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

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

Optical Isomer ⟶ Non-superimposable mirror images

 

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.

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.

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⁺.

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⁺​

Thus correct option is A

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.

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(NH_3​)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.

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.

The correct answer is option C
Cl⁻ is a weak ligand while CN⁻ is a strong ligand. 
In [MnCl_4​]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⁻
 

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.

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

The correct answer is option D
Two,tetrahedral

 

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.
 

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.

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:

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.
 

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
 

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.

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

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]