Coordination Compounds Test

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Coordination Compounds Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Primary and secondary valency of $$Pt$$ in $$[Pt(en)_2Cl_2]Cl_2$$ are:

A
2, 4

B
2, 6

C
6, 6

D
4, 2

Solution

Primary valency is the number of negative ions attached to the central metal atom.

Secondary valency is the number of ions or molecules that are coordinated to the central metal ion; its co-ordination number

Platinum is bonded with two types of ligands,
Ethylenediamine which is a bidentate ligand is two in number hence $$2\times2$$ =4
Chloride which is an unindented ligand is two in number hence $$1\times2$$ = 2
Co-ordination no.= 6
Oxidation state of central metal atom :$$x+ 0-2$$ = $$x$$ = 2
Question 2
1 / -0

The total number of isomers shown by $$[Co(NH_3)_4(NO_2)_2](NO_3)$$ complex is:

A
10

B
6

C
4

D
12

Solution

The complex $$[Co(NH_3)_4(NO_2)_2](NO_3)]$$ has ten isomers. They include geometrical, ionization and linkage isomers.
$$NO_3$$ and $$NO_2$$ groups can be either inside or outside the coordination sphere. This gives ionization isomers. $$NO_2$$ group can coordinate through N or O atom. This gives linkage isomerism. cis and trans isomers give geometrical isomers.
Question 3
1 / -0

Consider Crystal field theory and state which of the following statements is incorrect?

A
The $$Ni^{2+}(aq)$$ cation is coloured because $$Ni^{2+}$$ ion can absorb light, which promotes electrons from the filled d-orbitals to the higher energy half filled d-orbitals.

B
The $$Zn^{2+}(aq)$$ cation is colourless because the d-orbitals are completely filled and no electrons can be promoted, so on light is absorbed.

C
A complex which has just one absorption band at 455 nm, must be red coloured.

D
A complex which has just one absorption band at 425 nm, must be yellow coloured.

Solution

The wavelength 455 nm corresponds to blue light. When a complex absorbs blue light, the colour observed is orange. Thus the complex appears orange. So, the statement in option $$C$$ is incorrect.

Hence, the correct option is $$C$$
Question 4
1 / -0

Consider the following statements and arrange in the order of true/ false as given in the codes.
$$S_1$$: Both $$[Co(ox)_3]^{3-}$$ and $$[CoF_6]^{3-}$$ are paramagnetic.
$$S_2: CoCl_33NH_3$$ complex is non-conducting.
$$S_3$$: The number of possible isomers for the complex $$[Pt(NO_2)(py)(OH)(NH_3)]$$ is six.
$$S_4$$: The oxidation state of iron in brown ring complex $$[Fe(H_2O)_5NO^+]SO_4$$ + II where NO is $$NO^+$$.

A
F T T T

B
T T T T

C
F T T F

D
T T T F

Solution

In solution, $$[Co(NH_3)_3Cl_3]$$ does not dissociate into ions. It exists as neutral molecule. Hence, its aqueous solution does not conduct electricity.
Both linkage and geometrical isomerism are possible in the complex $$[Pt(NO_2)(py)(OH)(NH_3)]$$. Hence it shows six isomers.
Hence both statements are correct.
The complex $$[CoF_6]_3$$ contains 5 unpaired electrons. Hence, it is paramagnetic. In the complex $$[Co(ox)_3]_3$$ all the electrons are paired. Hence it is diamagnetic.
The oxidation state of iron in the complex $$Fe(H_2O)5NO^+]SO_2$$ is +1.
Hence both the statements are false.
Question 5
1 / -0

$$ K: K_4[Mn(CN)_6], \\L: K_3[Co(NO_2)_6], \\ M: [Co(H_2O)_6]Cl_3, \\ N:[Cr(CO)_6], \\ O:[Ni(en)_3](NO_3)_3 $$
$$P:[Pt(NH_3)_2Cl_2]$$
The diamagnetic complexes are :

A
L, M, N, P

B
K, L, N, O

C
L, M, O, P

D
K, N, O, P

Solution

The complex K having structure $$K_4[Mn(CN)_6]$$ has $$3d^5$$ electron configuration. It undergoes $$d^2sp^3$$ hybridization. It contains one unpaired electron and is paramagnetic.
The complex L having structure $$K_3[Co(NO_2)_6]$$ has $$3d^6$$ electron configuration. It undergoes $$d^2sp^3$$ hybridization. It contains no unpaired electron and is diamagnetic.
The complex M having structure $$[Co(H_2O)_6]Cl_3$$ has $$3d^6$$ electron configuration. It undergoes $$d^2sp^3$$ hybridisation. It contains no unpaired electron and is diamagnetic.
The complex N having structure $$[Cr(CO)6]$$ has $$3d^6$$ electron configuration. It undergoes $$d^2sp^3$$ hybridization. It contains no unpaired electron and is diamagnetic.
The complex O having structure $$[Ni(en)_3](NO_3)_3$$ has $$3d^8$$ electron configuration. It undergoes $$sp^3d^2$$ hybridization. It contains two unpaired electron and is paramagnetic.
The complex P having structure $$[Pt(NH_3)_2Cl_2]$$ has $$5d^8$$ electron configuration. It undergoes $$dsp^3$$ hybridization. It contains no unpaired electron and is diamagnetic.
Question 6
1 / -0

Directions For Questions

Crystal field theory vies the bonding in complexes as arising from electrostatic interaction and considers the effect of the ligand charges on the energies of the metal ion d-orbitals.
In this theory, a ligand lone pair is modelled as a point negative charge that repels electrons in the d-orbitals of the central metal ion. The theory concentrated on the resulting splitting of the d-orbitals in two groups with different energies and used that splitting to rationalize and correlate the optical spectra, thermodynamic stability, and magnetic properties of complexes. This energy splitting between the two sets of d-orbitals is called the crystal field splitting $$\Delta$$.
In general, the crystal field splitting energy $$\Delta$$ corresponds to wavelengths of light in visible region of the spectrum, and colours of the complexes can therefore be attributed to electronic transition between the lower-and higher energy sets of d-orbitals.
In general, the colour that we see is complementry to the colour absorbed.
Different metal ions have different values of $$\Delta$$, which explains why their complexes with the same ligand have different colour.

Similarly the crystal field splitting also depends on the nature of ligands and as the ligand for the same metal varies from $$H_2O$$ to $$NH_3$$ to ethylenediamine, $$\Delta$$ for complexes increases. According, the electronic transition shifts to higher energy (shorter wavelength) as the ligand varies $$H_2O$$ to $$NH_3$$ to en, thus accounting for the variation in colour.
Crystal field theory accounts for the magnetic properties of complexes in terms of the relative values of $$\Delta$$ and the spin pairing energy P. Small $$\Delta$$ values favour high spin complexes, and large $$\Delta$$ value favour low spin complexes.

...view full instructions

The $$[Ti(NCS)_6]^{3-}$$ ion exhibits a single absorption band at 544 nm. What will be the crystal field splitting energy $$(in kJ mol^{-1})$$ of the complex? $$(h=6.626\times 10^{-34}J.s; C=3.0\times 10^8 m/s; N_A=6.02\times 10^{23} ions/mole$$.

A
240

B
220

C
270

D
250

Solution

The expression for the crystal field splitting energy is $$=hv=hc$$. Here, h is the Planck's constant, v is the frequency of light, c and are the speed and the wavelengths of light respectively.
Substitute values in the above expression.
$$= \frac {(6.626 \times 10^{34}J.s) \times (3.0 \times 108m/s)} {544 \times 10^{9}m}=3.65 \times 10^{19}J/ion$$
Convert the crystal field splitting energy from J/ion to kJ/mol.
$$=(3.65 \times10^{19}J/ion) \times (6.02 \times 10^{23}ions/mole)= \frac {21.973 \times 10^4} {1000}kJmol^{1}=219.73kJmol^{1}$$
Question 7
1 / -0

The Alsatian-Swiss chemist Alfred Werner pioneered the field of coordination chemistry in the late nineteenth century. At that time, a number of compounds of cobalt (III) chloride with ammonia were known, with general formula $$CoCl_3.xNH_3$$ $$x=3\ to\ 6$$
Treatment of these compounds with aqueous HCl acid did not remove the ammonia. However, treatment of the compounds with excess of $$AgNO_3$$ at 273 K gave white precipitate of AgCl in varying ratio depending on the structure. Also each complex had electrical conductivity of varying nature.

There is no reaction of the compound with aqueous HCl indicating that there is no free $$NH_3$$. Thus, $$NH_3$$ ligand is bonded to cobalt(III) by________ .

A
Primary valency.

B
Secondary valency.

C
Both of these.

D
None of these.

Solution

Option (B) is correct.

Thus, $$NH_3$$ ligand is bonded to cobalt$$(III)$$ by secondary valency. Ammonia is present inside the coordination sphere and does not ionises while chloride ions present outside the sphere reacts with $$AgNO_3$$

$$[Co(NH_3)_5Cl]Cl_2 + AgNO_3 \longrightarrow [Co(NH_3)_5Cl](NO_3)_2 + AgCl_2$$
Question 8
1 / -0

How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with $$ Ca^{2+} $$?

A
six

B
three

C
one

D
two

Solution

An octahedral complex has 6 bonds around the central atom . One EDTA moledule complexes with one molecule of calcium cation to form octahedral complex.
Question 9
1 / -0

Which one of the following complexes will have four different isomers?

A
$$[Co(en)_2Cl_2]Br$$

B
$$[Co(en)(NH_3)_2Cl_2]Cl$$

C
$$[Co(PPh_3)_2(NH_3)Cl]Cl$$

D
$$[Co(en)_3]Cl_3$$
Question 10
1 / -0

Isomerisms exhibited by $$[Cr(NH_3)_2(H_2O)_2Cl_2]^+$$ are:

A
ionisation, optical

B
hydrate, optical

C
geometrical, optical

D
coordinate, geometrical

Solution

$$[Cr(NH_3)_2(H_2O)_2Cl_2]^+$$ can exhibit geometrical isomerism and because it's not symmetric and it will also exhibit optical isomerism.

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Coordination Compounds Test - 47

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Coordination Compounds Test - 47

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Question 1

2, 4

2, 6

6, 6

4, 2

Solution

Question 2

10

6

4

12

Solution

Question 3

The $$Ni^{2+}(aq)$$ cation is coloured because $$Ni^{2+}$$ ion can absorb light, which promotes electrons from the filled d-orbitals to the higher energy half filled d-orbitals.

The $$Zn^{2+}(aq)$$ cation is colourless because the d-orbitals are completely filled and no electrons can be promoted, so on light is absorbed.

A complex which has just one absorption band at 455 nm, must be red coloured.

A complex which has just one absorption band at 425 nm, must be yellow coloured.

Solution

Question 4

F T T T

T T T T

F T T F

T T T F

Solution

Question 5

L, M, N, P

K, L, N, O

L, M, O, P

K, N, O, P

Solution

Question 6

240

220

270

250

Solution

Question 7

Primary valency.

Secondary valency.

Both of these.

None of these.

Solution

Question 8

six

three

one

two

Solution

Question 9

$$[Co(en)_2Cl_2]Br$$

$$[Co(en)(NH_3)_2Cl_2]Cl$$

$$[Co(PPh_3)_2(NH_3)Cl]Cl$$

$$[Co(en)_3]Cl_3$$

Question 10

ionisation, optical

hydrate, optical

geometrical, optical

coordinate, geometrical

Solution

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