c.) I and III (d.) I and II
14 N

(11.) Which of the following compounds show linkage isomerism [Page: 244]
74 IE

(a.) Co ( NH3 )5 ( NO 2 )  Cl2 (b.) Co ( NH3 )6  Cr ( CN )6 

(c.) Co ( NH3 )5 SO 4  Br (d.) Cr ( H 2O )6  Cl3

B- C
O S

(12.) What are the coordination number of d 2sp3 and sp3d 2 hybridised complex respectively?
M C

(a.) 2 and 6 (b.) 4 and 6

(c.) 6 and 6 (d.) 6 and 4
PC

3−
(13.) What is the coordination number of the complex  Fe ( C2O4 )3  ?
(a.) 4 (b.) 6
(c.) 2 (d.) 0

(14.) IUPAC name of the complex Cr ( NH3 )3 ( H 2O )3  Cl3 is [Page: 24l]
(a.) triamminetriaquachromium(II) chloride (b.) hexaamminerriaquachromium(III) chloride
(c.) diamminetriamminechromium(II) chloride (d.) triamminetriaquachromium(III) chloride

(15.) 1 mol of CoCl3  6NH 3 gave how many mol of AgCl precipitate when it is treated with excess silver
nitrate solution? [Page: 238]

9
(a.) 3 mol of AgCl (b.) 2 mol of AgCl
(c.) 1 mol of AgCl (d.) 4 mol of AgCl

(16.) Match the metal ions given in column I with the spin magnetic moments of the given in column II and
assign the correct code: [NEET‐2O18, Page: 252]
Column I Column II

(P) Co3+ (i) 8B.M.

(Q) Cr 3+ (ii) 35B.M.

(R) Fe3+ (iii) 3B.M.

(S) Ni 2+ (iv) 24B.M.

Y
EM
(v) 15B.M.

D
P Q R S
(a.) (b.)

06 CA
(iv) (v) (ii) (i) (i) (ii) (iii) (iv)
(c.) (iv) (i) (ii) (iii) (d.) (iii) (v) (i) (ii)
66 A
(17.) Which of the following ligands form a chelate? [Page: 24O]
(a.) Acetate (b.) Oxalate
93 CE

(c.) Cyanide (d.) Ammonia

(18.) Which of the following statement is correct with respect to  Ni ( CO )4  and  NiCl4 
2−
[Page: 248]
14 N

 Ni ( CO )4  and  NiCl4   Ni ( CO )4  and  NiCl4 

2− 2−
(a.) both are square (b.) both are
74 IE

planar tetrahedral
B- C

 Ni ( CO )4  is tetrahedral while  NiCl4   Ni ( CO )4  is square planar while  NiCl4 

2− 2−
(c.) is (d.)
O S

square planar is tetrahedral

M C

(19.) Which of the following ligands is also considered as ambidentate Iigand)

(a.) C2O42− (b.) NO−2
PC

(c.) NH 3 (d.) H 2O

(20.) Select the incorrect statement [Page: 246]

(a.) Stereoisomerism have the same chemical (b.)  Pt ( NH 3 )2 Cl 2  has cis and trans isomerism.
formula and chemical bonds bur they have
different spatial arrangement.
+
(c.) Co ( NH3 )4 Cl2  has only cis isomerism. (d.) Ionisation isomerism is a type of structural
isomerism.

(21.) Crystal field theory explains

(a.) electrostatic model which considers the metal (b.) electrostatic model which considers the metal‐
− ligand bond to be ionic. ligand bond to be covalent.

10
(c.) electrostatic model which considers the metal‐ (d.) none of these.
ligand bond to be ionic and covalent.

(22.) The correct IUPAC name of  Pt ( NH 3 )2 Cl 2  is [QR code, NCERT Exemplar, Page: 243, HOTS]
(a.) diamminedichloridoplatinum (II) (b.) diamminedichloridoplatinum (IV)
(c.) diamminedichloridoplatinum (0) (d.) dichloridodiammineplatinum (IV)

(23.) Which of the following arrangement for spectrochemical series is incorrect? [Page: 251]
(a.) I−  Br −  SCN −  Cl− (b.) S2−  F−  OH−  C2O24−
(c.) en  CN−  CO  NCS− (d.) edta 4−  NH3  en  CN−

(24.) What is the correct IUPAC name of  Ag ( NH3 )2   Ag ( CN )2  ?

Y
(a.) Diammine silver(II) dicyanoargentate(II) (b.) Diammine silver(I) dicyanoargentate(I)

EM
(c.) Diammine silver(0) dicyanosilver(II) (d.) Diammine silver(II) dicyanoargentate(0)
2+
(25.) Assertion: Cu ( H2O )4 

D
absorb green light.
3+
Reason: Co ( NH3 )5 ( H 2O )

06 CA
complementary colour is yellow. [Page: 252]
(a.) Both Assertion and Reason are true and (b.) Both Assertion and Reason are true but
Reason is the best explanation of Assertion. Reason is not the correct explanation of
66 A
Assertion.
(c.) Assertion is true but Reason is false. (d.) Both Assertion and Reason are false.
93 CE

(26.) How many ions are produced from the complex Co ( NH3 )6 Cl2  in solution?
14 N

(a.) 6 (b.) 4
74 IE

(c.) 3 (d.) 2

(27.) The colour of the coordination compounds depends on the crystal field splitting. What will be the correct
B- C

order of absorption of wavelength of light in the visible region, for the complexes,
O S

3+ 3− 3+
Co ( NH3 )6  , Co ( CN )6  , Co ( H 2O )6 
M C

[QR code, NCERT Exemplar, Page: 252, HOTS]

(a.) (b.)
PC

3− 3+ 3+ 3+ 3+ 3−
Co ( CN )6   Co ( NH3 )6   Co ( H 2O )6  Co ( NH3 )6   Co ( H 2O )6   Co ( CN )6 
(c.) (d.)
3+ 3+ 3− 3+ 3− 3+
Co ( H 2O )6   Co ( NH3 )6   Co ( CN )6  Co ( NH3 )6   Co ( CN )6   Co ( H 2O )6 

(28.) Which of the following will not show geometrical isomerism? [Page: 244]
(a.) Cr ( NH3 )4 Cl2  Cl (b.) Co ( en )2 Cl2  Cl

(c.) Co ( NH 3 )5 NO 2  Cl 2 (d.)  Pt ( NH 3 )2 C  ]

2

11
(29.) Given below some coordinate complexes with IUPAC name. Identify the incorrect name with their
complex. [Page: 242]
(I)  NiC12 ( PPh 3 )2  : Dichloridobis(triphenyl phosphine)nickel(III)
(II) Cr ( NH3 )3 ( H 2O )3  Cl3 : Triamminetriaqua chromium(III) chloride
(III) Hg Co ( SCN )4  : Mercury tetrathiocynato cobaltate(II)
(a.) I and II (b.) I and III
(c.) II and III (d.) I, II and III

(30.) Which one of the following ions exhibits d‐d transition and paramagnetism as well? [NEET‐2018, Page:
252]
(a.) CrO24− (b.) Cr2O72−

Y
(c.) MnO4− (d.) MnO24−

EM
(31.) Which of the following option is correct with respect to the inner and outer orbital complex?
3+ 2+
(a.) Co ( NH3 )6  is inner orbital complex and (b.)  Ni ( NH3 )6  is inner orbital complex and

D
2+ 3+
 Ni ( NH3 )6  is outer orbital complex. Co ( NH3 )6  is outer orbital complex.

06 CA
3+ 2+ 3+ 2+
(c.) Both Co ( NH3 )6  and  Ni ( NH3 )6  are (d.) Both Co ( NH3 )6  and  Ni ( NH3 )6  are
inner orbital complex. outer orbital complex.
66 A
(32.) The spin only magnetic moment of  MnBr4 
2−
is 5.9 B.M. What is the geometry of the complexion?
93 CE

[Page: 249]
(a.) Tetrahedral (b.) Square planar
14 N

(c.) Trigonal planar (d.) None of these

74 IE

2+
(33.) Which isomer in the molecule  PtCl2 ( en )2  shows optical activity? [Page: 245]
B- C

(a.) Both cis and trans (b.) Neither cis nor trans
O S

(c.) Only cis (d.) Only trans

M C

(34.) Chlorophyll, hemoglobin and vitamin B12 are coordination compounds of

PC

(a.) cobalt, iron and magnesium respectively (b.) magnesium, cobalt and iron respectively
(c.) magnesium, iron and cobalt respectively (d.) iron, cobalt and magnesium respectively.

(35.) When 1 mol CrCl3  6H 2O is treated with excess of AgNO3 ,3 mol of AgCl are obtained. The formula of
the complex is [QR code, NCERT Exemplar, Page: 24O, HOTS]
(a.) CrCl3 ( H 2O )3   3H 2O (b.) CrCl2 ( H 2O )4   2H 2O

(c.) CrCl ( H 2O )5  Cl2  H 2O (d.) Cr ( H 2O )6  Cl3

(36.) Which of the following conditions is necessary for t 3e1 configuration and what is the nature of ligands?
(a.) Δ o  P , weak field ligand (b.) Δ o  P , strong field ligand
(c.) Δ o  P , weak field ligand (d.) Δ o  P , strong field ligand

12
(37.) Assertion: IUPAC name of Co ( NH3 )4 ( H 2O ) Cl  Cl2 is tetraammineaquachloridocobalt(III) chloride.
Reason: Chlorine is the central element in the compound Co ( NH3 )4 ( H 2O ) Cl  Cl2 . [Page: 242]
(a.) Both Assertion and Reason are true and (b.) Both Assertion and Reason are true but
Reason is the best explanation of Assertion. Reason is not the correct explanation of
Assertion.
(c.) Assertion is true bur Reason is false. (d.) Both Assertion and Reason are false.

(38.) Which of the following complex is also known as Wilkinson catalyst? [Page: 252]
(a.) Ag (S2O3 )2 
3−
(b.) ( Ph 3P )3 RhCl 

(c.)  Ni ( CO )4  (d.)
3−
Co ( C2O4 )3 

Y
(39.) What are the geometry of  Ni ( CO )4  , Fe ( CO )5 ] and [Cr ( CO)6  respectively? [Page: 255]

EM
(a.) Trigonal bipyramidal, tetrahedral, octahedral (b.) Tetrahedral, trigonal bipyramidal, octahedral
(c.) Octahedral, trigonal bipyramidal, tetrahedral (d.) Tetrahedral, octahedral, trigonal bipyramidal

D
(40.) The type of isomerism shown by the complex CoCl2 ( en )2  is [NEET‐2018, Page: 245]

06 CA
(a.) geometrical isomerism (b.) coordination isomerism
(c.) ionisation isomerism (d.) linkage isomerism
66 A
(41.) Co ( NH3 )6  Cr ( CN )6  , which type of isomerism arises in the given complex?
93 CE

(a.) Linkage isomerism (b.) Coordination isomerism

(c.) Ionisation isomerism (d.) Solvate isomerism
14 N

(42.) Which of the following is the correct chemical formula of Mohr’s salt? [Page: 239]
74 IE

(a.) KAl ( SO4 )2 12H 2O (b.) KCl  MgCl2  6H 2O

B- C

(c.) FeSO4  ( NH 4 )2 SO4  6H 2O (d.) None of these

(43.) Which of the following ligands are didentate ligand?
O S

(a.) C2O42− (b.) H 2 NCH 2CH 2 NH 2

M C

(c.) Both (a) and (b) (d.) Neither (a) nor (b)
PC

(44.) Which of the following statements regarding primary and secondary valences is correct?
[Page: 238]
(a.) Primary valences are ionisable while (b.) Primary valences are non‐ionisable while
secondary valences are non‐ionisable. secondary valences are ionisable.
(c.) Both primary and secondary valences are non‐ (d.) Both primary and secondary valences are
ionisable. ionisable.

(45.) Select the compound which shows ionization isomerism. [Page: 246]
(a.) Co ( NH3 )5 ( SCN )  Cl2 (b.) Co ( NH3 )5 Cl  SO 4

(c.) Co ( NH3 )6  Co ( CN )6  (d.) Co ( NH3 )5 ( NO2 )  ( NO3 )2

13
 (46.) Hexaamminenickel(II) hexanitrocobaltate(III) can be written as [Page: 242]
(a.)  Ni ( NH 3 )6  Co ( NO 2 )6  (b.)  Ni ( NH 3 )6  Co ( NO 2 )6 
2 3 3 2

(c.)  Ni ( NH 3 )6  Co ( NO 2 )6  (d.)  Ni ( NH3 )6 ( NO 2 )6  Co

(47.) Out of the following two coordination entities, which is chiral (optically active) . [Page: 244]
3−
(I) cis − CrCl2 ( Ox )2 
3−
(II) trans − CrCl2 ( Ox )2 
(a.) Only I (b.) Only II
(c.) Both I and II (d.) None of these

(48.) Which of the following has longest C‐O bond length? [NEET‐2O16, Page: 252]

Y
( Free C −O bond length in CO is 1.128Å )

EM
(a.)  Ni ( CO )4  (b.) Co ( CO )4 
−

2−
 Fe ( CO )4 
+
(c.) (d.)  Mn ( CO )6 

D
(49.) What is the primary valency of the given compounds CrCl3 ,CoCl 2 and PdCl 2 respectively? [Page: 238]

06 CA
(a.)
2, 2 and 3 (b.) 3, 2 and 2
(c.)
2, 3 and 2 (d.) 3,2 and 3
66 A
(50.) Select the correct statement
(a.) The secondary valences are ionisable. (b.) The primary valences are ionisable.
93 CE

(c.) The primary and secondary valences both are (d.) None of these.
non‐ionisable.
14 N
74 IE

TOPIC WISE PRACTICE QUESTIONS

B- C
O S

TOPIC 1: Coordination Number, Nomenclature and Isomerism

1. Select the correct IUPAC name for
M C

[Pt(C5H5N)4] both same [PtCl4] complex:

1) Tetrapyridineplatinate (II) tetrachloridoplantinate(II)
PC

2) Tetrapyridineplatinum (II) tetrachloridoplantinum(II)

3) Tetrapyridineplatinate (II) tetrachloridoplantinum(II)
4) Tetrapyridineplatinum (II) tetrachloridoplantinate(II)
2. [Pt(NH3)4Cl2] Br2 complex can show :
1) Hydrated as well as ionization isomerism 2) Ionization as well as geometrical isomerism
3) Linkage as well as geometrical isomerism 4) Ionization as well as optical isomerism
3. The complex ion [Pt (NO2)(Py) (NH3) (NH2OH)]+ will give
1) 2 isomers (Geometrical) 2) 3 isomers (Geometrical)
3) 6 isomers (Geometrical) 4) 4 isomers (Geometrical)
4. The geometry of Ni(CO)4 and Ni(PPh3)2Cl2 are
1) both square planar 2) tetrahedral and square planar
3) both tetrahedral 4) None of these
5. In the reactions

14
 [CoCl2 (NH3)4 ]++ Cl- → [CoCl3(NH3)3] +NH3 two isomers of the product are obtained. The initial
complex is
1) cis isomer 2) trans isomer 3) cis or trans isomers 4) None of these
6. Possible isomerism in complexes [Co(NH3)3(NO2)3] and [Co(NH3)5(NO2)]Cl2, respectively are:
1) Linkage and optical 2) Geometrical and linkage
3) Optical and ionization 4) Linkage and geometrical
7. Both geometrical and optical isomerisms are shown by
1) [Co(en)2Cl2]+ 2) [Co(NH3)5Cl]2+ 3) [Co(NH3)4Cl2]+ 4) [Cr(ox)3]3–
8. The complex given is

Y
EM
(i) non-superimposable on its mirror images (ii) optically inactive
(iii) rotate plane polarised light (iv) planar
1) (i) and (ii) 2) (i) and (iv) 3) (i), (ii) and (iii) 4) (ii) only

D
9. An example of double salt is
1) Bleaching powder 2) K4[Fe(CN)6] 3) Hypo 4) Potash alum

06 CA
10. Which of the following type of isomerism is shown by given complex compound?
66 A
93 CE

1) Facial 2) Meridional 3) Cis 4) Both b and c

11. The tetrahedral complex [M1)2)(X)(Y)], where A,B,X and Y are different ligands and M is a metal ion
14 N

is
74 IE

1) optically inactive 2) rotate plane polarized light

3) incomplete information 4) can’t be said
B- C

12. Which of the following will give maximum number of isomers?

1) [Ni (C2O4) (en)2]2– 2) [Ni (en) (NH3)4]2+ 3) [Cr (SCN)2 (NH3)4]+
O S

4) [Co (NH3)4 Cl2]

13. Which of the following compounds shows optical isomerism?
M C

3− 3− 2+
1) Co ( CN )6  2) Cr ( C2O4 )3  3)  ZnCl4  4) Cu ( NH3 )4 
2−
PC

14. The ionisation isomer of [Cr(H2O)4Cl(NO2)]Cl is

1) [Cr(H2O)4(O2N)]Cl2 2) [Cr(H2O)4Cl2](NO2)
3) [Cr(H2O)4Cl(ONO)]Cl 4) [Cr(H2O)4Cl2(NO2)].H2O
15. A bidenate ligand always
1) has bonds formed to two metals ions 2) has a charge of +2 or – 2
3) forms complex ions with a charge of +2 or –2
4) has two donor atoms forming simultaneously two sigma (s) bonds.
16. Which one of the following complexes is not expected to exhibit isomerism?
2+ 2+
1)  Ni ( en )3  2)  Ni ( NH3 )4 ( H 2O )2  3)  Pt ( NH 3 )2 Cl 2  4)  Ni ( NH 3 )2 Cl 2 
17. The complexes [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] are the examples of which type of
isomerism?
1) Linkage isomerism 2) Ionization isomerism

15
 3) Coordination isomerism 4) Geometrical isomerism
18. For which value of the x, and y, the following square planar compound shows geometrical isomers [Pt
(Cl)x (Br)y]2–
1) 1, 3 2) 3, 1 3) 2, 2 4) 1, 1
19. The structure of which of the following chloro species can be explained on the basis of dsp2
hybridization?
1) PdCl 24− 2) FeCl 24 − 3) CoCl 24− 4) NiCl 24 −
20. Which of the following is not chelating agent?
1) thiosulphato 2) oxalate 3) glycinato 4) ethylene diamine
21. A similarity between optical and geometrical isomerism is that
1) each gives equal number of isomers for a given compound
2) if in a compound one is present then so is the other
3) both are included in stereoisomerism

Y
4) they have no similarity

EM
22. Which is the pair of ambidentate ligand?
1) CN − , NO 2− 2) NO3− ,SCN− 3) N3− , NO2− 4) NCS− , C 2 O 42 −
23. Number of water molecules acting as ligands in CuSO4.5H2O, ZnSO4 .5H2O,FeSO4 .7H2O

D
respectively are

06 CA
1) 5, 5, 7 2) 4, 5, 4 3) 4, 4, 6 4) 4, 4, 7
24. The number of ions formed on dissolving one molecule of FeSO4(NH4)2SO4.6H2O in water is:
1) 4 2) 5 3) 3 4) 6
66 A
25. In octaamine-μ-dihydroxodiiron(III) sulphate, the number of bridging ligands is
1) 2 2) 1 3) 3 4) None of these
93 CE

TOPIC 2: Magnetic Moment, Valence Bond Theory and Crystal Field Theory
26. The crystal field splitting energy for octahedral (  0 ) and tetrahedral (  t ) complexes is related as
14 N

1 4 3 2
1)  t = −  0 2)  t = −  0 3)  t = −  0 4)  t = −  0
74 IE

2 9 4 5
27. Ammonia will not form complex with
B- C

1) Ag2+ 2) Pb2+ 3) Cu2+ 4) Cd2+

28. Which of the following complex compound is low spin, inner orbital, diamagnetic complex?
O S

1) [Ni(NH3)6]Cl2 2) K3[Fe(CN)6] 3) K2[PtCl6] 4) [Cr(H2O)6]Cl3

29. As aqueous solution of titanium bromide shows zero magnetic moment. Assuming the complex as
M C

octahedral in aqueous solution, the formula of the complex is

PC

1) [TiBr6]3– 2) [Ti(H2O)6]Br4 3) [TiBr6]2– 4) [Ti(H2O)4Br2]

30. Which of the following option is having maximum number of unpaired electrons –
1) A tetrahedral d6 ion 2) [Co(NH3)6]3+
3) A square planar d7 ion 4) A co-ordination compound with magnetic moment of 5.92 B.M.
31. The degeneracy of d-orbitals is lost under :
(I) Strong field ligand (II) Weak field ligand
(III) Mixed field ligand (IV) Chelated ligand field
1) I, II and IV 2) I and II 3) I, II, III and IV 4) I, II and III
32. Relative to the average energy in the spherical crystal field, the t2g orbitals in tetrahedral field is
1) raised by (2/5)  t 2) lowered by (2/5)  t
3) raised by (3/5)  t 4) lowered by (1/5)  t
33. Which of the following outer orbital complex has highest magnetic moment?
1) [Mn(NH3)6]Cl2 2) [Cr(NH3)6]Cl3 3) [Ni(NH3)6]Cl2 4) [Co(NH3)6]Cl3

16
34. Which of the following are inner orbital complex (i.e., involving d2sp3 hybridisation) and is
paramagnetic in nature?
1) [Mn(CN)6]3– , [Fe(CN)6]3–, [Co(C2O4)3]3– 2) [MnCl6]3–, [FeF6]3–, [CoF6]3–
3) [Mn(CN)6]3–, [Fe(CN)6]3– 4) [MnCl6]3– , [Fe(CN)6]3–, [Co(C2O4)3]3–
35. Mn2+ forms a complex with Br- ion. The magnetic moment of the complex is 5.92 B.M. What would be
the probable formula and geometry of the complex?
1) [MnBr6]4–, octahedral 2) [MnBr4]2–, square planar
3) [MnBr4]2–, tetrahedral 4) [MnBr5]3–, trigonal bipyramidal

36. Which of the following hydrate is diamagnetic?

1) [Mn(H2O)6]2+ 2) [Cu(H2O)6]3+ 3) [Co(NH3)6]3+ 4) [Co(H2O)6]2+
37. Which one of the following will show paramagnetism corresponding to 2 unpaired electrons?
(Atomic numbers : Ni = 28, Fe = 26)

Y
1) [FeF6]3– 2) [NiCl4]2– 3) [Fe (CN)6]3– 4) [Ni (CN)4]2–

EM
38. CN– is a strong field ligand. This is due to the fact that
1) it carries negative charge 2) it is a pseudohalide
3) it can accept electrons from metal species 4) it forms high spin complexes with metal species

D
39. When pink complex, [Co(H2O)6]2+ is dehydrated the colour changes to blue. The correct explanation for
the change is :

06 CA
1) The octahedral complex becomes square planar. 2) A tetrahedral complex is formed.
3) Distorted octahedral structure is obtained.
66 A
4) Dehydration results in the formation of polymeric species.
40. The crystal field stabilization energy (CFSE) is the highest for
93 CE

1) [CoF4]2– 2) [Co(NCS)4]2– 3) [Co(NH3)6]3+ 4) [CoCl4]2–

41. Which of the following complex ion is not expected to absorb visible light ?
2− 3+ 2+ 2+
1)  Ni ( CN )4  2) Cr ( NH3 )6  3)  Fe ( H 2 O )6  4)  Ni ( H 2O )6 
14 N

42. Crystal field stabilization energy for high spin d4 octahedral complex is:
74 IE

1) – 1.8  0 2) – 1.6  0 + P 3) – 1.2  0 4) – 0.6  0

43. Which of the following is incorrect regarding spectrochemical series?
B- C

1) NH3 > H2O 2) F−  C 2 O 42− 3) NCS−  SCN − 4) en > edta4–

O S

44. Of the following complex ions, which is diamagnetic in nature?

1) [NiCl4]2– 2) [Ni(CN)4]2– 3) [CuCl4]2– 4) [CoF6]3–
M C

45. The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d4, d5, d6 and d7, respectively. Which one
PC

of the following will exhibit minimum paramagnetic behaviour?

1) [Mn(H2O)6]2+ 2) [Fe(H2O)6]2+ 3) [Co(H2O)6]2+ 4) [Cr(H2O)6]2+
(At, nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
46. Which of the following complex compounds will exhibit highest paramagnetic behaviour?
(At. No. : Ti = 22, Cr = 24, Co = 27, Zn = 30)
1) [Ti (NH3)6]3+ 2) [Cr (NH3)6]3+ 3) [Co (NH3)6]3+ 4) [Zn (NH3)6]2+
47. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour ?
1) [Ni(NH3)6]2+ 2) [Zn(NH3)6)]2+ 3) [Cr(NH3)6]3+ 4) [Co(NH3)6]3+
48. Which of the following is the limitation of valence bond theory?
1) It does not distinguish between weak and strong ligands.
2) It does not give quantitative interpretation of magnetic data.
3) It does not explain the colour exhibited by coordination compounds
4) All of these

17
49. Low spin complex of d6-cation in an octahedral field will have the following energy :
−12 −12 −2 −2
1) 0 + P 2)  0 + 3P 3)  0 + 2P 4) 0 + P
5 5 5 5
(  0 = Crystal Field Splitting Energy in an octahedral field, P = Electron pairing energy)
50. Among the following complexes the one which shows zero crystal field stabilization energy (CFSE):
1) [Mn(H2O)6]3+ 2) [Fe(H2O)6]3+ 3) [Co(H2O)6]2+ 4) None of these
3+
51. [Sc(H2O)6] ion is
1) colourless and diamagnetic 2) coloured and octahedral
3) colourless and paramagnetic 4) coloured and paramagnetic
52. One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One
mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl (s). The
formula of the complex is
1) [Co(NH3)3Cl3]. 2 NH3 2) [Co(NH3)4Cl2] Cl . NH3

Y
3) [Co(NH3)4Cl] Cl2. NH3 4) [Co(NH3)5Cl] Cl2

EM
53. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral
complex with a Ca2+ ion?
1) One 2) Two 3) Six 4) Three

D
54. The molar ionic conductances of the octahedral complexes:

06 CA
(I) PtCl4.5NH3 (II) PtCl4.4NH3 (III) PtCl4.3NH3 (IV) PtCl4.2NH3
Follow the order
1) I < II < III < IV 2) IV < III < II < I 3) III < IV < II < I 4) IV < III < I < II
66 A
55. The most stable complex among the following is
1) K3[Al(C2O4)3] 2) [Pt(en)2]Cl 3) [Ag(NH3)2]Cl 4) K2[Ni(EDTA)]
93 CE

56. Consider the following complex [Co(NH3)5CO3]ClO4. The coordination number, oxidation number,
number of d-electrons and number of unpaired d-electrons on the metal are respectively.
1) 6, 3, 6, 0 2) 7, 2, 7, 1 3) 7, 1, 6, 4 4) 6, 2, 7, 3
14 N

57. Nickel (Z = 28) combines with a uninegative monodentate ligand to form a diamagnetic complex
74 IE

[NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are
respectively:
B- C

1) sp3, two 2) dsp2, zero 3) dsp2, one 4) sp3, zero

TOPIC 3: Organometallic Compounds
O S

58. The formula of ferrocene is

1) [Fe(CN)6]4– 2) [Fe(CN)6]3–
M C

3) [Fe(CO)5] 4) [(C5H5)2Fe]
59. The organometallic compound is :
PC

1) Ti(OCOCH3 )4 2) Ti(C2H4 )4 3) Ti(OC6H5 )4 4) Ti(OC2H5 )4

60. Which of the following does not have a metal- carbon bond?
1) Al(OC2H5 )3 2) C2H5MgBr 3) K[Pt(C2H4)Cl3] 4) Ni(CO)4
61. Which of the following is an organometallic compound?
1) Lithium methoxide 2) Lithium acetate 3) Lithium dimethylamide 4) Methyl lithium
62. In Fe(CO)5, the Fe – C bond possesses
1) ionic character 2)  -character only 3)  -character 4) both  and  characters
63. Which of the following may be considered to be an organometallic compound?
1) Nickel tetracarbonyl 2) Chlorophyll 3) K3 [Fe (C2O4)3] 4) [Co (en)3] Cl3
64. CH3 – Mg – Br is an organometallic compound due to
1) Mg – Br bond 2) C – Mg bond 3) C – Br bond 4) C – H bond.
65. Oxidation state of “V” in Rb4K[HV10O28] is
1) + 5 2) + 6 3)+7/5 4) + 4
66. For [Co2(CO)8], what is the total number of metal – carbon bonds and number of metal–metal bonds.
1) 10 ,1 2) 8, 2 3) 8, 1 4) 10, 0

18
67. Coordination compounds have great importance in biological systems. In this context which of the
following statements is incorrect?
1) Cyanocobalamin is B12 and contains cobalt
2) Haemoglobin is the red pigment of blood and contains iron
3) Chlorophylls are green pigments in plants and contain calcium
4) Carboxypeptidase - A is an exzyme and contains zinc.
68. An organometallic compound amongst the following is
1) Ferrocene 2) Diethyl zinc3) Tetraethyl lead (TEL) 4) All of these
69. Which of the following carbonyls will have the strongest C – O bond ?
1) [Mn (CO)6]+ 2) [Cr (CO)6] 3) [V (CO)6]– 4) [Fe (CO)5]
70. An example of a sigma bonded organometallic compound is :
1) Grignard's reagent 2) Ferrocene 3) Cobaltocene 4) Ruthenocene

NEET PREVIOUS YEARS QUESTIONS

Y
1. Iron carbonyl, Fe(CO)5 is [2018]

EM
1) Tetranuclear 2) Mononuclear 3) Dinuclear 4) Trinuclear
2. The type of isomerism shown by the complex [CoCl2(en)2] is [2018]
1) Geometrical isomerism 2) Coordination isomerism

D
3) Linkage isomerism 4) Ionization isomerism
3. The geometry and magnetic behaviour of the complex [Ni(CO)4] are [2018]

06 CA
1) Square planar geometry and diamagnetic 2) Tetrahedral geometry and diamagnetic
3) Tetrahedral geometry and paramagnetic 4) Square planar geometry and paramagnetic
4. An example of a sigma bonded organometallic compound is : [2017]
66 A
1) Grignard's reagent 2) Ferrocene 3) Cobaltocene 4) Ruthenocene
5. The correct order of the stoichiometries of AgCl formed when AgNO3 in excess is treated with the
93 CE

complexes : CoCl3.6NH3, CoCl3.5NH3, CoCl3.4NH3 respectively is : [2017]

1) 3 AgCl, 1 AgCl, 2 AgCl 2) 3 AgCl, 2 AgCl, 1 AgCl
3) 2 AgCl, 3 AgCl, 1 AgCl 4) 1 AgCl, 3 AgCl, 2 AgCl
14 N

6. Correct increasing order for the wavelengths of absorption in the visible region for the complexes of
Co3+ is : [2017]
74 IE

3+ 3+ 3+ 3+ 3+ 3+
1) [Co(H2O)6] , [Co(en)3] , [Co(NH3)6] 2) [Co(H2O)6] , [Co(NH3)6] , [Co(en)3]
3) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+ 4) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+
B- C

7. Pick out the correct statement with respect to [Mn(CN)6]3– [2017]

1) It is sp3d2 hybridised and tetrahedral 2) It is d2sp3 hybridised and octahedral
O S

3) It is dsp2 hybridised and square planar 4) It is sp3d2 hybridised and octahedral

8. Which of the following has longest C–O bond length? (Free C–O bond length in CO is 1.128Å) [2016]
M C

1) Ni(CO)4 2) [Co(CO)4]– 3) [Fe(CO)4]2– 4) [Mn(CO)6]+

9. The sum of coordination number and oxidation number of the metal M in the complex
PC

[M(en)2(C2O4)]Cl (where en is ethylenediamine) is: [2015]

1) 9 2) 6 3) 7 4) 8
10. The name of complex ion, [Fe(CN)6]3– is : [2015]
1) Hexacyanoiron (III) ion 2) Hexacyanitoferrate (III) ion
3) Tricyanoferrate (III) ion 4) Hexacyanidoferrate (III) ion
11. The hybridization involved in complex [Ni(CN)4]2– is (At. No. Ni = 28) [2015]
1) dsp2 2) sp3 3) d2sp2 4) d2sp3
12. Number of possible isomers for the complex [Co(en)2Cl2]Cl will be (en = ethylenediamine) [2015]
1) 2 2) 1 3) 3 4) 4
13. Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will
not give test of chloride ions with silver nitrate at 25 °C? [2015]
1) CoCl3·4NH3 2) CoCl3·5NH3 3) CoCl3·6NH3 4) CoCl3·3NH3
14. Which of these statements about [Co(CN)6]3– is true? [2015]
3–
1) [Co(CN)6] has four unpaired electrons and will be in a low-spin configuration.

19
 2) [Co(CN)6]3– has four unpaired electrons and will be in a high spin configuration.
3) [Co(CN)6]3– has no unpaired electrons and will be in a high-spin configuration.
4) [Co(CN)6]3– has no unpaired electrons and will be in a low-spin configuration.
15. Among the following complexes the one which shows zero crystal field stabilization energy (CFSE):
[2014]
3+ 3+ 2+ 3+
1) [Mn(H2O)6] 2) [Fe(H2O)6] 3) [Co(H2O)6] 4) [Co(H2O)6]
16. Which of the following complexes is used as an anti-cancer agent: [2014]
1) mer-[Co(NH3)3Cl3] 2) cis-[PtCl2(NH3)2] 3) cis-K2[PtCl2Br2] 4) Na2CoCl4
17. What is the correct electronic configuration of the central atom in K4[Fe(CN)6] based on crystal field
theory? [2019]
(1) t 2g eg
4 2
(2) t 2g eg
6 0
(3) e t 2
3 3
(4) e t 2
4 2

18. The Crystal Field Stabilisation Energy (CFSE) for [CoCl6]4– is 18000 cm–1. The CFSE for [CoCl4]2–
will be- [2019- ODISSA]
–1 –1 –1 –1
(1) 6000 cm (2) 16000 cm (3) 18000 cm (4) 8000 cm

Y
19. Match the coordination number and type of hybridisation with distribution of hybrid orbitals in space
based on Valence bond theory. [2020-COVID-19]

EM
Coordination Distribution
number and of hybrid
type of orbitals

D
hybridisation in space
1) 4, sp3 (i) trigonal bipyramidal

06 CA
2) 4, dsp2 (ii) octahedral
3
3) 5, sp d (iii) tetrahedral
4) 6, d2sp3 (iv) square planar
66 A
Select the correct option :
(1) 1)-(ii) 2)-(iii) 3)-(iv) 4)-(i) (2) 1)-(iii) 2)-(iv) 3)-(i) 4)-(ii)
93 CE

(3) 1)-(iv) 2)-(i) 3)-(ii) 4)-(iii) (4) 1)-(iii) 2)-(i) 3)-(iv) 4)-(ii)
20. Which of the following is the correct order of increasing field strength of ligands to form coordination
compounds? [2020]
14 N

1. CN −  C2O42−  SCN −  F − 2. SCN −  F −  C2O42−  CN −

74 IE

3. SCN −  F −  CN −  C2O42− 4. F −  SCN −  C2O42−  CN −

B- C

21. Ethylene diaminetetraacetate (EDTA) ion is [NEET-2021]

1) Unidentate ligand 2) Bidentate ligand with two “N” donor atoms
O S

3) Tridentate ligand with three “N” donor atoms

4) Hexadentate ligand with four “O” and two “N” donor atoms
M C

22. Match List – I with List – II. [NEET-2021]

List – I
PC

List – II
3−
a)  Fe ( CN )6  i) 5.92 BM
3+
b)  Fe ( H 2O )6  ii) 0 BM
4−
c)  Fe ( CN )6  iii) 4.90 BM
2+
d)  Fe ( H 2O )6  iv) 1.73 BM
23. The IUPAC name of the complex – [NEET-2022]
[Ag (H2O)2] [Ag(CN)2] is:
1) dicyanidosilver(II) diaquaargentate(II)
2) diaquasilver(II) dicyanidoargentate(II)
3) dicyanidosilver(I) diaquaargentate(I)
4) diaquasilver(I) dicyanidoargentate(I)

20
24. The order of energy absorbed which is responsible for the color of complexes [NEET-2022]
2+ 2+

(
A)  Ni H O
2 )2 ( en )2  (
B)  Ni H O
 2 )4 ( en ) and c)  Ni ( en )3 
2+
is

1)A > B > C 2) C > B > A 3) C > A > B 4) B > A > C

Y
EM
D
06 CA
66 A
93 CE
14 N
74 IE
B- C
O S
M C
PC

21
 NCERT LINE BY LINE QUESTIONS – ANSWERS
(1.) b (2.) d (3.) c (4.) a (5.) b
(6.) a (7.) a (8.) a (9.) d (10.) a
(11.) a (12.) c (13.) b (14.) d (15.) a
(16.) a (17.) b (18.) b (19.) b (20.) c
(21.) a (22.) a (23.) c (24.) b (25.) d
(26.) c (27.) c (28.) c (29.) b (30.) d
(31.) a (32.) a (33.) c (34.) c (35.) d
(36.) a (37.) c (38.) b (39.) b (40.) a
(41.) b (42.) c (43.) c (44.) a (45.) b
(46.) b (47.) a (48.) c (49.) b (50.) b
TOPIC WISE PRACTICE QUESTIONS - ANSWERS

Y
EM
1) 4 2) 2 3) 2 4) 3 5) 1 6) 2 7) 1 8) 3 9) 4 10) 2
11) 2 12) 3 13) 2 14) 2 15) 4 16) 4 17) 3 18) 3 19) 1 20) 1
21) 3 22) 1 23) 3 24) 2 25) 1 26) 2 27) 2 28) 3 29) 2 30) 4
31) 3 32) 1 33) 1 34) 3 35) 3 36) 3 37) 2 38) 2 39) 2 40) 3

D
41) 1 42) 4 43) 44) 2 45) 2 46) 3 47) 2 48) 1 49) 4 50) 2

06 CA
51) 4 52) 1 53) 4 54) 1 55) 2 56) 4 57) 1 58) 1 59) 4 60) 2
61) 1 62) 4 63) 4 64) 2 65) 1 66) 1 67) 3 68) 4 69) 1 70) 1
NEET PREVIOUS YEARS QUESTIONS-ANSWERS
66 A
1) 2 2) 1 3) 2 4) 1 5) 2 6) 4 7) 2 8) 3 9) 1 10) 4
11) 1 12) 3 13) 4 14) 4 15) 2 16) 2 17) 2 18) 4 19) 2 20) 2
93 CE

21) 4 22) 3 23) 4 24) 3

NCERT LINE BY LINE QUESTIONS – SOLUTIONS
+
(b) Complexes in which a metal is bound to more than one kind of donor groups, e.g., Co ( NH3 )4 Cl2 
14 N

(1.)
are known as heteroleptic. Complexes in which a metal is bound to only one kind of donor groups, e.g.,
74 IE

3+
Co ( NH3 )6  are known as homoleptic.
B- C

4
(2.) (d) Δ tetrahedra1 =  Δoctahedral
O S

9
4
= 18000 = 8000cm−1
M C

9
(3.) (c) Correct match is, P-(ii),Q-(i),R-(iii),S-(iv) .
PC

(4.) (a) Linkage isomers

(6.) (a)
Coordination number Type of hybridisation Distribution of hybrid orbitals in
space
4 sp3 Tetrahedral

4 dsp 2 Square planar

5 sp 3d Trigonal bipyramidal

6 sp3d 2 Octahedral

22
 6 d 2sp3 Octahedral

(a) The given compound is ethylenediaminetetraacetate ion ( EDTA )

4−
(7.) is hexadentate ligand.
(8.) (a)  Ni ( CO )4  − Terracarbonylnicke1(0) .
 MnCl6  ,  FeF6  and  CoF6 
3− 3− 3−
(10.) (a) areouterorbital complexes involving sp3d 2 hybridisation and
paramagnetic corresponding to four, five and four unpaired electrons.
(11.) (a) Co ( NH3 )5 ( NO 2 )  Cl2 shows linkage isomerism. Co ( NH3 )5 ( NO 2 )  Cl2 , which is obtained as the
red form, in which the nitrite ligands is bound through oxygen ( −ONO ) , and as the yellow form, in
which the nitrite ligand is bound through nitrogen ( − NO2 ) .
(14.) (d) Cr ( NH3 )3 ( H 2O )3  Cl3 : Triamminetriaqua chromium(III) chloride.
(15.) (a) 1mo1 CoCl3  6NH 3 (yellow) gave 3 mol of AgCl when it is treated with excess silver nitrate

Y
solution.

EM
(16.) (a) Co3+ , n = 4,μ = 4 ( 4 + 2 ) = 24B.M.
Cr 3+ , n = 3,μ = 3 ( 3 + 2 ) = 15B.M.

D
Fe3+ , n = 5,μ = 5 ( 5 + 2 ) = 35B.M.

06 CA
Ni3+ , n = 2,μ = 2 ( 2 + 2 ) = 8B.M.
(17.) (b) Among the given ligands oxalate is a bidentate ligand hence forms a chelate.
66 A
93 CE

(b) Both  Ni ( CO )4  and  NiCl4 

2−
(18.) are tetrahedral geometry.
14 N

+
(20.) (c) Co ( NH3 )4 Cl2  has geometrical isomers, cis and trans.
74 IE
B- C
O S
M C
PC

(22.) (a) The correct IUPAC name of  Pr ( NH 3 )2 Cl 2  is diamminedichloridoplatinum(II).

(23.) (c) In general, ligands can be arranged in a series in the order of increasing field strength as given below:
I−  Br −  SCN−  Cl−  S2−  F  OH−  C2O42−  H2O  NCS−  edra 4−  NH3  en  CN −  CO
2+ 3+
(25.) (d) Cu ( H2O )4  absorb red colour and its complementary colour is blue. Co ( NH3 )5 H 2O  absorb
blue green colour and its complementary colour is red.
(27.) (c) The CPSE of the ligands is in the order: H2O  NH3  CN− Therefore, excitation energies are in the
order:
3+ 3+ 3−
Co ( H 2O )6   Co ( NH3 )6   Co ( CN )6 
hc
From the reaction E =
λ

23
 1
Hence, E 
λ
The order of absorption of wavelength of light in the visible region is:
3+ 3+ 3−
Co ( H 2O )6   Co ( NH3 )6   Co ( CN )6 
(28.) (c) Octahedral complex of type  MA5B2  cannot show geometrical isomerism.
(29.) (b)  NiCl2 ( PPh 3 )2  : Dichloridobis (triphenyl phosphine) nickel (II)
Hg Co ( SCN )4  : Mercury tetrathiocynato cobaltate(III)
(30.) (d)
CrO24− Cr 6+ − Diamagnetic
CrO72− Cr 6+ − Diamagnetic
MnO−4 Mn 7+ − Diamagnetic

Y
MnO24− Mn 6+ − Paramagnetic

EM
In MnO24− unpaired electron is present, therefore d‐d transition is possible.
(32.) (a) Since the coordination number of Mn 2+ ion in the complex ion is 4, it will be either tetrahedral (sp3

D
hybridisation) or square planar ( dsp 2 hybridisation). But the fact that the magnetic moment of the
complex ion is 5.9 B.M., it should be tetrahedral in shape rather than square planar because of the

06 CA
presence of five unpaired electrons in the d‐orbitals.
2+
(33.) (c) In a coordination entity of the type  PtCl2 ( en )2  , only the cis‐isomer shows optical activity.
66 A
(35.) (d) As 3 moles of AgCl are obtained when 1 mol of CrCl3  6H 2O is treated with excess of AgNO3
which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of
93 CE

the complex is Cr ( H 2O )6  Cl3.

(37.) (c) Cobalt (Co) is the central metal in the complex Co ( NH3 )4 ( H 2O ) Cl  Cl2 .
14 N

(38.) (b) Coordination compounds are used as catalysts for many industrial processes. Examples include
rhodium complex, ( Ph 3P )3 RhCl  a Wilkinson catalyst is used for the hydrogenation of alkenes.
74 IE

(39.) (b) Tetracarbonyl nickel(O) is tetrahedral, pentacarbonyl iron (0) is trigonal bipyramidal while
B- C

hexacarbonyl chromium (0) is octahedral.

(40.) (a) CoCl2 ( en )2  has geometrical isomerism.
O S
M C
PC

(42.) (c) Mohr’s salt: FeSO4  ( NH 4 )2 SO4  6H 2O

(44.) (a) The primary valences are normally ionisable and are satisfied by negative ions while secondary
valences are non‐ionisable. These are satisfied by neutral molecules or negative ions. The secondary
valence is equal to the coordination number and is fixed a metal.
(45.) (b) Ionisation isomerism arises when the counter ion in a complex salt is itself a potential ligand and can
displace a ligand which can then become the counter ion. An example is provided by the ionisation
isomers Co ( NH3 )5 Cl  SO 4 and Co ( NH 3 )5 SO 4  Cl.
(46.) (b) Hexaamminenickel(II) hexanitrocobaltate(III) can be written as  Ni ( NH 3 )6  Co ( NO 2 )6  .
3 2

24
 3− 3−
(47.) (a) Out of cis − CrCl2 ( Ox )2  and trans ‐ CrCl2 ( Ox )2  , cis‐isomer is chiral (optically active).
2−
(48.) (c)  Fe ( CO )4 
Since metal atom is carrying maximum negative change therefore it would show maximum synergic
bonding as a result C − O bond length would be maximum.
(49.) (b) Primary compounds such as CrCl3 ,CoCl 2 or PdCl 2 have primary valency of 3, 2 and 2 respectively.

TOPIC WISE PRACTICE QUESTIONS - SOLUTIONS

1. 4)

2. 2)  Pt ( NH3 )4 Cl2  Br2

Y
Ionization isomer

EM
 Pt ( NH3 )4 ClBr  Cl.Br

Geometrical isomers

D
3. 2)

06 CA
4. 3) 66 A
93 CE

5. 1) The initial complex must have two Cl- ions in cis position as the third Cl- ion could replace an
14 N

ammonia cis to both or trans to one giving two isomers. If two Cl – ions were in the trans positions, the
74 IE

other four positions which could be replaced are equivalent and only one isomer of the product would
be obtained.
B- C

6. 2) Co ( NH3 )3 ( NO2 )3  Cl2 complex exhibits geometrical isomerism (G.I.)

O S

Geometrical isomers = 2 (1 cis + 1 trans)

M C

Optical isomers = 0
PC

Space isomers = 2
Co ( NH3 )5 ( NO 2 )  Cl2 complex shows linkage and ionization isomerism.

7. 1) The compounds of the type M ( AA )2 B2 exhibit both geometrical and optical isomerism

8. (3) Complex is not superimposable on its mirror image hence optically active i.e., rotate plane polarized
light.

25
9. 4)
10. 2)Given compound shows meridional isomerism.
11. (2) Non –superimposable mirror images are optically active, hence rotate plane polarized light.

Y
EM
D
+
12. 3) Cr (SCN )2 ( NH3 )4  shows linkage, geometrical and optical isomerism. Hence produces maximum

06 CA
no. of isomers.
13. 2)
66 A
14. 2) Ionisation isomer of Cr ( H 2 O )4 Cl ( NO 2 )  Cl is Cr ( H 2O )4 Cl2  NO 2
93 CE

15. 4)

16. 4)  Ni ( NH 3 )2 Cl2  , Ni 2+ is in sp3 hybridisation, thus tetrahedral in shape. Hence the four ligands are not
14 N

different to exhibit optical isomerism. In tetrahedral geometry all the positions are adjacent to one
74 IE

another
B- C

 geometrical isomerism also is not possible.

O S

17. (3) Coordination isomerism occurs when cationic and anionic complexes of different metal ions are
present in a salt. The two isomers differ in the distribution of ligands in cation and anion e.g.
M C

Co ( NH3 )6  Cr ( CN )6  is an isomer of Co ( CN )6  Cr ( NH 3 )6 

PC

18. 3) Geometrical isomers of following type of square planar complexes is possible. Ma2b2 type, Ma2bc
type and Mabcd type.
19. (1) [PdCl4]2– is dsp2 hybridized and square planar in shape.
20. (1) S2 O32− is a monodentate ligand whereas other ligands are bidentate.

21. (3) Similarity between optical and geometrical isomerism is that both are included in stereo isomerism.
22. (1)
23. 3) Coordination number of Cu2+, Zn2+ and Fe2+ are 4, 4 and 6 respectively.
i. [Cu(H2O)4]SO4.H2O ; ii. [Zn(H2O)4]SO4.H2O ; iii. [Fe(H2O)6]SO4.H2O

24. (2) It is a double salt: FeSO4 . ( NH 4 )2 SO4 .6H 2O → Fe2+ + 2SO42− + 2NH 4+

26
25. 1)
26. 2)
26. (2) The crystal field splitting in tetrahedral complexes is lower than that in octahedral complexes, and
4
t = − 0
9
27. (2) The complex formation is a characteristic of d-block elements. Lead is a p-block element hence does
not forms complex compounds.
28. (3) [Ni(NH3)6]Cl2 paramagnetic, outer orbital complex, sp3d2

K 3  Fe ( CN )6  paramagnetic, inner orbital complex, d2 sp3

K2 [PtCl6 ] diamagnetic, inner orbital complex, d2 sp3

Y
Cr ( H 2 O )6  Cl3 paramagnetic, inner orbital, d2 sp3

EM
29. (2)

D
30. (4)

06 CA
66 A
93 CE
14 N
74 IE
B- C

31. 3) Degenerate d-orbitals undergo splitting under ligand field created by strong, weak or mixed ligands.
O S

32. 1) t 2g orbitals have higher energy by ( 2 / 5)  t

M C

33. 1)  Mn ( NH 3 )6  Cl 2
PC

It is a outer orbital complex ; Mn 2+ ( d 5 )

M.M = 35 B.M
3− 3− 3−
34. 3)  Mn ( CN )6  and  Fe ( CN )6  are inner orbital complexes and paramagnetic while Co ( C2O4 )3 
is diamagnetic in nature

35. 3) Magnetic moment, n ( n + 2 ) = 5.92 B.M  n (unpaired electrons) = 5

27
36. 3)

Paramagnetic Paramagnetic Diamagnetic Paramagnetic

2) In  NiCl4  chloride ion, being a weak ligand, is not able to pair the electrons in d orbital.
−2
37.

38. 2) CN − is a strong field ligand as it is a pseudohalide ion. These ions are strong coordinating ligands and
hence have the tendency to form  -bond (from the pseudo halide to the metal) and  -bond (from the
metal to pseudo halide)
2+
39. 2) Hydrated CoCl2 .6H 2O is pink coloured and contains octahedral Co ( H 2O )6  ions. If this is partially

Y
2+
dehydrated by heating, then blue coloured tetrahedral ions Co ( H2O )4  are formed

EM
D
40. 3) Higher the oxidation state of the metal, greater the crystal field splitting energy. In options 1), 2) and
4), Co is present in + 2 oxidation state and in 3) it is present in + 3 oxidation state and hence has a

06 CA
higher value of CFSE.
41. 1)
66 A
42. 4)
93 CE
14 N
74 IE
B- C
O S

43. 2) According to spectrochemical series C 2 O 42−  F−

M C

44. 2)
PC

45. 3)

46. 2)
2+
47. 1)  Ni ( NH3 )6 

28
 Ni 2+ = 3d8 , according to CFT = t 62eg2 therefore, hybridisation is sp3d 2 and complex is paramagnetic.

48. 4)

49. 2) d 6 : t 2g
2,2,2
eg
0,0
( in low spin )
12
C.F.S.E = −0.4  6 0 + 3P = −  0 + 3P
5
50. 4) Due to d5 configuration CFSE is zero

51. 1) Sc =  Ar  3d1 , 4s 2
3+
Oxidation state of Sc in Sc ( H 2O )6  is Sc3+

Y
Sc3+ =  Ar  3d 0 , 4s0

EM
Thus Sc3+ does not have unpaired electron, hence it is diamagnetic and colourless.
52. 4)

D
06 CA
66 A
53. 1) EDTA has hexadentate four donor O atoms and 2 donor N atoms and for the formation of octahedral
complex one molecule is required
93 CE

54. 2)

55. 4) K 2  Ni ( EDTA )  Since EDTA is hexadentate and chelating and coordinates from six sides forming
14 N

more stable complex.

74 IE

56. 1) Co ( NH3 )5 CO3  ClO4 Six monodentate ligands are attached to Co hence C.N. of Co = 6; O.N. =
B- C

x + 5  ( 0 ) + 1 ( −2 ) + 1 ( −1) = 0  x = +3; electronic configuration of Co3+  Ar  3d 6 4s0 hence number of

O S

d electrons is 6. All d electrons are paired due to strong ligand hence unpaired electron is zero.
57. 1)
M C
PC

58. 4)
59. 2) Organometallic compound is 2) as it contains metalcarbon bonds. In others, direct link of carbon with
metal is not present
60. 1) Triethoxyaluminium has no Al – C linkage

29
61. (4) Compounds that contain at least one carbon metal bond are called organometallic compounds.
Hence, based on above definition methyl lithium is an organometallic compound. In other chemical
compounds, carbon is not linked with metal.
62. (4)
63. (2) Organometallic compounds are those compounds in which a metal is bonded directly to a carbon
atom of a molecule. In chlorophyll there is bond between carbon and Mg.
64. (2) Compounds that contain at least one carbon metal bond are known as organometallic compounds. In
CH3-Mg-Br (Grignard's reagent) a bond is present between carbon and Mg (Metal) hence it is an

Y
organometallic compound.

EM
65. (1)

66. (1) Structure of Co 2 ( CO )8 

D
06 CA
66 A
93 CE

67. (3) The chlorophyll molecule plays an important role in photosynthesis, contain porphyrin ring and the
metal Mg not Ca.
14 N

68. (4)
74 IE

69. 1) As positive charge on the central metal atom increases, the less readily the metal can donate electron
density into the  * orbitals of CO ligand (donation of electron density into  * orbitals of CO result in
B- C

weakening of C – O bond). Hence, the C – O bond would be strongest in [Mn(CO)6]+.

O S

70. 1) Grignard's reagent (RMgX) is a  -bonded organometallic compound.

NEET PREVIOUS YEARS QUESTIONS-EXPLANATIONS

M C

1. 2) Fe ( CO )5
PC

EAN = Z – O.N.+2(C.N) = 26-0+2(5) = 26 + 10 = 36

Only one central metal atom/ion is present and it follows EAN rule, so it is mononuclear.
2. 1) In the given complex, the CN of Co is 6, and the complex has octahedral geometry.

3. 2) Ni ( 28) :  Ar  3d8 4s 2
CO is a strong field ligand, so unpaired electrons get paired. Hence, configuration would be:

30
 For, four ‘CO’ligands hybridisation would be sp3 and thus the complex would be diamagnetic and of
tetrahedral geometry.
4. 1) Grignard's reagent (RMgX) is a s-bonded organometallic compound.
5. 2)

Y
EM
6. 4) The order of the ligand in the spectrochemical series H2O < NH3 < en Hence, the wavelength of the
light observed will be in the order
[Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(en)3]3+

D
Thus, wavelength absorbed will be in the opposite order

06 CA
i.e., [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+
3−
7. 2) In the complex  Mn ( CN )6  , O.S. of Mn is +3 E.C. of Mn+3 → 3d4
66 A
93 CE

The presence of a strong field ligand CN– causes pairing of electrons.

14 N
74 IE
B- C
O S

As, coordination number of Mn = 6, so it will form an octahedral complex.

3−
  Mn ( CN )6 
M C
PC

2−
8. 3)  Fe ( CO )4  Since metal atom is carrying maximum –ve charge therefore it would show maximum
synergic bonding as a resultant C—O bond length would be maximum.

9.  M ( en )2 ( C2O 4 )  Cl

C 2 O 4 = bidentate ligand, carry – 2 charge

en = bidentate ligand, carry 0 charge

 M carry + 3 charge;

31
 coordination number = 6
 Sum = + 3 + 6 = 9
10. (4) Carbocation Hexacyanidoerrate (III) ion.

1) Ni 2+ =  Ar  4s0 3d8
18
11.

Valence bond theory can be used to predict shape.

(In presence of ligand, pairing of electron occurs)

Y
 Square planar.

EM
12. 3)

D
06 CA
66 A
93 CE
14 N

13. 4) CoCl3 .3NH 3 will not give test for chloride ions with silver nitrate due to absence of ionisable chloride
74 IE

atoms.
B- C

CoCl3 .3NH3  Co ( NH3 )3 Cl3 

O S

Co ( NH3 )3 Cl3  ⎯⎯⎯

AgNO3
→ no ppt
M C
PC

−3
14. 4) In Co ( CN )6  O.N. of Co is +3

 Co+3 = 3d 6 4s0

CN − is a strong field ligand

 Pairing of electrons occurs so in this complex no unpaired electron is present and it is low spin
complex.
15. 2) Due to d5 configuration CFSE is zero.
16. 2)
17. 2)

32
18. 4)

Y
19. 2)sp3- tetrahedral, dsp2 - square planar

EM
sp3d - trigonal bipyramidal, d2sp3-octahedral
20. 2) According to spectro chemical series . SCN −  F −  C2O42−  CN −

D
21. 4)

06 CA
CH2COO
CH2 N
CH2COO
66 A
CH2COO
93 CE

CH2 N
CH2COO
3−
3)  Fe ( CN )6  Fe3+ − d 5 Sn = 1  = n ( n + 2 ) = 1.73 BM
14 N

22.
74 IE
B- C

3+
 Fe ( H 2O )6  Fe3+ − d 5 n=5  = n ( n + 2 ) = 5.92 BM
O S
M C

4−
 Fe ( CN )6  Fe2+ − d 6 n=0  = n ( n + 2 ) = 0 BM
PC

2+
 Fe ( H 2O )6  Fe2+ − d 6 n=4  = n ( n + 2 ) = 4.90 BM

23. IUPAC name of cpx  Ag ( H 2O )2   Ag ( CN )2  diaquasilver(1) dicyanidoargentate(1)

24. Energy absorbed by complex is directly proportional to strength of ligand

33