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Test: JEE Previous Year Questions- Coordination Compounds - NEET MCQ


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22 Questions MCQ Test Chemistry Practice Tests: CUET Preparation - Test: JEE Previous Year Questions- Coordination Compounds

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Test: JEE Previous Year Questions- Coordination Compounds - Question 1

In [Cr(C2O4)3]3_ , the isomerism shown is

[AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 1

The correct answer is B.
[Cr(Ox)3​]3−  Ox = C2O42−

Optical Isomer ⟶ Non-superimposable mirror images

 

Test: JEE Previous Year Questions- Coordination Compounds - Question 2

In the complexes [Fe(H2O)6]3+, [Fe(CN)6]3+, [Fe(C2O4)3]3_ and [FeCl6]3_, more stability is shown by -

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 2

The correct answer is Option C.

More basic and chelate forming ligand stabilises the complex to more extent C2O42- is a bidentate chelating ligand thus stabilises the complex [Fe(C2O4)3]3- to a high extent.

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Test: JEE Previous Year Questions- Coordination Compounds - Question 3

One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3solution to yield two moles of AgCl(s). The structure of the complex is – 

  [AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 3

Test: JEE Previous Year Questions- Coordination Compounds - Question 4

 In the coordination compound K4[Ni(CN)4], the oxidation state of nickel is – 

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 4

The correct answer is option A
Consider the compound K4​[Ni(CN)4​]
Let oxidation state of Nickel be x
(1×4)+x+(4×−1)=0
⇒ 4+x−4=0
⇒ x=0
Hence, in the given compound, the oxidation state of Nickel is zero.

Test: JEE Previous Year Questions- Coordination Compounds - Question 5

 The number of 3d-electrons remained in Fe2+ (At. no. of Fe = 26) is – 

 [AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 5

The correct answer is option C
Fe (26) 
Electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d5 3p6 4s2 
Fe2+  Electronic configuration: 1s22s22p6 3s23p63d53p64s0 
 6  d-electrons are retained in Fe2+.

Test: JEE Previous Year Questions- Coordination Compounds - Question 6

Ammonia forms the complex ion [Cu(NH3)4]2+ ion with copper ions in alkaline solutions but not in acidic solution. What is the reason for it – 

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 6

The correct answer is option B 
In acidic solutions, ammonia reacts with proton and forms ammonium ion which further cannot act as a ligand. Also, no free ammonia molecules are available for co-ordination. Hence, it does not form a complex with copper ion.
The reaction is as follows:
NH3​ + H+ → NH4+

Test: JEE Previous Year Questions- Coordination Compounds - Question 7

Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by CN_ ion towards metal species is –  

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 7

Thus correct option is A

Test: JEE Previous Year Questions- Coordination Compounds - Question 8

The coordination number of a central metal atom in a complex is determined by – 

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 8

The correct answer is Option C.
Coordination number is the maximum covalency shown by a metal or metal ion. lt is the maximum number of ligands attached to the metal by sigma bonds or coordinate bonds.
Here, the ligand is monodentate and attached to the metal via sigma bond.

Test: JEE Previous Year Questions- Coordination Compounds - Question 9

Which one of the following complexes in an outer orbitals complex–  

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 9

The correct answer is option D
Generally weaker field ligands form outer orbital complexes.
CN− is a very strong ligand and NH3​ is a weak ligand.
So [Co(NH3​)6​]3+ and  [Ni(NH3​)6​]2+ can form an outer orbital complex.
Electronic configuration of Co+3 is [Ar]3d64s0
Electronic configuration of Ni+2 is [Ar]3d84s0
Nickel complexes can form due to presence of d8 configuration.

Test: JEE Previous Year Questions- Coordination Compounds - Question 10

Coordination compounds have great importance in biological systems. In this contect which of the following statements is incorrect ?  

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 10

Chlorophylls are green pigments in plants and contain magnesium.
They are useful during photosynthesis to store energy in the form of glucose from carbon dioxide and water in presence of sunlight.

Test: JEE Previous Year Questions- Coordination Compounds - Question 11

The correct order of magnetic moments (spin only values in B.M. among is) –  

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 11

The correct answer is option C
Cl is a weak ligand while CN is a strong ligand. 
In [MnCl4​]2: d5 configuration and 5 unpaired electrons
In [CoCl4​]2: d7 configuration and 3 unpaired electrons
In [Fe(CN)6​]4: d6 configuration, low spin complex and 0 unpaired electrons
More the number of unpaired electrons, more the value of magnetic moment. Hence, the order is: [MnCl4​]2−>[CoCl4​]2−>[Fe(CN)6​]4−
 

Test: JEE Previous Year Questions- Coordination Compounds - Question 12

The value of the 'spin only' magnetic moment for one of the following configurations is 2.84 BM . The correct one is –

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 12

Spin only magnetic moment = √(n(n+2)) B.M.

Where n = number of unpaired electron

Given, √(n(n+2)) = 2.84

n(n + 2) = 8.0656

n = 2

In an octahedral complex, for a d4 configuration in a strong field ligand, number of unpaired electrons = 2

Hence A is correct.

Test: JEE Previous Year Questions- Coordination Compounds - Question 13

The IUPAC name for the complex [Co(NO2)(NH3)5] Cl2 is – 

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 13

The IUPAC name of the given coordination complex according to the nomenclature rules is Pentamminenitrito-N-cobalt(II)chloride.

Test: JEE Previous Year Questions- Coordination Compounds - Question 14

Nickel (Z = 28) combines with a ninegative monodentate ligand X_ to form a paramagnetic complex [NiX4]2_. The number of unpaired electron in the nickel and geometry of this complex ion are respectively –

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 14

The correct answer is option D
Two,tetrahedral

 

Test: JEE Previous Year Questions- Coordination Compounds - Question 15

In Fe(CO)5, the Fe–C bond possesses –  

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 15

The correct answer is option D
(Both s and p character)
It's called “synergic bonding”. The ligand (CO) donates it's lone pair of electrons to the vacant orbitals of the iron atom and forms the sigma-bond. Since the iron atom also possesses some electrons in its d-orbitals, it back donates those electrons to the molecular orbitals of the ligand forming a π-bond. In this way, the metal-carbon bond length is reduced and the complex gets more stability. One important thing to keep in mind is that the metal atom donates its electron pairs to the antibonding MO of CO, so the C-O bond is weakened by this synergic bonding, leading to a larger C-O bond length in the complex (as opposed to a free CO molecule).
Metal-C bond length reduces, C-O bond length increases, complex gets stability.
 

Test: JEE Previous Year Questions- Coordination Compounds - Question 16

 How many EDTA (ethylenediaminetetraacetate ion) molecules are required to make an octahedral complex with a Ca2+ ion ?  

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 16

The correct answer is option A

An octahedral complex has 6 bonds around the central atom . One EDTA moledule complexes with one molecule of calcium cation to form an octahedral complex.

Test: JEE Previous Year Questions- Coordination Compounds - Question 17

The ''spin-only'' magnetic moment [in units of Bohr magneton] of Ni2+ in aqueous solution would be (At. No. Ni = 28) –  

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 17

The correct answer is option C
In aqueous solution, as aqua is poor ligand so high spin complex forms.
Electronic configuration of Ni=[Ar] 4s23d8
Electronic configuration of Ni2+=[Ar] 4s0 3d8
Ni+2,d8 configuration, have 2 unpaired electrons so the magnetic moment is:

Test: JEE Previous Year Questions- Coordination Compounds - Question 18

Which one of the following has a square planar geometry - (Co = 27, Ni = 28, Fe = 28, Fe = 26, Pt = 78) –  

[AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 18

The correct answer is Option D.
Cl- is weak field ligand and so chloride complexes are mostly high spin, but CFSE for transition metals is in the order 3d<4d<5d , so the metal belonging to higher series will prefer low spin complex. Among the given complexes, Pt forms a low spin complex while others form a high spin complex. Square planar geometry of [PtCl4]2− is also supported by its diamagnetic nature.
Note: All the complexes of Pd (II) and Pt (II) are square planar.
 

Test: JEE Previous Year Questions- Coordination Compounds - Question 19

The coordination number and the oxidation state of the element 'E' in the complex [E(en)2(C2O4)] NO2 (where (en) is ethylene diamine) are, respectively –

 [AIEEE-2008]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 19

The correct answer is option D
The coordination number is the number of coordinate covalent bonds formed with a central metal atom. ligand en is a bidentate ligand that has two donor atoms in one molecule. So, when two en ligands attach to the central atom they will make four covalent coordinate bonds. Also, Oxalate (C2​O4​) is bidentate ligand making two coordinate covalent bonds at a time. Hence overall the coordination number of the element E is 4+2=6.
 The oxidation number can be calculated by adding the charges of all the ligands and counter ions (ions present outside the central atom).
Let the oxidation state of element E be x.
Ligand en is neutral. Oxalate carries−2charge NO2​ (counter ion) carries −1charge.
 
So,  x + (2 × 0) + (−2) + (−1) = 0
x − 2 − 1 = 0
x = +3
 

Test: JEE Previous Year Questions- Coordination Compounds - Question 20

In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of Dbe the highest ?

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 20

The correct answer is Option A.
∆о is the magnitude of splitting of d-orbitals in the octahedral field. Strong field ligands cause large splitting of d-orbitals into t2g and eg orbitals where t2g has lower energy than d-orbitals and eg has higher energy than d-orbitals.
Spectrochemical series is the series in which ligands are arranged in the order of their increasing splitting ability. The CN- ligands are at the right most end indicating the high splitting power.

Test: JEE Previous Year Questions- Coordination Compounds - Question 21

 Which of the following has an optical isomer ?

[AIEEE-2009]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 21

The correct answer is Option C.
[CO(en)2(NH3)2]3+ doesn’t have a plane of symmetry and shows optical isomerism.

Test: JEE Previous Year Questions- Coordination Compounds - Question 22

 Which of the following pairs represents linkage isomers ?

[AIEEE-2009]

Detailed Solution for Test: JEE Previous Year Questions- Coordination Compounds - Question 22

The correct answer is Option A
Linkage isomerism is shown by ambidentate ligands like NCS and SCN.
It can  be linked through N( or ) S
∴[Pd(PPh3)(NCS2)] and [Pd(PPh3)2(SCN)2

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