Coordination Compounds Test

Result

Coordination Compounds Test - 73

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which is correct statement?
As the s-character of a hybrid orbital decreases
(I) The bond angle decreases (II) The bond strength increases
(III) The bond length increases (IV) Size of orbital increases

A
(I), (III) and (IV)

B
(II), (III) and (IV)

C
(I) and (II)

D
All are correct

Solution

The statements (I), (III) and (IV) are correct.

As the s-character of a hybrid orbital decreases,

(I) the bond angle decreases. The bond angles for $$ sp $$, $$sp^{2}$$ and $$ sp^{3} $$ hybrid orbitals are $$180^{0}$$ , $$ 120^{0} $$ and $$109^{0}$$ respectively.

(II) the bond strength decreases. Triple bond ($$sp$$ hybridisation) is stronger than double bond ($$sp^{2}$$ hybridisation) which in turn is stronger than single bond ($$sp^{3}$$ hybridisation).

(III) the bond length increases. Triple bond ($$ sp $$ hybridisation) is shorter than double bond ($$sp^{2}$$ hybridisation) which in turn is shorter than single bond ($$sp^{3}$$ hybridisation).

(IV) the size of orbital increases. s orbital is smaller than p orbital. Hence, $$ sp $$ hybrid orbitals are smaller than $$sp^{2}$$ hybrid orbitals which in turn are smaller than $$sp^{3}$$ hybrid orbitals.

Hence , option A is correct .
Question 2
1 / -0

$$[CoCl_{2}(NH_{3})_{4}]^{+} + Cl^{-}\rightarrow [CoCl_{3}(NH_{3})_{3}] + NH_{3}$$. If in this reaction two isomers of the product are obtained, which is true for the initial (reactant) complex:

A
Compound is in cis-form

B
Compound is in trans-form

C
Compound is in both (cis and trans) form

D
Can't be predicted

Solution
Question 3
1 / -0

A complex which is formed by chloride and nitrate ligand gives two moles of precipitate of $$ AgCl $$ with $$ AgNO_3 $$. Its molecular formula will be

A
$$ [CO(NH_3)_5NO_3]Cl_2 $$

B
$$ [CO(NH_3)_5Cl]Cl NO_3Cl $$

C
$$ [CO(NH_3)_5Cl]NO_2 $$

D
None of the above

Solution
Question 4
1 / -0

Directions For Questions

Hybridisation involves the mixing of orbitals having comparable energies of same atom. Hybridised orbitals perform efficient overlapping than overlapping by pure s,p or d orbitals .

...view full instructions

Consider the following compounds and select the incorrect statement from the following :
$$NH_{3}, PH_{3}, H_{2}S, SO_{2}, SO_{3}, BF_{3}, PCl_{3}, IF_{7}, P_{4}, H_{2}$$

A
Six molecules out of given compound involves hybridisation

B
Three molecules are hypervalent compounds

C
Six molecules out of above compounds are non-planar in structure

D
Two molecules out of given compounds involves $$(d \pi- p \pi)$$ bonding as well as also involves $$(p \pi - p \pi)$$ bonding
Question 5
1 / -0

Directions For Questions

The crystal field theory assumes interaction between metal ion and the ligands as a purely electrostatic and ligands are supposed to be point charges.

...view full instructions

Which of the following match are incorrect?

A
$$[VCl_{3}(NMe_{3})_{3}], \sqrt {8} BM$$

B
$$[CrCl_{3}(NMe_{3})_{3}], \sqrt {15} BM$$

C
$$[Cu(CN)(NO_{2}) (NH_{3})(Py)], \sqrt {3} BM$$

D
$$[Co(ox) (H_{2}O)_{4}]^{+}, \sqrt {24} BM$$

Solution
Question 6
1 / -0

Chlorophyll contains

A
cobalt

B
magnesium

C
iron

D
nickel

Solution
Question 7
1 / -0

Mark the incorrect match.

A
Insulin- Zinc

B
Haemoglobin- Iron

C
Vitamin $$B_{12}$$- Cobalt

D
Chlorophyll- chromium

Solution

Insulin contains Zinc, Haemoglobin contains Iron, Vitamin $$\mathrm{B_{12}}$$ contains Cobalt and Chlorophyll contains Magnesium.

Hence, Option "D" is the correct answer.
Question 8
1 / -0

Which of the following isomers will give while precipitate with $$BaCl_{2}$$ solution?

A
$$[Co(NH_{3})_{5}SO_{4}]Br$$

B
$$[Co(NH_{3})_{5}Br]SO_{4}$$

C
$$[Co(NH_{3})_{4}(SO_{4})_{2}]Br$$

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

Solution

Reaction:
$$\mathrm{[Co(NH_{3})_{5}Br]SO_{4}\rightleftharpoons [Co(NH_{3})_{5}Br]^{2+}+SO_{4}^{2-}}$$

$$\mathrm{SO_{4}^{2-}+BaCl_{2}\rightarrow \underset{white\ ppt.}{BaSO_{4}}+2Cl^{-}}$$

So, the isomer which will give white ppt with $$\mathrm{BaCl_2}$$ solution is $$\mathrm{[Co(NH_3)_5Br]SO_4}$$
Hence, Option "B" is the correct answer.
Question 9
1 / -0

Which among the following will be named as dibromobis(ethylenediamine) chromium $$(III)$$ bromide?

A
$$[Cr(en)_2Br_2]Br$$

B
$$[Cr(en)Br_4]^-$$

C
$$[Cr(en)Br_2]Br$$

D
$$[Cr(en)_3]Br_3$$
Question 10
1 / -0

Geometrical shapes of the complexes formed by the reaction of $$Ni^{2+}$$ with $$Cl^-, CN^-$$ and $$H_2O$$, respectively, are:

A
Octahedral, tetrahedral and square planar

B
Tetrahedral, square planar and octahedral

C
square planar, tetrahedral and octahedral

D
Octahedral, square planar and octahedral

Solution

$$[NiCl_4]^{2-} = 3d^8$$
$$Cl^-$$ being a weak ligand paring of electrons does not take place.
Ref. image 1
Co-ordination number $$= 4$$, Hybridisation $$sp^3$$
Thus geometry is tetrahedral

$$[Ni(CN)_4]^{2-} = 3d^8$$
$$CN^-$$ being a strong ligand pairing of electrons takes place.
Ref. image 2
Co-ordination number $$=4$$, Hybridisation $$dsp^2$$
Thus geometry is square planar

$$[Ni(H_2O)_6]^{2+} = 3d^8$$
$$H_2O$$ is a weak filed ligand so pairing of electrons does not take place.
Ref. image 3
Co-ordination number $$= 6$$, Hybridisation $$sp^3d^2$$
Thus geometry is octahedral.