Coordination Compounds Test

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Coordination Compounds Test - 44

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Weekly Quiz Competition

Question 1
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In which of the following combination hybridisation of central atom (*) does not change ?

A
$$H_{2}O + CO_{2}$$

B
$$H_{3}BO_{3} + OH^{-}$$

C
$$BF_{3} + NH_{3}$$

D
None of these

Solution
Question 2
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Least stable hybride is :

A
stannane

B
silane

C
plumbane

D
germane

Solution

The stability of the hydrides of the particular group decreases down the group as the size of the element in that particular group increases which makes the bond between the element and hydrogen becomes weak, thereby it becomes unstable.
Stability order of carbon family hybride Silane > German > Stannane > Plumbane
Question 3
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Which of the following isomerism is not possible for complexes having molecular formulae?
(I) $$Pt(SCN)_{2}.3PEt_{3}$$, (II) $$CoBr . SO_{4} . 5NH_{3}$$ (III) $$FeCl_{2} . 6H_{2}O$$.

A
Optical

B
Linkage

C
Ionisation

D
Hydrate

Solution

(I) $$[Pt(SCN)(3PEt_{3})](SCN)$$: Exhibits only linkage isomerism.
as the thiocyanate group could be connected to the metal atom by either the S atom or the N atom. i.e thiocyanate, $$SCN^⁻$$ and isothiocyanate, $$NCS^⁻$$

(II) $$[CoBr(NH_{3})_{5}]SO_{4}$$: Exhibits only ionization isomerism.
as it can give different ion in solution i.e $$[CoBr(NH_3)_5]^{+2},SO_4^{2-} \ and\ [Co(SO_4)(NH_3)_5]^{+1},Br^-$$

(III) $$[Fe(H_{2}O)_{6}]Cl_{2}$$: Exhibits only hydrate isomerism.
Hydrate isomers are the type of isomers that have. similar composition but differ in the presence of the number of water molecules. ligands.

thus, the correct answer is A.
Question 4
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Directions For Questions

When hybridisation involving d-orbitals are considered then all the five d-orbitals are not degenerate, rather $$d_{x^{2} - y^{2}}, d_{x^{2}}$$ and $$d_{xy}, d_{yx}, d_{zx}$$form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation.

...view full instructions

Which of the following orbitals can not undergo hybridisation amongst themselves.
(I) 3d, 4s (II) 3d, 4d
(III) 3d,4s and 4p (IV) 3s, 3p and 4s

A
only II

B
II and III

C
I, II and IV

D
II and IV

Solution

The image shows the order of electron filling. The lower energy orbital is filled first followed by the next energy level orbit. The order s, p, d and f orbital is filled depends on their energy levels.

Hence from the image, we see that 3d, 4d and 3s, 3p, 4s can't undergo hybridisation among themselves, as these orbitals are not in increasing energy sequence or their energy difference is more for hybridisation as per the hybridisation concept.

The answer is option 2 and option 4.
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

Which of the following is different among structural isomers?

A
Oxidisation state

B
Co-ordination number

C
IUPAC name

D
None of these

Solution

IUPAC name is the Unique feature or concept for each and every complex compound
No two complexes have same IUPAC names
Question 6
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Directions For Questions

The magnetic property, dipole moment, plane of symmetry, colour and absorption band can be helpful in structure elucidation of complex compounds.

...view full instructions

Which of the following complex ion is expected to absorb light in $$4000\overset {\circ}{A}$$ to $$7800\overset {\circ}{A}$$ region>

A
$$[Ti(en)_{3}]^{4+}$$

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

C
$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$

D
$$[Zn(en)_{2}(NH_{3})_{2}]^{2+}$$

Solution

$$[\overset {+IV}{Ti}(en)_{3}]^{4+}$$ : No. of unpaired $$\overline {e}s = 0$$

$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$ : No. of unpaired $$\overline {e}s = 0$$

These complexes do not absorb light in visible range due to absence of unpaired electron and hence they are colourless.
While, in complex $$[Cr(H_{2}O)_{6}]^{3+}$$ there are unpaired $$\overline {e}s$$ and it is coloured due to absorption of light in visible range.
thus, the correct answer is B.
Question 7
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Directions For Questions

Magnetic moment, ionic conductance and colligative properties are useful in deciding structure. constitution of a given unknown complex compound.

...view full instructions

A metal $$M$$ having electronic configuration $$(n - 1)d^{8}ns^{2}$$ forms complexes with co-ordination No. $$= 4$$ and $$6$$, if it forms diamagnetic complexes then permissible oxidation states of metal cation and geometry is

A
$$+2$$, octahedral

B
$$+4$$, octahedral

C
$$+2$$, square planar

D
(b) and (c) both
Solution
Metal, M can be Nickel.
So the permissible oxidation states of metal cation and geometry are
$$+4$$, octahedral
$$+2$$, square planar
$$+2$$, tetrahedral

Hence, Option "D" is the correct answer.
Question 8
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Directions For Questions

Magnetic moment, ionic conductance and colligative properties are useful in deciding structure. constitution of a given unknown complex compound.

...view full instructions

If molar conductivity of complex is almost equal to that of $$NaCl$$ and it does not exhibits stereoisomerism then the complex will be

A
$$[Co(CO_{3})(en)_{2}]Br$$

B
$$[Co(CO_{3})(H_{2}O)_{2}(NH_{3})_{2}]Br$$

C
$$[Co(CN)(NH_{3})_{5}]Br_{2}$$

D
$$[Co(CO_{3})(NH_{3})_{4}]Br$$

Solution

The complex $$[Co(CO_{3})(NH_{3})_{4}]Br$$ has a molar conductivity similar to as that of NaCl and it also does not show stereo Isomerism.
Hence, Option "D" is the correct answer.
Question 9
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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

Types of isomerism exhibited by $$[CrCl_{2}(NO_{2})_{2}(NH_{3})_{2}]^{-}$$ complex ion are

A
Ionisation, optical

B
Hydrate, optical

C
Geometrical, optical

D
Co-ordinate, geometrical

Solution
Question 10
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Which of the following will not show geometrical isomerism?

A
$$[Cr(NH_{3})_{4}Cl_{2}]Cl$$

B
$$[Co(en)_{2}Cl_{2}]Cl$$

C
$$[Co(NH_{3})_{5}NO_{2}]Cl_{2}$$

D
$$[Pt(NH_{3})_{2}Cl_{2}]$$

Solution

Octahedral complex of type $$\mathrm{[MA_{5}B] \ \ \& \ \ [MA_6]}$$ cannot show geometrical isomerism.
Here, $$\mathrm{[Co(NH_3)_5NO_2]Cl}$$ is of $$\mathrm{[MA_5B]}$$ type. So, it does not show geometrical isomerism.

Hence, Option "C" is the correct answer.

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Coordination Compounds Test - 44

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Coordination Compounds Test - 44

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Question 1

$$H_{2}O + CO_{2}$$

$$H_{3}BO_{3} + OH^{-}$$

$$BF_{3} + NH_{3}$$

None of these

Solution

Question 2

stannane

silane

plumbane

germane

Solution

Question 3

Optical

Linkage

Ionisation

Hydrate

Solution

Question 4

only II

II and III

I, II and IV

II and IV

Solution

Question 5

Oxidisation state

Co-ordination number

IUPAC name

None of these

Solution

Question 6

$$[Ti(en)_{3}]^{4+}$$

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

$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$

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

Solution

Question 7

$$+2$$, octahedral

$$+4$$, octahedral

$$+2$$, square planar

(b) and (c) both

Solution

Question 8

$$[Co(CO_{3})(en)_{2}]Br$$

$$[Co(CO_{3})(H_{2}O)_{2}(NH_{3})_{2}]Br$$

$$[Co(CN)(NH_{3})_{5}]Br_{2}$$

$$[Co(CO_{3})(NH_{3})_{4}]Br$$

Solution

Question 9

Ionisation, optical

Hydrate, optical

Geometrical, optical

Co-ordinate, geometrical

Solution

Question 10

$$[Cr(NH_{3})_{4}Cl_{2}]Cl$$

$$[Co(en)_{2}Cl_{2}]Cl$$

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

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

Solution

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