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MCQ (Practice) - Coordination Cmpd (Level 2) - Class 12 MCQ


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25 Questions MCQ Test - MCQ (Practice) - Coordination Cmpd (Level 2)

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MCQ (Practice) - Coordination Cmpd (Level 2) - Question 1

The formula of the complex hydridotrimethoxidoborate (III) ion is :

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 1

Refer theory

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 2

The complex ion which has no `d' electrons in the central metal atom is;

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 2

[Co3+(NH3)6]3+ Co3+    d6
[Fe3+(CN)6]3– Fe3+    d5
[Cr3+(H2O)6]3+Cr3+    d3
[Mn7+O4] Mn7+    d0

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MCQ (Practice) - Coordination Cmpd (Level 2) - Question 3

Oxidation number of Fe in violet coloured complex Na4[Fe(CN)5(NOS)] is :

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 3

Let the O. N. of Fe = x
4 + x + (–1) × 5 + (–1) = 0
x = + 2

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 4

Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4 can be distinguished by :

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 4

Br + Ba2+ → x
SO42– + Ba2+ → BaSO4(s) (white ppt)
Br+ Ag+ → AgBr(s) Pale yellow ppt
SO42– + Ag+ → x
2 - ions form both the complex but magnitude of chene is different, second complex is more electrically conducting.

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 5

Amongst the following ions, which one has the highest paramagnetism?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 5

[Cr(H2O)6]3+
has maximum 3 unpaired electrons
[Fe(H2O)6]2+
has maximum 4 unpaired electrons
[Zn(H2O)6]2+
has maximum 0 unpaired electrons
[Cu(H2O)6]2+
has maximum 2 unpaired electrons H2O is weak field ligand.

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 6

Which of the following will produce a white preciptiate upon reacting with AgNO3 ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 6

Complex contains Cl out side the coordination sphere gives white ppt. AgNO3

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 7

Which isomer of CrCl3.6H2O is dark green in colour and forms one mole of AgCl with excess of AgNO3 solution

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 7

Complex contains only one Cl out side the coordination sphere which gives one mole white ppt. AgNO3

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 8

Which of the following are π-bonded organometallic compounds ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 8


*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 9

Which of the following is /are inner orbital complex (es) as well as diamagnetic in nature.

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 9

[Ir(H2O)6]3+ d2SP3 It is Inner orbital complex, dimagnetic
[Ni (NH3)6]2+ SP3d2 Outer orbital complex, Para Magnetic
[Cr (NH3)6]3+ d2SP3  Inner orbital complex, Para Magnetic
[CO(NH3)6]3+ d2SP3 Inner orbital complex, di Magnetic

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 10

Which of the following is /are correct about [Cu(NH3)4]SO4

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 10

[Cu2+(NH3)4] SO4
SO42– +Ba2+ → BaSO4(s)
                       (White ppt)
Conductivity corresponds to two ions

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 11

Which of the followng isomerism is /are shown by the complex [CoCl2(OH)2(NH3)2]Br ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 11

[COCl2(OH)2(NH3)2] Br & [CoClBr(OH)2 (NH3)2]Cl are ionisation isomers.

NO POS so optically active

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 12

Both geometrical and optical isomerism are shown by

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 12



both are optically inactive
[CO(OX)3]3– No geometrical isomers only optical isomers

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 13

WHich of the following complexes have tetrahedral shape ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 13

[Cu(NH3)4]2+ dsp2 square planner paramagnetic
[Ni(CO)4] sp3 Tetrahedral Dimagnetic
[Ni(CO)4] sp3 Tetrahedral Dimagnetic
[Zn(NH3)]42+sp3 Tetrahedral Dimagnetic

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 14

Which of the following is /are paramagnetic

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 14

[Ru (H2O)6]3+ d2sp3 1 unpaired electron (upe) octahedral
[Mn(CO)s] dsp3 0 upe TBP
[Fe(NH3)6]2+ sp3d24 upe octahedral
Cr2O72– sp3 0 upe dimagnetic

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 15

Co-ordination number of Cr in CrCl3.5H2O is six. The volume of 0.1 N AgNO3needed to ppt. the chlorine in outer sphere in 200 ml of 0.01 M solution of the complex is /are :

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 15

Applying law of equivalence eq of complex = eq of Ag NO3
[Cr(H2O)5Cl]Cl2
200×0.01×2=0.1×v
= 40 mL
[Cr(H2O)4Cl2]Cl. H2O
200 × 0.01 × 1 = 0.1 xV
V = 20 ml

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 16

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3 + 2). Three degenerate orbitals of lower energy (dxy, dyz, dzx) and a set of degenerate orbitals of higher energy (dx2 – y2 and dz2). The orbitals with lower energy are called t2g orbitals and those with higher energy are called eg orbitals.

In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of  and  lie along the axes, i.e., along the ligands the repulsions are more and so high is the energy. The lobes of the remaining three d-orbitals lie between the axes. i.e., between the ligands. The repulsion between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3 then for pairing them the option are

(i) Pairing may start with 4th electron in t2g orbitals.
(ii) Pairing may start normally with 6th electrons when t2g and eg orbitals are singly filled.

Q.

In which of the following configuration hybridisation and magnetic moment of octahedral complexes are independent of nature of ligands.

(i) d3 configuration of any metal cation.
(ii) d6 configuration of IIIrd transition series metal cation.
(iii) d8 configuration of Ist transition series metal cation.
(iv) d7 configuration of any metal cation

Select the correct code :

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 17

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3 + 2). Three degenerate orbitals of lower energy (dxy, dyz, dzx) and a set of degenerate orbitals of higher energy (dx2 – y2 and dz2). The orbitals with lower energy are called t2g orbitals and those with higher energy are called eg orbitals.

In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of  and  lie along the axes, i.e., along the ligands the repulsions are more and so high is the energy. The lobes of the remaining three d-orbitals lie between the axes. i.e., between the ligands. The repulsion between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3 then for pairing them the option are

(i) Pairing may start with 4th electron in t2g orbitals.
(ii) Pairing may start normally with 6th electrons when t2g and eg orbitals are singly filled.

Q.

Which of the following electronic arrangement is /are possible for inner orbital oct complex.

(I)  (II)  (III)  (IV) 

Select the correct code :

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 17

t32ge2g  outer orbital
t32ge0g  inner orbital
t42ge2g  outer orbital
t2g6e1g  inner  orbital by excitation of electron.

*Multiple options can be correct
MCQ (Practice) - Coordination Cmpd (Level 2) - Question 18

When a transition metal ion (usually) is involved in octahedral complex formation, the five degenerate d-orbitals split into two set of degenerate orbitals (3 + 2). Three degenerate orbitals of lower energy (dxy, dyz, dzx) and a set of degenerate orbitals of higher energy (dx2 – y2 and dz2). The orbitals with lower energy are called t2g orbitals and those with higher energy are called eg orbitals.

In octahedral complexes, positive metal ion may be considered to be present at the centre and negative ligands at the corner of a regular octahedron. As lobes of  and  lie along the axes, i.e., along the ligands the repulsions are more and so high is the energy. The lobes of the remaining three d-orbitals lie between the axes. i.e., between the ligands. The repulsion between them are less, so lesser the energy. In the octahedral complexes, if metal ion has electrons more than 3 then for pairing them the option are

(i) Pairing may start with 4th electron in t2g orbitals.
(ii) Pairing may start normally with 6th electrons when t2g and eg orbitals are singly filled.

Q.

Select incorrect match for the following complexes.

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 18

[Co(H2O)6]3+ Δ> P as for
Co3+  Cl, I, Br, F are only weak field ligands, rest are strong field ligands.

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 19

(where AB  Unsym. bidentate ligand, a,b,c,d & e  monodentate ligands)

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 19

(A)–S ; (B)–P ; (C)–Q

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 20

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 20

(A)–R,S ;  (B)–P,Q ;  (C)–P ;  (D)–P,Q




CN & NH3 are strong field ligand
CO3+d6


[PtCl2(NH3)4]Br2  d2sp3 No unpair e

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 21

The pair of compounds having metals in their highest oxidation state is

[JEE-2004]

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 21

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 22

The compound having tetrahedral geometry is

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 22

Cl is weak field ligand & CN is strong field ligand & for transition metal belonging to 2nd & 3rd transition series all lignads acts as strong field ligand.
[Ni(CN)4]2– dsp2 [PdCl4]2– dsp2
[Pd(CN)4]2– dsp2 [NiCl4]2– sp3

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 23

Spin only magnetic moment of the compound Hg[Co(SCN)4] is

[JEE-2005]

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 23


3 unpaired electron
 4sp3 hybrid orbital =

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 24

Which of the following pair is expected to exhibit same colour in solution ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 24

The chemical unit having same no. of unpaired electron has nearly same colour due to nearly same value of crystal field spliting energy Cu2+ in CuCl2 V4+ in VOCl2 has one unpaired electron each.

MCQ (Practice) - Coordination Cmpd (Level 2) - Question 25

Which type of isomerism is shown by [Co(NH3)4Br2] Cl ?

Detailed Solution for MCQ (Practice) - Coordination Cmpd (Level 2) - Question 25



n = 1
O.N. of Cr in the complex

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