Coordination Compounds Test

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Coordination Compounds Test - 39

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Weekly Quiz Competition

Question 1
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The primary valency and secondary valency of central metal ion and the no. of total ions produced in aqueous solution for $$K[Co(OX)_{2}(NH_{3})_{2}]$$ complex respectively is ________.

A
$$3, 4, 2$$

B
$$4, 4, 2$$

C
$$3, 6, 2$$

D
$$3, 6, 1$$

Solution

$$k[Co(OX)_2(NH_3)_2] \rightarrow k^+ + [Co(OX)_2(NH_3)_2]^-$$
So the number of total ions produced are 2
The oxidation state of $$K$$ is +1 and that of $$(OX)_2$$ is -2
Let the oxidation state of $$Co$$ be $$x$$
$$\therefore +1+x+2(-2)+2(0)=0$$
$$\therefore x=3$$
The primary valency of the $$Co$$ is 3
The secondary valency of the $$Co$$ is 6
Question 2
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The conductivity measurement of a coordination compound of Cobalt (III) shows that it dissociates into 3 ions in solution. The compound is:

A
Hexaamminecobalt (III) chloride

B
Pentaamminesulphatocobalt (III) chloride

C
Pentaamminechloridocobalt (III) sulphate

D
Pentaamminechloridocobalt (III) chloride

Solution

Pentaaminechloridocobalt (III) chloride

$$[Co(NH_3)_5 Cl] Cl_2 \rightarrow \, \underbrace { [Co(NH_ 3)_ 5Cl]^{ 2+} +2Cl^-] }_{ 3 \, ions } $$
Question 3
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After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.
$$AgBr+2Na_2S_2O_3\rightarrow 4Na^++Br^-+[Ag(S_2O_3)_2]^{3-}$$
What is the function of the thiosulfate ion?

A
To make the silver ions soluble

B
To oxidise the silver ions

C
To reduce the bromine

D
To reduce the silver ions

Solution

$$AgBr + 2 {Na}_{2}{S}_{2}{O}_{3} \longrightarrow 4 {Na}^{+} + {Br}^{-} + {\left[ Ag {\left( {S}_{2}{O}_{3} \right)}_{2} \right]}^{3-}$$
In the above reaction, thiosulfate ion $$\left( {{S}_{2}{O}_{3}}^{2-} \right)$$ is use to make the silver ion $$\left( {Ag}^{+} \right)$$ soluble.
Question 4
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Which of the above is a pair of geometrical isomers?

A
$$I,II$$

B
$$I,III$$

C
$$II,IV$$

D
$$III,IV$$

Solution

Geometric isomers are two or more coordination compounds which contain the same number and types of atoms, and bonds (i.e., the connectivity between atoms is the same), $$\text{but which have different spatial arrangements}$$ $$\text{of the}$$ $$\text{atoms.}$$

II and IV are geometrical isomers.
Option C is correct.
Question 5
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The hybrid states of carbon atoms in $${(\overset{B}CN)_4}\overset{A}{C_2}$$ are A and B and number of $$\pi$$ bonds in compound is C. Then?

A
$$A=sp, B=sp^2$$ and $$C=9$$

B
$$A=sp^2, B=sp$$, and $$C=9$$

C
$$A=sp^3, B=sp$$ and $$C=9$$

D
$$A=sp^2, B=sp^2$$ and $$C=9$$

Solution

From the above shown strucutre we can observe that-
The hybrid state B is $$sp$$ and that of A is $$sp^2$$ and the total no. of $$\pi$$ bond, $$C=9$$.

Option B is correct.
Question 6
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The compound $$[CoCl_2(NH_3)_2(en)]$$ can form:

A
Linkage isomers

B
Coordination isomers

C
Optical isomers

D
Linkage as well as optical isomers

Solution

The compound $$[CoCl_{2}(NH_{3})_{2}(en)]$$ has two geometrical isomers. One is the cis-isomer (with respect to $$NH_{3}$$) and the other is the trans- isomer (with respect to $$NH_{3}$$).

The cis-isomer has two optically active forms, $$d-$$cis isomer and $$l-$$cis isomer. So, the compound $$[CoCl_{2}(NH_{3})_{2}(en)]$$ can form geometrical isomers as well as optical isomers.

Hence, the correct answer is option (C).
Question 7
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In the hydrocarbon, $$H_3\underset{6}{C}-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}H_2-\underset{2}{C}\equiv \underset{1}{C}H$$

The state of hybridization of carbons $$1, 3, 5$$ are in the following sequence is:

A
$$sp, sp^2, sp^3$$

B
$$sp^3, sp^2, sp$$

C
$$sp^2, sp, sp^3$$

D
$$sp, sp^3, sp^2$$

Solution

$$H_3\underset{6}{\overset{sp^3}C}-\underset{5}{\overset{sp^2}C}H=\underset{4}{\overset{sp^2}C}H-\underset{3}{\overset{sp^3}C}H_2-\underset{2}{\overset{sp}C}\equiv \underset{1}{\overset{sp}C}H$$
The state of hybridisation of carbon in $$1, 3$$ and $$5$$ position are $$sp, sp^3, sp^2$$
Question 8
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On hybridization of one s-and one p-orbitals, we get:

A
Two mutually perpendicular orbitals

B
Two orbitals at $$ 180^o $$

C
four orbitals directed tetrahedrally

D
Three orbitals in the plane

Solution

On hybridization of one s-and one p-orbital, we get two orbitals at $$180^o$$. The number of hybrid orbitals formed is equal to the number of participating atomic orbitals, which in present case is 2. The angle between these two is equal to $$360^o$$ divided by the number of orbitals. $$\dfrac {360^o}{2} =180^o$$
Question 9
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On hybridization of one s- and one p-orbital we get :

A
Two mutually perpendicular orbitals

B
Two orbitals at $$ 180^0 $$

C
Four orbitals directed tetrahedrally

D
Three orbitals in a plane

Solution
Question 10
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Directions For Questions

One cationic complex has two isomers A and B. Each has one $$Co^{3+}$$, five $$NH_3$$, one $$Br$$, and one $$SO^{2-}_4$$ stoichiometrically. A gives white ppt with $$BaCl_2$$ while B give yellow ppt with $$AgNO_3$$.

...view full instructions

Complexes A and B have similarity in which type of isomerism?

A
Hydrate

B
Ionization

C
Linkage

D
Coordination

Solution