Coordination Compounds — Page 2 | JEE Chemistry

Q11

Which of the following cannot be explained by crystal field theory?

A The order of spectrochemical series

B Stability of metal complexes

C Magnetic properties of transition metal complexes

D Colour of metal complexes

Correct Answer

Option A

Solution

CFT does not explain the order of spectrochemical series because as per CFT, anionic ligands should exert greatest splitting effect.

However, they lie on lower end of the spectrochemical series.

Q12

The complex that dissolves in water is :

A

\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}

B

\left(\mathrm{NH}_{4}\right)_{3}\left[\mathrm{As}\left(\mathrm{Mo}_{3} \mathrm{O}_{10}\right)_{4}\right]

C

\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]

D

\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}(\mathrm{OAc})_{6}\right] \mathrm{Cl}

Correct Answer

Option D

Solution

Option A :

\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}

(Prussian Blue) Prussian blue is a very stable and intensely colored compound, but it is not soluble in water.

The individual cyanide ligands and the iron ions are bonded together strongly, which results in a crystalline structure that does not easily dissociate into its components in water.

Option B :

\left(\mathrm{NH}_{4}\right)_{3}\left[\mathrm{As}\left(\mathrm{Mo}_{3} \mathrm{O}_{10}\right)_{4}\right]

(Ammonium Arseno Molybdate) In this compound, the arsenic and molybdenum atoms are coordinated in a complex polyatomic anion.

Despite the presence of ammonium ions, which are generally soluble, the overall complex has a low solubility due to the strong interactions between arsenic, molybdenum, and oxygen within the anion.

Option C :

\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]

This is a potassium salt of an anionic complex.

Although potassium salts are generally soluble in water, the nitrite ligands attached to the cobalt in the anion form a tightly bonded structure, resulting in very poor solubility in water.

Option D :

\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}(\mathrm{OAc})_{6}\right] \mathrm{Cl}

This complex is composed of iron atoms coordinated with hydroxide and acetate ligands, and a chloride counterion.

Hydroxide and acetate ions are both polar and capable of forming hydrogen bonds with water molecules, which is a key factor in solubility.

The chloride ion is also soluble in water.

The presence of these ions, combined with the likely overall charge of the complex (since it's paired with a counterion), makes it soluble in water.

Q13

The correct sequence of ligands in the order of decreasing field strength is :

A

\mathrm{NCS}^{-}>\mathrm{EDTA}^{4-}>\mathrm{CN}^{-}>\mathrm{CO}

B

\mathrm{CO}>\mathrm{H}_2 \mathrm{O}>\mathrm{F}^{-}>\mathrm{S}^{2-}

C

\mathrm{S}^{2-}>{ }^{-} \mathrm{OH}>\mathrm{EDTA}^{4-}>\mathrm{CO}

D

{ }^{-} \mathrm{OH}>\mathrm{F}^{-}>\mathrm{NH}_3>\mathrm{CN}^{-}

Correct Answer

Option B

Solution

\text{ Field strength order : } \mathrm{CO}>\mathrm{H}_2 \mathrm{O}>\mathrm{F}^{-}>\mathrm{S}^{2-}

Q14

If an iron (III) complex with the formula

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_x(\mathrm{CN})_y\right]^-

has no electron in its

e_g

orbital, then the value of

x+y

is

A 6

B 4

C 3

D 5

Correct Answer

Option A

Solution

Balancing charges,

\begin{aligned} & 3-y=-1 \\ & \Rightarrow y=4 \\ & x=2 \\ & x+y=6 \\ & \end{aligned}

Q15

The correct IUPAC name of

[\mathrm{PtBr}_2(\mathrm{PMe}_3)_2]

is :

A bis(trimethylphosphine)dibromoplatinum(II)

B dibromodi(trimethylphosphine)platinum(II)

C dibromobis(trimethylphosphine)platinum(II)

D bis[bromo(trimethylphosphine)]platinum(II)

Correct Answer

Option C

Solution

The correct IUPAC name of

[\mathrm{PtBr}_2(\mathrm{PMe}_3)_2]

is dibromobis(trimethylphosphine)platinum(II).

Q16

Consider the following complexes (A)

\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}

, (B)

\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}

, (C)

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+}

, (D)

\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+}

The correct order of A, B, C and D in terms of wavenumber of light absorbed is :

A D < A < C < B

B B < C < A < D

C C < D < A < B

D A < C < B < D

Correct Answer

Option A

Solution

Wavenumber

=\frac{1}{\lambda} \propto

Frequency

\propto \Delta_0

Wavenumber order : $$\mathrm{D}

Q17

In which of the following complexes the CFSE,

\Delta_o

will be equal to zero?

A

\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{SCN})_6\right]

B

\left[\mathrm{Fe}(\mathrm{en})_3\right] \mathrm{Cl}_3

C

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Br}_2

D

\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]

Correct Answer

Option A

Solution

For complex $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{SCN})_6\right]$ Calculation of CFSE

\begin{aligned} & =(-0.4 \times 3+0.6 \times 2) \Delta_0 \\\\ & =0 \Delta_0 \end{aligned}

Q18

An octahedral complex having molecular composition

\mathrm{Co} \cdot 5 \mathrm{NH}_3 \cdot \mathrm{Cl}^2 . \mathrm{SO}_4

has two isomers A and B. The solution of A gives a white precipitate with

\mathrm{AgNO}_3

solution and the solution of B gives white precipitate with

\mathrm{BaCl}_2

solution. The type of isomerism exhibited by the complex is,

A Co-ordinate isomerism

B Ionisation isomerism

C Geometrical isomerism

D Linkage isomerism

Correct Answer

Option B

Solution

In this octahedral complex with the molecular composition $\mathrm{Co} \cdot 5 \mathrm{NH}_3 \cdot \mathrm{Cl}^2 \cdot \mathrm{SO}_4$ , two isomers, A and B, are present.

Isomer A is represented as $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{Cl}$ .

When this isomer is dissolved in solution, it reacts with $\mathrm{AgNO}_3$ , forming a white precipitate of silver chloride ( $\mathrm{AgCl}$ ).

Isomer B is written as $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$ .

In solution, it reacts with $\mathrm{BaCl}_2$ , forming a white precipitate of barium sulfate ( $\mathrm{BaSO}_4$ ).

These two isomers demonstrate ionisation isomerism.

This type of isomerism occurs when the exchange of ions inside and outside the coordination sphere of a complex leads to the formation of isomers with different chemical behaviors in solution.

Q19

'

X

' is the number of electrons in

t_{2 g}

orbitals of the most stable complex ion among

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{FeCl}_6\right]^{3-}, \quad\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}

and

\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}

. The nature of oxide of vanadium of the type

\mathrm{V}_2 \mathrm{O}_{\mathrm{X}}

is :

A Amphoteric

B Acidic

C Basic

D Neutral

Correct Answer

Option A

Solution

Most stable is $\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ due to Chelation effect. $\mathrm{V}_2 \mathrm{O}_5$ is amphoteric.

Q20

'X' is the number of acidic oxides among VO2, V2O3, CrO3, V2O5 and Mn2O7. The primary valency of cobalt in [Co(H2NCH2CH2NH2)3]2(SO4)3 is Y. The value of X + Y is _________.

A 3

B 4

C 2

D 5

Correct Answer

Option D

Solution

In the given compounds, we evaluate which are acidic oxides: CrO3 is acidic.

Mn2O7 is acidic.

Therefore, the number of acidic oxides, $X$ , is 2.

Next, for the complex $[Co(H2NCH2CH2NH2)3]2(SO4)3$ , we consider its dissociation: ${[\mathop {Co}\limits^{III} {({H_2}NC{H_2}C{H_2}N{H_2})_3}]_2}{(SO_4)_3} \rightleftharpoons 2{[\mathop {Co}\limits^{III} {({H_2}NC{H_2}C{H_2}N{H_2})_3}]^{3+}} + 3SO_4^{2-}$ The primary valency of cobalt (Y) in this context is 3.

Thus, the value of $X + Y$ is: $X + Y = 2 + 3 = 5$