Coordination Compounds Test

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Coordination Compounds Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The formula of prussian blue is:

A
$$Na_{4}[Fe(CN)_{6}]$$

B
$$Fe(CN)_{2}$$

C
$$Fe(CNS)_{3}$$

D
$$Fe_{4}[Fe(CN)_{6}]_{3}$$

Solution

Ferric ferrocyanide /Potassium ferrocyanide/ Sodium ferrocyanide is prussian blue.
$$Fe_{4}[Fe(CN)_{6}]_{3}$$
Option D is correct,
Question 2
1 / -0

In which of the following coordination entities the magnitude of $$\Delta _{0}$$ (CFSE in the octahedral field) will be maximum? (Atomic number of $$Co=27$$)

A
$$[Co(CN)_{6}]^{3-}$$

B
$$[Co(C_{2}O_{4})_{3}]^{3-}$$

C
$$[Co(H_{2}O)_{6}]^{3+}$$

D
$$[Co(NH_{3})_{6}]^{3+}$$

Solution

Since CFSE depends on strength of the ligand which depends on the electronegativity of the donor atom, therefore the strength of the ligand is in order of CN^->NH₃>C₂O₄^2->H₂O. [Lesser electronegativity means easy electron-donating capacity and hence stronger ligand].
Question 3
1 / -0

$$[Co(NH_3)_4Cl_2]$$ possesses :

A
Square planar geometry

B
Tetrahedral geometry

C
Tetrahedral nature

D
Octahedral geometry

Solution
Question 4
1 / -0

$$[Fe(H_2O)_4(CN)_2]$$ is the empirical formula of a compound which has a magnetic moment corresponding to $$2\frac{2}{3}$$ unpaired electrons per iron. The best possible formula of the compound is

A
$$[Fe(H_2O)_6]2[Fe(CN)_6]$$

B
$$[Fe(H_2O_6)][Fe(H_2O)_2(CN)_4]
$$

C
$$[Fe(H_2O)_4(CN)_2]_2[Fe(H_2O)_4(CN)_2]$$

D
None of these

Solution

Magnetic moment $$M=\sqrt {n(n+2)}$$ magneton And here $$M=2\frac {2}{3}=\frac {8}{3}$$ unpaired $$e^-$$ per iron.
best formula of compound is $$[Fe(H_2O)_6]_2[Fe(CN)_6]$$
Question 5
1 / -0

Directions For Questions

In the valence bond theory, hybridisation of orbitals is an integral part of bond formation. Hybridisation consists

ofmixing or linear combination of the "pure" atomic orbitals in such a way as to form new hybrid orbitals such
as $$sp, sp^{2} , sp^{3} ,sp^{3}d, sp^{3}d^{2}$$ etc.

...view full instructions

Which one of the following molecular geometries (i.e. shapes) is not possible for the $$sp^{3}d^{2}$$ hybridization?

A
Capped octahedral

B
Octahedral

C
Square planar

D
Square pyramidal

Solution
Question 6
1 / -0

The correct order regarding the electronegativity of hybrid orbitals of carbon is:

A
$$sp > sp^{2}< sp^{3}$$

B
$$sp > sp^{2} > sp^{3}$$

C
$$sp < sp^{2}> sp^{3}$$

D
$$sp < sp^{2}< sp^{3}$$

Solution

$$sp>s{ p }^{ 2 }>s{ p }^{ 3 }$$
$$s$$ orbitals hold electrons more tightly to nucleus than $$p$$ orbitals. That is the $$s$$ orbitals are effectively more electronegative.
In $$s{ p }^{ 2 }$$ carbon, the character of each orbital has $$33$$% of s character and $$25$$% s character in $$s{ p }^{ 3 }$$ carbon, whereas in $$sp$$, $$s$$ character has $$50$$% and therefore the electronegativity order goes this way
$$sp>s{ p }^{ 2 }>s{ p }^{ 3 }$$
Question 7
1 / -0

Which one of the following complexes is not expected to exhibit isomerism?

A
$$[Pt(NH_3)_2Cl_2]$$

B
$$[Ni(NH_3)_2Cl_2]$$

C
$$[Ni(en)_3]^{2+}$$

D
$$[Ni(NH_3)_4(H_2O)_2]^{2+}$$
Question 8
1 / -0

Coordination number of Ni in $$[Ni(C_2O_4)_3]^{4-}$$ is:

A
3

B
6

C
4

D
2

Solution

Co-ordination number of $$Ni$$ in $$[Ni(C_2O_4)]^{-4}$$ is $$6.$$. Because $$C_2O_4^{-3}$$ is a bidentate ligand and these ligands are $$3$$ in number. So $$2 \times 3=6$$ be the co-ordination number.
Question 9
1 / -0

The two compounds $$\displaystyle \left [ Co\left ( SO_{4} \right )\left ( NH_{3} \right )_{5} \right ]Br$$ and $$\displaystyle \left [ Co\left ( SO_{4} \right )\left ( NH_{3} \right )_{5} \right ]Cl$$ represent

A
Linkage is isomerism

B
Ionisation isomerism

C
Co $$\displaystyle -$$ ordination isomerism

D
No isomerism

Solution

Even formula of compounds is not same so no isomerism.
Question 10
1 / -0

The total number of possible isomers of the compound $$\displaystyle \left [ Cu^{II}\left ( NH_{3} \right )_{4} \right ]\left [ Pt^{II}Cl_{4} \right ]$$ are :

A
3

B
5

C
4

D
6

Solution

$$\displaystyle \left [ Cu^{II}\left ( NH_{3} \right )_{4} \right ]\left [ Pt^{II}Cl_{4} \right ]$$ will show 4 isomers (2 coordinate isomers and 2 geometrical isomer of coordinate isomer).

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Coordination Compounds Test - 17

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Coordination Compounds Test - 17

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Question 1

$$Na_{4}[Fe(CN)_{6}]$$

$$Fe(CN)_{2}$$

$$Fe(CNS)_{3}$$

$$Fe_{4}[Fe(CN)_{6}]_{3}$$

Solution

Question 2

$$[Co(CN)_{6}]^{3-}$$

$$[Co(C_{2}O_{4})_{3}]^{3-}$$

$$[Co(H_{2}O)_{6}]^{3+}$$

$$[Co(NH_{3})_{6}]^{3+}$$

Solution

Question 3

Square planar geometry

Tetrahedral geometry

Tetrahedral nature

Octahedral geometry

Solution

Question 4

$$[Fe(H_2O)_6]2[Fe(CN)_6]$$

$$[Fe(H_2O_6)][Fe(H_2O)_2(CN)_4]$$

$$[Fe(H_2O)_4(CN)_2]_2[Fe(H_2O)_4(CN)_2]$$

None of these

Solution

Question 5

Capped octahedral

Octahedral

Square planar

Square pyramidal

Solution

Question 6

$$sp > sp^{2}< sp^{3}$$

$$sp > sp^{2} > sp^{3}$$

$$sp < sp^{2}> sp^{3}$$

$$sp < sp^{2}< sp^{3}$$

Solution

Question 7

$$[Pt(NH_3)_2Cl_2]$$

$$[Ni(NH_3)_2Cl_2]$$

$$[Ni(en)_3]^{2+}$$

$$[Ni(NH_3)_4(H_2O)_2]^{2+}$$

Question 8

3

6

4

2

Solution

Question 9

Linkage is isomerism

Ionisation isomerism

Co $$\displaystyle -$$ ordination isomerism

No isomerism

Solution

Question 10

3

5

4

6

Solution

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$$[Fe(H_2O_6)][Fe(H_2O)_2(CN)_4]
$$