Coordination Compounds Test

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Coordination Compounds Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

S in $$\displaystyle SF_{6}$$ is ______________ hybridized.

A
$$\displaystyle sp^{3}d^{3}$$

B
$$\displaystyle sp^{3}d^{4}$$

C
$$\displaystyle sp^{3}d^{2}$$

D
None of these

Solution

In $$SF_6$$, sulphur has six valence electrons and there are six fluorine atoms that are monovalent. So, by applying the hybridization formula answer comes for 6 hybrid orbitals, which is $$sp^3d^2$$ hybridisation.
Question 2
1 / -0

True structure is predicted by :

A
valence bond approach

B
Sidgwick approach

C
hybrid orbital formation

D
None of the above

Solution

Hybrid formation was introduced to explain molecular structure when the other theories failed to correctly predict them.

Hence the correct option is C.
Question 3
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Directions For Questions

An isomer of the complex $$Co(en)_2(H_2O)ICl_2$$, on reaction with concentrated $$H_2SO_4$$ it suffers loss in weight and on reaction with $$AgNO_3$$ solution gives a yellow precipitate, which is insoluble in $$NH_3$$ solution.

...view full instructions

Total number of space isomers of the formula of the above complex is :

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution

Loss of water on treatment with sulfuric acid indicates presence of water molecule in the ionization sphere.
Yellow precipitate of AgI on treatment with silver nitrate indicates presence of the iodide ion in the ionization sphere.
Thus the complex is $$[Co(en)_2Cl_2](H_2O)I$$.
Two geometrical isomers cis and trans are possible.
Cis isomer is optically active and exists in d and l forms.
Trans isomer is optically inactive .
Thus total three steroisomers are possible.
Question 4
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Identify the type of isomerism shown by $$[Co(NH_3)_4Cl_2]^+$$ :

A
$$geometrical$$ $$cis$$, $$trans$$

B
$$geometrical $$ $$fac-$$, $$mer-$$

C
$$optical$$

D
None of the above

Solution

$$[Co(NH_3)_4Cl_2]^+$$ will show cis, trans geometrical isomerism.
Question 5
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Directions For Questions

Complex compounds that have the same molecular formula but have different structural/ space arrangements of ligands around central metal atom/ ion are called isomers, these are of two types namely structural and stereoisomers.

...view full instructions

Complex species that exhibits isomerism is :

A
$$[Ag(NH_{3})_{2}]^{+}$$

B
$$[Co(NO_{2})(NH_{3})_{5}]^{2+}$$

C
$$[PtCl_{2}(en)]$$

D
$$[CoCl(NH_{3})_{5}]^{2+}$$

Solution

None of the given complexes can exhibit either geometrical or optical isomerism. But the complex $$[Co(NO_2)(NH_3)_5]^{2+}$$ shows linkage isomerism as it contains nitro group which is ambidentate ligand and can be attached through either N atom or O atom.
Question 6
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Chrome green is _____

A
$$PbCrO_4 + Fe^{III} [Fe^{II} (CN)_6]^-$$

B
$$Cr_2O_5$$

C
$$PbCrO_4$$

D
$$FeCrO_4$$

Solution

Chrome green is $$PbCrO_4 + Fe^{III} [Fe^{II} (CN)_6]^-$$.

It is a composite pigment consisting of a combination of chrome yellow and prussian blue.

Hence, option A is correct.
Question 7
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If the metal ion of $$[Fe(CN)_6]^{3-}$$ has $${ d }^{ x }$$ electronic configuration then value of $$x$$ is:

A
2

B
4

C
5

D
3

Solution

$$[Fe(CN)_6]^{3-}$$
$$x+(6x+1)+3=0$$
$$x=3$$
$$Fe^{3+}=d^5$$
$$x=5$$
Question 8
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When $$KCN$$ is added to $$CuS{ O }_{ 4 }$$ solution, there is formation of the stable water soluble complex. This complex is :

A
$${ K }_{ 4 }\left[ Cu{ \left( CN \right) }_{ 6 } \right] $$

B
$${ K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$

C
$${ K }_{ 2 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$

D
$${ K }\left[ Cu{ \left( CN \right) }_{ 3 } \right] $$

Solution

When $$KCN$$ is added to $$CuS{ O }_{ 4 }$$ solution, the following reaction occurs.

$$4KCN+CuSO_4 \rightarrow { K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$

Hence, the complex formed is $${ K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$.
Question 9
1 / -0

True structure is predicted by:

A
Valence-bond approach

B
Sidgwick approach

C
Hybrid orbital

D
None of the above

Solution

Hybridisation helps to explain molecular shape, since the angles between bonds are (approximately) equal to the angles between hybrid orbitals
Question 10
1 / -0

Turnbull's blue is :

A
$$Fe_3[Fe(CN)_6]_2$$

B
$$K_4Fe(CN)_6$$

C
$$K_3Fe(CN)_6$$

D
$$Na_4Fe(CN)_6$$

Solution

Turnbull's blue is $$Fe_3[Fe(CN)_6]_2$$.
The structure is as follows:

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Coordination Compounds Test - 19

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Coordination Compounds Test - 19

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Question 1

$$\displaystyle sp^{3}d^{3}$$

$$\displaystyle sp^{3}d^{4}$$

$$\displaystyle sp^{3}d^{2}$$

None of these

Solution

Question 2

valence bond approach

Sidgwick approach

hybrid orbital formation

None of the above

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$1$$

Solution

Question 4

$$geometrical$$ $$cis$$, $$trans$$

$$geometrical $$ $$fac-$$, $$mer-$$

$$optical$$

None of the above

Solution

Question 5

$$[Ag(NH_{3})_{2}]^{+}$$

$$[Co(NO_{2})(NH_{3})_{5}]^{2+}$$

$$[PtCl_{2}(en)]$$

$$[CoCl(NH_{3})_{5}]^{2+}$$

Solution

Question 6

$$PbCrO_4 + Fe^{III} [Fe^{II} (CN)_6]^-$$

$$Cr_2O_5$$

$$PbCrO_4$$

$$FeCrO_4$$

Solution

Question 7

2

4

5

3

Solution

Question 8

$${ K }_{ 4 }\left[ Cu{ \left( CN \right) }_{ 6 } \right] $$

$${ K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$

$${ K }_{ 2 }\left[ Cu{ \left( CN \right) }_{ 4 } \right] $$

$${ K }\left[ Cu{ \left( CN \right) }_{ 3 } \right] $$

Solution

Question 9

Valence-bond approach

Sidgwick approach

Hybrid orbital

None of the above

Solution

Question 10

$$Fe_3[Fe(CN)_6]_2$$

$$K_4Fe(CN)_6$$

$$K_3Fe(CN)_6$$

$$Na_4Fe(CN)_6$$

Solution

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