Coordination Compounds — Page 16 | JEE Chemistry

Q151

Which of the following complex is octahedral, diamagnetic and the most stable?

A

\mathrm{Na}_{3}\left[\mathrm{CoCl}_{6}\right]

B

\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}

C

\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]

D

\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}

Correct Answer

Option C

Solution

Option A :

\mathrm{Na}_{3}\left[\mathrm{CoCl}_{6}\right]

This is an octahedral complex, but it is not diamagnetic.

Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6 e_{g}^0$ , leading to two unpaired electrons.

Option B :

\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}

This is also an octahedral complex, but it is not diamagnetic.

The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5 e_{g}^2$ , leading to 3 unpaired electrons.

Option C :

\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]

This complex is both octahedral and diamagnetic.

Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6 e_{g}^0$ , which means no unpaired electrons.

This complex is more stable due to a strong field ligand (CN-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

Option D :

\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}

This is an octahedral complex, and it is diamagnetic.

The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8 e_{g}^0$ , leading to no unpaired electrons.

However, NH3 is a weaker field ligand compared to CN-, so the complex is less stable than Option C.

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is

\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]

.

Q152

The correct order of spin only magnetic moments for the following complex ions is

A

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}

B

\left[\mathrm{MnBr}_{4}\right]^{2-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}

C

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}

D

\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}<\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}

Correct Answer

Option C

Solution

The spin-only magnetic moment of a complex ion is calculated using the formula $\mu = \sqrt{n(n+2)}$ , where $n$ is the number of unpaired electrons in the complex ion.

In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

Here, CN $^-$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F $^-$ and Br $^-$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ : In this complex, Fe is in +3 oxidation state and has 5 d-electrons.

Due to the strong field ligand, CN $^-$ , all the 5 electrons are paired, and there are 0 unpaired electrons.

The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$ .

$\left[\mathrm{CoF}_{6}\right]^{3-}$ : Here, Co is in +3 oxidation state and has 6 d-electrons.

As F $^-$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons.

The spin-only magnetic moment is $\mu = \sqrt{4(4+2)} = \sqrt{24}$ .

$\left[\mathrm{MnBr}_{4}\right]^{2-}$ : Here, Mn is in +2 oxidation state and has 5 d-electrons.

Since Br $^-$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons.

The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35}$ .

$\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ : In this complex, Mn is in +3 oxidation state and has 4 d-electrons.

Due to the strong field ligand, CN $^-$ , all the 4 electrons are paired, and there are 0 unpaired electrons.

The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$ .

So the correct order of spin-only magnetic moments is:

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}

Q153

Given below are two statements, one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

. Assertion A: The spin only magnetic moment value for

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}

is

1.74 \mathrm{BM}

, whereas for

\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}

is

5.92 \mathrm{BM}

. Reason

\mathbf{R}

: In both complexes,

\mathrm{Fe}

is present in +3 oxidation state. In the light of the above statements, choose the correct answer from the options given below:

A Both A and R are true and R is the correct explanation of A

B A is true but R is false

C A is false but R is true

D Both A and R are true but R is NOT the correct explanation of A

Correct Answer

Option D

Solution

Unpaired electron $=1$

\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1 \times 3}=1.74 \text{ B.M. }

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ No pairing because $\mathrm{H}_2 \mathrm{O}$ is WFL Number of unpaired electrons $=5, \mu=5.92 \mathrm{BM}$ Assertion is true, Reason is true but not correct explanation.

Q154

The IUPAC name of

\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]

is:-

A Potassium tris(oxalato)cobalt(III)

B Potassium tris(oxalato)cobaltate(III)

C Potassium trioxalatocobaltate(III)

D Potassium trioxalatocobalt(III)

Correct Answer

Option C

Solution

To determine the IUPAC name of the compound K₃[Co(C₂O₄)₃], let's follow the nomenclature rules for coordination compounds: Identify the Cation and Anion: The compound consists of potassium ions (K⁺) and a complex anion [Co(C₂O₄)₃]³⁻.

Name the Cation First: The cation is named first.

So, we start with "Potassium."

Determine the Ligand Name: The ligand C₂O₄²⁻ is called "oxalato."

Count the Number of Ligands: There are three oxalato ligands.

Since "oxalato" does not contain any numerical prefixes (like di-, tri-), we use the prefixes "di-", "tri-", etc., for simple ligands.

So, we use "trioxalato."

Name the Central Metal Atom: Since the complex ion is an anion, the metal name ends with the suffix "-ate."

Therefore, "cobalt" becomes "cobaltate."

Specify the Oxidation State: The oxidation state of cobalt in this complex can be calculated: Let the oxidation state of Co be x.

Each oxalato ligand has a charge of -2.

The overall charge of the complex ion is -3.

So, x + 3(-2) = -3 ⇒ x -6 = -3 ⇒ x = +3.

Therefore, we indicate the oxidation state as (III).

Combining all these, the IUPAC name is Potassium trioxalatocobaltate(III).

Answer: Option C Potassium trioxalatocobaltate(III)

Q155

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}

and

\left[\mathrm{CoF}_6\right]^{3-}

are respectively known as :

A Inner orbital Complex, Spin paired Complex

B Spin paired Complex, Spin free Complex

C Spin free Complex, Spin paired Complex

D Outer orbital Complex, Inner orbital Complex

Correct Answer

Option B

Solution

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}

$\mathrm{Co}^{3+}($ strong field ligand $) \Rightarrow 3 \mathrm{~d}^6\left(\mathrm{t}_{2 \mathrm{~g}}^6, \mathrm{e}_{\mathrm{g}}^0\right)$ , Hybridisation : $\mathrm{d}^2 \mathrm{sp}^3$ Inner obital complex(spin paired complex) Pairing will take place.

\left[\mathrm{CoF}_6\right]^{3-}

$\mathrm{Co}^{3+}($ weak field ligand $) \Rightarrow 3 \mathrm{~d}^6\left(\mathrm{t}_{2 \mathrm{~g}}^4, \mathrm{e}_{\mathrm{g}}^2\right)$ Hybridisation : $\mathrm{sp}^3 \mathrm{~d}^2$ Outer orbital complex (spin free complex) No pairing will take place

Q156

Which of the following complex is homoleptic?

A

\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]

B

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

C

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

D

\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

Correct Answer

Option D

Solution

A homoleptic complex is one in which a central metal atom or ion is surrounded by only one kind of donor groups, ligands or ions.

On the other hand, a heteroleptic complex contains a central metal surrounded by more than one kind of donor groups, ligands or ions.

Let's examine the given options : Option A: $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ This complex contains two different ligands, ammonia ( $\mathrm{NH}_3$ ) and chloride (Cl).

Therefore, it is a heteroleptic complex.

Option B: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$ Just like Option A, this complex has two types of ligands, ammonia ( $\mathrm{NH}_3$ ) and chloride (Cl), making it a heteroleptic complex.

Option C: $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$ Again, this complex has ammonia ( $\mathrm{NH}_3$ ) and chloride (Cl) ligands, so it is a heteroleptic complex.

Option D: $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ This complex has only cyanide (CN) ligands surrounding the nickel ion, with no other types of ligands present.

Thus, this is a homoleptic complex because it is coordinated by a single type of ligand.

Therefore, Option D is the correct answer as it represents a homoleptic complex.

Q157

Given below are two statements : Statement (I) : A solution of

\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}

is green in colour. Statement (II) : A solution of

\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

is colourless. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Both Statement I and Statement II are incorrect

D Statement I is correct but Statement II is incorrect

Correct Answer

Option B

Solution

In order to determine the correctness of the given statements, we need to analyze the nature of the mentioned complexes.

Statement (I) asserts that a solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green in color.

This complex involves a nickel(II) cation coordinated with six water molecules.

Nickel(II) has the electronic configuration $[Ar]3d^8$ .

In an octahedral field created by water ligands, the d-orbitals split into two sets, $t_{2g}$ and $e_g$ , due to crystal field splitting.

The presence of unpaired electrons allows for d-d transitions when light is absorbed, which is responsible for the characteristic color of the complex.

Since typical $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ complexes are indeed green in color due to such transitions, Statement (I) is correct.

Statement (II) states that a solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colorless.

In this complex, nickel(II) is surrounded by four cyanide ligands.

Cyanide is a strong field ligand, which leads to a large crystal field splitting that exceeds the pairing energy of the electrons.

Hence, all the electrons in the nickel(II) ion are paired, and there are no unpaired electrons available for d-d transitions.

As there are no d-d transitions to absorb light in the visible region, the complex appears colorless.

Therefore, Statement (II) is also correct.

Given that both statements are correct, the most appropriate answer would be : Option B : Both Statement I and Statement II are correct.

Q158

Consider the following complex ions

\begin{aligned} & \mathrm{P}=\left[\mathrm{FeF}_6\right]^{3-} \\ & \mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \\ & \mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \end{aligned}

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

A R < Q < P

B R < P < Q

C Q < R < P

D Q < P < R

Correct Answer

Option C

Solution

\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^5

F : Weak field Ligand No. of unpaired electron's

=5

\begin{aligned} & \mu=\sqrt{5(5+2)} \\ & \mu=\sqrt{35} \mathrm{~BM} \\ & {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^3} \end{aligned}

No. of unpaired electron's

=3

\begin{aligned} & \mu=\sqrt{3(3+2)} \\ & \mu=\sqrt{15} \mathrm{~BM} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6} \end{aligned}

H

_2

O : Weak field Ligand No. of unpaired electron's

=4

\begin{aligned} & \mu=\sqrt{4(4+2)} \\ & \mu=\sqrt{24} \mathrm{~BM} \end{aligned}

Q159

Yellow compound of lead chromate gets dissolved on treatment with hot

\mathrm{NaOH}

solution. The product of lead formed is a :

A Tetraanionic complex with coordination number six

B Neutral complex with coordination number four

C Dianionic complex with coordination number six

D Dianionic complex with coordination number four

Correct Answer

Option D

Solution

Lead chromate (

\mathrm{PbCrO}_4

) dissolves in hot NaOH solution to form products due to a chemical reaction.

The reaction involves the formation of a compound where lead (

\mathrm{Pb}

) is coordinated by hydroxide ions (

\mathrm{OH}^-

). The balanced chemical reaction is:

\mathrm{PbCrO}_4 + 4\mathrm{NaOH} (hot, excess) \rightarrow \left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-} + \mathrm{Na}_2\mathrm{CrO}_4

In this reaction, lead forms a dianionic complex (

\left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-}

) with hydroxide ions.

This indicates that the lead is in the center of a complex with a coordination number of four, meaning it is directly bound to four hydroxide ions.

Therefore, the correct description of the product formed with lead is a dianionic complex with a coordination number of four.

Q160

Identity the incorrect pair from the following :

A Haber process - Iron

B Polythene preparation -

\mathrm{TiCl}_4, \mathrm{Al}\left(\mathrm{CH}_3\right)_3

C Photography - AgBr

D Wacker process -

\mathrm{Pt} \mathrm{Cl}_2

Correct Answer

Option D

Solution

The catalyst used in Wacker's process is

\mathrm{PdCl_2}