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1

All India Test Series/IIT/FLT 1/ Paper : 1 - SOLUTIONS 1

THE ACE TUTORIALSFLT - 1 (PAPER - 1) : SOLUTIONS

(1) (B). CD = h,h h

tan , tan2x L x

θ = θ =−

x

L

D L–x

h

C

A B

θ 2θ

Now, x tan θ = (L – x) tan 2θx (tanθ + tan 2θ) = L tan 2θ

tan2x L

tan tan2θ = θ + θ ;

tan22

2x Ltan tan2

22

θ θ θ= θ θ

θ + θ θ θ

0

2Ll imx

3θ→=

(2) (B). For 0 < a ≤ 1 the line always cuts y = ax.

O x

y

y=a , a (0, 1]x ?

For a > 1, say a = e

Consider f (x) = ex – x

Ox

y

a = 1

f ' (x) = ex – 1f ' (x) > 0 for x > 0 and f ' (x) < 0 for x < 0∴ f (x) is increasing ( ↑ ) for x > 0and decreasing ( ↓ ) for x < 0

x=0x

II

y = ex always lies above y = x i.e. ex – x ≥ 1 for a > 1Hence never cuts = a = (0, 1] ⇒ B

(3) (A). SS' = 2ae, where a and e are length of semi-major axisand eccentricity respectively.

2 2(9 3) (12 4) 2ae− + − = ∴ ae = 5

∴ Centre is mid point of SS'∴ Center ≡ (6, 8)Let the equation of auxiliary circle be (x – 6)2 + (y – 8)2 = a2

We know that the foot of the perpendicular from the focuson any tangent lies on the auxiliary circle

∴ (1, – 4) lies on auxiliary circle.i.e. (1 – 6)2 + (– 4 – 8)2 = a2 ⇒ a = 13Q ae = 5 ⇒ e = 5/13

Q 1 2 3 4 5 6 7 8 9 10 11

A B B A D A C BD AC BC CD A

Q 12 13 14 15 16 17 18 19 20 21 22

A C B D B A C A D C B B

Q 23 24 25 26 27 28 29 30 31 32 33

A A C C D C A A BCD ABC ABD AC

Q 34 35 36 37 38 39 40 41 42 43 44

A A C C C D A A C A A C

Q 45 46 47 48 49 50 51 52 53 54 55

A A B B B B B D B AD AC ACD

Q 56 57 58 59 60 61 62 63 64 65 66

A ABD C D B C C B D A C D

Q 67 68 69

A B C B

STANDARD ANSWER KEY

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2

2 All India Test Series/IIT/FLT 1/ Paper : 1 - SOLUTIONS

(4) (D). A is non singular ∴ det A ≠ 0Given AB – BA = A, hence AB = A + BA = A(I + B)det. A · det. B = det. A · det. (I + B)det. B = det. (I + B) ....(1) (as A is non singular)again AB – A = BA

A(B – I) = BA(det. A) · det.(B – I) = det. B · det. A⇒ det. (B – I) = det. (B) ....(2)From (1) and (2)det. (B – I) = det. (B + I)

(5) (A). Solving line and circlem2x2 + (δ – lx)2 = m2c2

(m2 + l2) x2 – 2lδx + (δ2 – m2c2) = 0x

1

x2

(l2 + m2 = 1)

Given2 2 2

1 2

1 2

2 x x 2 ( m c )H

x x 2δ −

= =+ δl

2 2 2m cH

δ −=

δl ......... (1)

A

B

(x ,y )11

( x , y )22

δx+my =

Similarly2 2 2c

mKδ −

=δ

l......... (2)

(1) + (2) ⇒2 2 2 2 22 c ( m ) c

H mK 2δ − +

+ = = δ −δ δl

l

where (l2 + m2 = 1)(6) (C).

Let z = a + i b ⇒ z a ib= −

Hence, we have 2008z z=

∴ 2008| z | | z | | z |= =

2007| z | [ | z | 1] 0− =| z | = 0 or | z | = 1, if | z | = 0⇒ z = 0 ⇒ (a, b) = (0, 0)

if | z | = 1 ; 2009 2z zz | z | 1= = =⇒ 2009 values of z ⇒Total number of ordered pairs = 2010

(7) (BD). As a > 2 hence a2 > 2a > a > 2Now (x – a)(x – 2a)(x – a2) < 0 ⇒ the solution set is asshown

0 a 2a a2

Between (0, a) there are (a – 1) positive integersBetween (2a, a2) there are (a2 – 2a – 1) positive integers∴ a2 – 2a – 1 + a – 1 = 18 ⇒ a2 – a – 20 = 0

(a – 5)(a + 4) = 0∴ a = 5 and a = –4 but a > 2∴ a = 5⇒ (B) and (D)

(8) (AC). Clearly a.c 0 & b.c 0= =rr r r . Also a.b 0 (A)= ⇒

rr

Again,| a | | b | | c | | a | | c |

| a | | c | & | b | 1| c | | a || b | | c | | a |

= ⇒ = ⇒ = ==

rr r r r rr rr rr r r

⇒ 2 2a b.c | a | | b | | c | | a | | c |× = = =r rr r r r r r

(Student will assume ˆ ˆ ˆa i , b j,c k= = =rr r but in this case all

the four will be correct which will be wrong)

(9) (BC).

n1/2 1/41

x x2

− + ;n r

n r/42r 1 r r

1T C x . .x

2

−−

+ =

Coefficient of the first 3 terms aren n n

0 1 2 2

1 1C , C , C .

2 2

n n n0 2 1

1 1C C . 2. C .4 2+ =

n(n 1)1 n

8−

+ = ;n(n 1)

(n 1) n 8 (a s n 1)8−

= − ⇒ = ≠

3r8 r 48 r/4 8 42

r 1 r rr r

1 1T C x . .x C . .x

2 2

− − −+ = =

Number of terms with integer power of x occur when r = 0,4, 8⇒ 3 terms. Hence B and C are correct.

(10) (CD). Given : (f '(x))2 + (g (x))2 = 1

x

0f(x) g( t )d t sinx(cosx sinx)+ = −∫

Differentiating both sidesf ' (x) + g (x) = cos 2x – sin 2x ....... (1)

Squaring (1)(f '(x))2 + (g (x))2 + 2 f '(x). g (x) = 1 – sin 4x1 + 2 f '(x) . g (x) = 1 – sin 4x

∴ 2 f ' (x) g (x) = – sin 4x

Now, substituting g (x) =sin4x2f (x )

−′ in eq. (1)

sin4xf (x) cos2x sin2x

2f ( x)− = −′ ′Put f ' (x) = t

2t2 – 2 (cos 2x – sin 2x) t – sin 4x = 0

⇒2(cos2x sin2x) 4 (1 sin4x) 8sin4x

t4

− ± − +=

∴ 4t = 2 (cos 2x – sin 2x) ± 4(1 sin4x) 8sin4x− +

⇒ 2t = (cos 2x – sin 2x) ± 1 sin4x+taking (+) ve sign, 2t = cos 2x – sin 2x + cos 2x + sin 2x⇒ t = cos 2xtaking (–) ve sign, t = – sin 2xHence f ' (x) = cos 2x or f ' (x) = – sin 2x





3


Integrating, 11

f(x) sin2x C2

= + or 2cos2x

f(x) C2

= +

f (0) = 0 ⇒ C1 = 0 and C2 = –1/2

∴ f (x) =

1

sin2x2 or

cos2x 1

f(x) 2

−

=If f ' (x) = cos 2x then g (x) = – sin 2xIf f ' (x) = – sin 2x then g (x) = cos 2x

i.e.1

f(x) sin2x2

= and g (x) = – sin 2x ⇒ (C)

cos2x 1

f(x)2−

= and g (x) = cos 2x ⇒ (D)

(11) (A). y = | ln x | not differentiable at x = 1

y = | cos | x | | is not differentiable at x =3

,2 2

π π

y = cos–1 (sgn x) = cos–1 (1) = 0 differentiable x∈ (0, 2π)

(12) (C). Let1

0

f (t)dt k=∫ , so f (x) = kx + 1, now1

0

(kt 1) dt k+ =∫

⇒k

1 k,2+ = so k = 2

∴ f (x) = 2x + 1,

Also3

0

f(x)dx 12=∫ ⇒ Option (C) is correct

(13) (B). Let x1, x2, x3 ∈ R be the roots of f (x) = 0∴ f (x) = (x – x1) (x – x2) (x – x3)f (i) = (i – x1) (i – x2) (i – x3)

| f (i) | = | x1 – i | | x2 – i | | x3 – i | = 1

∴ 2 2 21 2 3x 1 x 1 x 1 1+ + + =

This is possible only if x1 = x2 = x3 = 0⇒ f (x) = x3 ⇒ a = 0 = b = c ⇒ a + b + c = 0⇒/ all roots are zero.

(14) (D). L1 and L2 are obviously non-parallelConsider the determinant

2 4 1

D 2 4 31 3 2

−

= − = 2 (8 + 9) + 4 (4 + 3) + 1 (6 – 4)

= 34 + 28 + 2 = 64 ⇒ D ≠ 0⇒ skew linesHence Statement-1 is false.

(15) (B)., (16) (A)., (17) (C).

Urn-I5R

1B; Urn-II

2R

4BA : first two draws resulted in a blue ball.

B1 : urn-I is used 11

P(B )2

=

B2 : urn-II is used 21P(B )2

=

11 1 1

P ( A / B ) .6 6 36

= = , 24 4 16 4

P ( A/ B ) .6 6 36 9

= = =

1

E1

1 1.

12 36P(B / A) 1 1 1 16 17. .2 36 2 36

= =+

14243

2

E2

1 16. 162 36P(B /A)

1 16 1 1 17. .2 36 2 36

= =+

14243 E

E1

E2

E : third ball drawn is redP (E) = P (E ∩ E1) + P (E ∩ E2)

=1 5 16 2 5 32 37

. .

17 6 17 6 102 102 102

+ = + =

(18) (A). f (0) = 2,

{{

x xx x

I0 0 II

f(x) (e e )cosx 2x x f (t)dt t f (t)dt− = + − − −′ ′ ∫ ∫

x xf(x) (e e )cosx 2x−= + −

xx0

0

xf(x) xf(0) t.f(t) f (t)dt − − − −

∫

x xf(x) (e e )cosx 2x x f ( x ) 2x−= + − − +x

0

x f ( x ) f (t)dt

+ −

∫

xx x

0

f(x) (e e )cosx f(t)dt−= + − ∫ ........... (1)

Differentiating eq. (1)f ' (x) + f (x) = cos x (e x – e–x) – (ex + e–x) sin x ........ (2)

Hence

x xdyy e (cosx sinx) e (cosx sinx)

dx

−

+ = − − +

(19) (D). f ' (0) + f (0) = –2 + 2 = 0(20) (C). I.F. of DE (1) is ex.

x 2xy.e e (cosx sinx)dx (cosx sinx)dx= − − +∫ ∫ x 2xy.e e (cosx sinx)dx (sinx cosx) C= − − − +∫ ∫

Let 2x 2xI e (cosx sinx)dx e (Acosx Bsinx)= − = +∫ Solving A = 3/5 and B = –1/5 and C = 2/5

∴ x x x3 1 2y e cosx sinx (sinx cosx)e e5 5 5

− − = − − − +





4


(21) (B). Equation of normal in terms of slope ism3x + (4 – y) m2 + 2 = 0

point P(h, k) satisfies this equation

∴ m3

h + (4 – k) m2

+ 2 = 0

m1

m2

m3....... (1)

algebraic sum of slopes is m1 + m2 + m3 =k 4

h−

(22) (B).

If two normals are perpendicular then

m1m2 = – 1 and m1m2m3 =2

h−

substituting m3 =2

h= m in (1) we get

2 28 4(4 k) 2 0h h

−+ + =

4 + 2 (4 – k) + h 2 = 0x2 = 2 (y – 12)

∴ latus rectum = 2(23) (A).

If the slopes are complementary then m1m2 = 1

∴ m3 = –2

h

using m3

= –2

hin (1) we get

2 2

8 4(4 k) 2 0

h h− + − + =

⇒ – 4 + 2 (4 – k) + h2 = 0⇒ h2 + 4 – 2k = 0∴ x2 = 2 (y – 2)x2 = 2Y

where Y = y – 2⇒ a = 1/2∴ directrix is Y = –1/2 or y – 2 = –1/2

y= –a

∴ 2y – 3 = 0(24) (C).

1 04S

P P2R

− = ......... (1)

2 14S

P PR

− = ......... (2)

P0

P1

P2

Add (1) and (2)

2 0 6SP P R− =

(25) (C). Least count =1(main scale)

N

0.1 mm =1(main scale)

20; 1 Main scale division = 2mm.

(26) (D).dN N

50dt 0.5

= − ;

N t

0 0

dNdt

50 2N=

−∫ ∫ N = (100 (1 – e–t/2)) ; N = 25 ; t = 2 ln (4/3)

(27) (C). For loop B, µ 0 (2i – i) = B.d∫ ur uur

l

For loop C, µ 0 (i – 2i) = B.d∫ ur uur

l

For loop A, µ 0 (3i – 3i) = B.d∫ ur uur

l

For loop D, µ 0 (0 – i) = B.d∫ ur uur

l

So, B > A > C = D(28) (A). At resonance XL = XC and Z = Zmin = R

XL =ωL and C1

XC=

ωIf ‘f’ is decreased then ωwill decrease and hence XC willincrease therefore at f < f r, circuit behaves as capacitative.VL and VC always differ in phase by 180° at any frequency

(29) (A).

L m

1 2 1

F F F= +

/ / / / / / / / / / / / / / /

/

/ / / / / / /

//

/ / / /

/

/ /

/

/

/ /

Fm = ∞ mR

F2

=

L

1 n 1

F R−

=L

1 n 1

F R−

=

1R

F2 ( n 1)

=− 2

RF

2n=

Ratio = 1

2

F n

F n 1=

−

(30) (BCD). Intensity increases no. of photon coming on platestherefore it wouldn’t change stopping potential, work func-

tion and maximum K.E.(31) (ABC). When return to same point displacement is zero

therefore average velocity is also zero.

Average speed =2Total distance 2 (u / 2g) u

Totaltime 2 ( u / g ) 2= =





5


(32) (ABD). Equivalent circuit

VK1

K2

C

C

C

C

When K1 is closed energy stored = 21CV

2

When K2 is also closed equivalent capacitance =5C

3

Energy stored

22 2

eq1 1 5 C 5CV

E C V V2 2 3 6= = =

Change =2 2

2 25CV CV 2 1CV CV

6 2 6 3− = =

(33) (AC).

(1) (2)

Deviation, δ= i + e – A

dispersion, θ = (δV – δR) = (nV – nR) A

δ1 = –δ2δ1 + δ2 = 0

∴ Final ray will be parallel to initial ray.

But θ1 ≠θ 2 ⇒ θ1+ θ 2 ≠ 0

(34) (A). Statement 1 : I ∝A2 and I ∝ f 2

2 2 2 21 1 1

2 2 2

I A f 2 1

1I A f 1 2

= = =

Statement 1 : Velocity amplitude : A Iω∝

therefore if velocity amplitude is same then intensity willalso be same.

(35) (C). Increase in R deceases i, magnetic flux through ring,which is inside paper, decreases and by Lenz’s law, currentflows to produce flux in inward direction i.e. clockwise.

(36) (C). Electric field also depends upon charge distribution.

(37) (C). In free expansion pressure outside is zero. So no workis done by ideal gas but in real gas work may be doneagainst internal force existing between molecules.

(38) (D).V1 + V2 = V0 (remains same st. line)

2 2 21 2 1 0

1 1 1MV MV C MV

2 2 2+ + =

C1

is loss in kinetic energy

On solving, 2 2 20 1 2 2V V V C= + +

2 2 21 2 0 2V V V C+ = − ; ( )

22 2 2

1 2 0 2V V V C+ = −

20 2 0R V C ( V )= − <

Therefore it yields circle of smaller radius.

(39) (A). In inelastic collision radius will be less but straight lineof momentum conservation will be same.

(40) (A). Velocity of object I will always be less than object II,therefore from above two solutions (0.5, 1.5) is correct.

V1

V2

2

2 (0.5, 1.5)

(1.5, 0.5)

A

B

(41) (C). Just after closing, capacitor behaves as short circuitand all current flows through it hence ammeter reads zero.

(42) (A). After long time capacitor behaves as open circuit andno current flows through it.

Therefore i = 0

1 2

V 302mA

R R 10 5= =

+ +

(43) (A). Just after reopening, potential difference across R2remains same initially as charge on capacitor does notchange initially, hence current remains same.

(44) (C).

a

E

mgmg

22m

mg mg m2 3

τ = + = +

l ll l α

9g

8α =

l

9g 9g

a 8 8= × =ll







7


(52) (B).

To convert FeO into Fe2O3 otherwise FeO can combinewith SiO2 present as gangue in the haematite ore.

FeO + SiO2 → FeSiO3gangue slag

(53) (AD).4

m

m

200 100.1

0.2

−

∞∧ ×α = = =∧

Acid is 10% ionised so enthalpy of ionisation of

XOH =2.7

3kJ/mol0.9

=

XOH (aq) X+(aq) + OH– (aq)c – x x x

2

b(0.1C)

K 0.0110.9C

= =

1413

a10

K 9.1 100.011

−−= = ×

Due to cationic hydrolysis resultant solution is acidic(54) (AC). XY (aq)→ X+ (aq) + Y–(aq)

X+ (aq) + H2O (l) XOH (aq) + H+(aq)At equilibrium :

C (1 –α) – Cα Cα

[H+] = Cα =w

b

KC

K

× ;14

5 25

10 80 1(10 )

M 24 10

−−

−= × ×

×

M = 100 gm/mol; 3hK% 100 2.5 10

C−α = × ⇒ ×

a for XY : 3a 2( r r )−= +

1.732 × a = 2 (1.6 + 1.864)a = 4Å = 4 × 10–8 cm.

Effective formula unit of XY = 1

3 8 3 23A

Z.M 1 100 1 1000d 2.6gm/cc

384a N (4 10 ) 6 10−× ×

= = = =× × × ×

(55) (ACD). (A) At T1 and T2 the value of ∆Gf for MgO be-comes more and more positive due to decrease in entropyin the formation of MgO.

2 Mg (l) + O2 (g) T1 → MgO(s)

2 Mg (g) + O2 (g) T2 → 2MgO(s)

(C) Beyond T3, ∆Gf of CO is more negative than that of MgO

(D) MgO (s) + C(s)T3

gaseousproduct

RapidcoolingMg(s)

Mg(g) CO

↓

+ ↑1442443

(56) (ABD).CH3 – CH2 – CH3It has 6 primary H = 44% given product and 2 secondaryH = 56% given product.

So44

7.33

6

= ;28.0 3.82 reactivitywith 2 H

7.33 1 reactivitywith1 H

°= =

°In 3 3

3

CH CH CH|

CH

− −

It has 9 primary H = 66 % given productIt has 1 tertiary H= 33% given product

so66

7.339= ;

33 4.5 reactivitywith3 H

7.33 1 reactivitywith1 H

°= =

°

3333

1=

∴Reactivity ratio for monochlorination with1°H : 2°H : 3°H = 1 : 3.82 : 4.5

(A) 3 2 3

2 3

H|

CH CH CH CH CH|

CH CH

− − − −

−

→ 3 2 3

2 3

Cl|

CH CH CH CH CH|

CH CH

− − − −

−

(Major product)

1° H = 9 × 1 = 92° H = 6 × 3.82 = 22.92 (major) % of 2-chloro-3-ethyl pen-tane

3° H = 1 × 4.5 = 4.5 (minor) =22.92

62.93%36.42

=

Total ⇒ 36.42 ≈ 63% ∴ X = 63

(B) 3 2 2 3

2 3

Cl|

CH CH C CH CH (minor)|

CH CH

− − − −

−

% of 3-chloro-3-ethyl pentane

=4.5

100 12.36% 1236.42

× = ≈ ∴ Y = 12

(C) (i) O3

(ii) Zn, H O2H – C – C

O O

CH3

2 +

CHO

CHO

(i) O3

(ii) Zn, HO2CH3 – C – C

O O

CH3

+

CHO

CHO

2





8


2 +CHO

CHO

CH – C3 – C H

O O

(i) O3

(ii) Zn, H O2

2 + CHO

CHO

CH – C3 – C H

O O

(i) O3

(ii) Zn, H O2

+CHO

CHO

C H2 5 – C H– C

O O

(i) O3

(ii) Zn, H O2

C H2 5

2

So we obtain four types of product (given in bracket)∴ Z = 4(D) X + Y + Z = 63 + 12 + 4 = 79

(55) (A) p, q, r, (B) – p, q, s, (C) – r, (D) – p, q, r

For 2s : Number of radial node = 1, number of maxima inradial probability distribution curve = 2, No. of angularnode = 0For 4d : Number of radial node = 1, number of maxima inradial probability distribution curve = 2, No. of angularnode = 0

(56) (A) r, s; (B) – r ; (C) p, r, s ; (D) p, q

(A) 2 4CuS 2O CuSO∆+ →

4 2CuS CuSO 2Cu 2SO∆+ → + ↑

2 4PbS 2O PbSO∆+ →

4 2PbSO PbS 2Pb 2SO∆+ → + ↑(B) During smelting step of extraction of Cu copper (I)sulphide is not reduced, while other metal oxides can bereduced to respective metals during carbon reduction i.e.smelting process.(C) Main ores of Ag, Pb and Cu occur as sulphides hence

they are concentrated by froth floatation process.

(D) (p) 2 2 21

2Ag 4CN O H O 2[Ag(CN) ] 2OH2

(Impure) [Soluble]

− − −+ + + +

22 42[Ag(CN) ] Zn [Zn(CN) ] 2Ag− −+ → + ↓

(q) 2 2 21

2Au 4CN O H O 2[Au(CN) ] 2OH2

(Impure) [Soluble]

− − −+ + + +

22 42[Au(CN) ] Zn [Zn(CN) ] 2Au− −+ → + ↓

(57) (C). Victor Meyer test is a identification test.(58) (D). Nitrobenzene is a deactivated ring & deactivated rings

do not give F.C. reaction.

(59) (B). 4S + 6NaOH → 2Na2S + Na2S2O3 + 3H2Oi.e. Initially H2S and H2S2O3 are formed which react withNaOH and final products are Na2S and Na2S2O3.

(60) (C). (CO)5 Mn – Mn (CO)5Total number of electrons in both Mn atoms

= 2 [Atomic number of Mn + 5 × number of electrons con-tributed by CO] = 2 [25 + 5 × 2] = 70It can not act as an oxidising agent.

(61) (C).

(NH ) C r O(A)

Orange solid

74 2 2 Cr O (s)(B)

Green solid

2 3 + N + H O (g)(C) (D)

2 2

Mg

CrCl(F)

Green sol.

3

Conc.HCl

Mg N(I)

3 2

CoCl.6H OPink

(E)

2 2

NH(J)

3

H O /OH2 2

CrO42–

(G)Yellow sol.

H O /H2 2+Et O2

CrO.OEt(H)

Blue

5 2

H O2

NiClsol.

2

[Ni(NH ) ]Deep blue

(K)

3 62+

CoCl2

2

3 2 2 4Mn NH 2H O Mn(OH) 2NH+

+ + → ↓ +

H O2 2

2 2MnO.(OH) orMnOBrownishblack

↓

(62) (B). Ni2+ in [Ni(NH3)6]2+

4p

3d

4s 4d

Hyb. sp d23

µ = 2.8 B.M.eff

(63) (D).

CrO42–

+ 2H O + 2H2 2+

Cr

OO

O

O

O

+ H O3 2

deep blue(unstable in the absence

of organic solvent)→ CrO5 is stabilized in presence of ether, Amyl alcohol,amylacetate.





9


(64) (A), (65) (C), (66) (D).

Et

NBS

CH

BrCH3

(A)

alc. KOH

CH = CH2

Br

(B)

Br2

CCl4

CH – CH2Br

(C)

alc. KOH

NaNH2

CH CH

(D)

HgSO4

dil. HS O2 4

COCH3

(E)

HOClexcess

(F)

C – CH

OHOH Cl

ClC – CHCl2

O

CO – CHO

KOH

CH– COOK

OH

Intramolecular

Cannizaro(G)

(67) (B).

(I) eg dx – y2 2 dz

2

dxy

t2g

dyz dzx

0

In d1, d2, d3 system for octahedral ligand field, electronic distribution of any metal cation does not change, whatever may beligands.

(II) eg dx – y2 2 dz2

dxyt2g

dyz dzx

0

5d6

Due to increased Zeff value for 3rd transition series metal cation, ∆0 value gets increased, and P < ∆0, therefore electronicdistribution for d6 will not change, whatever may be the ligands.

(III) egdx – y

2 2 dz2

dxy

t2g

dyz dzx

0

Electronic distribution for d8 system for any octahedral ligand field does not change, hence, hybridization and magnetic

moment do not change.(IV)

4p4s 4d

egdx – y

2 2 dz2

dxy

t2g

dyz dzx

0

4p4s 4d

egdx – y

2 2 dz2

dxy

t2g

dyz dzx

0

P >0

d7

Hybridisation : sp d

µ : 3.89 B.M.eff

3 2

P <0

d7

Hybridisation : d

µ : B.M.eff

2sp

1.732

3

Hence for d7 system hybridisation and µ eff value changes according to strong or weak field ligand.





10


(68) (C).

(I) egdx – y

2 2 dz2

dxy

t2g

dyz dzx

0

d5

For given P > ∆0Hybridisation : sp3d2

µ eff : 5.9 B.M.Outer orbital complex/high spin complex.

(II)

4p4s 4d

egdx – y2 2 dz2

dxy

t2g

dyz dzx

0

d7

For given P < ∆0Hybridisation : d2sp3

µ eff : 1.732 B.M.Inner orbital complex/low spin complex/spin paired com-plex

(III) eg dx – y2 2 dz2

dxy

t2g

dyz dzx

0

d3

For given P < ∆0Hybridisation : d2sp3

Inner orbital complexµ eff : 3.89 B.M.

(IV) egdx – y

2 2 dz2

dxy

t2g

dyz dzx

0

d6

For given P > ∆0Hybridisation : sp3d2Outer orbital/high spin/spin free complex.

(69) (B).

(A) Due to higher oxidation state of Cobalt in

III3

2 6[Co(H O) ] , P < ∆0, also H2O lies towards stronger

ligand side in spectrochemical series.(B) Due to high value of Zeff in Ir+3, which belongs to IIIrd

transition series (P < ∆0)(C) CO is a strong field ligand hence P < ∆0(D) Due to high value of Zeff of Pd2+, which belongs to

2nd

transition series, [P < ∆0]

