MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 14

The number of isomers possible for distributed borazine,

$B_3N_3H_4X_2$ is:

0%
$3$
0%
$4$
0%
$6$
0%
$5$

$[Co(NH_3)_4(NO_2)_2]Cl$ exhibits:

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.

1575254_1730992_ans_137f3f12ea714201b414c73ee167757e.jpg

The primary and secondary valency of copper in the complex

$Cu(NH_3)_4SO_4$ are :

0%
$2, 4$
0%
$4, 2$
0%
$0, 4$
0%
$1, 4$

$[Fe(CN)_6]^{4-}$ ion is:

0%
square planar
0%
octahedral
0%
tetrahedral
0%
none of these

Two complexes given below are :

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

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.

1864909_1746476_ans_cd74cd66f449400abeb05350f1ef1b2b.png

Which of the following is/are outer orbital complex(es)?

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?

0%
$I_3^-$
0%
$H_2O_2$
0%
$B_2H_6$
0%
$[Ni(CN)_4]^{2-}$

Turnbull's blue is a compound having formula:

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:

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:

0%
$4$
0%
$2$
0%
$3$
0%
$1$

The formulae of

$A$ and

$B$ are:

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:

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 $1.73\ B.M$

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

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

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

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?

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

0%
Rn
0%
Ni
0%
Fe
0%
Co

The pair of complex compounds

$[Cr(H_2O)_6]Cl_3$ and

$[Cr(H_2O)_5Cl]Cl_2H_2O$ are an example of

0%
Linkage isomerism
0%
Ionization isomerism
0%
Coordination isomerism
0%
Hydrate isomerism

The formula of dichlorobis(urea)copper(II) is:

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

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?

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

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.
1672574_1760661_ans_1c2fde1ba6764143b65bf1fc41c130d2.png

Heamoglobin is a complex of

0%
$Fe^{3+}$
0%
$Fe^{2+}$
0%
$Fe%_4++$
0%
$Cu^{2+}$

The formula of alum is

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?

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$H-C\equiv C-C\equiv-H$
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?

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.
1672646_1760673_ans_6a207a42ba3444009da71e45f7506f5d.png

1672646_1760673_ans_6a207a42ba3444009da71e45f7506f5d.png

Which of the following reacts with haemoglobin in the blood to form carboxyhaemoglobin

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

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?

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

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:

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?

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 :

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:

Which of the following does not represent the isostructural pair?

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

0%
Geometrical, ionization
0%
Ionization, linkage
0%
Linkage, optical
0%
Geometrical, optical

In which compound vacant hybride orbital take part in bonding :

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 :

$NH_{3}, PH_{3}, H_{2}S, SO_{2}, SO_{3}, BF_{3}, PCl_{3}, IF_{7}, P_{4}, H_{2}$

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?

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

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

0%
cobalt
0%
magnesium
0%
iron
0%
nickel

Mark the incorrect match.

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

$Ni = 28$ )

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

$[Ni(CO)_4]$ , 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

$sp^3$ .

2036504_1964276_ans_f70f56faeb434f1e82ac205f0bf06651.png

Which among the following will be named as dibromobis(ethylenediamine) chromium

$(III)$ bromide?

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?

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:

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.

2036497_1964267_ans_9759e6d80b2c4afbb494de9d4644564e.png

The formula of the complex diamminechloro-(ethylenediamine)nitroplatinum

$(IV)$ chloride is:

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

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:

$Diamminetetraaquacobalt\ (III) \ chloride$

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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