Coordination Compounds Test

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Coordination Compounds Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The loss or reduction of chlorophyll in the leaves is termed:

A
necrosis

B
chlorosis

C
epinasty

D
lichen

Solution
Question 2
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Directions For Questions

An aqueous solution of an inorganic compound (X) gives the following reactions:
(i) With an aqueous solution of $$BaCl_2$$, a precipitate insoluble in dilute HCl is obtained.
(ii) Addition of excess of KI give a born precipitate which turns white on addition of excess hypo solution
(iii) With an aqueous solution of potassium ferrocyanide a chocolate colored ppt. is obtained.

...view full instructions

What is the formula of choclate colored ppt.?

A
$$Fe_4[Fe(CN)_6]$$

B
$$Cu_2[Fe(CN)_6]$$

C
$$Cu_2[Fe(CN)_4]$$

D
$$Cu(CN)_2$$

Solution
Question 3
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The formulae of $$(A)$$ and $$(B)$$ with colours are:

A
$$Na_2CrO_4$$, yellow and $$PbCrO_4$$, yellow

B
$$Na_2Cr_2O_7$$, orange and $$PbCrO_4$$ , yellow

C
$$Cr(OH)_3$$, green and $$(CH_3COO)_3Cr$$, green

D
$$CrO_2Cl_2$$, red and $$(CH_3COO)_3 Cr$$, green

Solution
Question 4
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Hybridisation involves:

A
Addition of an electron pair

B
Mixing up of atomic orbital

C
Removal of an electron pair

D
Separation of orbitals

Solution

Merging (mixing) of dissimilar orbitals of different energies to form new orbitals is known as hybridisation and the new orbital formed are known as hybrid orbitals, They have similar energy.
Question 5
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In which one of the following changes there are transfer of five electrons

A
$$Mn{O}_{4}^{-}\rightarrow {Mn}^{2+}$$

B
$$Cr{O}_{4}^{2-}{Cr}^{3+}$$

C
$$Mn{O}_{4}^{2-}Mn{O}_{2}$$

D
$${Cr}_{2}{O}_{7}^{2-} 2{Cr}^{3+}$$

Solution

$$\overset{+7}{Mn}{O}_{4}^{-}\rightarrow {Mn}^{2+}+5{e}^{-}$$

Option A.
Question 6
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Directions For Questions

In coordination chemistry, there are a variety of methods applied to find out the structure of complexes. One method involves treating the complex with know regents and from the nature of the reaction, the formula of the complex can be predicted. An isomer of the complex $$ Co(en)_{2} $$ $$ (H_{2}O) $$ $$ Cl_2 Br$$, on reaction with concentrated the
$$ H_{2}SO_{4} $$ (dehydrating agent ) it suffers loss in weight and on reaction with $$ AgNO_{3} $$ solution it gives a while precipitate which is solution in $$ NH_{3}(eq) $$.

...view full instructions

The correct formula of the complex is

A
$$ \left [ CoClBr(en)_{2} \right ]H_{2}O $$

B
$$ \left [ CoCl(en)_{2}(H_{2}O) \right ]BrCl $$

C
$$ \left [ CoBr(en)_{2} (H_{2}O)\right ] Cl_{2} $$

D
$$ \left [ CoBrCl(en)_{2} \right ]Cl.H_{2}O $$

Solution

Solution:

As per given conditions: Complex suffer loss in weight with conc.$$H_2SO_4$$ as it is dehydrating agent so ionization sphere must contain $$H_2O$$ molecules and on reaction with $$AgNO_3$$ solution it gives a white $$ppt$$ the ionization sphere should contain $$Cl^-$$ ion. So formula may be $$[CoBrCl(en)_2]Cl.H_2O$$

Option D is correct
Question 7
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Which of the following is the odd one out?

A
Potassium ferrocyanide

B
Tetramminezinc(II) chloride

C
Ferrous ammonium sulphate

D
Potassium ferricyanide

Solution

All are the complexes other than ferrous ammonium sulphate are complexes with a single metal salt. But $$\text{ferrous}$$ $$\text{ammonium sulphate is a double salt with two different cations.}$$ Note that double salts are different from coordination compounds. All others are coordination compounds, whereas ferrous ammonium sulphate is not a coordination compound and it ionizes into ferrous, ammonium and sulphate ions.

Option C is correct.
Question 8
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One mole of the complex compound $$Co\left(NH_{3}\right)_{5}Cl_{3}$$, gives $$3$$ moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $$AgNO_{3}$$ solution to yield two moles of $$AgCl\left(s\right)$$. The structure of the complex is

A
$$[Co(NH_{3})_{3}Cl_{3}].2NH_{3}$$

B
$$\left[Co\left(NH_{3}\right)_{4}Cl_{2}\right]Cl.NH_{3}$$

C
$$\left[Co\left(NH_{3}\right)_{4}Cl\right]Cl_{2}.NH_{3}$$

D
$$\left[Co\left(NH_{3}\right)_{5}Cl\right]Cl_{2}$$

Solution

As complex gives 2 moles of AgCl on reaction with water. This means that there are 2 moles of Chloride ions out of 3 moles of total ions.

$$\left[Co\left(NH_{3}\right)_{5}Cl\right]Cl_{2}$$ yields three moles of ions out of which two mole of ions are Chloride ions.

Hence, Option "D" is the correct answer.
Question 9
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Which one of the following compounds helps in achieving equilibrium between $$O_2$$ and $$CO_2$$ in the atmosphere?

A
Chlorophyll

B
Vitamin - B$$_{12}$$

C
Porphyrin

D
Acetyl salicylic acid

Solution

Chlorophyll is a green substance in producers that traps light energy from the sun, which is then used to combine carbon dioxide and water into sugars in the process of photosynthesis. Chlorophyll is vital for photosynthesis, which helps plants get energy from light.

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (A)}$$
Question 10
1 / -0

$$MgSO_{4}$$ on reaction with $$NH_{4}OH$$ and $$Na_{2}HPO_{4}$$ forms a white crystalline precipitate. What is its formula?

A
$$Mg(NH_{4})PO_{4}$$

B
$$Mg_{3}(PO_{4})_{2}$$

C
$$MgCl_{2}-MgSO_{4}$$

D
$$MgSO_{4}$$

Solution

Magnesium is Group VI radical.

The preliminary test of Mg involves following reaction, $$MgSO_4$$ reacts with $$NH_4OH$$ and $$Na_2HPO_4$$ forming a white crystalline precipitate of Magnesium ammonium phosphate.

$$MgSO_4+NH_4OH+Na_2HPO_4\rightarrow Mg(NH_4)PO_4\downarrow +Na_2SO_4+H_2O$$
white ppt

Hence the correct option is $$(A)$$.

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Coordination Compounds Test - 41

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Coordination Compounds Test - 41

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Question 1

necrosis

chlorosis

epinasty

lichen

Solution

Question 2

$$Fe_4[Fe(CN)_6]$$

$$Cu_2[Fe(CN)_6]$$

$$Cu_2[Fe(CN)_4]$$

$$Cu(CN)_2$$

Solution

Question 3

$$Na_2CrO_4$$, yellow and $$PbCrO_4$$, yellow

$$Na_2Cr_2O_7$$, orange and $$PbCrO_4$$ , yellow

$$Cr(OH)_3$$, green and $$(CH_3COO)_3Cr$$, green

$$CrO_2Cl_2$$, red and $$(CH_3COO)_3 Cr$$, green

Solution

Question 4

Addition of an electron pair

Mixing up of atomic orbital

Removal of an electron pair

Separation of orbitals

Solution

Question 5

$$Mn{O}_{4}^{-}\rightarrow {Mn}^{2+}$$

$$Cr{O}_{4}^{2-}{Cr}^{3+}$$

$$Mn{O}_{4}^{2-}Mn{O}_{2}$$

$${Cr}_{2}{O}_{7}^{2-} 2{Cr}^{3+}$$

Solution

Question 6

$$ \left [ CoClBr(en)_{2} \right ]H_{2}O $$

$$ \left [ CoCl(en)_{2}(H_{2}O) \right ]BrCl $$

$$ \left [ CoBr(en)_{2} (H_{2}O)\right ] Cl_{2} $$

$$ \left [ CoBrCl(en)_{2} \right ]Cl.H_{2}O $$

Solution

Question 7

Potassium ferrocyanide

Tetramminezinc(II) chloride

Ferrous ammonium sulphate

Potassium ferricyanide

Solution

Question 8

$$[Co(NH_{3})_{3}Cl_{3}].2NH_{3}$$

$$\left[Co\left(NH_{3}\right)_{4}Cl_{2}\right]Cl.NH_{3}$$

$$\left[Co\left(NH_{3}\right)_{4}Cl\right]Cl_{2}.NH_{3}$$

$$\left[Co\left(NH_{3}\right)_{5}Cl\right]Cl_{2}$$

Solution

Question 9

Chlorophyll

Vitamin - B$$_{12}$$

Porphyrin

Acetyl salicylic acid

Solution

Question 10

$$Mg(NH_{4})PO_{4}$$

$$Mg_{3}(PO_{4})_{2}$$

$$MgCl_{2}-MgSO_{4}$$

$$MgSO_{4}$$

Solution

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