5 Marks Questions

Question 15 Marks

What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.

Answer

The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant. $\text{M}+3\text{L}\leftarrow\rightarrow\text{ML}_{3}$ Stability constant, $\beta=\frac{[\text{ML}_{3}]}{[\text{M}][\text{L}]^{3}}$ For this reaction, the greater the value of the stability constant, the greater is the proportion of $ML^3$ in the solution. Stability can be of two types:

Thermodynamic stability:

The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.

Kinetic stability:

This helps in determining the speed with which the transformation will occur to attain the state of equilibrium.
Factors that affect the stability of a complex are:

Charge on the central metal ion: Thegreater the charge on the central metal ion, the greater is the stability of the complex.
Basic nature of the ligand: A more basic ligand will form a more stable complex.
Presence of chelate rings: Chelation increases the stability of complexes.

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Question 25 Marks

$\mathrm{CoSO}_4 \mathrm{Cl} .5 \mathrm{NH}_3$ exists in two isomeric forms ' A ' and ' B '. Isomer ' A ' reacts with $\mathrm{AgNO}_3$ to give white precipitate, but does not react with $\mathrm{BaCl}_2$. Isomer 'B' gives white precipitate with $\mathrm{BaCl}_2$ but does not react with $\mathrm{AgNO}_3$. Answer the following questions.

Identify ‘A’ and ‘B’ and write their structural formulas.
Name the type of isomerism involved.
Give the IUPAC name of ‘A’ and ‘B’.

Answer

$\mathrm{CoSO}_4 \mathrm{Cl}_{.} 5 \mathrm{NH}_3:$
Isomer A reacts with $\mathrm{AgNO}_3$ but not with $\mathrm{BaCl}_2$, it shows it has CP ion outside the coordination sphere.
Hence, $A=\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl}$
Isomer B reacts with $\mathrm{BaCl}_2$ but not with $\mathrm{AgNO}_3$, it shows it has $\mathrm{SO}_4{ }^{-}$outside the coordination sphere.

Hence, $B=\left[\mathrm{CO}\left(\mathrm{NH}_3\right) 5 \mathrm{Cl}\right] \mathrm{SO}_4$

Ionisation isomerism.
A = Pentaamminesulphatocobalt (III) chloride and.

B = Pentaamminesulphatocobalt (III) sulphate.

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Question 35 Marks

Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code:

	Column I (Complex ion)		Column II (Hybridisation, number of unpaired electrons)
$a.$	$[Cr(H_2O)_6]^{3+}$	$1.$	$dsp^2, 1$
$b.$	$[Co(CN)_4]^{2-}$	$2.$	$sp^3d^2, 5$
$c.$	$[Ni(NH_3)_6]^{2+}$	$3.$	$d^2sp^3, 3$
$d.$	$[MnF_6]^{4-}$	$4.$	$sp^3, 4$
		$5.$	$sp^3d^2, 2$

Code:

$A (3), B (1), C (5), D (2).$
$A (4), B (3), C (2), D (1).$
$A (3), B (2), C (4), D (1).$
$A (4), B (1), C (2), D (3).$

Answer

$A (4), B (3), C (2), D (1).$

	Column I (Complex ion)		Column II (Hybridisation, number of unpaired electrons)
$a.$	$[Cr(H_2O)_6]^{3+}$	$4.$	$sp^3, 4$
$b.$	$[Co(CN)_4]^{2-}$	$3.$	$d^2sp^3, 3$
$c.$	$[Ni(NH_3)_6]^{2+}$	$2.$	$sp^3d^2, 5$
$d.$	$[MnF_6]^{4-}$	$1.$	$dsp^2, 1$

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Question 45 Marks

Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory. $[CoF_6]^{3-}, [Fe(CN)_6]^{4-}$ and $[Cu(NH_3)_6]^{2+}$

Answer

According to spectrochemical series related ligands the given complexes can be arranged in a series in the order of increasing field strength as- $CN^- > NH_3> F^-$
Thus $CN^-$ and $NH_3$ are strong field ligand pair up the $t_{2g}$ electrons before filling $e_g$ set.
$[\text{CoF}_6]^{3-},\text{Co}^{3+}(\text{d}^6)\text{t}^4_{2\text{g}}\text{e}^2_\text{g},$

$[\text{Fe}(\text{CN})_6]^{4-},\text{Fe}^{2+}(\text{d}^6)\text{t}^6_{2\text{g}}\text{e}^3_\text{g},$

$[\text{Cu}(\text{NH}_3)_6]^{2+},\ \text{Cu}^{2+}(\text{d}^9)\text{t}^6_{2\text{g}}\text{e}^3_\text{g},$

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Question 55 Marks

Match the complex ions given in Column I with the colours given in Column II and assign the correct code:

	Column I (Complex ion)		Column II (Colour)
$a.$	$[Co(NH_3)_6]^{3+}$	$1.$	Violet
$b.$	$[Ti(H_2O)_6]^{3+}$	$2.$	Green
$c.$	$[Ni(H_2O)_6]^{2+}$	$3.$	Pale blue
$d.$	$[Ni(H_2O)_4(en)]^{2+} (aq)$	$4.$	Yellowish orange
		$5.$	Blue

Code:

$A (1), B (2), C (4), D (5).$
$A (4), B (3), C (2), D (1).$
$A (3), B (2), C (4), D (1).$
$A (4), B (1), C (2), D (3).$

Answer

$A (4), B (1), C (2), D (3).$

	Column I (Complex ion)		Column II (Colour)
$a.$	$[Co(NH_3)_6]^{3+}$	$4.$	Yellowish orange
$b.$	$[Ti(H_2O)_6]^{3+}$	$1.$	Violet
$c.$	$[Ni(H_2O)_6]^{2+}$	$2.$	Green
$d.$	$[Ni(H_2O)_4(en)]^{2+} (aq)$	$3.$	Pale blue

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Question 65 Marks

Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.

	Column I (Compound)		Column II (Oxidation state of Co)
$a.$	$[Co(NCS)(NH_3)_5](SO_3)$	$1.$	$+4$
$b.$	$[Co(NH_3)_4Cl_2]SO_4$	$2.$	$0$
$c.$	$Na_4[Co(S_2O_3)_3]$	$3.$	$+1$
$d.$	$[Co_2(CO)_8]$	$4.$	$+2$
		$5.$	$+3$

Code:

$A (1), B (2), C (4), D (5).$
$A (4), B (3), C (2), D (1).$
$A (5), B (1), C (4), D (2).$
$A (4), B (1), C (2), D (3).$

Answer

$A (5), B (1), C (4), D (2).$

	Column I (Compound)		Column II (Oxidation state of Co)
$a.$	$[Co(NCS)(NH_3)_5](SO_3)$	$5.$	$+3$
$b.$	$[Co(NH_3)_4Cl_2]SO_4$	$1.$	$+4$
$c.$	$Na_4[Co(S_2O_3)_3]$	$4.$	$+2$
$d.$	$[Co_2(CO)_8]$	$2.$	$0$

Explanation:
Oxdiation state of CMI (central metal ion) can be calculated by considering the oxidation state of whole molecule is equal to charge present on coordination sphere.

$[Co(NCS)(NH_3)_5]SO_3$

Let oxidation state of Co be $x.$
$x - 1 + \times 0 = +2$
$x = +2 + 1 = +3$

$[Co(NH_3)_4Cl_2]SO_4$

Let oxidation state of $Co = x$
$\Rightarrow x + 4 \times 0 + 2 \times (-1) = +2$
$\Rightarrow x - 2 = +2$
$x = 4$

$Na_4[Co(S_2O_3)_3]$

Let oxidation state of Co $= x$
$x + 3 \times (-2) = -4$
$x - 6 = -4$
$x = -4 + 6 = +2$

$[Co(CO)_8]$

Let oxidation state of Co$ = x$
$x - 8 \times 0 = 0$
$x = 0$
Hence, correct choice is $(d).$

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MCQ 75 Marks

Match the complex species given in Column $I$ with the possible isomerism given in Column $II$ and assign the correct code:

	Column $I \ ($Complex species$)$		Column $II \ ($Isomerism$)$
$a.$	$\ce{[Co(NH_3)_4C_{l2}]^+}$	$1.$	Optical
$b.$	$\ce{cis-[Co(en)_2Cl_2]^+}$	$2.$	Ionisation
$c.$	$\ce{[Co(NH_3)_5 (NO_2)]Cl_2}$	$3.$	Coordination
$d.$	$\ce{[Co(NH_3)_6][Cr(CN)_6]}$	$4.$	Geometrical
		$5.$	Linkage

A
$A (1), B (2), C (4), D (5).$
B
$A (4), B (3), C (2), D (1).$
C
$A (4), B (1), C (5), D (3).$
✓
$A (4), B (1), C (2), D (3).$

Answer

Correct option: D.

$A (4), B (1), C (2), D (3).$

	Column $I \ ($Complex species$)$		Column $II \ ($Isomerism$)$
$a.$	$\ce{[Co(NH_3)_4C_{l2}]^+}$	$4.$	Geometrical
$b.$	$\ce{cis-[Co(en)_2Cl_2]^+}$	$1.$	Optical
$c.$	$\ce{[Co(NH_3)_5 (NO_2)]Cl_2}$	$2.$	Ionisation
$d.$	$\ce{[Co(NH_3)_6][Cr(CN)_6]}$	$3.$	Coordination

Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangement of ligands.

$\ce{[Co(NH_3)_4Cl_2]+}$ shows geometrical isomerism due to presence of two types of ligand whose $\ce{[Co(NH_3)_4Cl_2]+}$ arrangement around central metal ion.

$\ce{cis-[Co(en)_2Cl_2]+}$ shows optical isomer due to its non-superimposable mirror image relationship.

$\ce{[Co(NH_3)_5(NO_2)]Cl_2}$ shows ionization isomer due to its interchanging ligand from outside the ionization sphere.
$\ce{[Co(NH_3)_6][Cr(CN)_6]}$ shows coordination isomer due to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.

Hence, correct choice is $(d).$

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Question 85 Marks

Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code:

	Column I (Coordination Compound)		Column II (Central metal atom)
a.	Chlorophyll	1.	Rhodium
b.	Blood pigment	2.	Cobalt
c.	Wilkinson catalyst	3.	Calcium
d.	Vitamin $B_{12}$	4.	Iron
		5.	Magnesium

Code:

A (5), B (4), C (1), D (2).
A (3), B (4), C (5), D (1).
A (4), B (3), C (2), D (1).
A (3), B (4), C (1), D (2).

Answer

A (5), B (4), C (1), D (2).

	Column I (Coordination Compound)		Column II (Central metal atom)
a.	Chlorophyll	5.	Magnesium
b.	Blood pigment	4.	Iron
c.	Wilkinson catalyst	1.	Rhodium
d.	Vitamin $B_{12}$	2.	Cobalt

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Question 95 Marks

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

$[\text{CoF}_6]^{3-}, [\text{Co(H}_2\text{O})_6]^{2+}, [\text{Co(CN)}_6]^{3-}$
$[\text{FeF}_6]^{3-}, [\text{Fe(H}_2\text{O})_6]^{2+}, [\text{Fe(CN})_6]^{4-}$

Answer

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 4
Magnetic moment
$=\sqrt{\text{n}(\text{n}+2)}=\sqrt{4(4+2)}=4.9\text{B.M.}$
$[\text{Co(H}_2\text{O})_6]^{2+}:$

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 3
Magnetic moment = $\sqrt{3(3+2)}=3.87\text{B.M.}$
$[\text{Co(CN)}_6]^{3-}:$

$\text{Co}^{3+}=3\text{d}^6$
Number of unpaired electrons = 0
Diamagnetic.

$\text{FeF}^{3-}_6:$

$\text{Fe}^{3+}=3\text{d}^5$
Number of unpaired electrons = 5
Magnetic moment = $\sqrt{5(5+2})=5.92\text{B.M.}$
$[\text{Fe(H}_2\text{O})_6]^{2+}:$

$\text{Fe}^{2+}=3\text{d}^6$
Number of unpaired electrons = 4
Magnetic moment = $\sqrt{4(4+2})=4.9\text{B.M.}$
$[\text{Fe(CN)}_6]^{4-}:$

$\text{Fe}^{2+}=3\text{d}^6$
Diamagnetic.

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