Coordination Compounds — Page 14 | JEE Chemistry

Q131

The complex cation which has two isomers is :

A

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{+}

B

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

C

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}

D

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}

Correct Answer

Option C

Solution

Complex

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}

will have two isomer one linked through N (Nitro) and one through O (Nitrite).

Q132

Which of the following complex will show largest splitting of d-orbitals?

A

[\mathrm{Fe(C_2O_4)_3]^{3-}}

B

[\mathrm{Fe(CN)_6]^{3-}}

C

[\mathrm{Fe(NH_3)_6]^{3+}}

D

[\mathrm{FeF_6]^{3-}}

Correct Answer

Option B

Solution

CN– is strongest field ligand among given ligands. So maximum splitting in d-orbitals take place.

Q133

Which of the following are the example of double salt? A.

\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}

B.

\mathrm{CuSO}_{4}\cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}

C.

\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}

D.

\mathrm{Fe}(\mathrm{CN})_{2}\cdot4 \mathrm{KCN}

Choose the correct answer :

A A, B and D only

B B and D only

C A and B only

D A and C only

Correct Answer

Option D

Solution

$A=\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - double salt B. $\mathrm{CuSO}_{4} \cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ $=\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$ - complex salt C. $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$ - double salt D. $\mathrm{Fe}(\mathrm{CN})_{2} \cdot 4 \mathrm{KCN}$ = $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ - complex salt Option A, $\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ , is a double salt because it is formed by the combination of two different salts: iron(II) sulfate ( $\mathrm{FeSO}_{4}$ ) and ammonium sulfate ( $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ ).

Option B, $\mathrm{CuSO}_{4}\cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ , is a complex salt because it contains a central copper ion coordinated to several ammonia molecules.

In this compound, the copper(II) ion (Cu2+) acts as the central metal ion, while the four ammonia molecules (NH3) act as ligands, coordinating to the copper ion through their lone pairs of electrons.

The sulfate ion (SO42-) and water molecules (H2O) are not involved in the coordination sphere and simply crystallize along with the complex.

Option C, $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$ , is also a double salt, formed by the combination of two different salts: potassium sulfate ( $\mathrm{K}_{2} \mathrm{SO}_{4}$ ) and aluminum sulfate ( $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ ).

The 24 water molecules in the formula are not directly involved in the salt formation but instead function to stabilize the crystal lattice by forming hydrogen bonds with the sulfate ions and other water molecules.

Option D, $\mathrm{Fe}(\mathrm{CN})_{2}\cdot4 \mathrm{KCN}$ , is not an example of a double salt.

It is a complex salt, formed by the coordination of iron(II) ions to cyanide ligands and potassium ions.

Q134

The

\mathrm{Cl}-\mathrm{Co}-\mathrm{Cl}

bond angle values in a fac-

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]

complex is/are :

A

90^{\circ} ~\&~ 120^{\circ}

B

90^{\circ}

C

180^{\circ}

D

90^{\circ} ~\& ~180^{\circ}

Correct Answer

Option B

Solution

All the Cl $-$ Co $-$ Cl bond angles are of 90

^\circ

.

Q135

Correct order of spin only magnetic moment of the following complex ions is : (Given At.no. Fe : 26, Co : 27)

A

\mathrm{[CoF_6]^{3-} > [FeF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}

B

\mathrm{[FeF_6]^{3-} > [Co(C_2O_4)_3]^{3-} > [CoF_6]^{3-}}

C

\mathrm{[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}

D

\mathrm{[Co(C_2O_4)_3]^{3-} > [CoF_6]^{3-} > [FeF_6]^{3-}}

Correct Answer

Option C

Solution

{\left[ {Fe{F_6}} \right]^{3 - }}\overset{{}}\longrightarrow 5

unpaired electrons

{\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}\overset{{}}\longrightarrow 0

unpaired electron

{\left[ {Co{F_6}} \right]^{3 - }}\overset{{}}\longrightarrow 4

unpaired electrons So, correct answer is : (3)

Q136

Chiral complex from the following is : Here en = ethylene diamine

A

\mathrm{cis-[PtCl_2(en)_2]^{2+}}

B

\mathrm{trans-[Co(NH_3)_4Cl_2]^{+}}

C

\mathrm{trans-[PtCl_2(en)_2]^{2+}}

D

\mathrm{cis-[PtCl_2(NH_3)_2]}

Correct Answer

Option A

Solution

This is chiral complex form.

Q137

The hybridization and magnetic behaviour of cobalt ion in

\mathrm{[Co(NH_3)_6]^{3+}}

complex, respectively is :

A

\mathrm{sp^3d^2}

and paramagnetic

B

\mathrm{sp^3d^2}

and diamagnetic

C

\mathrm{d^2sp^3}

and diamagnetic

D

\mathrm{d^2sp^3}

and paramagnetic

Correct Answer

Option C

Solution

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is diamagnetic with $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation of $\mathrm{Co}^{+3}$ .

This is because $\mathrm{NH}_{3}$ is a strong field ligand and forces electrons to pair up in a $\mathrm{d}^{6}$ configuration.

Q138

The primary and secondary valencies of cobalt respectively in

\mathrm{[Co(NH_3)_5Cl]Cl_2}

are :

A 2 and 8

B 2 and 6

C 3 and 5

D 3 and 6

Correct Answer

Option D

Solution

\mathrm{[Co(NH_3)_5Cl]Cl_2}

x + 5 (0) – 1 = + 2 $\Rightarrow$ x = + 3 Secondary valency is the number of ligands in the complex which is equal to 6.

Q139

The complex with highest magnitude of crystal field splitting energy

\left(\Delta_{0}\right)

is :

A

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

B

\left[\mathrm{Mn}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}

C

\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}

D

\left[\mathrm{Ti}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}

Correct Answer

Option C

Solution

The crystal field splitting energy ( $\Delta_0$ ) is proportional to the electrostatic interaction between the metal ion and the ligands.

This interaction is affected by the size of the metal ion, as well as the size of the ligands.

The size of the metal ion can be estimated by its ionic radius, which is the distance from the nucleus to the outermost electron shell.

A smaller ionic radius corresponds to a greater electrostatic interaction with the ligands, resulting in a larger $\Delta_0$ value.

Out of the given options, the ionic radii of the metal ions are: - $\mathrm{Ti}^{3+}$ : 67 pm - $\mathrm{Cr}^{3+}$ : 62 pm - $\mathrm{Mn}^{3+}$ : 65 pm - $\mathrm{Fe}^{3+}$ : 65 pm The smaller ionic radius of $\mathrm{Cr}^{3+}$ compared to the other metal ions indicates a stronger electrostatic interaction with the ligands, resulting in a higher crystal field splitting energy ( $\Delta_0$ ).

Therefore, the correct answer is option (C) $\left[\mathrm{Cr}\left(\mathrm{OH}_2\right)_6\right]^{3+}$ , as it has the highest tendency to attract ligands due to its smaller ionic radius.

Q140

The total number of stereoisomers for the complex

\left[\mathrm{Cr}(o x)_{2} \mathrm{ClBr}\right]^{3-}

(where

o x=

oxalate) is :

A 1

B 3

C 2

D 4

Correct Answer

Option B

Solution

$\left[\mathrm{Cr}(\mathrm{Ox})_2 \mathrm{ClBr}\right]^{-3}$ - No. of isomers - - This structure has plane of symmetry, So no optical isomerism will be shown. - This structure does not contain plane of symmetry, So two forms $d$ as well as 1 will be shown.