Inorganic Chemistry - Topic Wise Multiple Choice By Malik Xufyan

O N L Y C H E M I S T R Y
D I S C U S S I O N

9 2 3 1 3 7 3 5 5 7 2 7
0 3 1 3 7 3 5 5 7 2 7
1 / 1 / 2 0 2 0
Malik Xufyan
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CHEMICAL PERIODICITY
1.The size of the d orbitals in Si, P, S and Cl follows the order.
(a) Cl> S > P > Si
(b) Cl> P > S > Si
(c) P > S > Si >Cl
(d) Si > P > S >Cl
Answer
Ans (d)
Solution:
Size of d-orbitals decrease with decrease in size of element. The decreasing order of size is Si >
P > S >Cl.
Therefore decreasing order of size of d-orbitals is Si > P > S >Cl
2.The least basic among the following is:
(a) Al (OH)3
(b) La (OH)3
(c) Ce (OH)3
(d) Lu (OH)3
Answer
Ans (a)
Solution:
Al (OH)3→Amphoteric whereas hydroxides of lanthanoids are all base.
3.For an odd nucleon in ‘g’ nuclear orbital and parallel to I, spin and parity are
(a) 9/2 and (+)
(b) 7/2 and (+)
(c) 9/2 and (–)

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(d) 7/2 and (–)
Answer
Ans (a)
Solution:
In g nuclear orbital total subshell is nine.
+½
One odd nucleon finding in g-subshell
Total spin = 9/2
Parity=(-1)l
= (-1)4
= (+)
4.The electronegativity differences is the highest for the pair
(1) Li, Cl
(b) K, F
(c) Na, Cl
(d) Li, F
Answer
Ans (b)
Solution.
Among these elements K is least electronegative and F is most electronegative. Therefore
electronegativitydifference is highest for the pair K, F.
5. Among F–, Na+, O2-and Mg2+ions, those having the highest and the lowest ionic radii
respectively are
(a) O2-
andNa+
(b) F–
and Mg2+
(c) O2-
andMg2+

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(d) Mg2+
andO2-
Answer
6. 12-Crown-4 binds with the alkali metal ions in the following order:Li+>> Na+> K+>
Cs+ .It is due to the
(a) Right size of cation
(b) Change in entropy being positive
(c) Conformational flexibility of crown ether
(d) Hydrophobicity of crown ether
Answer
Ans (a)
Solution:
Each crown ether binds different ions, depending on the size of the cavity.
(Reference: Advance Organic Cheimistry, Jerry March, M. Smith)
7. The correct order of decreasing electronegativity of the following atoms is,
(a) As > Al > Ca > S
(b) S > As > Al > Ca
(c) Al > Ca > S > As
(d) S > Ca > As > Al
Answer
Ans (b)
Solution:
The electronegativities of elements are
Ca Al As S
1.0 1.5 2.0 2.5
8.The correct order of the size of S, S2–, S2+ and S4+ species is,

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(a) S>S2–>S2+> S4+
(b) S2+>S4+>S> S2–
(c) S2- >S>S2+> S4+
(d) S4+>S2–>S> S2+
Answer
Ans (c)
Solution:
As positive charge increases the size decreases while with increase in negative charge increase
the size. This is due to increase in Zeff in former case while decrease in Zeff in later case.
Hence, order of size is S2-
>S> S2+
> S4+
9. Which of the following pairs has the highest difference in their first ionization energy?
(a) Xe, Cs
(b) Kr, Rb
(c) Ar, K
(d) Ne, Na
Answer
Ans (d)
Solution:
First ionization potential of Ne – 2080
First ionization potential of Na – 495
——–
I.E. (Ne–Na) 1585 eV
So, 1585 eV is the largest difference in given pairs. The reason being as we move down the
group number of electrons and proton increases simultaneously with addition of new energy
shells so increase in distance from Nucleus to electron is more pronounced as that of increases in
electron and proton resultantly Zeff (effective nuclear charge) decreases and first ionization
potential also decreases down the group.

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10. Among the following pairs
(1) oxygen-sulfur (2) nitrogen -phosphorus
(3) phosphorus arsenic (4) chlorine- iodine
Those in which the first ionization energies differ by more than 300kJ mole-1are :
(a) (1) and (3) only
(b) (1) and (2) only
(c) (2) and (3) only
(d) (3) and (4) only
Answer
Ans (b)
11.The formation constant for the complexation of M+
(M = Li, Na, K and Cs) with
cryptant, C222 follows the order
(1.) Li+
< Cs+
< Na+
< K+
(2.) Li+
< Na+
< K+
< Cs+
(3.) K+
< Cs+
< Li+
< Na+
(4.) Cs+
< K+
< Li+
< Na+
Answer
Ans (2)
Solution:
The ability of cryptant to trap an alkali metal cation depends on size of both cage and metal ion,
the better the math between these; the more effectively the ions can be trapped.

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STRUCTURE AND BONDING
1. The total number of lone pairs of electrons in I3
–is:
(a) Zero
(b) Three
(c) Six
(d) Nine
Answer
Ans (d)
Solution:
Number of lone pair of electrons = 9
2. The strength of pΠ-dΠ bonding in E–O (E = Si, P, S and C) follows the order
(a) Si – O > P – O > S – O >Cl – O
(b) P – O > Si – O > S – O >Cl – O
(c) S – O >Cl – O > P – O > Si – O
(d) Cl – O > S – O > P – O > Si – O
Answer
Ans(d)

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Solution:
Smaller the inter nuclear distance between two atoms, stronger will be Π-the bonds. The
increasing or due of inter nuclear distance is Cl – O < S – O < P – O < Si – 0
Therefore strength of pn—dΠ
bonding is Cl – O > S –O > P – O > Si – O
3.The decreasing order of dipole moment of molecules is
(a) NF3 > NH3 > H2O
(b) NH3 > NF3 > H2O
(c) H2O > NH3 > NF3
(d) H2O > NF3 > NH3
Answer
Ans(c)
Solution:
Therefore, order of dipole moment is H2O > NH3 > NF3.
4.Which ones among CO3
2-, SO3, XeO3 and NO3
– have planar structure?
(1) CO3
2–
, SO3 and XeO3
(b) NO3
–
, SO3 and XeO3

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(c) CO3
2-
, NO3
–
and XeO3
(d) CO3
2-
, SO3 and NO3
–
Answer
Ans (d)
Solution
sp2
-hybridization sp2
-hybridization sp3
-hybridization Sp3
-hybridization
(Trigonal planar) (Trigonal planar) (Trigonal pyramidal) (Trigonal pyramidal)
NO3
–
,CO3
2-
and SO3have planar structure.
5.The correct schematic molecular energy diagram for SF6 molecule is

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Answer
Ans (a)
Solution:

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Consider SF6, which has six S-F bonds and hence 12 electrons involved in forming bonds and is
therefore hypervalent. The simple basis set of atomic orbitals that are used to construct the
molecular orbitals consists of the valence shell s and p orbitals of the S atom and one p orbital of
each of the six F atoms and pointing towards the S atom. We use the F2p orbitals rather than the
F2s orbitals because they match the S orbitals more closely in energy. From these ten atomic
orbitals it is possible to construct ten molecular orbitals. Calculations indicate that four of the
orbitals are bonding and four are antibonding; the two remaining orbitals are nonbonding.
(Reference : Inorganic Chemistry, Shriver Atkin)
6.The correct non-linear and iso-structural pair is
(a) SCl2and I3
–
(b) SCl2and I3
–
(c) SCl2and ClF2
(d) I3
+
and ClF2
Answer
Ans (b)
7.The oxidation state of Ni and the number of metal-metal bonds in [Ni CO6]2-that are
consistent with the18 electron rule are
(a) Ni(–II), 1 bond
(b) Ni(IV), 2 bonds
(c) Ni(–I), 1 bond
(d) Ni(IV), 3 bonds
Answer
Ans (c)
Solution:

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8.In metal-olefin interaction, the extent of increase in metal ® olefin p-back-donation
would
(a) lead to a decrease in C = C bond length
(b) change the formal oxidation state of the metal
(c) change the hybridisation of the olefin carbon from sp2
to sp3
.
(d) increase with the presence of electron donating substituent on the olefin.
Answer
Ans (c)
Solution:
9.The compound that will behave as an acid in H2SO4 is
(a) CH3COOH
(b) HNO3
(c) HClO4
(d) H2O

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Answer
Ans (c)
Solution:
HClO4behave as an acid in H2SO4. Because HClO4contain higher dissociation constant than that
of H2SO4.HClO4pKa value –9,
H2SO4pKa value –3, HNO3pKa value –1, CH3COOH pKa value 4.7.pKa value of HClO4> pKa
value of H2SO4.
10.Electron change in reduction of Ce(SO4)2, KMnO4, HNO2 and I2 with hydrazine in
acidic medium, respectively is
(a) 1e, 1e, 2e and 4e
(b) 1e, 3e, 2e and 4e
(c) 2e, 3e, 1e and 4e
(d) 2e, 4e, 1e and 3e
Answer
Ans (a)
Solution:
Ce(SO4)2and KMnO4gives one electron on reduction with hydrazine in acidic medium and
HNO2 and I2gives two electron and four electron on reduction with hydrazine in acidic medium.
11.The reason for the chemical inertness of gaseous nitrogen at room temperature is best
given by its
(a) high bonding energy only
(b) electronic configuration
(c) HOMO-LUMO gap only
(d) high bond energy and HOMO-LUMO gap
Answer
Ans (d)

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Solution:
High bond energy and HOMO-LUMO gap.
12.The orbital interactions n below represent
(a) CH3–Al interactions in Al2(CH3)6
(b) B–H interactions in B2H6
(c) CH3–Li interaction in Li4(CH3)4
(d) CH3CH2–Mg interactions in EtMgBr.(OEt2)2
Answer

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Ans (c)
Solution
The diagram clearly indicates the four centered-two electron interaction (4c-2e). This takes place
in Liu(CH3)4.
The sp3
hybrid orbital is of carbon while the three s-orbitals are of three surrounding lithium
atoms.
13.Correct combination for π and π* orbital’s in B2 molecules is…..
π π*
(1) Gerade Ungerade
(2) Ungerade Gerade
(3) Gerade Gerade
(4) Ungerade Ungerade
Answer
14.The correct shape of [TeF5]–ion on the basis of VSEPR theory is ……
(1) trigonalbipyramidal
(2) square pyramidal

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(3) pentagonal planar
(4) see-saw
Answer
Ans (2)
Solution:
Shape of [TeF5]–
ion on the basis of VSEPR theory is square pyramidal, as n in the figure.
15.Choose the correct option for carbonyl fluoride with respect to bond angle and bond
length
(1) ÐF-C-F >ÐF-C-O and C-F > C-O
(2) ÐF-C-F >ÐF-C-O and C-F < C-O
(3) ÐF-C-F <ÐF-C-O and C-F > C-O
(4) ÐF-C-F <ÐF-C-O and C-F < C-O
Answer
Ans (3)
Solution:
Carbonyl flouride is written in the following form

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From the structure it is clear that q1 is greater than q2 due to more bp-bp repulsion in q1So, ÐF-C-
O is greater than ÐF-C-F and bond length of C-F is greater than C-O bond.
16.In the molecules H2O, NH3 and CH4.
(a) The bond angles are same
(b) The bond distances are same.
(c) The hybridizations are same
(d) The shapes are same.
Answer
Ans (c)
Solution:
All are sp3
hybradized molecules.
17.According to VSEPR theory, the molecule/ion having ideal tetrahedral shape is:
(a) SF4

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(b) SO3
-2
(c) S2Cl2
(d) SO2Cl2
Answer
Ans (b)
18.The highest occupied MO in N2 and O2
+respectively are (take x-axis as internuclear
axis)
(a) σ2px, π*2py
(b) π2py ,π2pz
(c) σ*2px , σ2px
(d) π*2pz , π*2py
Answer
Ans (a)
19.The molecule with highest number of lone-pairs and has a linear shape based on VSEPR
theory is:
(a) CO2
(b) I3
–
(c) N02
–
(d) N02
+
Answer
Ans (b)
20.The number of antibonding electrons in NO and CO according to MO theory are
respectively.
(a) 1, 0
(b) 2, 2
(c) 3, 2

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(d) 2, 3
Answer
Ans (a)
21.Among the following pairs, those in which both species have similar structures are:
(A) N3
–,XeF2 (B)[ICl4]– , [PtCl4]2- (C)[ClF2]+ , [ICl]– (D) XeO3 , SO3
(a) (A) and (B) only
(b) (A) and (C) only
(c) (A), (B) and (C) only
(d) (B), (C) and (D) only
Answer
Ans (b)
22.Match list I (compounds) with list II (structures), and select the correct answer using the
codes given below.
List-I List-II
(A) XeO4 (i) square planar
(B) BrF4
–
(ii) tetrahedral
(C) SeCl4 (iii) distorted tetrahedral.
(a) (A–ii) (B–iii) (C–i)
(b) (A–iii) (B–i) (C–ii)
(c) (A–ii) (B–i) (C–iii)
(d) (A–i) (B–ii) (C–iii)
Answer
Ans (c)
23. The molecule C3O2 has a linear structure. This compound has
(1) 4σ and 4π bonds

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(3) 2σ and 3π bonds bonds
Answer
Ans (1)
Solution:
Structure of carbon sub oxide C3O2
24.The structures of XeF2 and XeO2F2respectively are
(a) bent, tetrahedral
(b) linear, square planar
(c) linear, see-saw
(d) bent, see-saw
Answer
Ans (c)
Solution:
XeF2 : linear molecule in vapor state

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The three 5p electrons are promoted to higher energy 5d sub-level. The 5s three 5p and one 5d
orbitals hybridize to give five sp3d hybrid orbitals. The four singly occupied orbitals are used for
bond formation to two fluorine and two oxygen atoms. The fifth hybrid orbital contains the lone
pair. The other 5d electrons of xenon which do not take part in the hybridization scheme are
involved in π-bond formation to two oxygen atoms. The structure of XeO2F2is represented in
Fig.
25.The number of lone pair(s) of electrons on the central atom in [BrF4]–, XeF6and
[SbCl6]3-are, respectively,
(a) 2, 0 and 1
(b) 1, 0 and 0
(c) 2, 1 and 1

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(d) 2, 1 and 0
Answer
Ans (c)
Solution:
Formula to calculate Bond Pair and Lone Pair
1. calculate TVE (total valence electron)
2. Divide TVE by 8 if 8 < TVE < 56
3. The quotient will give number of bond pair
4. while dividing the remainder by 2 will give lone pair on central atom
26.The ground state electronic configuration of C2 using all electrons is
Answer
Ans (4)
Solution:
C – 1s2
2s2
p2
MOT diagram for C2
1s2
2s2
has been omitted for clarity

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CONCEPTS OF ACIDS AND BASES
1.In the reactions (A) and (B),
nH2O + Cl–→ [Cl(H2O)n]– … (A)
6H2O + Mg2+→ [Mg (H2O)6] 2+ … (B)
water behaves as
(a) An acid in both (A) and (B)
(b) An acid in (A) and a base in (B)
(c) A base in (A) an acid in (B)
(d) A base in both (A) and (B)
Answer
Ans (b)
Solution:
nH2O + Cl–
→ [Cl(H2O)n]–
If negative end of Oδ-
– Hδ+
dipole approaches the Cl–
ion, then there well be repulsion
ion dipole interaction positive and of Oδ-
– Hδ+
dipole approaches the Cl–
ion and form [Cl(BO)n]–
Not stabilized. Thus H2O does not behave as a base
6H2O + Mg2+
→ [Mg (H2O)6] 2+
In this reaction Mg2+
behave as lewis acid and H2O behave as a ligand or lewis base.
2.In the following reactions carried out in liquid NH3.
Zn(NH2)2+ 2KNH2 → K2[Zn( NH2)4]
K2[Zn( NH2)4]+ 2NH4NO3 → Zn(NH2)2 + 2KNO3+4NH3
KNH2 and NH4NO3 act respectively as
(a) Solvo-acid and solvo-base
(b) Solvo-base and solvo-acid

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(c) Conjugate acid and conjugate base
(d) Conjugate base and conjugate acid
Answer
Ans(a)
3.Match each item from the List-I (compound in solvent) with that from the List-II (its
behaviour) and select thecorrect combination using the codes given below.
List-I List-II
1. CH3COOH in pyridine (i) strong acid.
2. CH3COOH in H2SO4 (ii) weak acid
3. HClO4in H2SO4 (iii) strong base
4. SbF5 in HF (iv) weak base
(1) (A–i), (B–ii), (C–iii), (D–iv)
(b) (A–ii), (B–i), (C–iii), (D–iv)
(c) (A–iii), (B–iv), (C–ii), (D–i)
(d) (A–iv), (B–ii), (C–iii), (D–i)
Answer
Ans (c)
Solution:
CH3COOH + C5H5N : → CH3COO–
+ C5H5NH+
(solvocation)
CH3COOH + H2SO4→ CH3COOH2
+
+ HSO4 (solvent anion)
HClO4 + H2SO4→ClO4 + H3SO4 (solventcation weak acid)
SbF5 +2HF→[SbF6]H2F (strong acid)
4.The reaction between SbF5 and two equivalents of HF leads to the formation of
(a) H2SbF3+ 2F2
(b) HSbF2+ 3F2

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(c) SbF3+ H2 + 2F2
(d) [SbF6 ]–
+[H2F]+
Answer
Ans (d)
Solution:
This is a reaction for the production of fluoroantimonic acid.
2HF ⇌ H2F+
+ F–
SbF5 + F–
→ SbF6
–
Overall reaction is
SbF5 + 2 HF → SbF6
–
+ H2F+
5.Among the following species, (A) Ni(II) as dimethylglyoximate, (B) Al(III) as oxinate, (C)
Ag(I) as chloride,those that precipitate with the urea hydrolysis method are
(a) A, B and C
(b) A and B
(c) A and C
(d) B and C
Answer
Ans (b)
6.The particles postulated to always accompany the positron emission among
(A) neutrino, (B) anti-neutrino, (C) electron,are
(a) A, B and C
(b) A and B
(c) A and C
(d) B and C

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Answer
Ans (c)
Solution:
Particles that are always to emit during position emission
1P1
→ 0n1
+ 1e0
+ v̅ (antineutrino)
7.Among the compounds A-D, those which hydrolyse easily are
(a) NCl3
(b) NF3
(c) BiCl3
(d) PCl3
Answer
Ans (c)
Solution:
BiCl3is readily hydrolysed by water to give BiOCl
BiCl3+H2O —> BiOCl
But BiOCl redissolve in conc. HCl to produce BCl3 after evaporation. It has quasi molecular
structure. SbCl3 has similar behaviour to BCl3.
8.The reaction of FeCl3.6H2O with SOCl2yields.
(a) FeCl2(s), SO2(g) and HCl (g)
(b) FeCl3(s), SO2(g) and HCl(l)
(c) FeCl2(s), SO3(s) and HCl (g)
(d) FeCl3(s), SO2(g) and HCl(g)
Answer
Ans (d)

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9.Among the following, the correct acid strength trend is represented by
(a) [Al(H2O)6]3+
<[Fe(H2O)6]3+
<[Fe(H2O)6]2+
(b) [Fe(H2O)6]3+
<[Al(H2O)6]3+
<[Fe(H2O)6]2+
(c) [Fe(H2O)6]2+
<[Fe(H2O)6]3+
<[Al(H2O)6]3+
(d) [Fe(H2O)6]2+
<[Al(H2O)6]3+
<[Fe(H2O)6]3+
Answer
Ans (c)
Solution
As the size of metal ion decreases, acidic character increases.
Order of size is Fe2+
>Al3+
>Fe3+
[Fe(H2O)6]2+
<[Fe(H2O)6]3+
<[Al(H2O)6]3+
Reference – Shriver Atkin (P.No. 123, Bronsted axis base concept)
10.Among the following, an example of a hypervalent species is
(a) BF3.OEt2
(b) SF4
(c) [PF6]–
(d) Sb2S3
Answer
Ans (c)
Solution
More than 8 electrons in the valence shell of P in [PF6
–
]
11.Two tautomeric forms of phosphorus acid are

31
Answer
Ans (a)
Solution
H3PO3 is a dibasic and reducing in nature.
12.Xenon forms several fluorides and oxofluorides which exihibit acidic behaviour. The
correct sequence of descending Lewis acidity among the given species is represented by
(a) XeF6> XeOF4> XeF4> XeO2F2
(b) XeOF4 > XeO2F2> XeOF4 > XeF6
(c) XeF4> XeO2F2> XeOF4> XeF6
(d) XeF4 > XeF6> XeOF4> XeO2F2
Answer
Ans (a)
Solution
Relative acidic strength of xenon fluorides follows order
XeF6> XeO3> F2> XeOF4> XeF4 > XeO2F2 > XeO3> XeF2
This order depends upon
(i) number of lone pair (ii) number of ‘F’ atoms
13.The gases SO2 and SO3 were reacted separately with ClF gas under ambient conditions.
The major products expected from the two reactions respectively, are

32
(a) SOF2 and ClOSO2F
(b) SOF2and SO2F2
(c) SO2ClF and SO2F2
(d) SO2ClF and ClOSO2F
Answer
Ans (d)
Solution
SO2+ClF –>SO2ClF
ClF+SO3 –>ClOSO2 F
14.Among KF, SnF4 and SbF5, solute(s) that increases(s) the concentration of BrF4
– in BrF–
3,is are
(1) KF only
(2) KF and SnF4
(3) SnF4 and SbF5
(4) KF, SnF4 and SbF5
Answer
Ans (1)
Solution:
The cation resulting formautodissociation of the solvent in the acid and the anion is the base
solutes that increase the concentration of cation of the solvent are considered acids and solutes
that increases the concentration of the anion are considered base. BrF3 also undergoes
autodissociation
2BF3–> BrF2
+
+ BrF4
–
Solutes such as KF that increase the concentration BrF4
–
are considered bases.
F–
+ BrF3–>BrF4
–

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15.Among SF4 ,BF4,XeF4 and ICl4
–the number of species having two lone pair of electrons
on the central atom according to VSEPR theory is:
(a) 2
(b) 3
(c) 4
(d) 0
Answer
Ans (a)
16.Lewis acidity of BCl3, BPh3 ad BMe3 with respect to pyridine follows the order
(a) BCl3>BPh3>BMe3
(b) BPh3> BMe3>BCl3
(c) BCl3>BPh3>BMe3
(d) BCl3>BMe3>BPh3
Answer
Ans (b)
17.The acid-base indicator (HIn) s a colour change at pH 6.40 when 20% of it is ionized.
The dissociationonstant of the indicator is
(a) 9.95 x 10-8
(b) 3.95x 10-6
(c) 4.5x 10-8
(d) 6.0x 10-8
Answer
Ans (a)

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MAIN GROUP ELEMENTS AND THEIR COMPOUNDS
1.Among the halides, NCl3(A), PCl3(B) and AsCl3(C), those which produce two different
acids.
(a) A and B
(b) A and C
(c) B and C
(d) A, B and C
Answer
Ans(c)
Solution:
NCl3+4H2 O —> NH 4OH+3HOCl ; PCl3+ 3H2O —>H PO3+3HCl (two different Acids)
AsCl3+ 3H2 O –> H3 AsO3+3H3 (two different acids)
2.The correct structure of P4 S3 is:
Answer
Ans(a)
Solution:
Structure of P4 S3.

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3.The reaction that yields Li AlH4 is:
(a) HCl( excess)+ AlCl3+ Li—>
(b) H2+ Al+ Li—>
(c) LiH (excess)+ AlCl3—>
(d) LiH(excess) +Al—>
Answer
Ans(c)
Solution:
LiH(exless)+AlCl3—>Li (AlH4)+3HCl
4.Boric acid is a weak acid in aqueous solution. But its acidity increases significantly in the
presence of ethyleneglycol, because
(a) ethylene glycol releases additional H+
(b) B(OH)4-
is consumed in forming a compound with ethylene glycol.
(c) ethylene glycol neutralizes H+
released by boric acid.
(d) Boric acid dissociates better in the mixed-solvent.
Answer
Ans (b)
Solution:
5.For the deposition of Pb by electroplating, the best suited compound among the following
is
(a) PbCl2
(b) PbSO4
(c) Pb(Et)4
(d) Pb(BF4)2
Answer

36
Ans (d)
Solution.
PbCl2 and PbSO4 are ionic compounds and insoluble in cold water. Therefore, cannot be used for
depositionof Pb and Pb(Et4) is toxic.
6.The correct order of the retention of cations on a sulfonated cation exchange resin
column is
(a) Ag+
> K+
> Na+
> Li+
(b) K+
> Na+
> Ag+
> Li+
(c) Li+
> Na+
> K+
> Ag+
(d) Li+
> Na+
> Ag+
> K+
Answer
Ans (a)
Solution:
The retention of cation is measured in term of distribution constant K. In general, where K for an
ion is large,
there is strong tendency for the resin phase to retain that ion. Therefore, for a typically
sulphonated cation exchange resin ,
value of K for a univalent ion decreases in the order
Ag+
> Cs+
> Rb+
> K+
>NH4
+
> Na+
> H+
> Li +
7.The reaction between diphenyldichlorosilane and water in 1:2 molar ratio gives product
A which on heating above 100°C yields a cyclic or polymeric product B. The products A
and B respectively, are

37
Answer
Ans (c)
Solution:
Diphenyl dichlorosilanes, Ph2SiCl2, which, like other halosilanes, are easily hydrolyzed, initially
to silanediols, R2Si(OH)2.
This on heating eliminate water molecule to form polymeric structure
8.The main products of the reaction of equimolar quantities of XeF6 with NaNO3 are
(a) XeOF4, NaF and NO2F
(b) Xe2O2 F2, NaF, NOF and F
(c) XeOF4, NaNO2 and F2
(d) XeF4, NaNO2 and F2O
Answer

38
Ans (a)
Solution:
Nitrates react with XeF6to form XeOF4with stoichiometric deficiency of NaNO3
XeF6+NaNO3 —> NaF+XeoF4+FNO2
9.Ozone present in upper atmosphere protects people on the earth
(a) due to its diamagnetic nature
(b) due to its blue colour
(c) due to absorption of radiation of wavelength at 255nm
(d) by destroying chlorofluoro carbons
Answer
Ans (c)
Solution:
Ozone is a diamagnetic gas which is of dark blue coloured due to absorption of red light.
(λ = 557 and 602 nm)
Ozone depliction discovered by J.C. Farman over Halley Bay in Antarctica.
Ozone also strong absorption inλ =255 UV which is good for earth and living beings as this
‘UV-b’ is
most dangerous
λ = 557= UV- b
10.Amongst organolithium (A), Grignard (B) and organoaluminium (C) compounds, those
react with SiCl4 to give compound containing Si-C bond are
(a) A and B
(b) B and C
(c) A and C

39
(d) A, B and C
Answer
Ans (d)
Solution:
11.Which of the following is used as propellant for whipping creams?
(a) N2O
(b) NO
(c) N2O3
(d) N2O5
Answer
Ans (a)
Solution:
Nitrous acid (N2O) which is commonly known as laughing gas used as a propellent in whipping
cream.So, also known as whippits or nangs.
12.Flame proof fabrices contain
(a) H2NC(O)NH2 .Na2SO4
(b) H2NC(S)NH2 .Na2SO4

40
(c) H2NC (O) NH2 .H3PO4
(d) H2NC(S) NH2 .H3PO4
Answer
Ans (c)
Solution:
The Flame proof fabrices contain urea and phosphoric acid (H2NC(O)NH2.H3PO4)
13.Among the compounds A-D, those which hydrolyse easily are
(a) NCl3
(b) NF3
(c) BiCl3
(d) PCl3
Answer
Ans (c)
Solution:
BiCl3is readily hydrolysed by water to give BiOCl
BiCl3+H2O —> BiOCl
But BiOCl redissolve in conc. HCl to produce BCl3 after evaporation. It has quasi molecular
structure. SbCl3 has similar behaviour to BCl3.
14.The number of lone-pairs are identical in the pairs
(a) XeF4, ClF3
(b) XeO4, ICl4
–
(c) XeO2F2, ICl4
–
(d) XeO4, ClF3

41
Answer
Ans (a)
Solution:
15.Among the oxides of nitrogen, N2O3, N2O4 and N2O5, the compound(s) having N–N
bond is/are
(a) N2O4 and N2O5
(b) N2O3 and N2O5
(c) N2O3 and N2O4
(d) N2O5 only
Answer
Ans (c)
Solution:
(A) N2O3: HNO2+HNO2—> N2O3
-H2O

42
(B) N2O4: HNO3+HNO2 —> N2O
-H2O
16.Among the molten alkali metals, the example of an immiscible pair (in all proportions) is
(a) K and Na
(b) K and Cs
(c) Li and Cs
(d) Rb and Cs
Answer
Ans (c)
Solution
Large difference between the size of Li and Cs. So, it is difficult to get a solid solution of these
two metals.

43
17.Among the following, an example of a hypervalent species is
(a) BF3.OEt2
(b) SF4
(c) [PF6]–
(d) Sb2S3
Answer
Ans (c)
Solution
More than 8 electrons in the valence shell of P in [PF6
–
]
18.Two tautomeric forms of phosphorus acid are
Answer
Ans (a)
Solution
H3PO3 is a dibasic and reducing in nature.
19.In a specific reaction, hexachlorocyclophazene, N3P3Cl6 was reacted with a metal
fluoride to obtain mixed halo derivatives namely N3P3Cl5F(A), N3P3Cl4F2(B), N3P3Cl3F3(C),
N3P3Cl2F4(D), N3P3ClF5(E). Compositions among these
which can give isomeric products are
(a) A, B and C

44
(b) B, C and D
(c) C, D and E
(d) E, A and B
Answer
Ans (b)
Solution
(A) N3P3Cl5F = only one isomer
(B) N3P3Cl3F3= three isomer
(C) N3P3Cl3F4 = three isomer
(D)N3P3Cl2F4 = one isomer

45
20.Xenon forms several fluorides and oxofluorides which exihibit acidic behaviour. The
correct sequence of descending Lewis acidity among the given species is represented by
(a) XeF6> XeOF4> XeF4> XeO2F2
(b) XeOF4 > XeO2F2> XeOF4 > XeF6
(c) XeF4> XeO2F2> XeOF4> XeF6
(d) XeF4 > XeF6> XeOF4> XeO2F2
Answer
Ans (a)
Solution
Relative acidic strength of xenon fluorides follows order
XeF6> XeO3> F2> XeOF4> XeF4 > XeO2F2 > XeO3> XeF2
This order depends upon
(i) number of lone pair (ii) number of ‘F’ atoms
21.The correct statement for ozone is
(a) It absorbs radiations in wavelength region 290-320 nm.
(b) It is mostly destroyed by NO radical in atmosphere
(c) It is non toxic even at 100 ppm level
(d) Its concentration near poles is high due to its paramagnetic nature.
Answer
Ans (a)
Solution
(i) Ozone is diamagnetic in nature
(ii) It is non toxic even at 1PPm level

46
(iii) It is not destroyed by no radial in atmosphere
(iv) It absorbs radiations in wave length region 290-320 nm
22.The orbital interactions n below represent
(a) CH3–Al interactions in Al2(CH3)6
(b) B–H interactions in B2H6
(c) CH3–Li interaction in Li4(CH3)4
(d) CH3CH2–Mg interactions in EtMgBr.(OEt2)2
Answer
Ans (c)
Solution

47
The diagram clearly indicates the four centered-two electron interaction (4c-2e). This takes place
in Liu(CH3)4. The sp3
hybrid orbital is of carbon while the three s-orbitals are of three
surrounding lithium atoms.
23.The numbers of P-S and P-P bonds in the compounds P4S3are, respectively
(a)6 and 3
(b) 4 and 3
(c) 3 and 6
(d) 6 and 2
Answer
Ans (a)
Solution:

48
P4S3is a derivative of tetrahedral (P4) unit form insertion of sulphur into three P-P bonds. The P-
S and P-P bond distance are 2.235Å and 2.090Å respectively. P4S3is called
phosphrousesquisulfide. Number of P-S bond is 6 and number of P-P is 3
Point group C3V signal = 2
24.Choose the correct option for carbonyl fluoride with respect to bond angle and bond
length
(a) <F-C-F > <F-C-O and C-F > C-O
(b) <F-C-F > <F-C-O and C-F < C-O
(c) <F-C-F < <F-C-O and C-F > C-O
(d) <F-C-F ><F-C-O and C-F < C-O
Answer
Ans (c)
Solution:
Carbonyl flouride is written in the following form
From the structure it is clear that q1 is greater than q2 due to more bp-bp repulsion in q1So, ÐF-C-
O is greater than ÐF-C-F and bond length of C-F is greater than C-O bond.
25.Which of the following react(s) with AsF5 in liquid BrF3
(a) XeF6 only
(b) XeF6 and XeF4
(c) XeF6 and XeF2
(d) XeF4 and XeF2
Answer
Ans(c)

49
Solution
XeF6 is flouride donor and react with PF5,AsF5&SbF5 in the presence of covalent solvent such
as BrF3
26.Consider the following reactions:
Answer
Ans (2)
Solution:
For C: Bromintrifluoride is a strong fluorinating agent that is able to convert a metal to its
associated fluoride compound. The reaction of NOCl with BrF3 yields the monofluoride. If the
reaction is combined in BrF3 solution a mixed metal fluoride salt is formed.
NOCl+ BrF3 –>[NO]+
[BrF4Cl]–
For D: NOCl behave as an electrophilic and an oxidant in most of its reactions with halide
acceptor
NOCl+ SbCl5 –>[NO]+
[SbCl6]–
27.The number of 3c-2e bonds present in Al(BH4)3 is
(a) four
(b) three
(c) six
(d) zero
Answer
Ans (c)

50
Solution:
Aluminium borohydride contain six bond which is n in figure
3c-2e bonds are ‘B–H–Al’ which is ‘6’.
28.The correct order of stability of difluorides is:
(a) GeF2 > SiF2 > CF2
(b) CF2 > SiF2 >GeF2
(c) SiF2 >GeF2 >CF2
(d) CF2 >GeF2 >SiF2
Answer
Ans (a)
29.Alkali metal superoxides are obtained by the reaction of
(a) Oxygen with alkali metals in liquid ammonia.
(b) Water with alkali metals in liquid ammonia
(c) H2O2 with alkali metals.
(d) H2O2 with alkali metals in liquid ammonia.
Answer

51
Ans (a)
30.H2O2reduces
(A) [Fe (CN)6]4- (B) KIO4 (C) Ce (SO4)3 (D) S03
2-
(a) A and B only
(b) B and C only
(c) C and D only
(d) B and D only
Answer
Ans (a)
31.Match List-I (compounds) with List-II (application) and select the correct answer using
the codes given below the lists.
List-I List-II
(A) Trisodium phosphate (i) Plasticizer
(B) Triarylphosphates (ii) Water softener
(C) Triethylphosphate (iii) Toothpaste
(D) Calcium hydrogen phosphate (iv) Insecticides
(a) (A)-ii (B)-i (C)-iv (D)-iii
(b) (A)-i (B)-ii (C)-iv (D)-iii
(c) (A)-ii (B)-iii (C)-iv (D)-i
(d) (A)-iii (B)-i (C)-ii (D)-iv
Answer
Ans (a)
Solution:

52
32.Among the following, the number of anhydrides of acids are CO, NO, N2O, B2O3, N2O5,
SO3 and P4O10.
(a) 3
(b) 4
(c) 5
(d) 6
Answer
Ans (b)
33.Lewis acidity of BCl3, BPh3 ad BMe3 with respect to pyridine follows the order
(a) BCl3>BPh3>BMe3
(b) BPh3> BMe3>BCl3
(c) BCl3>BPh3>BMe3
(d) BCl3>BMe3>BPh3
Answer
Ans (b)
34.The reaction between NH4Br and Na metal in liquid ammonia (solvent) results in the
products
(a) NaBr,HBr
(b) NaBr, H2

53
(c) H2 ,HBr
(d) HBr, H2
Answer
Ans (b)
35.The material that exhibits the highest electrical conductivity among the following sulfur-
nitrogen compounds is
(a) S4N4
(b) S7NH
(c) S2N2
(d)(SN)x
Answer
Ans (d)
36.The final product (s) of the reaction P(OR)3+ R’X is/are
(a) R’PO (OR)2 and RX
(b) [R’PO (OR)2]X
(c) [R’RPO2 (OR)]X
(d) ROR’ and P(OR)2X
Answer
Ans (a)
37.The structures of XeF2 and XeO2F2respectively are
(1.) bent, tetrahedral
(2.) linear, square planar

54
(3.) linear, see-saw
(4.) bent, see-saw
Answer
Ans (3)
Solution:
XeF2 : linear molecule in vapor state
The three 5p electrons are promoted to higher energy 5d sub-level. The 5s three 5p and one 5d
orbitals hybridize to give five sp3d hybrid orbitals. The four singly occupied orbitals are used for
bond formation to two fluorine and two oxygen atoms. The fifth hybrid orbital contains the lone
pair. The other 5d electrons of xenon which do not take part in the hybridization scheme are
involved in π-bond formation to two oxygen atoms. The structure of XeO2F2is represented in
Fig.
38.The formation constant for the complexation of M+
(M = Li, Na, K and Cs) with
cryptant, C222 follows the order
(1.) Li+
< Cs+
< Na+
< K+

55
(2.) Li+
< Na+
< K+
< Cs+
(3.) K+
< Cs+
< Li+
< Na+
(4.) Cs+
< K+
< Li+
< Na+
Answer
Ans (2)
Solution:
The ability of cryptant to trap an alkali metal cation depends on size of both cage and metal ion,
the better the math between these; the more effectively the ions can be trapped.
39.The number of lone pair(s) of electrons on the central atom in [BrF4]–, XeF6and
[SbCl6]3-are, rzspectively,
(1.) 2, 0 and 1
(2.) 1, 0 and 0
(3.) 2, 1 and 1
(4.) 2, 1 and 0
Answer

56
Ans (3)
Solution:
Formula to calculate Bond Pair and Lone Pair
1. calculate TVE (total valence electron)
2. Divide TVE by 8 if 8 < TVE < 56
3. The quotient will give number of bond pair
4. while dividing the remainder by 2 will give lone pair on central atom
40.Consider the following reaction:
N3P3Cl6 + 6 HNMe2 → N3P3Cl3(NMe2)3 + 3Me2NH.HCl
[A]
The number of possible isomers for [A] is
(1.) 4
(2.) 3
(3.) 2
(4.) 5
Answer
Ans (2)
Solution:
The cyclophosphazenes, first discovered by Liebig in 1834, have aroused great interest and,
especially within the last twenty years, have been intensively investigated. Much has been
discovered about their chemical and physical properties.

57
Two opposite points of view were put forward, one based on a completely delocalized molecular
orbital covering the whole of the cycle , the other based on the three-centered P-N-P islands with
little or no interaction between them.

58
TRANSITION METALS
1.The number of microstates for d5 electron configuration is:
(a) 21×63
(b) 14×63
(c) 7×62
(d) 28×63
Answer
Ans(c)
Solution:
2.The number of possible isomers of [Ru(PPh3)2(acac)2] (acac=acetylacetonate) is:
(1) 2
(b) 3
(c) 4
(d) 5
Answer
Ans (b)
Solution:
3.The correct spinel structure of Co3O4 is:

59
Answer
4.In the solid state, the CuCl5
3-ion has two types of bonds. These are
(1) Three long and two short
(b) Two long and three short
(c) One long and four short
(d) Four long and one short
Answer
Ans (a)
Solution:
In trigonalbipyramidal complexes, the two ligands lie on z-axis and the three in xy plane
somewhere in between the axes. In xy plane, there are four electrons and on z-axis there is only
one electron in dz2 orbital
(E.C.
5.The platinum complex of NH3 and Cl– ligands is an anti-tumour agent. The correct
isomeric formula of the complex and its precursor are
(1) cis-Pt(NH3)2Cl2 and PtCl4
2-
(b) trans-Pt(NH3)2Cl2 and PtCl4
2-
(c) cis-Pt(NH3)2Cl2 and Pt(NH3)4
2+
(d) trans-Pt(NH3)2Cl2 and Pt(NH3)4
2-
Answer
Ans (a)
Solution:
Trans effect of Cl–
> NH3Antitumer agent is cis-[Pt(NH3)2Cl]

60
The precursor of this complex is [PtCl4]2-
6.Successive addition of NaCl, H3PO4, KSCN and NaF to a solutin of Fe(NO3)3.9H2O gives
yellow, colourless, red and again colorless solutions due to the respective formation of:
(1) [Fe (H2O)5Cl]2+
, [Fe (H2O)5 (PO4)] , [Fe (H2O)5 (SCN)]2+
, [Fe (H2O)5 F]2+
(b) [Fe (H2O)5Cl(OH)]+
, [Fe (H2O)5 (PO4)] , [Fe (H2O)5 (SCN)]2-
, [Fe (H2O)5 F]2+
(c) [Fe (H2O)5Cl]2+
, [Fe (H2O)6 ]3+
, [Fe (H2O)5 (SCN)]2+
, [Fe (H2O)5 F]2+
(d) [Fe (H2O)5Cl]2+
, [Fe (H2O)5 (PO4)] , [Fe (H2O)5 (SCN)]2+
, [Fe (H2O)4(SCN)F]+
Answer
Ans (a)
Solution:
When Fe (NO3)3.9H2O is dissolved in water, the complex ion [Fe (H2O)6 ]3+
is formed.
7. The rate of exchange of OH2 present in the coordination sphere by 18OH2 of,
(i) [Cu(OH)2)6]2+, (ii) [Mn(OH2)6]2+, (iii) Fe(OH2)6]2+, (iv) [Ni(OH2)6]2+, follows an order

61
(1) (i) > (ii) > (iii) > (iv)
(b) (i) > (iv) > (iii) > (ii)
(c) (ii) > (iii) > (iv) > (i)
(d) (iii) > (i) > (iv) > (ii)
Answer
Ans (a)
Solution:
The rate of water exchange in [Cu(H2O)6]2+
is fastest due to John Teller distortion. For other
three complexes of 3d-series dipositive metal cation. The rate of water exchange decreases with
increase in effective nuclear charge and decrease in size.
8.The δ-bond is formed via the overlap of
(a) dx
2
-y and dx
2
-y 2
orbitals
(b) dxz and dxz orbitals
(c) dxy and dxy orbitals
(d) dyz and dyz orbitals
Answer
Ans (a)
Solution:
If the molecular orbital has two nodal planes passing through both the atomic nuclei , it is called
δ bond. Two δ bond can be formed per pair of atom one using two𝑑and the other using two
𝑑𝑥2−𝑦2 which overlap in four region. Hence, (1) and (3) both seems to be correct.
9.In the following reaction compound B is
(a) trans-[PtCl2 (NO2 )(NH3)]–

62
(b) cis-[PtCl2 (NO2 )(NH3)]–
(c) trans-[PtCl2 (NH3)2]
(d) cis-[PtCl2 (NO2)2]2-
Answer
Ans (a)
Solution:
This is an example of trans effect.
In 1st step addition substitution of NO2 take place at any place since all are equivalent.
In 2nd Trans effect of NO2 > Cl hence it will direct NH3 trans to itself leading to the formation of
(A).
10.The number of stereoisomers of trans-[CoCl2(triethylenetetramine)]Br is
(a) One
(b) Two
(c) Three
(d) Four
Answer
Ans (c)
11.Base hydrolysis of [CoCl3(NH3)5] 2+is an overall second order reaction, whereas that of
[Co(CN)6 ]3- is of first order. The rates depend in both cases solely on the concentrations of
the cobalt complex. This may be dueto
(A) Presence of ionizable proton in [CoCl2 (NH3) 5] 2+
but not in [Co(CN)6]3-
(B)SN
1
CBmechanism in the case of [CoCl2 (NH3) 5] 2+
only
(C) S N
1
CBmechanism in the case of [Co(CN)6]3-
only

63
(D) S N
1
CBmechanism in both the complexes
Correct explanation(s) is/are
(a)A and B
(b) A and C
(c) B only
(d) A and D
Answer
Ans (a)
Solution:
Base Hydrolysis mechanism follows as
[CoCl3(NH3)5] 2+
+ OH–
→ [CoCl(NH2)(NH3)4]+
+ H2O
[CoCl(NH2)(NH3)4]+
→ [Co (NH2)(NH3)4]2+
+ Cl–
(slow)
[Co (NH2)(NH3)4]2+
+ H2O→ [Co(OH)(NH3)5] 2+
(fast)
In the first step, an NH3 ligand acts as a Brønsted acid, resulting in the formation of its conjugate
base, the NH2ion, as a ligand. Because the deprotonated form of the complex has a lower charge,
it will be able to lose a Cl ion more readily than the protonated form, thus accelerating the
reaction. In addition, according to the conjugate base mechanism, loss of a proton from an
NH3 ligand changes it from a pure σ -donor ligand to a strong σ and π donor (as NH2
–) and so
helps to stabilize the five-coordinate transition state, greatly accelerating the loss of the Cl ion.
12.The spin-only magnetic moment and the spectroscopic ground state term symbol of
manganese center in[MnF6]3-ion respectively, are
(a) 4.9 BM and 5
D
(b) 4.9 BM and 4
F
(c) 3.9 BM and 3
D
(d) 4.9 BM and 3
F

64
Answer
Ans (A)
Solution:
[MnF6]3-
= 𝑀𝑛3+
= 4𝑆0
3𝑑4
∴ n=4
↑ ↑ ↑ ↑
S = 2 ∴ 2S+1 = 5
L=2 ∴ D
Term symbol :5
D
𝜇 = √(𝑛 + 2)
= √𝟐𝟒 = 𝟒. 𝟗
13.The three dimensional structure of compound [Co(Co(NH3) 4 (OH)2)3]Br 6 has
(a) Twelve Co–O and twelve Co–N bonds
(b) Ten Co–O and ten Co–N bonds
(c) Fourteen Co–O and ten Co–N bonds
(d) Twelve Co–O and ten Co–N bonds
Answer
Ans (a)
Solution:
Werner first resolve this compound which was totally inorganic optically active compound.

65
14.The spin-only (µS) and spin plus orbital (µS+L) magnetic moments of [CrCl6]3- are
(a) 3.87 BM and 5.20 BM
(b) 2.84 BM and 5.20 BM
(c) 3.87 BM and 6.34 BM
(d) 2.84 BM and 6.34 BM
Answer
Ans (a)
Solution:
Spin only magnetic moment is given by
μ = √(n + 2)
For [CrCl6
3−
]
n = 3
μ = 3.87
Spin plus orbital magnetic moment is given by
μ = √(n + 2) + n(n + 1)
Value of L is 3

66
Hence
μeff =5.2
15.In bis(dimethylglyoximato)nickel(II), the number of Ni–N, Ni–O and intramolecular
hydrogen bond(s),respectively are
(a) 4, 0 and 2
(b) 2, 2 and 2
(c) 2, 2 and 0
(d) 4, 0 and 1
Answer
Ans (a)
Solution:
Ni2+
is tested by dmg.
Ni2+
+ 2dmg – → [Ni(dmg)2] (Tschugaeff test)
(Red)
The structure of complex is
Therefore,
Ni – N bonds → 4

67
Ni – O → Zero
Hydrogen bonds → two
16.NiBr2reacts with (Et)(Ph2)P at –78ºC in CS2to give red compound ‘A’, which upon
standing at room temperatureturns green to give compound, ‘B’ of the same formula. The
measured magnetic moments of ‘A’ and‘B’ are 0.0 and 3.2 BM, respectively. The
geometries of ‘A’ and ‘B’ are
(a) square planar and tetrahedral
(b) tetrahedral and square planar
(c) square planar and octahedral
(d) tetrahedral and octahedral
Answer
Ans (b)
Solution:
Therefore,
17.If L is a neutral monodentate ligand, the species,[AgL4]2+, [AgL6]2+and [AgL4]3+,
respectively are
(a) paramagnetic, paramagnetic and dimagnetic
(b) paramagnetic, diamagnetic and paramagnetic
(c) diamagnetic, paramagnetic and diamagnetic

68
(d) paramagnetic, diamagnetic and diamagnetic
Answer
Ans (a)
Solution:
In ,[AgL4]2+
and [AgL6]2
,Ag2+
has d9
configuration, hence have unpaired electron.
Hence paramagnetic.
[AgL4]3+
=>Ag3+
has d8
configuration and dsp2
hybridization (square planar so pairing of
electron).
Hence,diamagnetic in nature.
18.Chromite ore on fusion with sodium carbonate gives
(a) Na2CrO4and Fe2O3
(b) Na2Cr2O7and Fe2O3
(c) Cr2(CO)3 and Fe(OH)3
(d)Na2CrO4andFe2(CO)3
Answer
Ans (a)
Solution:
Chromium is extracted from chromite ore:
(I) 4FeCr2O4+ 8Na2CO3+ 7O2→ 8Na2CrO4+ 2Fe2O3+ 8CO2
(II) 2Na2CrO4+ H2SO4→ Na2SO4 + Na2Cr2O7+ H2O
(III) Na2Cr2O7+2C →Cr2O7+ Na2CO3+ CO
(IV) Cr2O3+ 2Al → Al2O3+ 2Cr

69
19.[CoL6]2+is red in colour whereas [CoL‘6]2+is green. L and L‘ respectively corresponds to,
(a) NH3and H2O
(b) NH3and 1, 10-phenanthroline
(c) NH3and 1, 10-phenanthroline
(d) H2O and NH3
Answer
Ans (a)
Solution:
[CoL6]3+
→red colour → absorbs green radiations.
[CoL’6]2+
→ green colour → absorbs red radiations.
Energy of green radiations > enegy of red radiations.
Therefore, L will be stronger ligand and than L’. Thus, L and L’ are NH3 and H2O respectively.
20.The typical electronic configurations of the transition metal centre for oxidative addition
(a) d0
and d8
(b) d6
and d8
(c) d8
and d10
(d) d5
and d10
Answer
Ans (c)
Solution:

70
Most commonly the metal in the complexes in their low oxidation state with d8 or d10
configuration undergooxidative addition.
21.Reaction of [Ru(NH3)5(isonicotinamide)]3+ with [Cr(H2O)6]2+ occurs by inner sphere
mechanism and rate of the reaction is determined by dissociation of the successor complex.
It is due to the
(a) Inert ruthenium birdged to inert chromium centre
(b) Inert ruthenium bridged to labile chromium centre
(c) Labile ruthenium bridged to inert chromium centre
(d) Labile ruthenium bridged to labile chromium centre
Answer
Ans (a)
Solution:
In the successor complex inert rutherm bridged to inert chromium.

71
22.Consider the second order rate constants for the following outer-sphere electron
transfer reactions:
[Fe (H2O)6]3+ / [Fe (H2O)6]2+4.0 M–1sec–1
[Fe (phen)3]3+/ [Fe (phen)3]2+3.0×107 M–1sec–1
(phen = 1, 10-phenanthroline)
The enhanced rate constant for the second reaction is due to the fact that
(a) The ‘phen’ is a π-acceptor ligand that allows mixing of electron donor and acceptor orbitals
that enhancesthe rate of electron transfer
(b) The ‘phen’ is a π-donor ligand that enhances the rate of electron transfer
(c) The ‘phen’ forms charge transfer complex with iron and facilitates the eletron transfer
(d) The ‘phen’ forms kinetically labile complex with iron and facilitates the electron transfer.

72
Answer
Ans (a)
Solution:
(Phen) is anΠ-accepter ligands hence there is mixing of donor and acceptor orbital having similar
symmetry this leads to fast transfer of electron leading to enhance rate of reaction.
23.Identify the correct statement about
[Ni(H2O)6]2+and [Cu(H2O)6]2+
(a) All Ni-O and Cu-O bond lengths of individual species are equal
(b) Ni-O(equatorial) and Cu-O(equatorial)
(c) All Ni-O bond lengths are equal whereas Cu-O (equatorial) bonds are shorter than Cu-
O(axial) bonds
(d) All Cu-O bond lengths are equal whereas Ni-O(equatorial) bonds are shorter than Ni-O(axial)
bonds.
Answer
Ans (c)
Solution:

73
24.A 1 : 2 mixture of Me2NCH2CH2CH2PPh2 and KSCN with K2[PdCl4] gives a square
planar complex A.Identify the correct pairs of donor atoms trans to each other in complex
A from the following combinations.
(a) P, N
(b) N, S
(c) P, S
(d) N, N
Answer
Ans (a)
Solution:

74
s and p both form Π -bonding with complex and Π-bonding capacity of sulphur is greater than
phosphorus due to smaller size of d-orbital of sulphur. Hence, in presence of sulphur trans to
phosphorus donor atom phosphorus- metal bond will be weak hence they do not lie trans to each
other in the complex. As nitrogen does not involvent in Π-bonding with complex hence when
nitrogen atom is trans to phosphorus, phosphorus become able to form efficient Π-bond with
metal hence become stable that’s why P and N are trans to each other
25.The reaction of [PtCl4]2- with two equivalents of NH3 produces
(a) cis-[Pt(NH3)2Cl2]
(b)trans-[Pt(NH3)2Cl2]
(c)both cis-[Pt(NH3)2Cl2] and trans-[Pt(NH3)2Cl2]
(d) cis-[Pt(NH3)2Cl4]2-
Answer
26.Which of the pairs will generally result in tetrahedral coordination complexes, when
ligands are Cl– or OH–
(A) Be(II), Ba(II) (B) Ba(II), Co(II) (C) Co(II), Zn(II) (D) Be(II), Zn(II)
(a) A and B
(b) B and C
(c) C and D
(d) A and D

75
Answer
27.Silica gel contains [CoCl4]2–as an indicator. When activated, silica gel becomes dark blue
while upon absorption of moisture, its colour changes to pale pink. This is because,
(a) Co(II) changes its coordination from tetrahedral to octahedral.
(b) Co(II) changes its oxidation state to Co(III)
(c) Tetrahedral crystal field splitting is NOT equal to octahedral crystal field splitting.
(d) Co(II) forms kinetically labile while Co(III) forms kinetically inert complexes
Answer
Ans (a)
Solution:
28.The Δt of the following complexes
(A) [CoCl4]2– (B) [CoBr4]2– (C) [Co(NCS)4]2–
(a) C > A > B
(b) A > B > C
(c) B > A > C
(d) C > B > A
Answer
Ans (a)
Solution:
From spectro chemical series order, Br–
<Cl–
<NCS–
Order of Δt

76
[Co(NCS)4]2–
> [CoCl4]2–
> [CoBr4]2–
29.Identify the chiral complexes from the following
(A) [Cr (EDTA)]–(B) [Ru (bipy)3]3+(C) [PtCl(diene)]+
(a) A only
(b) A and B only
(c) A and C only
(d) B and C only
Answer
Ans (b)
Solution:
[Cr (EDTA)]–
and [Ru (bipy)3]3+
are octahedral and have no plane of symmetry. Therefore
these arechiral complex [PtCl(diene)]+
is square planar and has plane of symmetry. Therefore, it
is achiral
30.Among the following, the correct acid strength trend is represented by
(a) [Al(H2O)6]3+
<[Fe(H2O)6]3+
<[Fe(H2O)6]2+
(b) [Fe(H2O)6]3+
<[Al(H2O)6]3+
<[Fe(H2O)6]2+
(c) [Fe(H2O)6]2+
<[Fe(H2O)6]3+
<[Al(H2O)6]3+
(d) [Fe(H2O)6]2+
<[Al(H2O)6]3+
<[Fe(H2O)6]3+
Answer
Ans (c)
Solution
As the size of metal ion decreases, acidic character increases.
Order of size is Fe2+
>Al3+
>Fe3+
[Fe(H2O)6]2+
<[Fe(H2O)6]3+
<[Al(H2O)6]3+

77
Reference – Shriver Atkin (P.No. 123,Bronsted axis base concept)
31.An octahedral metal ion M2+has magnetic moment of 4.0 B.M. The correct combination
of metal ion and delectronconfiguration is given by
(a) Co2+
, t2g5
e g
2
(b) Cr2+
,t2g4
eg
2
(c) Mn2+
, t2g3
eg
1
(d) Fe2+
,t2g4
eg
2
Answer
Ans (a)
Solution
Co2+
, t2g5
e g
2
μ=3.89 B.M.μ
Cr2+
,t2g4
eg
2
μ=4.9 B.M.μ
Mn2+
, t2g3
eg
1
μ=4.9 B.M.μ
Fe2+
,t2g4
eg
2 μ
=4.9 B.M.μ
32.According to VSEPR theory, the geometry (with lone pair) around the central iodine in
I3
+ and I3
–
ions respectivelyare
(a) tetrahedral and tetrahedal
(b) trigonalbipyramidal and trigonalbipyramidal
(c) tetrahedral and trigonalbipyramidal
(d) tetrahedral and octahedral
Answer
Ans (c)
Solution
Number of valence electrons in I3+
central atom are 8 while in I3–
are 10 and hence the geometries.

78
33.Among the following, the correct combination of complex and its color is
Answer
Ans (c)
Solution
Order of ligand is spectro chemical series
SCN–
<Cl–
< F–
< CN–
From above it is clear that CN–
will produce highest energy difference, followed by F–
, Cl–
, SCN–
.As colour of complex is complementary colour of absorbed light.
Thus,
[Co(CN)4] 2–
absorbs orange colour radiation which complementary colour is blue therefore
colour of [Co(CN)4] 2–
is blue.
The other combination according to energyare mismatched. Hence, only option (c) is correct.
34.MnCr2O4 is likely to have a normal spinel structure because
(a) Mn2+
will have a LFSE in the octahedral site whereas the Cr3+
will not
(b) Mn is +2 oxidation state and both the Cr are in +3 oxidation state.
(c) Mn is +3 oxidation state and 1 Cr is in +2 and the other is in +3 state.
(d) Cr3+
will have a LFSE in the octahedral site whereas the Mn2+
ion will not.
Answer

79
Ans (d)
Solution
Mn2+ —->
3d5 —->
t3
2geeg(high spin octahedral)
CFSE in oh fluid = (–0.4×3 + 0.6×2) = 0
Cr 3+
–>3d3
–>t3
2geg0
(octahedral )
CFSE in octahedral field. = –0.4×3D0 = -(0.4×3)Δ0=-1.2Δ0
35.Compounds K2Ba[Cu(NO2)6] (A) and Cs2Ba[Cu(NO2)6] (B) exhibit tetragonal elongation
and tetragonal compression, respectively. The unpaired electron in A and B are found
respectively, in orbitals,
(a) dz2
and dx2
-y2
(b) dx2
-y2
and dz2
(c) dz2
and dz2
(d) dx2
-y2
and dx2
-y2
Answer
Ans (b)
Solution:
36.Formonoionic complex [UO2(NO3)3]–,the correct coordination number and geometry
respectively are

80
(1) 8 and hexagonal bipyr90amidal
(2) 5 and square pyramidal
(3) 8 and squareantiprism
(4) 5 and trigonalbipyramidal
Answer
Ans (1)
Solution:
[UO2(NO3)3]–
NO3 =bidented ligand
U= Coordination number = 8
Geometry = Hexagonal bipyramidal
37.The complex that s orbital contribution to the magnetic moment, is
(1) [Cu(H2O)6]2+
(2) [Ni(H2O)6]2+
(3) [Co(H2O)6]2+
(4) [Cr(H2O)6]2+
Answer
Ans (3)
Solution:
If the electron is present in the t2g orbitals are unsymmetrical field then the orbital contribution to
the magnetic moment is large.
▪ The electronsare present in eg orbitals they do not contribute.
▪ Symmetrically field (t3
2g t6
2g ) they do not contribute.

81
38.Paramagnetic susceptibility of the order of 10–6 cm3mol–1 observed for KMnO4 is due to
(1) random spin alignment
(2) antiferromagnetic exchange interaction
(3) paramagnetic impurity
(4) temperature independent paramagnetism
Answer
Ans (4)
Solution:
KMnO4 the temperature independent paramagnetion due to Mn in O.S. +7 then they are
d0
orbital. They are diamagnetic in nature and magnetic moment is zero then they are
temperature independent paramagnetism.
39.According to crystal field theory, N+2can have two unpaired electrons in
(a) Octahedral geometry only
(b) Square-planar geometry only
(c) Tetrahedral geometry only
(d) Both octahedral and tetrahedral geometry.

82
Answer
Ans (d)
40.[Ni(CN)4]-2 and [NiCl4]-2
4complex ions are
(a) Both diamagnetic
(b) Both paramagnetic
(c) Diamagnetic and paramagnetic respectively
(d) Antiferromagnetic and diamagnetic respectively
Answer
Ans (c)
41.The number of possible isomers for [Ru (bpy)2 Cl2] is (bpy = 2,2 ‘−bipyridine )
(a) 2
(b) 3
(c) 4
(d) 5
Answer
Ans (d)
42.Cis and trans complexes of the type [PtA2X2] are distinguished by
(a) Chromyl chloride test
(b) Carbylamine test
(c) Kurnakov test
(d) Ring test
Answer

83
Ans (c)
43.In the H2Ru6( CO)18 cluster, containing 8-coordinated Ru centers, the hydrogen atoms
are
(a) Both terminal
(b) One terminal and the other bridging
(c) Both bridging between two Ru centers
(d) Both bridging between three Ru centers.
Answer
Ans (b)
44.The correct order of LMCT energies is:
(a) MnO4
–
< CrO4
2-
< VO4
3-
(b) MnO4
–
>CrO4
2-
> VO4
3-
(c) MnO4
–
> CrO4
2-
< VO4
3-
(d) MnO4
–
< CrO4
2-
> VO4
3-
Answer
Ans (a)
45.The complex [Mn (H2O)6] +2 has very light pink colour. The best reason for it is
(a) The complex does not have a charge transfer transition.
(b) d-d transitions here are orbital forbidden but spin allowed.
(c) d-d transitions here are orbital allowed but spin forbidden.
(d) d-d transitions here are both orbital forbidden and spin forbidden
Answer
Ans (d)

84
46.Consider the compounds, (A) SnF4, (B) SnCl4and (C) R3SnCl. The nuclaerquadrupole
splitting are observed for
(a) (A), (B) and (C)
(b) (A) and (B) only
(c) (B) an d (C) only
(d) (A) and (C) only
Answer
Ans (d)
47.Consider two redox pairs
(1) Cr(II)/Ru(III) (2) Cr(II)/Co(III)
The rate of acceleration in going from a outer-sphere to a inner-sphere mechanism is lower
for (1) relative to (2). Its correct explanation is:
(a) HOMO/LUMO are σ * and σ * respectively.
(b) HOMO/LUMO are σ * and π * respectively.
(c) HOMO/LUMO are π * and σ * respectively.
(d) HOMO/LUMO are π * and π * respectively.
Answer
Ans (a)
48.For the complexes
(A) [Ni (H2O)6]2+ ,(B) [Mn (H2O)6]2+ ,(C) [Cr (H2O)6]3+ ,(D) [Ti (H2O)6]3+, the ideal
octahedral geometry will not be observed in
(a) (A) and (D)
(b) (C) and (D)

85
(c) (B) only
(d) (D) only
Answer
Ans (d)
Solution:
49.The actual magnetic moment s a large deviation from the spin-only formula in the case
of
(a) Ti3+
(b) V3+
(c) Gd3+
(d) Sm3+
Answer
Ans (d)
50.The complex that absorbs light of shortest wavelength is
(a) [CoF6]3-
(b) [Co(H2O)6]3+
(c) [Co(NH3)6]3+
(d) [Co(OX)3]3-
(OX = C2O4
2-
)
Answer
Ans (c)
51.Green colouredNi(PPh2Et)2 Br2 , has a magnetic moment of 3.20 B.M. The geometry and
the number of isomers possible for the complex respectively, are
(a) square planar and one
(b) tatrahedral and one

86
(c) Square planer and two
(d) tetrahedral and two
Answer
Ans (b)
52.The correct order of acidity among the following species is
(a) [Na (H2O6)]+
> [Ni (H2O6)]2+
> [Mn (H2O6)] 2+
> [Sc (H2O6)]3+
(b) [Sc (H2O6)]3+
> [Ni (H2O6)]2+
> [Mn (H2O6)] 2+
> [Na (H2O6)]+
(c) [Mn (H2O6)] 2+
> [Ni (H2O6)]2+
> [Sc (H2O6)]3+
> [Na (H2O6)]+
(d) [Sc (H2O6)]3+
> [Na (H2O6)]+
> [Ni (H2O6)]2+
> [Mn (H2O6)] 2+
Answer
Ans (a)
53.A true statement about base hydrolysis of [Co (NH3)5Cl]2+ is:
(a) It is a first order reaction
(b) The rate determining step involves the dissociation of chloride in [Co (NH3)4(NH2)Cl]+
(c) The rate is independent of the concentration of the base
(d) The rate determining step involves the abstraction of a proton from [Co (NH3)5Cl]2+
Answer
Ans (b)
54.The correct order of d-orbital splitting in a trigonal bipyramidal geometry is:

87
Answer
Ans (d)
55.For the following outer sphere electron transfer reactions.
[Co(NH3)6]2+ + [Co*(NH3)6]3+→ [Co(NH3)6]3+ + [Co*(NH3)6]2+
[Ru(NH3)6]2+ + [Co*(NH3)6]3+→ [Co(NH3)6]3+ + [Co*(NH3)6]2+
the rate constants are 10–6 M–1 s–1 and 8.2×102 M–1s–1 respectively. This difference in the
rate constants is due to
(a) A change from high spin to low spin in Co* and high spin to low spin in Ru.
(b) A change from high spin to low spin in Co* and low spin to high spin Ru*.
(c) A change from low spin to high spin in Co* and the low spin state remains unchanged in Ru.
(d) A change from low spin to high spin in Co* and high spin to low spin in Ru*.
Answer
Ans (c)
56.Using crystal field theory, identify from the following complex ions that s the same
μeff(spin only) values
(A) [CoF6]3-
(B) [IrCl6] 3-
(C) [Fe(H2O) 6] 2+
(1) A and B
(2) B and C
(3) A and C
(4) A, B and C
Answer

88
Ans (3)
Solution:
All the complexes are d6
configured and contain weak field ligand but except Ir all form high
spin complex
*[4d and 5d metals never form high spin complex]
Hence spin only magnetic moment will be same for Co(III) and Fe(II).
57.Identify the complex ions in the sequential order when ferroin is used as an indicator in
the titration of iron(II) with potassium dichromate. (phen = 1, 10- phenathroline)
(1) [Fe(phen)3]2+
and [Fe(phen)3]3+
(2) [Fe(phen)3] 3+
and [Fe(phen)3] 2+
(3) [Fe(CN)6]4-
and [Fe(CN)6]3-
(4) [Fe(CN)6]3-
and [Fe(CN)6]4-
Answer
Ans (1)
Solution:
The color changes observed with redox indicators at the equivalence point is due to structural
changes in changes upon being oxidized or reduced. This can be illustrated with ferroin indicator
which is the iron(II) complex of the organic compound 1,10-phananthroline .
Each of the two nitrogen atoms in 1, 10-phenanthroline has an unshared pair of electrons that
can be shared with the Fe(II) ion. Three such ligand molecules get attached to one Fe(II) ion to
form a blood red complex ion. This complex can be oxidized to the corresponding light blue

89
Fe(III) complex with the same structure. Therefore, a sharp colour change from red to blue
occurs on oxidation of the complex.
[Ph3Fe]2+
→ [Ph3Fe]3+
+ e–
E0 = -1.06 V
The indicator is prepared by mixing equivalent quantities of iron(II) sulphate and 1, 10-
phananthroline. The resulting Fe(II) complex sulphate is called Ferroin. The Fe(III) complex
sulphate is called ferrin. The colour changes occur at about -1.11 V (not at -1.06), since the color
of ferroin is so much more intense than that of ferrin.
58.The correct statement for Mn-O bond lengths in [Mn(H2O)6] 2+ is
(1.) All bonds are equal
(2.) Four bonds are longer than two others
(3.) Two bonds are longer than four others
(4.) They are shorter than the Mn-O bond in [MnO4]-
Answer
Ans (1)
Solution:
Mn aqua complex do not allow any distortion because of symmetrical d5
distribution.
59.Among the following, species expected to fluxional behavior are
(A) [NiCl4]2- (tetrahedral), (B) IF7(pentagonal bipyramidal),
(C) [CoF6] 3- (octahedral), (D) Fe(CO)5(trigonal, bipyramidal)

90
(1.) B and C
(2.) B and D
(3.) C and D
(4.) A and D
Answer
Ans (2)
Solution:
Fe(CO)5 fluxional behavior by process called berry pseudo rotation. And molecule IF7 s similar
mechanism called Bartell mechanism which exchanges the axial atoms with one pair of the
equatorial atoms with an energy requirement of about 2.7 kcal/mol.
60.The oxidizing power of [CrO4]2-, [MnO4]2-, and [FeO4]2-follows the order
(1.) [MnO4]2-
< [MnO4]2-
<[FeO4]2-
(2.) [FeO4]2-
<[MnO4]2-
< [CrO4]2-
(3.) [MnO4]2-
<[FeO4]2-
< [CrO4]2-
(4.) [CrO4]2-
< [FeO4]2-
<[MnO4]2-
Answer

91
Ans (1)
Solution:
Since oxidation number is same (VI) in all three metals. Hence, in going from Cr to Mn to Fe,
size decrease due to which oxidizing power increase.
61.Aqueous Cr2+effects one electron reduction of [Co(NH3)5Cl]2+giving compound Y.
compound Y undergoes rapid hydrolysis. Y is,
(1.) [Co(NH3) 5Co] 2+
(2.) [Co(NH3) 5Co(OH)] 2+
(3.) [Co(NH3)4Co (OH) 2]
(4.) [Cr(H2O)5Cl] 2+
Answer
Ans (1)
Solution:
The reduction of cobalt(III) (in [Co(NH3)5Cl] 2+
) by chromium(II) (in [Cr(H2O)6] 2+
) and was
specifically chosen because (I) both CO(III) and Cr(III) form inert complexes and (2) the
complexes of Co(II) and Cr(III) are labile.
Under these circumstances the chlorine atom, while remaining firmly attached to the inert Co(III)
ion, can displace a water molecule from the labile Cr(III) complex to form a bridged
intermediate:
[Co(NH3)3Cl] 2+
+ [Cr(H2O)6] 2+
→ [(NH3)5Co-Cl-Cr(H2O) 5]4+
+ H2O
The redox reaction now takes place within this dinuclear complex with formation of reduced
Co(II) and oxidized Cr(III). The latter species forms an inert chloroaqua complex, but the
cobalt(II) is labile, so the intermediate dissociates with the chlorine atom remaining with the
chromium:
[(NH3)5Co-Cl-Cr(OH2) 5] 4+
→ [(NH3) 5Co] 2+
+ [ClCr(H2O)5]2+
62.The correct statement about the substitution reaction of [Co(CN) 5 (Cl)]3-with OH – to
give [Co(CN) 5(OH)] 3-is,

92
(1.) it obeys first order kinetics
(2.) its rate is proportional to the concentration of both the reactants
(3.) it follows the SN1 CB mechanism
(4.) its rate is dependent only on the concentration of [OH] –
Answer
Ans (1)
Solution:
Complexes such as [Co(CN) 5 (Cl)]3-[Co(py) 4Cl2] + would not be expected typical base
hydrolysis and reaction proceeds slowly without dependence of OH –ion.
(Reference : Inorganic Chemistry , Huheey)
63.Among the following complexes[Co(ox) 3] 3-, B. trans-[CoCl2 (en) 2] +, C. [Cr(EDTA)]–
the chiral one(s) is/are
(1.) A and B
(2.) C and B
(3.) C only
(4.) A and C
Answer
Ans (4)
Solution:
The essential condition to the chirality is the lack of any symmetry element and it have
“handedness”
→ Chiral complex is able to handedness properly (right or left handedness) by Δ and “Λ
”designation
→ However the complex trans-[CoCl2(en)2]+ contains plane of symmetry and hence are achiral

93
64.The room temperature magnetic moment (µeff in BM) for a monomeric Cu(II) complex
is greater than 1.73. This may be explained using the expression:
Answer
Ans (1)

94
ORGANOMETALLIC COMPOUNDS
1.The reactions of Ni(CO)4with the ligand (L = PMe3or P(OMe)3)L yields 3 Ni(CO)3L . The
reaction is
(a) Associative
(b) Dissociative
(c) Interchange (Ia)
(d) Interchange (Id)
Answer
Ans (b)
Solution:
Ni(CO)4 [L = PMe3,P(OMe)3]
Ni(CO)4 Total valence electron around
Ni = 10 + 4 x 2 = 18 electron
Substitution reactions at coordinatively saturated tetrahedral complexes with a 18 electron count
likeNi(CO)4 or Ni(CO)2(PR3)2 follows a simple first order kinetics.
Since rate does not depend upon the concentration and nature of the ligand suggesting a
dissociative mechanism.
2.The correct statement for the aggregating nature of alkyl lithium (RLi) reagent is:
(a) The carbanionnucleophilicity increases with aggregation.
(b) The observed aggregation arises from its electron deficient nature.
(c) Carbanionnucleophilicity does not depend on aggregation.
(d) The extent of aggregation is maximum in polar dative solvents.

95
Answer
Ans (b)
Solution:
CH3Li →the tendency of organolithium compounds to associate in the solid state as well as
in solution is due to the fact that in a single molecule (LiR), the number of valence electrons
is too low to use all the available Li valence orbitals for 2e–2c bonding. This electron
deficiency is compensated by the formation of multicentre bonds e.g. (2e – 4c) bonding.
In this structure methyl group of one (LiCH3)4unit interacts with the Li atoms of
neighbouring (Li)4tetrahedron.
3.For the reaction, trans-[IrCl (CO) (PPh3)2] + Cl2→ trans-[IrCl3(CO) (PPh3)2], the correct
observation
(a) VCO(product) > VCO(reactant)
(b) VCO(product) < VCO(reactant)
(c) VCO(product) = VCO(reactant)
(d) VCO(product) = VCO(free CO)
Answer
Ans (a)
Solution:

96
Since in product metal Ir is present in the higher oxidation state.So, there is less back
bonding to π*orbital of CO.
Hence, V(CO) product >V(CO) reactant
4.The nucleophilic attack on olefins under mild conditions:
(a) Is always facile
(b) Is more facile than electrophilic attack on olefins

97
(c) Is facile for electron-rich olefins
(d) Requires activation by coordination to metal.
Answer
Ans (d)
Solution:
Nucleophilic attack on olefins under mild conditions
So, it requires activation by coordination to metal.
5.The cluster having arachno type structure is:
(a) Os5(CO)16
(b) Os3(CO)12
(c) Ir4(CO)12
(d) Rh6(CO)16
Answer
Ans(b)
Solution:
Step-I: Calculate T.V.E. (total valency electron)
= 8×3 + 2 ×12 = 48
Step-II: Polyhedral electron count
= T.V. E. – (n×12) = 48–36=12 n = number of the metal atom
Step-III: = P.E.C./2 = 12/2 = 6 = (n + 3) = Archano.

98
6.The carbonyl resonance in 13C NMR spectrum of [(h5– C5 H5)Rh(CO)]3 (103Rh, nuclear
spin, I=1/2, 100%) s a triplet at –65º C owing to the presence of
(a) Terminal CO
(b) μ2– CO
(c) μ3– CO
(d) η5-C5H5
Answer
Ans(b)
Solution:
Doublet of carbonyl carbon indicates the each CO is attached with chemically equivalent
two Rh- atoms–that means μ2-CO complex.
Actual structure is:

99
7.Low oxidation state complexes are often air-sensitive, but are rarely water sensitive
because
(a) Air is reducing in nature while water is inert
(b) Both air and water are oxidizing in nature
(c) Both air and water are not π-acceptors
(d) Complexes with low oxidation states will easily lose electrons to O2but will not bind to p-
donor molecule like H2O.
Answer
Ans(d)
8.The complex that DOES NOT obey 18- electron rule is:
(a) [(η5
-C5 H5) RuCl (CO) (PPh)3]
(b) [W(CO)3(SiMe3) (Cl) (NCMe)2]
(c) [IrCl3(PPh3)2(AsPh2)]–
(d) [Os(N) Br2 (PMe3)NMe2]–
Answer
Ans(d)
Solution:
[η5
C5 H5
–
Ru+2
Cl– 1
(CO)(PPh)3]
T.V.E=6+6+2+2+2=18
[W+2
(CO)3(Si–1
Me3) Cl–1
(NCMe)2]
Total valence electron = 4 + 6 + 2 + 2 + 4 = 18
[IrCl3
–3
(PPh3)2(AsPh–
2)]–1
Total valence elecatron = 6 + 6 + 4 + 2 = 18

100
[Os+5
(N-3
) Br-2
2 (PMe3)N-1
Me2]–
Total valence electron = 3 + 2 + 4 + 2 + 2 = 13
9.The final product of the reaction [Mn (CO)6]++M eLi —> is:
(a) [Mn(CO)6]+
Me–
(b) [Mn(CO)5 Me]
(c) [Mn(CO)6]
(d) [(MeCO)Mn(CO)5]
Answer
Ans(d)
Solution:
[Mn(CO)6]+
+ MeLi
10.The reaction 3 [Rh4(CO)12] —> 2[Rh6(CO)16]+ 4CO [25ºC, 500 atm CO] is:
(a) Exothermic as more metal-metal bonds are formed.
(b) Endothermic as stronger metal-carbonyl bonds are cleaved while weaker metal-metal bonds
are formed.
(c) Is entropicallyfavorable but enthalpicallyunfavorable such that ΔG= 0
(d) Thermodynamically unfavourable ΔG >0

101
Answer
Ans(b)
Solution:
Because metal carbonyls are stronger bond then break this bond huge amount of energy are
required. But metal-metal bonds are weaker bond. So, reaction is endothermic and enthalpy
predominant and entropy unfavorable.
11.The substitution of η5–Cp group with nitric oxide is the easiest for
(a) η5
-Cp2Fe
(b)η5
-Cp2CoCl
(c)η5
– Cp2Ni
(d) η5
-Cp2Co
Answer
Ans (c)
Solution:

102
12.The molecule (OC)5M=C(OCH3)(C6H5)obeys 18 e rule. The two ‘M’ satisfying the
condition are
(a) Cr, Re+
(b) Mo, V
(c) V, Re+
(d) Cr, V
Answer
Ans (a)
Solution:
2 × 5 + M + 2 = 18, M = 18–12 = 6
So, M = Cr, and Re+
.
13.Complexes of general formula, fac-[Mo(CO)3(phosphite)3] have the C—O stretching
bands as given below.
Phosphines: PF3(A); PCl3(B); P(Cl)Ph2(C); PMe3(D)
v(CO), cm-1: 2090(i); 2040(ii); 1977(iii); 1945(iv)
The correct comibination of the phsphine and the streching frequency is,
(1) (A–i), (B–ii), (C–iii), (D–iv)
(b) (A–ii), (B–i), (C–iv), (D–iii)
(c) (A–iv), (B–iii), (C–ii), (D–i)
(d) (A–iii), (B–iv), (C–i), (D–ii)

103
Answer
Ans (a)
Solution:
As the π-accepting abilities of phosphine increases vC–O stretching frequency of the complex
increases.So,the order of π-accepting abilities among the given phosphine is:
PF3> PCl3> PClPh2> PMe3.
i.e. A > B > C > D.
So, the correct combination for vC–O streching is:
A →2090 C →1977
B →2040 D →1945
14.For the molecule below,
consider the following statements about its room temperature spectral data.
(1) 1H NMR has singlets at 5.48 and 3.18 ppm
(B) 1H NMR has multiplet at 5.48 and singlet at 3.18 ppm
(C) IR has CO stretching bands at 1950 and 1860 cm-1
(D) IR has only one CO stretching band at 1900 cm-1.
The correct pair of statement is,
(a) A and C

104
(b) B and C
(c) A and D
(d) B and D
Answer
Ans (a)
Solution:
all 5H proton gives on signal in
1
H NMR at 5.48
Since, the molecule is C2–symmetric. So, two CO’s will give two absorption band in IR
spectrum.
15.In the cluster [Co3(CH)(CO) 9] obeying 18e rule, the number of metal-metal bonds and
the bridgind ligands respectively, are
(a) 3 and 1 CH
(b) 0 and 3 CO
(c) 3 and 1 CO
(d) 6 and 1 CH
Answer

105
Ans (a)
Solution:
The structure of cluster [Co3(CH)(CO)9]
So, the number of M–M bond = 3 and bridging ligand = 1CH
16.Consider the catalyst in column-I and reactin in column-II
Column-I Column-II
1. [(R)-BINAP]Ru2-
(i) hydroformylation
2. [Rh(CO)2I2] –
(ii) asymmetric hydrogenation.
3. Pd(PPh3)4 (iii) asymmetric hydrogen transfer
(iv) heck coupling.
The best match of a catalyst of column-I with the reaction nuclear column-II is
(a) (A–ii), (B–i), (C–iv), (D–iii)
(b) (A–i), (B–ii), (C–iii), (D–iv)
(c) (A–iii), (B–i), (C–iv), (D–ii)
(d) (A–iv), (B–iii), (C–ii), (D–i)
Answer
Ans (a)

106
Solution:
[Rh(CO)2I2]–
→ Hydroformylation.
[Pd(PPh3)4] → Heck coupling (It is a catalyst for Heck coupling)
→ Asymmetry hydrogen transfer.
17.Which one of the following will NOT undergo oxidative addition by methyl iodide?
(1) [Rh (CO)2I2]–
(b) [Ir (PPh3) 2 (CO) Cl]
(c) [ŋ2
– CpRh (CO)2]
(d) [ŋ5
– Cp2Ti (Me) Cl]
Answer
Ans (d)
Solution:
[Rh (CO)2I2]–
→ 16e–
and a better cordidate for O.A.
[Ir (PPh3) 2 (CO) Cl] → 16e–
species (Vaska’s complex) O.A. is possible.
ŋ5
– Cp2Ti (Me) Cl] → d0
– system O.A. is not possible.
[ŋ5
– Cp2Ti (Me) Cl] → 18e– species as such O.A. is not possible but after the dissociation
ofligand CO’ O.A. is possible.
18.In hydrofomylation reaction using [Rh (PPh3)3 (CO) (H)]as the catalyst, addition of
excess PPh3 would
(1) increase the rate of reaction
(b) decrease the rate of reaction.
(c) not influence of the rate of reaction

107
(d) stop the reaction.
Answer
Ans (b)
Solution:
18e–
species in excess PPh3 it becomes 20e–
species but the active catalyst is 16e–
species. So, in
presence of excess PPh3. The rate of hydroformylation will be decreases.
19.Reactions A and B are, termed as respectively.
(1) Insertion, Metathesis
(b) Metathesis, insertion
(c) Oxidative, addition, metathesis
(d) Oxidative addition, insertion
Answer
Ans (a)
Solution:

108
Since SnCl2 will behaves as a carbene and it insert into the M–M bond. So, it is a kind of
insertion reaction.
20.Consider the following reaction mechanism. The steps A, B and C, respectively, are
(1) Oxidative addition; transmetallation; reductive elimination.
(b) Oxidative addition; carbopalladation; β-hydride elimination.
(c) Carbopalladation; transmetallation; reductive elimination.
(d) Metal halogen exchange; transmetallation; metal extrusion.
Answer
Ans (a)
Solution:
Step (1) is O.A. because the oxidation state of the metal is increased by two units.

109
Step(B) is transmetallation because the ligand is transferring from one metal to another metal.
Step(C) is reductive elimination because the oxidation state of the metal is increased by two
units.
21.The hapticities ‘x’ and ‘y’ of the arene moieties in the diamagnetic complex
[(ηx– C6H6 )Ru(ηy–C6H6)]respectively are
(a) 6 and 6
(b) 4 and 4
(c) 4 and 6
(d) 6 and 2
Answer
Ans (c)
Solution:
In the compound [𝜂−𝐶6 𝐻6 𝑅𝑢 𝜂𝑦−𝐶6𝐻6]
∴𝑇𝑜 𝑓𝑜𝑟𝑚 𝑎 𝑠𝑡𝑎𝑏𝑙𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑚𝑢𝑠𝑡 𝑓𝑜𝑙𝑙𝑜𝑤 18−𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑟𝑢𝑙𝑒.
TVE = 8 + x + y = 18
∴𝑥 & 𝑦 𝑚𝑢𝑠𝑡 𝑏𝑒 6 & 4.

110
22.The rate of the reaction depends on
(a) Concentration of both the reactants
(b) Concentration of Ni(CO)4only
(c) Concentration of PPh3only
(d) The steric bulk of PPh3
Answer
Ans (b)
Solution:
Ni(CO)4: This is 18 electron complex hence stable. In order to undergo substitution, it must
follow
dissociative mechanism (SN1). Hence, it depends only on concentration of reactant.
23.The product of the reaction of propene, CO and H2in the presence of CO2(CO) 8as a
catalyst is
(a) Butanoic acid
(b) Butanal
(c) 2-butanone
(d) Methylpropanoate
Answer
Ans (b)
Solution:
This is an example of Hydroformylation reaction (oxo) process. Conversion of alkene to
aldehyde.

111
24.Reductive elimination step in hydrogenation of alkenes by Wilkinson catalyst results in
(neglecting solvent incoordination sphere of Rh)
(a) T-shaped [Rh(PPh3)2 Cl]
(b) Trigonal-planar [Rh(PPh3)2 Cl]2+
(c) T-shaped [Rh(H)(PPh3)2 Cl]+
(d) Trigonal-planar [Rh(H)(PPh3)2]
Answer
Ans (a)
Solution:
Structure is T- shape.
25.Na[(η5– C5H5)Fe(CO)2 ] reacts with Br2 to give A. Reaction of A with LiAlH4 results in
B. The protonNMR spectrum of B consists of two singlets of relative intensity 5:1.
Compounds A and B, respectively, are
(a) (η5
– C5H5)Fe(CO)2Br and (η5
– C5H5)Fe(CO)2H
(b) (η4
– C5H5)Fe(CO)2Br2 and (η4
– C5H5)Fe(CO)2HBr
(c) (η5
– C5H5)Fe(CO)2(H)2
(d) (η5
– C5H5)Fe(CO)2HBr
Answer
Ans (a)
Solution:

112
First bromination occurs at Fe center to form stable 18 e–
complex. Then reaction with LAH will
substitute Br with H. In 1
H NMR singlet appears for Cp ring and 1
H in the ratio of 5:1 (5 protons
in Cp and for 1
H in Fe)
26.The compound that undergoes oxidative addition reaction in presence of H2is
(a) [Mn(CO)5]–
(b) [(η5
– C5H5)Mo(CO)]–
(c) [IrCl(CO)(PPh3)2]
(d) [(η5
– C5H5)2 ReH]
Answer
Ans (c)
Solution:
Oxidative addition is possible only in case of Iridium complex, commonly known as vaska’s
complex. Rest all are 18e–
complex hence stable.
27.1H NMR spectrum of free benzene s a peak at ~ 7.2 ppm. The expected chemical shift (in
ppm) of C6H6ligand in 1H NMR spectrum of [(η6 – C6H6)Cr(CO)3 ] and the reason for it, if
an, is/are
(a) 4.5 ; disruption of ring current
(b) 9.0 ; inductive effect
(c) 7.2
(d) 2.5 ; combination of inductive effect and disruption of ring current
Answer
Ans (a)
Solution:
There is change in chemical shift experience by particular proton on the formation of complex, bt
still depends on number of other factors including change in ring current brought about by
complex formation.

113
The calculated δ value experimentally obtained is 4.43. this work was carried out by Strohmeier
and Helhnanni which differ substantially for those in free ligand
(Ref : J. organometal. Chem, 5, 1966, 147-154)
28.The reaction of phosphorus trichloride with phenyllithium in 1:3 molar ratio yields
product ‘A’, which on furthertreatment with methyl iodide produces ‘B’. The reaction of B
with n-BuLi gives product ‘C’. The products A, Band C, respectively, are
(a) [PPh4]Cl, [Ph2P=CH2]I, Ph2P(nBu)
(b) PPh3, [Ph3PI]Me, Ph2P(nBu)3
(c) PPh3, [Ph3PMe]I, Ph3P=CH2
(d) [PPh4]Cl, [Ph3P=CH2]I, [Ph3P(nBu)]Li
Answer
Ans (c)
Solution:
PCl3is the precursor for the formation of PPh3
PCl3+ 3PhCl + 6Na → PPh3+ 6NaCl
PPh3combines with most alkyl halides to give phosphonium salts. The facility of the reaction
follows the usual pattern whereby alkyl iodides and benzylic and allylic halides are particularly
efficient reactants:
PPh3+ CH3I → [CH3PPh3]+
I−
These salts, which are readily isolated as crystalline solids, react with strong bases to form
ylides:

114
Such ylides are key reagents in the Wittig reactions, used to convert aldehydes and ketones into
alkenes. Nickel salts are required to react PPh3 with PhBr to give [PPh4]Br. The
tetraphenylphosphonium cation is widely used to prepare crystallizable lipophilic salts.
(References : Elschenbroich, C.; Salzer, Organometallic Chemsitry)
29.The final product in the reaction of [Cp2* ThH] with CO in an equimolar ratio is
Answer
Ans (d)
Solution:
In Lanthanides and Actinides, reaction with CO , the initial product undergoes a facile
conversion in which the acyl ligands appear to be dihapto bonded due to special affinity of these
metals towards oxygen
(𝜂5−𝐶5 𝐻5)2 (𝑇𝐻𝐹) 𝐿𝑛−𝐶𝑀𝑒3 + 𝐶→ (𝜂5− 𝐶5 𝐻5)2 𝐿𝑛−𝐶𝑂−𝐶𝑀𝑒3

115
(Ref : Organometallic Chemistry By R. C. Mehrotra)
30.Complexes HM(CO)5 and[(η5
-C5H5 )M’(CO)3]2 obey the 18-electron rule. Identify M
and M’ and their1H NMR chemical shifts relative to TMS.
(a) M =Mn, -7.5; M’=Cr, 4.10
(b) M =Cr, 4.10; M’=Mn, -7.5
(c) M =V, -7.5; M’=Cr, 4.10
(d) M =Mn,10.22; M’=Fe, 2.80
Answer
Ans (a)
Solution:
HM(CO)5
M + 1 + 10 = 18
M= 7 VE ∴ M = Mn
[(η5
-C5H5 )M’(CO)3]2
10 + 2M’ + 12 = 36
M’ = 7 VE Since it is dimer M’ must be Cr (6 VE) with one M-M bond

116
31.The ligand(s) that is (are) fluxional in [(η5– C5H5) (η1– C5H5)Fe(CO)2] in the
temperature range 221–298K, is (are)
(a) η5
– C5H5
(b) η1
– C5H5
(c) 5
– C5H5 and CO
(d) η1
– C5H5 and CO
Answer
Ans (b)
Solution:
32.The oxidation state of Ni and the number of metal-metal bonds in [Ni CO6]2-that are
consistent with the18 electron rule are
(a) Ni(–II), 1 bond
(b) Ni(IV), 2 bonds
(c) Ni(–I), 1 bond
(d) Ni(IV), 3 bonds
Answer
Ans (c)
Solution:

117
33.The compound [Re2(Me2PPh)4Cl4] (M) having a configuration of σ2π4δ2δ*2 can be
oxidized to M+ and M2+.The formal metal-metal order in M, M+ and M2+ respectively, are
(a) 3.0, 3.5 and 4.0
(b) 3.5, 4.0 and 3.0
(c) 4.0, 3.5 and 3.0
(d) 3.0, 4.0 and 3.5
Answer
Ans (a)
Solution:

118
34.Reaction of nitrosyl tetrafluoroborate to Vaska’s complex gives complex A with <M-N-
O = 124º . Thecomplex A and its N-O stretching frequency are, respectively
(a) [IrCl(CO)(NO)(PPh3)2]BF4, 1620 cm–1
(b) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1730 cm–1
(c) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1520 cm–1

119
(d) [IrCl(CO)(NO)(PPh3)2], 1820 cm–1
Answer
Ans (a)
Solution:
NO = 1650 cm–1
Bent
NO = 1520 cm–1
Linear
35.Amongst the following which is not isolobal pairs
(a) Mn(CO)5, CH3
(b) Fe(CO)4, O
(c) Co(CO)3, R2Si
(d) Mn(CO)5, RS
Answer
Ans (c)
Solution:
Total number of electrons
Co(CO)3 15 electrons not isolobal

120
R2Si 6 electrons
Mn(CO)5 17 electrons isolobal
CH3 7 electrons
Fe(CO)4 16 electrons isolobal
O 6 electrons
Mn(CO)5 17 electrons isolobal
RS 7 electrons
36.The ligand in uranocene is:
(a) C8 H8
2-
(b) C5 H5
2-
(c) C6 H6
(d) C4H4
2-
Answer
Ans (a)
Solution:
Uranocene U(C8H8)2 is the most notable cyclo octatetraenidl
of f-block elements and first organouranium compounds.
2K+C8H8 —> K2(C8H8)
2K2(C8H8) + UCl4—> U(C8H8)2+ 4KCl

121
37.In metal-olefin interaction, the extent of increase in metal —> olefin p-back-donation
would
(a) lead to a decrease in C = C bond length
(b) change the formal oxidation state of the metal
(c) change the hybridisation of the olefin carbon from sp2
to sp3
.
(d) increase with the presence of electron donating substituent on the olefin.
Answer
Ans (c)
Solution:
38.The oxidation state of molybdenum in [(η7–tropylium) Mo (CO)3]is :
(a) +2

122
(b) +1
(c) 0
(d) –1
Answer
Ans (c)
Solution:
Mo in zero oxidation state.
39.The number of metal-metal bonds in [W2(OPh)6] is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer
Ans (c)
Solution:
Total valency electron = 12 + 18 = 30 (A) [ (OPh) in bridging donate 3 electron]

123
B = (n×18–A) = 36–30 = 6
Metal-metal bond = B/2 = 6/2 = 3
40.The reaction,
[(CO)5 Mn( Me)+CO → [(CO)5 Mn{ C(O )Me}]is an example for
(a) oxidative addition
(b) electrophilic substitution
(c) nucleophilic substitution
(d) migratory insertion
Answer
Ans (d)
Solution:
41.The orders of reactivity of ligands, NMe3, PMe3and CO with complexes MeTiCl3and
(CO)5Mo(thf) are
(a) CO > PMe3> NMe3 and CO > NMe3> PMe3
(b) PMe3 > CO > NMe3 and NMe3> CO > PMe3
(c) NMe3 > PMe3> CO and CO > PMe3> NMe3
(d) NMe3 > CO > PMe3 and PMe3> NMe3> CO
Answer

124
Ans (c)
Solution:
42.Intense band at 15000 cm–1in the UV-visible spectrum of [Bu4N]2Re2Cl8is due to the
transition
(a) p-p*
(b) d-d*
(c) d-p*
(d) p-d*
Answer
Ans (b)
Solution:

125
43.The treatment of PhBr with n-BuLi yields:
(a) 2 n-BuPh + Br2 + Li2
(b) PhPh + octane + 2 LiBr
(c) n-BuPh + LiBr
(d) PhLi + n-BuBr
Answer
Ans (d)
Solution:

126
This follows electron transfer (radical) process
Hence, formation of 4-membered transition state model.
44.Though cyclobutadiene (C4H4) is highly unstable and readily polymerizes in its free
stae, its transition metal complexes could be isolated because
(a) it engages in long-range interaction with transition metals.
(b) it gains stability due to formation of C4H4
2–
on binding to transition metals.
(c) its polymerization ability reduces in presence of transition metal.
(d) it becomes stable in presence of transition metals due to formation of C4H4
2+.
Answer
Ans (c)
Solution:
Cyclobutadiene is
which is antiaromatic and unstable due to this reasons it dimerizes, but when it attached with
metal such as

127
The back p-bonding occurs from metal to cyclobutadienes due to this it gain electro from the
metal and converted, into cyclobutadiene anion which aromatic and hence stables.
45. Identify the order representing increasing p– acidity of the following ligands C2F4,
NEt3, CO and C2H4
(a) CO < C2F4< C2H4< NEt3
(b) C2F4< C2H4< NEt3< CO
(c) C2H4< NEt3< CO < C2F4
(d) NEt3< C2H4< C2F4< CO
Answer
Ans (d)
Solution:
Increasing p-acidity
NEt3< C2H4< C2F4< CO
Where, NEt3=Neitherp-donor nor p-acceptor
C2H4< C2F4< CO=Increasing order of p-acceptor ability (on the bases of LFT)
46.The species with highest magnetic moment (spin only value) is
(a) VCl6
4–
(b)[(h5
-C5H5)2Cr
(c) [Co (NO2)6]3–

128
(d)[Ni(EDTA)]2–
Answer
Ans (a)
Solution:
47.The number of metal-metal bonds in Ir4(CO)12is
(a) 4
(b) 6
(c) 10
(d) 12
Answer
Ans (b)
Solution:

129
A = TVE = 60
B = (n × 18)–TVE = 72–60 = 14
Metal metal bond = B/2 = 12/2 = 6
48.The bond order of the metal-metal bond in the dimericcomplex [Re2Cl4(PMe2)Ph4]+is
(a) 4.0
(b) 3.5
(c) 3.0
(d) 2.5
Answer
Ans (b)
Solution

130
49.Which of the following in NOT suitable as catalyst for hydroformylation?
(a) HCo(CO)4
(b) HCo(CO)3PBu3
(c) HRh(CO)(PPh3)3
(d) H2Rh(PPh3)2Cl
Answer
Ans (d)
Solution
HCo(CO)4, HCo(CO)3PBu3and HRb(CO)(PPh3)3. These all are the catalyst for hydroformylation
process.
50.In a cluster, H3CoRu3(CO)12, total number of electrons considered to be involved in its
formation is
(a) 57
(b) 60

131
(c) 63
(d) 72
Answer
Ans (b)
Solution
H3CoRu3(CO)12
3+a+8×3 + 12×2 = 60 electron.
51.The correct statement regarding terminal/bridging CO groups in solid Co4(CO)12 and
Ir4(CO)12 is
(a) both have equal number of bridging CO groups
(b) number of bridging CO groups in Co4(CO)12is 4
(c) the number of terminal CO groups in Co4(CO)12is 8
(d) the number of bridging CO groups in Ir4(CO)12is zero.
Answer
Ans (d)
Solution

132
52.On reducing Fe3(CO)12 with an excess of sodium, a carbonylate ion is formed. The iron
is isoelectronic with
(a) [Mn(CO)5]–
(b) [Ni(CO)4]
(c) [Mn(CO)5]+
(d) [V(CO)6]–
Answer
Ans (b)
Solution

133
53.Among the following clusters,
H is encapsulated in
(a) A only
(b) B only
(c) B and C only
(d) A and B only
Answer
Ans (a)
Solution
54.The electrophile Ph3C+ reacts with [(h5 – C5H5)Fe(CO)2(CDMe)2]+
to give a product A. The productA is formed because
(a) Fe is oxidised
(b) alkyl is susbtituted with Ph3C
(c) Fe-Ph bond is formed
(d) Alkyl is converted to alkene
Answer

134
Ans (d)
Solution
55.Subsitution of L with other ligands will be easiest for the species
Answer
Ans (c)
Solution

135
In case of (c) the chances of conversion of haptacityfrom h5
®h3
is more due to partial bond
fixation.
56.Among the following, the correct statement is
(a) CH is isolobal to Co(CO)3
(b) CH2 is isolobal to Ni(CO)2
(c) CH is isolobal to Fe(CO)4
(d) CH2 is isolobal to Mn(CO)4
Answer
Ans (a)
Solution
CH→4 +1=5 e-
Co (CO )3→9+ 6 =15 e-
But,
CH2→4+2 =6e-
Ni (CO) 2→10+4= 14 e-
i.eand so on
57.The most appropriate structure for the [Pt2(NH3)2(NCS)2(PPh3)2] is

136
Answer
Ans (c)
Solution
Pt -S→dπ, -dπ, bonding in weaker than that of Pt-P bonding. When Pt-S and Pt-P bond are trans
to each other then Pt-S become weaker therefore in such a situation SCM ligand tends to form
bonding through M atoms of ligand as M do not form dπ, -pπ, bonding.Hence the most propable
structure will be
58.CpM[Cpis (η5-C5H5)]fragmentisolobal with a BH fragment is……..
(1) CpGe
(2) CpMn

137
(3) CpRu
(4) CpCO
Answer
Ans (4)
Solution:
CpCoThe species which have same difference of electron to complete their valence shell
electronic configuration called isolobal fragment or electrochemical equivalent species
Total number of electron in BH = 4e–
Electron deficiency = 4e–
Total electron in M-Cp
x + 5 = 18
x = 13
For isolobal fragment, e–
deficiency = 4
So, number of electrons present in metal=13 – 4 = 9e–
Then electronic configuration of valence shell of metal is 3d7
4s2
, which is configuration of Co
Metal, therefore CpCo is isolobal fragment of BH
59.The number of metal-metal bonds in [Co2Fe2(CO)11(µ4-PPh)2] is……
(1) 3
(2) 4
(3) 5
(4) 6
Answer
Ans (2)
Solution:

138
Number of metal-metal bond is calculated by following formula
Number of M-M bond = ½[18 x No.of metals –Totalvalenceshellelectron]
Here, Number of M-M = ½[18 x 4 –(2 x 9 + 2 x 8 + 2 x 11 + 2 x 4)]
= ½[72 – 64]
=½ x 8
= 4
Number of M-M bond = 4
60.Correct order of M-C bond length of metallocenes (a-c)
1. [Fe(η5-Cp)2]
2. [Ni(η5-Cp)2]
3. [Co(η5-Cp)2] is
(1) a> b > c
(2) b> c > a
(3) c> b > a
(4) a > c > b
Answer
Ans (2)
Solution:
The value of M-C bond length of metallocenes are
Cp2Fe Cp2Co Cp2Ni
M-C Distance in (Å ) 2.06 2.12 2.20
61.For fluxional Fe(CO)5 (structure given below) in solution, the exchange of numbered CO
groups will be between

139
(1) 2 and 5; 3 and 4
(2) 2 and 3; 4 and 5
(3) 2 and 3; 1 and 5
(4) 1 and 2; 4 and 5
Answer
Ans (1)
Solution:
Fluxional molecules that undergo dynamics such that some or all of their atoms interchange
between symmetry equivalent positions. In Fe(CO)5 is the archetypal fluxional molecule due to
rapid interchange of the axial and equatorial CO group.
62.In the following reaction sequence

140
Answer
Ans (1)
Solution:
In the following reaction sequence the product is given below
63.Reaction of Cr(CO)6 with LiC6H5 gives A which reacts with [Me3O][BF4] to give B. The
structures of A and B respectively, are

141
Answer
Ans (1)
Solution:
Reaction of Cr(CO)6 with LiC6H5 gives A which react with [Me3O] [BF4] to give B. The
structure of A and B in the following reaction sequence are given below
64.Heating a sample of [(η5-C5H5)Mo(CO)3]2 results in the formation of [(η5-
C5H5)Mo(CO)2]2with elimination of 2 equivalents of CO. The Mo–Mo bond order in this
reaction changes from
(1) 2 to 3
(2) 1 to 2
(3) 1 to 3
(4) 2 to 4
Answer
Ans (3)
Solution:
65.A plausible intermediate involved in the self metathesis reaction of C6H5–C≡C-C6H4-p-
Me catalyzed by [(tBuO)3W≡C–tBu] is

142
Answer
Ans (2)
Solution:
66.In Ziegler-Natta catalysis the commonly used catalyst system is:
(a) TiCl4 ,Al (C2 H5)3
(b) η5
– Cp2TiCl2 ,Al (OEt)3
(c) VO(acac) , Al2 (CH3) 6
(d) TiCl4 ,BF3
Answer
Ans (a)
67.Oxidation occurs very easily in case of
(a)η5
-( C5H5)2 Fe
(b)η5
-( C5H5)2 Co
(c) η5
-( C5H5)2 Ru
(d)η5
-( C5H5)2 Co+
Answer

143
Ans (b)
68.Complex in which organic ligand is having only s–bond with metal is:
(a) W (CH3)6
(b) η5
– ( C5H5)2Fe Ru
(c) K [PtCl3(C2H4 )]
(d) η6
-( C6H6)2Ru
Answer
Ans (a)
69.The oxidative addition and reductive elimination steps are favoured by
(a) Electron rich metal centres.
(b) Electron deficient metal centers
(c) Electron deficient and electron rich metal centers respectively.
(d) Electron rich and electron deficient metal centers respectively
Answer
Ans (d)
70.Identify the order according to increasing stability of the following organometallic
compounds, TiMe4, Ti(CH2Ph)4, Ti(i-Pr)4and TiEt4.
(Me = methyl, Ph = phenyl, i-Pr = isopropyl, Et = ethyl)
(a) Ti(CH2Ph)4 < Ti(i-Pr)4< TiEt4 <TiMe4
(b) TiEt4 < TiMe4 < Ti(i-Pr)4 < Ti(CH2Ph)4
(c) Ti(i-Pr)4 < TiEt4< TiMe4 < Ti(CH2Ph)4
(d) TiMe4 < TiEt4 <Ti(i-Pr)4 < Ti(CH2Ph)4

144
Answer
Ans (c)
71.Among the metals, Mn, Fe, Co and Ni, the ones those would react in its native form
directly with CO givingmetal carbonyl compounds are:
(a) Co and Mn
(b) Mn and Fe
(c) Fe and Ni
(d) Ni and Co
Answer
Ans (c)
72.In the hydroformylation reaction, the intermediate CH3CH2CH2Co(CO)4:
(a) Forms are acyl intermediate CH3CH2CH2CO Co(CO)3
(b) Forms an adduct with an olefin reactant.
(c) Reacts with H2.
(d) Eliminates propane.
Answer
Ans (a)
73.The correct combination of metal, number of carbonyl ligands and the charge for a
metal carbonyl complex [M(CO)x]z-that satisfies the 18 electron rule is
(a) M = Ti, x = 6, z =1
(b) M = V, x = 6, z =1
(c) M = Co, x = 4, z=2
(d) M = Mo, x = 5, z = 1
Answer

145
Ans (b)
74.The stable cyclopentadienyl complex of beryllium is
(a) [Be(η2
-C5H5)2]
(b) [Be(η2
-C5H5)(η3
-C5H5)]
(c) [Be(η1
-C5H5)(η3
-C5H5)]
(d) [Be(η1
-C5H5)(η5
-C5H5)]
Answer
Ans (d)
75.Reaction of Fe (CO)5with OH– leads to complex A which on oxidation with MnO2gives
B. Compounds A and B respectively are
(a) [HFe(CO 4 ]–
and Fe3(CO)12
(b) [Fe(CO)5(OH)]–
and Fe2(CO)9
(c) [Fe(CO)4 ]2-
and Mn2(CO)10
(d) [HFe(CO)4 ]–
and Fe2O3
Answer
Ans (a)
76.The number of metal-metal bonds in the dimers, [CpFe (CO) (NO)]2 and
[CpMo(CO)3]2respectively, are
(a) two and two
(b) two and three
(c) one and two
(d) zero and one
Answer
Ans (d)

146
77.In the trans-PtCl2L(CO) complex, the CO stretching frequency for L = NH3, pyridine,
NMe3 decreases in theorder.
(a) pyridine>NH3> NMe3
(b) NH3> pyridine> NMe3
(c) NMe3>NH3> pyridine
(d) pyridine>NMe3> NH3
Answer
Ans (a)
78.The catalyst involved in carrying out the metathesis of 1-butene to give ethylene and 3-
hexene is:
(b) Na2 PdCl4
(c) Co2(CO)8 , H2
(d) RhCl(PPh3)3
Answer
Ans (a)

147
79.The greater stability of (( CH3)3C – CH2 –)4 Ti (A) compared to that of (( CH3)3C –
CH2 –)4 Ti (B)is due to
(a) Hyperconjugation present in complex (A)(b) β -hydride elimination is not possible in
complex (A)
(c) Steric protection of titanium from reactive species in complex (A)
(d) The stronger nature of Ti–C bond in complex (A)
Answer
Ans (b)
80.A compound A having the composition 9 8 3 Fe C H O s one signal at 2.5 ppm and
another one around 5.0ppm in its 1H NMR spectrum. The IR spectrum of this compound s
two bands around and 1680 cm–1.The compound follows the 18 electron rule of the
following statements for A, the correct one is/are
(A) It has η5 – Cp group. (B) It has a terminal CO ligand.
(C) It has a CH3ligand (D) It has Fe–H bond.
(a) (A) and (B) only
(b) (C) only
(c) (A) and (C) only
(d) (B) and (D) only.
Answer
Ans (a)
81.The W-W bond order in [W(η5-C5H5)(μ- Cl)(CO)2]2 is …….
(1.) three
(2.) two
(3.) one
(4.) zero

148
Answer
Ans (4)
Solution:
To calculate bond order
(a) Calculate TVE : a
(b) Calculate n x 18 : b [n = number of metal]
82.[(η3
-C3H5)Mn(CO)4] s fluxional behaviour. The 1
H NMR spectrum of this compound
when it is in the non-fluxional state s
(1.) one signal
(2.) two signals in the intensity ratio of 4 : 1
(3.) three signals in the intensity ratio of 2 : 2 : 1
(4.) five signals of equal intensity
Answer
Ans (3)
Solution:

149
Non fluxional behavior can be observed if NMR is carried out at low temperature. For this
complex, 3 signals are observed
→ one for H3
→ two for H1and H4
→ third for H2and H5
In the intensity ratio 2 : 2 : 1
83.Reaction of [Mn2(CO)10] with I2 results in A without loss of CO. Compound A, on
heating to 1200C loses a CO ligand to give B, which does not have a Mn-Mn bond.
Compound B reacts with pyridine to give 2 equivalents of C. Compounds A, B and C from
the following respectively, are
(1.) II, V and IV
(2.) II, III and IV
(3.) V, III and IV
(4.) II, V and III
Answer
Ans (1)
Solution:
(a) Dimanganesedecacarbonyl reacts with halogens to form carbonyl halides.

150
Mn2CO10+ I2→ 2Mn(Co)5I
84.The final product of the reaction of carbonyl metalates [V(CO)6]–and [Co(CO)4]– with
H3PO4, respectively, are
(1.) V(CO)6and HCo(CO)4
(2.) HV(CO)6and Co2(CO)8
(3.) [H2V(CO)6]+ and HCo(CO)4
(4.) V(CO)6and Co2(CO)8
Answer
Ans ()
Solution:
85.For, uranocene, the correct statement(s) is/are:
(1.) Oxidation use of uranium is ‘+4’.
(2.) it has cyclocotatetraenide ligands
(3.) it is a bend sandwich compound
(4.) it has ‘-2’ charge.
Correct answer is
(1.) A and B
(2.) B and C

Inorganic Chemistry - Topic Wise Multiple Choice By Malik Xufyan

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