MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 14 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 14
The number of isomers possible for distributed borazine, $$B_3N_3H_4X_2$$ is:
Report Question
0%
$$3$$
0%
$$4$$
0%
$$6$$
0%
$$5$$
$$[Co(NH_3)_4(NO_2)_2]Cl$$ exhibits:
Report Question
0%
ionisation isomerism, geometrical isomerism and optical isomerism
0%
linkage isomerism, geometrical isomerism and optical isomerism
0%
linkage isomerism, ionisation isomerism and optical isomerism
0%
linkage isomerism, geometrical isomerism and ionisation isomerism.
Explanation
It has linkage isomerism, geometrical isomerism and ionisation isomerism.
The given compound may have linkage isomerism due to presence of $$NO_2$$ group which may be in the form $$–NO_2$$ or $$–ONO$$.
$$[Co(NH_3)_4(NO_2)_2]Cl$$
$$[Co(NH_3)_4(ONO)_2]Cl$$
It may have ionisation isomerism due to presence of two ionisable group $$–NO_2$$ and $$–Cl$$.
$$[Co(NH_3)_4(NO_2)_2]Cl$$
$$[Co(NH_3)_4(NO_2)Cl]NO_2$$
It has two geometrical isomers are shown in figure.
The primary and secondary valency of copper in the complex $$Cu(NH_3)_4SO_4$$ are :
Report Question
0%
$$2, 4$$
0%
$$4, 2$$
0%
$$0, 4$$
0%
$$1, 4$$
$$[Fe(CN)_6]^{4-}$$ ion is:
Report Question
0%
square planar
0%
octahedral
0%
tetrahedral
0%
none of these
Two complexes given below are :
Report Question
0%
geometrical isomers
0%
position isomers
0%
optical isomers
0%
identical
$$S_{1}$$ : The species $$\left [ CuCl_{4} \right ]^{2-}$$ exists but $$\left [ Cul_{4} \right ]^{2-}$$ does not.
$$S_{2}: \left [ RhCl\left ( Ph_{3}P_{3} \right ) \right ]$$ and $$\left [ Ni(CO)_{4} \right ]$$ both are tetrahedral and diamagnetic.
$$S_{3}:N\left ( Me \right )_{3}$$ and $$N\left ( SiMe_{3} \right )_{3}$$ are isostructural
Report Question
0%
T T F
0%
T F F
0%
F T F
0%
F T T
Explanation
$$S_1: I^−\text{ ion is a stronger reducing agent than }Cl^−\text{ ion. It reduces }Cu^{2+} to \ Cu^0+ \ ion.$$
$$S_2:\text{ Both diamagnetic but }[Ni(CO)_4]\text{ is tetrahedral and }[RhCl(Ph_3P_3)]\text{ is a square planar.}$$
Option B is correct.
Which of the following is/are outer orbital complex(es)?
Report Question
0%
$$[Ni(NH_3)_6]^{2+}$$
0%
$$[Co(NH_3)_6]^{3+}$$
0%
$$[Fe(H_2O)_6]^{3+}$$
0%
$$[Co(CN)_6]^{3-}$$
Which of the following have planar structure?
Report Question
0%
$$I_3^-$$
0%
$$H_2O_2$$
0%
$$B_2H_6$$
0%
$$[Ni(CN)_4]^{2-}$$
Turnbull's blue is a compound having formula:
Report Question
0%
$$KFe(III)[Fe(II)(CN)_6]$$
0%
$$KFe(II)[Fe(III)(CN)_6]$$
0%
$$Na_2[Fe(CN)_5NO]$$
0%
$$Fe_4[Fe(CN)_6]_3$$
The geometry of $$[Ni(CN)_4]^{2-}$$ and $$[NiCl_4]^{2-}$$ are:
Report Question
0%
both square planar
0%
both tetrahedral
0%
tetrahedral and square planar respectively
0%
square planar and tetrahedral respectively
Number of unpaired electrons in $$[Co(F)_6]^{3-}$$ is:
Report Question
0%
$$4$$
0%
$$2$$
0%
$$3$$
0%
$$1$$
The formulae of $$A$$ and $$B$$ are:
Report Question
0%
$$K_{2}[Cu(CN)_{4}],[Cu(NH_{3})_{4}]SO_{4}$$
0%
$$K_{3}[Cu(CN)_{4}], [Cu(NH_{3})_{4}]SO_{4}$$
0%
$$K[Cu(CN)_{4}], [Cu(NH_{3})_{6}]SO_{4}$$
0%
$$K_{3}[Cu(CN)_{6}], [Cu(NH_{3})_{6}]SO_{4}$$
The formula of the compound which gives violet colour in Lassaigne's test for sulphur with sodium nitroprusside is:
Report Question
0%
$$Na_{4}[Fe(CN)_{6}S]$$
0%
$$Na_{4}[Fe(CN)_{5}NCS]$$
0%
$$Na_{4}[Fe(CN)_{5}NOS]$$
0%
$$Na_{2}[Fe(CN)_{5}NOS]$$
Explanation
While preparation of Lassaigne's extract, sulphur from the organic compound reacts with sodium to form sodium sulphide.
$$Na+S\rightarrow Na_2S$$
Sodium sulphide reacts with sodium nitroprusside to form violet colour compound, which confirms presence of Sulphur.
$$Na_2S+Na_2[Fe(CN)_5NO]\rightarrow Na_4[Fe(CN)_5NOS]$$
Hence the correct option is $$(C)$$.
Match the List $$I$$ and List $$II$$ and pick the correct matching from the codes given below:
List $$I$$
List $$II$$
$$[Ag(CN)_2]^-$$
Square planer and $$1.73\ B.M$$
$$[Cu(CN)_4]^{3-}$$
Linear and zero
$$[Cu(CN)_6]^{4-}$$
Octahedral and zero
$$[Cu(NH_3)_4]^{2+}$$
Tetrahedral and zero
$$[Fe(CN)_6]^{4-}$$
Octahedral and $$1.73\ B.M$$
Report Question
0%
$$2, 4, 5, 1, 3$$
0%
$$5, 4, 1, 3, 2$$
0%
$$1, 3, 4, 2, 5$$
0%
$$4, 5, 2, 1, 3$$
When $$ CuSO_4 $$ solution is added to $$ K_4[Fe(CN)_6] $$ the formula of the product formed is
Report Question
0%
$$ Cu_2Fe(CN)_6 $$
0%
$$ KCN $$
0%
$$ Cu(CN)_3 $$
0%
$$ Cu(CN)_2 $$
Explanation
When $$ CuSO_4 $$ solution is added to $$ K_4[Fe(CN)_6] $$ the formula of the product formed is $$ Cu_2[Fe(CN)_6]. $$
$$ 2CuSO_4 +K_4(Fe(CN)_6] \rightarrow Cu_2[Fe(CN)_6] + 2K_2SO_4 $$
Blood haemoglobin contains the metal
Report Question
0%
$$ Al $$
0%
$$ Mg $$
0%
$$ Cu $$
0%
$$ Fe $$
Explanation
Blood haemoglobin contains the metal
$$ Fe $$.
Hence, Option "D" is the correct answer.
$$[Co(NH_3)_5Br]SO_4$$ and $$[Co(NH_3)_5SO_4]Br$$ are examples of which type of isomerism
Report Question
0%
Linkage
0%
Geometrical
0%
Ionization
0%
Optical
Explanation
The two given compounds have the same composition but in solution, both will give different ions. The isomerism is known as ionisation isomerism.
$$[Co(NH_3)_5Br]SO_4\to [Co(NH_3)_5Br]^{2+}+SO_4^{2-}$$
$$[Co(NH_3)_5SO_4]Br\to [Co(NH_3)_5SO_4]^++Br^-$$
Which would exhibit co-ordination isomerism?
Report Question
0%
$$[Cr(NH_3)_6][Co(CN)_6]$$
0%
$$[Co(en)_2Cl_2]$$
0%
$$[Cr(NH_3)_6]Cl_3$$
0%
$$[Cr(en)_2Cl_2]^+$$
The substance used in cancer therapy is
Report Question
0%
Rn
0%
Ni
0%
Fe
0%
Co
Explanation
Cobalt-60 isotope is used in cancer therapy.
The pair of complex compounds $$[Cr(H_2O)_6]Cl_3$$ and $$[Cr(H_2O)_5Cl]Cl_2H_2O$$ are an example of
Report Question
0%
Linkage isomerism
0%
Ionization isomerism
0%
Coordination isomerism
0%
Hydrate isomerism
The formula of dichlorobis(urea)copper(II) is:
Report Question
0%
$$[CuCl_2({O=C(NH_2)_2})_2]$$
0%
$$[Cu(O=C(NH_2)_2)Cl]Cl$$
0%
$$[Cu({O=C(NH_2))_2}]Cl2$$
0%
$$[CuCl_2({O=C(NH_2))_2H_2}]$$
In solid $$ CuSO_5H_2O $$ copper is co-ordinated to
Report Question
0%
five water molecules
0%
four water molecules
0%
one sulphate anion
0%
one water molecule
Explanation
$$ CuSO_4.5H_2O $$ is a crystalline salt. Four $$ H_2 O $$ molecules attach to copper forming a square planar symmetry. The fifth water molecule is attached through hydrogen bonding between one of the co-ordinated water molecules and one of the sulphate ion.
In coordination compound $$[Co(en)_2Cl_2]Cl$$ which is false?
Report Question
0%
Show geometrical isomerism
0%
Show optical isomerism
0%
Shown ionic isomerism
0%
A octahedral complex
0%
A cationic complex
Explanation
When coordinate compounds give different ions in solution then it produces ionic isomerism while this situation is not present in $$[Co(en)_2Cl_2]Cl$$.
$$[Co(en)_2Cl_2]Cl\to [Co(en)_2Cl_2]^++Cl^-$$
Complex ion shows geometrical isomerism and cis form shows optical isomerism. An octahedral complex since there are six donor atom and complex part is cation.
$$[Pt(NH_{3})_{4}]Cl_{2}$$ is
Report Question
0%
Square planar
0%
Tetrahedral
0%
Pyramidal
0%
Pentagonal
Explanation
$$[Pt(NH_3)_4]Cl_2\to Pt^{2+}\to [Xe]5d^86s^0$$. Due to the presence of the ligand pairing of electrons take place.
Since hybridisation is $$dsp^{2}$$ so it is square planar.
Heamoglobin is a complex of
Report Question
0%
$$Fe^{3+}$$
0%
$$Fe^{2+}$$
0%
$$Fe%_4++$$
0%
$$Cu^{2+}$$
The formula of alum is
Report Question
0%
$$K_2SO_4.Al_2(SO_4)_3. 24H_2O$$
0%
$$K_2[Fe(CN)_6]$$
0%
$$K_2SO_4.Al_2(SO_4)_3. 6H_2O$$
0%
$$Na_2CO_3. 10H_2O$$
Explanation
General formula for alum is $$M_2SO_4. R_2(SO_4)_3 .24H_2O$$
$$M=$$ mono valent cation $$(K^+ , Na^+,....)$$
$$R=$$ Trivalent cation $$(Al^{3+} , Fe^{3+},......)$$
Hence, $$K_2SO_4.Al_2(SO_4)_2.24H_2O$$ represent an alum.
Which of the following compounds contain all the carbon atoms in the same hybridization state?
Report Question
0%
$$H-C\equiv C-C\equiv-H$$
0%
$$CH_3-C\equiv C-CH_3$$
0%
$$CH_2=C=CH_2$$
0%
$$CH_2=CH-CH=CH_2$$
Explanation
Only in $$H-C\equiv C-C\equiv-H$$ and $$CH_2=CH-CH=CH_2$$ compounds all carbon atoms are in same hybridization state i.e., $$sp$$ and $$sp_2$$ hybridized, respectively.
$$H-\overset{sp}C\equiv \overset{sp}C-\overset{sp}C\equiv-H$$
$$\overset{sp^2}CH_2=\overset{sp^2}CH-\overset{sp^2}CH=\overset{sp^2}CH_2$$
Which reagent can be used to identity nickel ion?
Report Question
0%
Resorcinol
0%
Dimethyl glyoxime $$[DMG]$$
0%
Diphenyl benzidine
0%
Potassium ferrocyanide
Explanation
$$Ni$$ reacts with dimethylglyoxime to give red precipitate of nickel dimethyl glyoxime complex.
Which of the following reacts with haemoglobin in the blood to
form carboxyhaemoglobin
Report Question
0%
$$CO$$
0%
$$CO_2$$
0%
$$HCOOH$$
0%
$$H_2CO_3$$
Explanation
$$\text{Carboxyhemoglobin or carboxyhaemoglobin is a stable complex of carbon monoxide and hemoglobin}$$ that forms in red blood cells upon contact with carbon monoxide. Carboxyhemoglobin is often mistaken for the compound formed by the combination of carbon dioxide and hemoglobin, which is actually carbaminohemoglobin.
Which is the example of branch isomerization
Report Question
0%
$$C-\overset{C}{\overset{|}{C}}-C-C-C$$ and $$C-\overset{C}{\overset{|}{\underset{C}{\underset{|}{C}}}}-C$$
0%
$$C-\underset{C}{\underset{|}{C}}-\overset{C}{\overset{|\\}{C}}$$ and $$\overset{C}{\overset{|}{C}}-\underset{C}{\underset{|}{C}}-C$$
0%
0%
$$C-C-C-C$$ and $$C-C-\underset{C}{\underset{|}{C}}$$
In which of the following species maximum atom can lie in same plane?
Report Question
0%
$$XeF_{2}O_{2}$$
0%
$$PCl_{5}$$
0%
$$AsH_{4}^{+}$$
0%
$$XeF_{4}$$
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
Report Question
0%
(I), (III) and (IV)
0%
(II), (III) and (IV)
0%
(I) and (II)
0%
All are correct
Explanation
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 .
$$[CoCl_{2}(NH_{3})_{4}]^{+} + Cl^{-}\rightarrow [CoCl_{3}(NH_{3})_{3}] + NH_{3}$$. If in this reaction two isomers of the product are obtained, which is true for the initial (reactant) complex:
Report Question
0%
Compound is in cis-form
0%
Compound is in trans-form
0%
Compound is in both (cis and trans) form
0%
Can't be predicted
Which of the following is incorrectly matched?
Report Question
0%
$$[FeF_{6}]^{3-}, 5$$
0%
$$[Cr(en)_{3}]^{2+}, 2$$
0%
$$[Co(NH_{3})_{6}]^{3+}, 4$$
0%
$$[Mn(H_{2}O)_{6}]^{2+}, 5$$
Select the correct statement :
Report Question
0%
$${HSO_5}^-$$ ion has one $$S-O-H$$ linkage
0%
Number of $$B-O-B$$ linkages in borax is equal to number of $$P-O-P$$ linkages in $$P_4O_{10}$$
0%
Hybridization of both sulphur in $$H_2S_2O_5$$ (pyrosulphurous acid) is same but oxidation state of both sulphur are different
0%
Tetra-polyphosphoric acid has four $$P-O-P$$ and no $$P-P$$ linkage
Select the incorrect match:
Report Question
0%
0%
0%
0%
Explanation
In see-saw structure, as lone pair lies in equatorial position, hence there is more reduction in equatorial bonds than in axial bonds.
Which of the following does not represent the isostructural pair?
Report Question
0%
$$SF_{5}^{-}$$ and $$IF_{5}$$
0%
$$ClO_{2}F_{3}$$ and $$SOF_{4}$$
0%
$$SeF_{3}^{+}$$ and $$XeO_{3}$$
0%
None
The types of isomerism shown by the complex compound is
Report Question
0%
Geometrical, ionization
0%
Ionization, linkage
0%
Linkage, optical
0%
Geometrical, optical
In which compound vacant hybride orbital take part in bonding :
Report Question
0%
$$B_2H_6$$
0%
$$Al_2Cl_6$$
0%
$$C_2H_5Cl$$
0%
$$H_3BO_3$$
Consider the following compounds and select the incorrect statement from the following :
$$NH_{3}, PH_{3}, H_{2}S, SO_{2}, SO_{3}, BF_{3}, PCl_{3}, IF_{7}, P_{4}, H_{2}$$
Report Question
0%
Six molecules out of given compound involves hybridisation
0%
Three molecules are hypervalent compounds
0%
Six molecules out of above compounds are non-planar in structure
0%
Two molecules out of given compounds involves $$(d \pi- p \pi)$$ bonding as well as also involves $$(p \pi - p \pi)$$ bonding
Which of the following match are incorrect?
Report Question
0%
$$[VCl_{3}(NMe_{3})_{3}], \sqrt {8} BM$$
0%
$$[CrCl_{3}(NMe_{3})_{3}], \sqrt {15} BM$$
0%
$$[Cu(CN)(NO_{2}) (NH_{3})(Py)], \sqrt {3} BM$$
0%
$$[Co(ox) (H_{2}O)_{4}]^{+}, \sqrt {24} BM$$
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
Report Question
0%
$$ [CO(NH_3)_5NO_3]Cl_2 $$
0%
$$ [CO(NH_3)_5Cl]Cl NO_3Cl $$
0%
$$ [CO(NH_3)_5Cl]NO_2 $$
0%
None of the above
Chlorophyll contains
Report Question
0%
cobalt
0%
magnesium
0%
iron
0%
nickel
Mark the incorrect match.
Report Question
0%
Insulin- Zinc
0%
Haemoglobin- Iron
0%
Vitamin $$B_{12}$$- Cobalt
0%
Chlorophyll- chromium
Explanation
Insulin contains Zinc, Haemoglobin contains Iron, Vitamin $$\mathrm{B_{12}}$$ contains Cobalt and Chlorophyll contains Magnesium.
Hence, Option "D" is the correct answer.
Among $$[Ni(CO)_4], [Ni(CN)_4]^{2-}, [NiCl_4]^{2-}$$ species, the hybridization states at the $$Ni$$ atom are, respectively: (Atomic number of $$Ni = 28$$)
Report Question
0%
$$sp^3, dsp^2, sp^3$$
0%
$$sp^3, sp^3, dsp^3$$
0%
$$dsp^2 , sp^3, sp^3$$
0%
$$sp^3, dsp^2, dsp^2$$
Explanation
In $$[Ni(CO)_4]$$, oxidation state of $$'CO'$$ is $$0$$. So oxidation state of $$'Ni'$$ is also $$0$$. $$4s$$ and $$4p$$ orbitals are involved in hybridization and so its hybridization is $$sp^3$$.
Which among the following will be named as dibromobis(ethylenediamine) chromium $$(III)$$ bromide?
Report Question
0%
$$[Cr(en)_2Br_2]Br$$
0%
$$[Cr(en)Br_4]^-$$
0%
$$[Cr(en)Br_2]Br$$
0%
$$[Cr(en)_3]Br_3$$
Which of the following isomers will give while precipitate with $$BaCl_{2}$$ solution?
Report Question
0%
$$[Co(NH_{3})_{5}SO_{4}]Br$$
0%
$$[Co(NH_{3})_{5}Br]SO_{4}$$
0%
$$[Co(NH_{3})_{4}(SO_{4})_{2}]Br$$
0%
$$[Co(NH_{3})_{4}Br(SO_{4})]$$
Explanation
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.
Geometrical shapes of the complexes formed by the reaction of $$Ni^{2+}$$ with $$Cl^-, CN^-$$ and $$H_2O$$, respectively, are:
Report Question
0%
Octahedral, tetrahedral and square planar
0%
Tetrahedral, square planar and octahedral
0%
square planar, tetrahedral and octahedral
0%
Octahedral, square planar and octahedral
Explanation
$$[NiCl_4]^{2-} = 3d^8$$
$$Cl^-$$ 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.
The formula of the complex diamminechloro-(ethylenediamine)nitroplatinum $$(IV)$$chloride is:
Report Question
0%
$$[Pt(NH_3)_2Cl(en)NO_2]Cl_2$$
0%
$$Pt[Pt(NH_3)_2(en)Cl_2NO_2]$$
0%
$$Pt[(NH_3)_2(en)NO_2]Cl_2$$
0%
$$Pt[(NH_3)_2(en)NO_2Cl_2]$$
The IUPAC name of the complex $$[Co(NH_3)_2(H_2O)_4]Cl_3$$ is
Report Question
0%
Diamminetetraaquacobalt (III) trichloride
0%
Diamminetetraaquacobalt (II) chloride
0%
Diamminetetraaquacobalt (III) chloride
0%
Tetraaquadiamminecobalt (III) trichloride
0%
Tetraaquadiamminecobalt (II) chloride
Explanation
First of all, the compound has complex positive part therefore, according to IUPAC conventions, positive part will be named first. Secondly, in writing name of complex, ligands are named first in alphabetical order, irrespective of its charge, hence -ammine will be written prior to "aqua.
Oxidation state of $$Co$$ be "x"
$$x+2\times0+4\times0=+3$$
$$x=+3$$
Therefore, name of the complex is:
$$Diamminetetraaquacobalt\ (III) \ chloride$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Engineering Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
