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Coordination Compounds

JEE Chemistry · 225 questions · Page 7 of 23 · Click an option or "Show Solution" to reveal answer

Q61
Which of the following pairs represents linkage isomers ?
A [Pd (P Ph3)2 (NCS)2] and [Pd (P Ph3)2 (SCN)2]
B [Co(NH3)5 NO3]SO4 and [Co(NH3)5 SO4]NO3
C [Pt Cl2(NH3)4] Br2 and [Pt Br2(NH3)4] Cl2
D [Cu (NH3)4] [Pt Cl4] and [Pt (NH3)4] [CuCl4]
Correct Answer
Option A
Solution

The

SCN−SC{N^ - }

ion can coordinate through

SS

or

NN

atom giving rise to linkage isomerism

M←SCN   M \leftarrow SCN\,\,\,

thiocyanto

M←NCS   M \leftarrow NCS\,\,\,

isothiocyanato.

Q62
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+
Correct Answer
Option B
Solution

For a substance to be optical isomers following conditions should be fulfilled

(a)   (a)\,\,\,

A coordination compound which can rotate the plane of polarized light is said to be optically active.

(b)   (b)\,\,\,

When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarized light are said to exhibit optical isomerism and the molecules are optical isomers.

The optical isomers are pair of molecules which are non-super-imposable mirror images of each other.

(c)   (c)\,\,\,

This is due to the absence of elements of symmetry in the complex.

(d)   (d)\,\,\,

Optical isomerism is expected in tetrahedral complexes of the type

Mabcd.Mabcd.

Based on this only option

(2)(2)

shows optical isomerism

[Co(en)3]3+{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}

Complexes of

Zn++  Z{n^{ + + }}\,\,

cannot show optical isomerism as they are tetrahedral complexes with plane of symmetry.

[Co(H2O)4(en)]3+  {\left[ {Co{{\left( {{H_2}O} \right)}_4}\left( {en} \right)} \right]^{3 + }}\,\,

have two planes of symmetry hence it is also optically inactive. Hence the formula of the complex is

[Co(NH3)6]Cl3\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}
Q63
A solution containing 2.675g of CoCl3. 6NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is : (At. Mass of Ag = 108 u)
A [Co(NH3)6]Cl3
B [CoCl2(NH3)4]Cl
C [CoCl3(NH3)3]
D [CoCl(NH3)5]Cl2
Correct Answer
Option A
Solution
CoCl3.6NH32.675g⟶ xCl−\mathop {CoC{l_3}.6N{H_3}}\limits_{2.675g} \overset{\,}\longrightarrow xC{l^ - }
xCl−+AgNO3⟶ x AgCl↓4.78gxC{l^ - } + AgN{O_3}\overset{\,}\longrightarrow \mathop {x\,AgCl \downarrow }\limits_{4.78g}

Number of moles of the complex

=2.675267.5=0.01= {{2.675} \over {267.5}} = 0.01

moles Number of moles of

AgClAgCl

obtained

=4.78143.5=0.03= {{4.78} \over {143.5}} = 0.03

moles ∴\therefore

   \,\,\,

Number of moles of

AgClAgCl

obtained

=3×  = 3 \times \,\,

No. of moles of complex ∴\therefore

   \,\,\,
n=0.030.01=3n = {{0.03} \over {0.01}} = 3
Q64
The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is :
A L3 < L2 < L4 < L1
B L1 < L2 < L4 < L3
C L4 < L3 < L2 < L1
D L1 < L3 < L2 < L4
Correct Answer
Option D
Solution

For a given metal ion, weak filed ligands create a complex with smaller

Δ\Delta

, which will absorbs light of longer λ\lambda and thus lower frequency.

Conservely, stronger filled ligands create a larger

Δ,\Delta ,

absorb light of shorter λ\lambda and thus higher

v.i.e.v.i.e.

higher energy.

Redλ=650 nm  <Yellow570  nm  <Green490 nm<Blue450 nm\mathop {{\mathop{\rm Re}\nolimits} d}\limits_{\lambda = 650\,nm} \,\,\mathop { < Yellow}\limits_{570\,\,nm} \,\,\mathop { < Green}\limits_{490\,nm} < \mathop {Blue}\limits_{450\,nm}

So order of ligand strength is

L1<L3<L2<L4{L_1} < {L_3} < {L_2} < {L_4}
Q65
The equation which is balanced and represents the correct product(s) is :
A [Mg (H2O)6 ]2+ + (EDTA)4- ⟶excess NaOH\overset{{excess\,NaOH}}\longrightarrow [Mg (EDTA) ]2+ + 6H2O
B CuSO4 + KCN →\to K2 [Cu (CN)4] + K2SO4
C Li2O + 2KCl →\to 2LiCl + K2O
D [CoCl(NH3)5]+ + 5H+ →\to Co2+ + 5NH4+5NH_4^+ + Cl-
Correct Answer
Option D
Solution

(A) [Mg (H2O)6 ]2+ + (EDTA)4-

⟶excess NaOH\overset{{excess\,NaOH}}\longrightarrow

[Mg (EDTA) ]2+ + 6H2O The above equation is incorrect because the product formed would be [Mg (EDTA)]2− .

(B) CuSO4 + KCN →\to K2[Cu (CN)4] + K2SO4 The above equation is incorrect.

Thus, the correct equation is 2KCN + CuSO4 →\to K2SO4 + Cu + (CN)2(Cyanogen Gas) (C) Li2O + 2KCl →\to 2LiCl + K2O The above equation is incorrect because K2O, a stronger base cannot be generated by a weaker base, Li2O.

(D) [CoCl(NH3)5]+ + 5H+ →\to Co2+ +

5NH4+5NH_4^+

+ Cl- The above equation is correct because amine complexes decomposes under acidic medium.

Thus, the complex [CoCl(NH3)5] decomposes under acidic conditions to give ammonium ions.

Q66
The pair having the same magnetic moment is : [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
A [Cr(H2O)6]2+ and [Fe(H2O)6]2+
B [Mn(H2O)6]2+ and [Cr(H2O)6]2+
C [CoCl4]2– and [Fe(H2O)6]2+
D [Cr(H2O)6]2+ and [CoCl4]2–
Correct Answer
Option A
Solution

Since

(a)(a)

and

(b),(b),

each has

44

unpaired electrons, they will have same magnetic moment.

Q67
Identify the correct trend given below : (Atomic No.=Ti : 22, Cr : 24 and Mo : 42)
A Δ\Delta o of [Cr(H2O)6]2+ > [Mo(H2O)6]2+ and Δ\Delta o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
B Δ\Delta o of [Cr(H2O)6]2+ > [Mo(H2O)6]2+ and Δ\Delta o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
C Δ\Delta o of [Cr(H2O)6]2+ < [Mo(H2O)6]2+ and Δ\Delta o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
D Δ\Delta o of [Cr(H2O)6]2+ < [Mo(H2O)6]2+ and Δ\Delta o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
Correct Answer
Option C
Solution
Δ\Delta

0 of complex [Mo(H2O)6]2+ is greater than that of [Cr(H2O)6]2+.

It is due to the fact that Mo2+ belongs to second row transition element while Cr2+ is first row transition element.

Magnitude of crystal field splitting energy (

Δ\Delta

0) depends on the charge on the metal ion.

Greater the charge, greater is the crystal field splitting energy.

Therefore, [Ti(H2O)6]3+ > [Ti(H2O)6]2+.

Q68
On treatment of 100 mL of 0.1 M solution of CoCl3. 6H2O with excess AgNO3; 1.2 ×\times 1022 ions are precipitated. The complex is :
A [Co(H2O)3Cl3].3H2O
B [Co(H2O)6]Cl3
C [Co(H2O)5Cl]Cl2.H2O
D [Co(H2O)4Cl2]Cl.2H2O
Correct Answer
Option C
Solution

Moles of complex =

Molarity×Volume(mL)1000{{Molarity \times Volume\left( {mL} \right)} \over {1000}}

=

100×0.11000{{100 \times 0.1} \over {1000}}

= 0.01 mole Moles of ions precipitated with excess of AgNO3 =

1.2×10226.023×1023{{1.2 \times {{10}^{22}}} \over {6.023 \times {{10}^{23}}}}

= 0.02 mole So 0.01 × n = 0.02 ⇒\Rightarrow n = 2 It means 2Cl– ions present in ionization sphere.

∴\therefore The formula of complex is [Co(H2O)5Cl]Cl2.

H2O

Q69
[Co2(CO)8] displays :
A one Co−Co bond, six terminal CO and two bridging CO
B one Co−Co bond, four terminal CO and four bridging CO
C no Co−Co bond, six terminal CO and two bridging CO
D no Co−Co bond, four terminal CO and four bridging CO
Correct Answer
Option A
Solution

The structure of [Co2(CO)8] is From the structure, we can see that there is one CO−CO bond, six terminals CO and two bridging CO.

Q70
The correct combination is :
A [Ni(CN)4]2- −- tetrahedral ; [Ni(CO)4] −- paramagnetic
B [Ni(Cl)4]2- −- paramagnetic ; [Ni(CO)4] −- tetrahedral
C [Ni(Cl)4]2- −- square-planar ; [Ni(CN)4] 2-−- paramagnetic
D [Ni(Cl)4]2- −- diamagnetic ; [Ni(CO)4] −-square-planar
Correct Answer
Option B
Solution

[NiCl4]2– : Oxidation state of Ni in [NiCl4]2– = + 2 Cl– is a weak field ligand and cannot take part in pairing of electrons.

Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.

[Ni(CO)4] : Oxidation state of Ni in [Ni(CO)4] is zero.

CO is a strong field ligand thus pairing of electrons takes place in d-orbitals.

Hence, the complex is tetrahedral and diamagnetic.

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