1. science
2. chemistry

MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
1. An organic compound undergoes first order decomposition. The time taken for its
decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What
is the value of
[t1 / 8 ]
X 10 ? ( log102=0.3)
[t1 / 8 ]
(a) 2
(b) 3
(c) 6
(d) 9
Solution :
For a first order process,
kt = ln
[ A]0
where [A]0 = initial concentration, [A] = concentration of reactant remaining at
[ A]
time t.
kt1/8 = ln
[ A] 0
[ A]0
 ln 8 , kt1/10 = ln
 ln 10 therefore,
[ A]0 / 8
[ A]0 / 10
t1 / 8
t
t
ln 8

 log 8  3 log 2 1 / 8  3 X 0.3  0.9 1 / 8 X 10  0.9 X 10  9
t1 / 10 ln 10
t1 / 10
t1 / 10
2.
For a first order reaction, (A) products the concentration of A charges from 0.1 M to 0.025
M in 40 min. The rate of reaction when the concentration of A is 0.01 M is……
(a) 1.73 X 10-5 M/min
(b) 3.47 X 10-4- M/min
(c) 3.47 X 10-5 M/min (d) 1.73 X 10-4 M/min
Solution :
By first order kinetic, rate constant
K=
k=
2.303
 a 
log 
, a=0.1 M , (a-x)= 0.025 M , t = 40 min
t
 a  x 
2.303
0.1
 dx 
log
 0.0347 min 1 Rate =    k[ A]1 = 0.0347 X 0.01 = 3.47 X 10-4 M min-1
40
0.025M
 dt 
3. The rate of a chemical reaction doubles for every 100 C rise of temperature. if the
temperature is raised by 500 C, the rate of the reaction increases by about….
(a) 10 times
(b) 24 times
(c) 32 times
(d) 64 times
Solution :
For every 100 C rise of temperature, rate is doubled. Thus temperature coefficient of the reaction
= 2.
When temperature is increased by 50, rate becomes
 50 
 
 10 
= 2
= 25 times = 32 times.
4. The time for half life period of a certain reaction A products is 1h. when the initial
concentration of the reactant ‘A’ is 2.0 mol L-1. How much time does it takes for its
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
concentration to come from 0.50 to 0.25 mol L-1 , if it is a zero order reaction?
(a) 4 h
(b) 0.5 h
(c) 0.25 h
(d) 1 h
Solution :
For a zero order reaction,
k0 =
[ A] 0
2t 1 / 2
Since,
[A]0 = 2 M, t1/2 = 1 h
So,
k0 = 1
and
k0 =
or
X
t
0.50  0.25
t=
 0.25 h
1
5. Consider the reaction
Cl2(aq) + H2S(aq)  S(s) + 2H+(aq) + 2Cl-(aq) the rate equation for this reaction is rate = K[Cl2][H2S]
which of these mechanism is/are consists with this rate reaction?
(a) Cl2+ H2S H++ Cl-+ Cl+ + HS- (Slow)
Cl+ + HS-  H+ + Cl- + S(Fast)
(b) H2S  H+ + HS- (Fast equilibrium)
Cl2 + HS-  2Cl- + H+ + S(Slow)
(a) (B) only
(c) Neither (A ) nor (B)
(b) Both (A) and ( B) only
(d) (A) only
Solution :
Slowest step is the rate determining step. Thus, in case (A) rate law is given as rate
= k[Cl2] [H2S]
While for the reaction given in case (B), rate law is given as rate = k [H2S] [Cl2] [H+]-1.
Hence, only mechanism (A) is consistent with the given rate law.
6. The half life period of a first order chemical reaction is 6.93 min. the time required for the
completion of 99% of the chemical reaction will be (log 2 – 0.301)
(a) 230.3 min
(b) 23.03 min
(c) 46.06 min
(d) 460.6 min
Solution :
Half-life = 6.93 min k1 =
0.693
 0.1
6.93
We know, k1 for per cent completion.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
2.303
 100 
log 

t
 100 x 
k1 =
2.303
100
 log
t
1
2.303
0.1 =
 log 10 2
t
2.303  2
t=
 46.06 min .
0.1
0.1 =
7. For reaction ½ A 2B , rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’
by the expression
(a) 
d [ A] d [ B]

dt
dt
(b) 
d [ A]
d [ B]
4
dt
dt
(c) 
d [ A] 1 d [ B]
d [ A] 1 d [ B]
(d) 


dt
2 dt
dt
4 dt
Solution :
2d [ A] 1 d [ B]

dt
2 dt
d [ A] 1 d [ B]


dt
4 dt
Rate of reaction = 

8. Under the same reaction conditions, initial concentration of1.386 mol dm-3 of a substance
becomes half in 40 S and 20 S through first order and zero order kinetics respectively.
Ration(
k1
) of the rate constant for first order (K1) and zero (K0) of the reaction is….
k0
(a) 0.5 mol-1dm3
(b) 1.0 mol dm-3
(c) 1.5 mol dm-3
(d) 2.0 mol-1dm3
Solution :
First order kinetics,
k1 =
0.693 0.693 1

s
t1 / 2
40
Zero order kinetics,
k1 =
Hence,
C0
1.386

2t1 / 2 2  20
k1 0.693

 0 .5
k 0 1.386
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
9. Consider the reaction aG + bH  products when the concentration of both the reactants G
and H is doubled . the rate increases by eight times. However, when concentration of G is
doubled keeping the concentration of H Fixed, the rate is doubled. The overall order of the
reaction is…..
(a) 2
(b) 1
(c) 2
(d)3
Solution :
aG + bH  product
Suppose order of reaction = n
When concentration of both G and H is doubled, the rate increased by eight times.
Rate = k (reactants)n
(8) = k(2)n
(2)3 = k(2)n
n=3
When concentration of G is doubled keeping the concentration of H fixed, the rate is double.
Rate  [G]1
then,
Rate  [G]1 [H]2
10. The energies of activation for forward and reverses reaction for A2 + B2  2AB are 180 kJ
mol -1 and 200 KJ mol-1 respectively. The presence of a catalyst lower the activation energy
of both ( forward and reverse) reaction by 100 KJ mol-1. The enthalpy change of the reaction
(A2 + B2 2AB) in the presence of a catalyst will be (in kJ mol-1)
(a) 300
(b) 120
(c) 280
(d) 20
Solution :
A2 + B2  2AB
Ea (forward) = 180 kJ mol-1
Ea (backward) = 200kJ mol-1
In the presence of catalyst
Ea (forward) = 180 kJ mol-1
Ea(backward) 200 – 100 = 100 kJ mol-1
H = Ea (forward) – Ea (backward)
= 80 – 100
= -20 kJ mol-1
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
11. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr
NO(g) + Br2(g)  NOBr2
NOBr2 (g) + NO(g)  2NOBr(g)
if the second step is the rate determining step, the order of the reaction with respect to
NO(g) is
(a) 1
(b) 0
(c) 3
(d) 2
Solution :
Rate = k[NOBr2] [NO]
But NOBr2 is in equilibrium Keq =
[ NOBr 2 ]
[ NO ][ Br2 ]
[NOBr2] = Kaq [NO] [Br2]
Putting the [NOBr2] in Eq (i)
Rate = k . Keq [NO][Br2] [NO]
Hence Rate = k’ [NO]2 [Br2]
Where,
k’ = k . Keq
12. A reaction involving two different reactants can never be
(a) bimolecular reaction
(b) Second order Reaction
(c) First Order Reaction
(d) unimolecular Reaction
Solution :
There are two different reactants (say A and B)
A + B  product
Thus, it is a bimolecular reaction,
If
dx
 k[ A][B],
dt
It is second order reaction.
13. (A) follows first order reaction (A)  product
concentration of A, changes from 0.1 M to 0.025 M in 40 min. find the rate of reaction of A
when concentration of A is 0.01M.
(a) 3.47 X 10-4 M min-1
(b) 3.47 X 10-5 M min-1 (c) 1.73 X 10-4 M min-1 (d) 1.73 X 10-6 M min-1
Solution :
A  Product (first order reaction)
For first order reaction,
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
Rate constant (k) =
[ A] 0
2.303
log 10
t
[ A]t
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
At, t = 40 min,
2.303
0.1
2.303
log 10

log 10 4
40
0.025
40
2.303
2.303
=
 2 log 10 2 
 2  0.3010  0.0347 min 1
40
40
=
14. In a first order reaction , the concentration of reactant decreases from 800 mol/dm3 to 50
mol/dm3 in 2X104 s. the rate constant of reaction is S-1is
(a) 2 X 104
(b) 3.45 X 10-5
(c) 1.386 X 10-4
(d) 2 X 10-4
Solution :
For 1st order reaction,
k=
=
A
2.303
log 0
t
At
2.303
800
log
4
50
2  10
= 1.386  10-4 mol L-1 min-1
15. For the reaction system
2NO(g) + O2(g) 2NO2(g)
volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction
is of first order with respect to O2 and second order with respect to NO. the rate of reaction
will
(a) diminish to one fourth of its initial value
(b) diminish to one eight of its initial value
(c) increase to eight times of its initial value
(d) increase to four times of its initial value
Solution :
 dx 
2
   k[ NO ] [O2 ]
 dt 
 n   nO 
= k  NO   2 
 V   V 
k
 dx 
2
 '  3 (n NO ) N O 2 
dt
V
 
2
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
2
 dx  k (n NO ) N O 2 
'

 
3
 dt 
V 
 
2
 dx 

 dt 
=8 
16. According to freundlich adsorption isotherm which of the following is correct?
(a)
x
 p0
m
(b)
x
 p1
m
(c)
x
 p1/n
m
(d) all of these are correct for different range of pressure
Solution :
By Freundlich adsorption isotherm
x
 kp1 / n
m
x
when n = 1,
 p1
m
x
When n is large,
k
m
x
Thus,
 p0
m
(in between low and high pressure)
(in lower pressure range)
(independent of pressure)
(in high pressure range when saturation point is reached)
17. Which of the following statements is incorrect regarding physisorptions?
(a) It occurs because of van der waals forces
(b) more easily liquefiable gases are adsorbed readily
(c) under high pressure it results into multi molecular layer on adsorbent surface.
(d) enthalpy of adsorption (Adsorption) is slow and positive
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
Solution :
Adsorption is an exothermic process i.e., energy is released van der Waals force of attraction
(physisorptions). Hence,  H is always negative.
18. In langmuir’s model of adsorption of a gas on a solid surface.
(a) the rate of dissociation of adsorbed molecules from the surface does not depend on the
surface covered.
(b)the adsorption at a single site on the surface may involve multiple molecule at the same
time
(c) the mass of gas striking a given area of surface is proportional to the pressure of a gas.
(d) the mass of gas striking a given area of surface is independent of the pressure of the gas
Solution :
The adsorption of a gas is directly proportional to the pressure of the gas.
19. Rate of physisorption increases with.
(a) decreases in temperature
(b) increase in temperature
(c) decreases in pressure
(d) decrease in surface area
Solution :
The physical adsorption is an exothermic process, so on the basis of Le-Chatelier’s principle
increasing temperature is favourable towards a direction in which heat is absorbed, so the rate
of physisorption increases with decrease in temperature.
20. Which of the following characteristics is not correct for physical adsorption?
(a) Adsorption on solids is reversible
(b) adsorption increases with increase in temperature
(C) adsorption is spontaneous
(d) both enthalpy and entropy of adsorption are negative
Solution :
As temperature increases desorption increases. Adsorbent +
Adsorbate  adsorbed state + E
Adsorption is exothermic process (forward direction) desorption is endothermic process
(backward direction).
According to Le-Chatelier’s principle increase in temperature favours endothermic process.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
21. Gold number of protective colloids A, B,C and D are 0.50, 0.01, 0.10 and 0.005 respectively.
The correct order of their protective power is..
(a) D < A < C < B (b) C < B < D < A
(c) A < C < B < D
(d) B < D < A < C
Solution :
Higher the gold number, less will be the protective power of colloid.
22. Among the following the surfactant that will form micelles in aqueous solution at the lower
molar concentration at ambient condition is
(a) CH3(CH2)15N+(CH3)3Br-
(b)
CH3(CH2)11OSO3-Na+
(c) CH3(CH2)6COO-Na+
(d)
CH3(CH2)11N+(CH3)3Br-
Solution :
Sodium docecyl
Hexadecyltrimethyl ammonium
Sulphate (SDS)
bromide (CTAB)
CMC (mm) > -10
1
23. Lyophilic sols are
(a) irreversible sols
(b) prepared from inorganic compounds
(c) Coagulated by adding electrolyte
(d) self-stabilizing
Solution :
Lyophilic sols are self stabilizing because these sols are reversible and are highly hydrated in
the solution.
24. The disperse phase in colloidal ion (III) hydroxide and colloidal gold is positively and
negatively charged respectively, which of the following statements is not correct?
(a) coagulation in both sols can be brought about by electrophoresis
(b) Mixing the sols has no effect
(c) sodium sulphate solution causes coagulation in both sols
(d) magnesium chloride solution coagulates the gold sol more readily than the iron (III)
hydroxide sols
Solution :
Mixing the soles together can cause coagulation, since, the charges are neutralised.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
25. The volume of a colloidal particle Vc as compared to the volume of a solute particle in a true
solution Vs. could be
(a)
Vc
 10 3
Vs
(b)
Vc
 10 3
Vs
(c)
Vc
 10 23
Vs
(d)
Vc
1
Vs
Solution :
Size of colloidal particles = 1 to 100 nm
(say 10 nm)
4 3
r
3
4
=  (10)3
3
VC =
Size of true solution particles  1 nm
Vs =
4
 (1) 3
3
VC
 10 3 .
VS
Thus,
26. NiCl2 [P(C2H5)2 (C2H5)]2 exhibit temperature dependent magnetic behavior
(paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and
diamagnetic states are respectively.
(a) Tetrahedral and tetrahedral
(b)Square planar and square planar
(c) etrahedral and square planar
(d)square planar and tetrahedral
Solution :
-
In the given complex, NiCl2 {P(C2H5)2 (C2H5)}2, nickel is in + 2 oxidation state.
-
For the given four-coordinated complex to be paramagnetic, it must possess unpaired
electrons in the valence shell. Therefore, the four sp3 hybrid orbitals of metal occupies
four lone pairs.
-
The geometry of paramagnetic complex must be tetrahedral.
-
On the other hand, for complex to be diamagnetic, there should not be any unpaired
electrons in the valence shell.
-
This condition can be fulfilled by pairing electrons of 3d orbitals.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
-
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
The geometry of diamagnetic complex is square planer.
27. Which among the following will be named as dibromidobis – (ethylenediamine) chromium
(III) bromide ?
(a)[Cr(en)3]Br3
(b) [Cr(en)2Br2]Br
(c) [Cr(en)Br4]-
(d)[Cr(en)Br2]Br
Solution :
-
Two Br, two (en) and one Cr are parts of complex.
Charge on the complex is :
2(Br) = -2
2(en) = 0
=+1
1(Cr) = + 3
-
Thus, complex ion is [Cr(en)2Br2]+.
-
Since, anion is bromide thus, complex is [Cr(en)2Br2]Br.
28. The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions
present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:
(a)3
(b)4
(c)5
(d)6
Solution :
m mol of complex = 30  0.01 = 0.3
1 mole of complex [Cr(H2O)5Cl] Cl2 gives only two moles of chloride ion when dissolved in
solution .
[Cr(H2O)5 Cl]Cl2  [Cr(H2O5Cl)2+ + 2Clm mol of Cl- ion produced from its 0.3 m mol = 0.6 Hence, 0.6 m mol of Ag+ would be required
for precipitation.
0.6 m mol of Ag+ = 0.1 M  V (in mL)
V = 6 mL
29. Which of the following facts about the complex [Cr(NH3)6Cl3 is wrong ?
(a) The complex involves d2sp3 hybridization and is octahedral in shape
(b) The complex is paramagnetic
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
(c) The complex is an outer orbital complex
(d) The complex gives white precipitate with silver nitrate solution
Solution :
Cl3 

[Cr ( NH 3 ) 6 ]

coordinate sphere
ionisable
AgNO3
[Cr(NH3)6]3+ + 3Cl- 
 AgCl 
white precipitate
Cr(24) = [Ar] 3d5 4s-1
Cr3+ = [Ar] 3d3 4s0
[Cr(NH3)63+ =
Indicates that lone – pair of NH3 are donated to Cr.
(a) d2 sp3 hybridisation, octahedral. Thus, correct
(b) there are three unpaired electrons, hence paramagnetic. Thus, correct
(c) d2 sp3 inner orbital complex, thus incorrect
(d) due to ionizable Cl- ions, white precipitate with AgNO3, thus correct.
30. The magnetic moment (spin only) of [NiCl4]2- is :
(a)1.82 BM
(b)5.46 BM
(c)2.82 BM
(d)1.41 BM
Solution :
[NiCl4]2-; oxidation number of Ni, x – 4 = -2
x = +2
Ni(28) = [Ar] 3, 4s2d
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
Sp3 hybrid orbitals, tetrahedral Cl- is a weak field ligand and thus unpaired electrons are not
paired.
Lone pairs from 4 Cl- are accommodated in four sp3 hybrid orbitals.
N = unpaired electrons = 2, paramagnetic
Magnetic moment (spin only)
=
n(n  2) BM
=
8
= 2.828 BM
31. Which one of the following has an optical isomer ? [en = ethylenediamine]
(a)[Zn(en)(NH3)2]2+
(b)[Co (en)3]3+
(c)[Co(H2O)4(en)]3+
(d)[Zn(en)2]2+
Solution :
Complex [Co(en)3]3+ has no plane of symmetry and centre of symmetry that’s why it is optically
active.
32. Which of the following pairs represents linkage isomers ?
(a) [Cu(NH3)4 [PtCl4] and {pt(NH3)4] [CuCl4]
(b) [Pd(PPh3)2 (NCS)2] and [Pd(PPh3)2(SCN)2]
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
(c) [Co(NH3)5NO3]SO4 and [Co(NH3)5SO4] NO3
(d) [Pt(Cl2(NH3)4 Br2 and [Pt(Br2(NH3)4]Cl2
Solution :
Linkage isomers are formed due to the presence of ambidentate ligands. [Pd(PPh3)2 (NCS)] and
[Pd(PPh3)2 (SCN)2] are linkage isomers due to SCN. Ambidentate ligand.
33. The coordination number and the oxidation state of the element ‘E’ in the complex
[E(en)2(C2O4)]NO2 are respectively, [where, en is ethylene diamine].
(a)6 and 2
(b)4 and 2
(c)4 and 3
(d)
6 and 3
Solution :
CH2 – NH2 is a bidentate ligand C2O42- is also a bidentate
|
CH2 – NH2
ligand, Hence coordination number = 6
Complex can be ionized as
[E(en)2 (C2O4)]NO2  [E(en)2(C2O4)]+ + NO2Oxidation number = x + 0 + (-2) = 1
 x= 3.
34. In which of the following octahedral complexes of Co(at. No 27), will the magnitude of o be
the highest?
(a)[Co(CN)6]3-
(b)[Co(C2O4)3]3-
(c)[Co(H2O)6]3- (d)[Co(NH3)5]3+
Solution :
CFSE (Crystal field splitting energy) for octahedral complex, 0 depends on the strength of
negative ligand. Spectrochemically, it has been found that the strength of splitting is as follows :
CO > CN- > NO2- > en > NH3 > py >
NCS- > H2O > O2- > OC2- > OH- > F- >
Cl- > SCN- > S2- > Br- > I35. Both [Ni(CO)4] and [Ni(CN)4]2- are diamagnetic. The hybridizations of nickel in these
complexes are respectively,
(a)sp3, sp3
(b)sp3, dsp3
(c)dsp2, sp3
(d)dsp2, dsp2
Solution :
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
36. Which one of the following has a square planar geometry ?
(At. No. Co = 27, Ni = 28, Fe = 26, Pt = 78)
(a)[CoCl4]2-
(b)[FeCl4]2-
(c)[NiCl4]2-
(d)[PtCl4]2-
Solution :
Cl- is a weak field ligand but Cl- causes pairing of electrons with large Pt2+ and consequently
give dsp2 hybridization and square planar geometry.
37. Among the following metal carbonyls, the C – O bond order is lowest in :
(a)[Mn(CO)6]+
(b)[Fe(CO)5]
(c)[Cr(CO)6]
(d)[V(CO)6]2+
Solution :
(a) Mn+ = 3d5, 4s1; in the presence of CO, effective configuration = 3d6 4s0.
Three lone pairs for back bonding with vacant orbital of C in CO.
(b) Fe = 3d6, 4s2; in the presence of CO, effective configuration = 3d8, four lone pairs of back
bonding with CO.
(c) Cr = 3d5, 4s1 effective configuration = 3d6. Three lone pairs for back bonding with CO.
(d) V- = 3d4 , 4s2; effective configuration = 3d6
Three lone pairs for back bonding with CO. Maximum back bonding in Fe(CO)5,
therefore, CO bond order id lowest here.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6
MERIDIAN INSTITUTE
Date : 19/02/2015
Sub : Chemistry
JEE Chemical kinetics/ surface
chemistry / complex compounds
Batch : 12th GSEB
38. Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in the alkaline solutions but
not in acidic solutions. What is the reason for it ?
(a)In acidic solutions hydration protects copper ions
(b)In acidic solutions protons coordinate with ammonia molecules forming NH4+ ions and
NH3 molecules are not available
(c)In alkaline solutions insoluble Cu(OH)2 is precipitated which is soluble in excess of any
alkali
(d)Copper hydroxide is an amphoteric substance
Solution :
The pair of electrons present with nitrogen will be available to be donated as H+ will consume
that one.
2nd F l o o r , R a j c a s t l e , N e a r K a l p a t a r u , E l l o r a P a r k , B a r o d a – 3 9 0 0 0 7 P h : +9 1 – 2 6 5 – 3 0 5 3 1 6 6