MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 2

The formula of prussian blue is:

0%
$Na_{4}[Fe(CN)_{6}]$
0%
$Fe(CN)_{2}$
0%
$Fe(CNS)_{3}$
0%
$Fe_{4}[Fe(CN)_{6}]_{3}$

Explanation

Ferric ferrocyanide /Potassium ferrocyanide/ Sodium ferrocyanide is prussian blue.

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

Option D is correct,

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
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Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
0%
Assertion is correct, but Reason is incorrect.
0%
Assertion is incorrect, but Reason is correct.

Explanation

Assertion : It is correct statement because all three p-orbitals are involved in sp

$^3$ hybridisation and thus no p-orbital is left for side wise overlapping.
Reason: XeO

$_3$ is sp

$^3$ hybridisaed and Xe form three double through p

$\pi$ -d

$\pi$ overlapping.

Which of the following complexes is(are) paramagnetic ?

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$[Mn(CN_{6})]^{3-}$
0%
$[Cr(NH_{3})_{6}]^{3+}$
0%
$[Fe(CN)_{6}]^{4-}$
0%
$[CO(CN_{6})]^{3-}$

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

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Capped octahedral
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Octahedral
0%
Square planar
0%
Square pyramidal

$[Co(NH_3)_4Cl_2]$ possesses :

0%
Square planar geometry
0%
Tetrahedral geometry
0%
Tetrahedral nature
0%
Octahedral geometry

Which of the following is considered to be an anticancer species ?

Which of the following are isostructural ?

(I) $NO_{3}^{-}$

(II) $CO_{3}^{2-}$

(III) $CIO_{3}^{-}$

(IV) $SO_{3}$

(V) $XeO_{3}$

0%
(I) and (IV)
0%
(II) and (V)
0%
(III) and (IV)
0%
(IV) and (V)

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

0%
$sp > sp^{2}< sp^{3}$
0%
$sp > sp^{2} > sp^{3}$
0%
$sp < sp^{2}> sp^{3}$
0%
$sp < sp^{2}< sp^{3}$

Explanation

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

$s$ orbitals hold electrons more tightly to nucleus than

$p$ orbitals. That is the

$s$ orbitals are effectively more electronegative.

$s{ p }^{ 2 }$ 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

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

Which complex compound will give three isomers?

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$\left[ Fe(en)_{ 3 } \right] C{ l }_{ 3 }$
0%
$\left[ Co(en)_{ 2 }C{ l }_{ 2 } \right] C{ l }$
0%
$\left[ Fe(PP{ h }_{ 3 })_{ 3 }N{ H }_{ 3 }ClBr \right] C{ l }$
0%
$\left[ Co(PP{ h }_{ 3 })_{ 3 }Cl \right] C{ { l }_{ 3 } }$

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

0%
Linkage is isomerism
0%
Ionisation isomerism
0%
Co $\displaystyle -$ ordination isomerism
0%
No isomerism

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 :

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

Explanation

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

$\displaystyle Ti^{3+}\left ( aq \right )$ is violet while

$\displaystyle Ti^{4+}\left ( aq \right )$ is colourless because

$\displaystyle -$

0%
There is no crystal field effect in $\displaystyle Ti^{4+}$
0%
There energy difference between $\displaystyle t_{2g}$ and $\displaystyle e_{g}$ of $\displaystyle Ti^{4+}$ is quite high and does not fall in the visible region.
0%
$\displaystyle Ti^{4+}$ has $\displaystyle d^{0}$ configuration.
0%
$\displaystyle Ti^{4+}$ is very small in comparison to $\displaystyle Ti^{3+}$ and hance does not absorb any radiation.

Explanation

$Ti^{+3}$ :

$d^1$ configuration have one unpaired d-electron so it is colourful but

$Ti^{+4}$ :

$d^0$ configuration have zero unpaired d-electron so it is colourless.

Coordination compounds are mainly known for transition metals.

0%
True
0%
False

The IUPAC nomenclature for the complex

$Na[PtBrCl(NO_{2})(NH_{3})]$ is

0%
Sodium amminechlorobromonitro-N-platinum(II)
0%
Sodium nitrochlorobromoammine-N-platinate(II)
0%
Sodium amminebromochloronitro-N-platinate(II)
0%
Sodium amminebromochloronitro-N-platinum(II)

Explanation

The IUPAC name of the complex

$Na[PtBrCl(NO_2)(NH_3)]$ is sodium amminebromochloronitro-N-platinate(II).
The cation is sodium ion and the anion is the complex ion.
The ligands present are ammine, bromo, chloro, and nitro-N. Here N represents that the nitro group coordinates through the N atom.
The names of the ligands are written in the alphabetical order.
The oxidation state of platinum is +2. The oxidation state is written in Roman numerals inside the parenthesis.

The IUPAC name of

$[PtCl(NH_2CH_3)(NH_3)_2]Cl$ is :

0%
diamminechloro(methylamine)platinum(II)chloride
0%
(dimethylamine)chlorodiamminoplatinum(II)chloride
0%
bis(ammine)chloro(methylamine)platinate(II)chloride
0%
diaminechloro(mehylamine)platinum(II)chloride

Explanation

The IUPAC name of

$[PtCl(NH_2CH_3)(NH_3)_2]Cl$ is diamminechloro(methylamine)platinum(II)chloride.
The ligands present are ammine, chloro and methyl amine. The ligands are named according to the alphabetical order.
The prefix di indicates two.
The oxidation state of platinum is +2. The oxidation state is written in roman numerals inside parenthesis.

Type of isomerism exhibited by

$\displaystyle \left [ Cr\left ( NCS \right )\left ( NH_{3} \right )_{5} \right ]\left [ ZnCl_{4} \right ]$ is :

0%
coordination isomerism.
0%
linkage isomerism.
0%
ionization isomerism.
0%
both coordination and linkage isomerism.

Which one of the following platinum complexes is used in cancer chemotherapy?

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$Cis-[PtCl_{2}(NH_{3})_{2}]$
0%
$Trans-[PtCl_{2}(NH_{3})]$
0%
$[Pt(NH_{3})_{4}]^{2+}$
0%
$[PtCl_{4}]^{2-}$

Explanation

$cis-[PtCl_2(NH_3)_2]$ known as cis-platin.

Cisplatin is a chemotherapy medication used to treat several cancers. These include testicular cancer, ovarian cancer, cervical cancer, breast cancer, bladder cancer, head and neck cancer, oesophagal cancer, lung cancer, mesothelioma, brain tumours and neuroblastoma. It is given by injection into a vein.

Hence, the correct option is

$\text{A}$

The IUPAC name of

$\displaystyle \left [ CuCl_{4} \right ]^{2-}$ is tetrachlorido cuprate (II).

0%
True
0%
False

Explanation

True.

$[CuCl_4]^{-2}$

Here, chlorine is the ligand and there are

$4$ chlorine molecules. Chlorine is written as chlorido.

Oxidation state is

$x+4(-1)=-2$

$x-4=-2$

$\boxed {x=+2}$

So, IUPAC name is tetrachlorido cuprate (II)

Total number of space isomers of the formula of the above complex is :

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

Explanation

Loss of water on treatment with sulfuric acid indicates presence of water molecule in the ionization sphere.
Yellow precipitate of AgI on treatment with silver nitrate indicates presence of the iodide ion in the ionization sphere.
Thus the complex is

$[Co(en)_2Cl_2](H_2O)I$ .
Two geometrical isomers cis and trans are possible.
Cis isomer is optically active and exists in d and l forms.
Trans isomer is optically inactive .
Thus total three steroisomers are possible.

Four one–litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same room temperature and pressure. The ratio of total number of atoms of these gases present in the different flasks would be:

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1:1:1:1
0%
1:2:2:3
0%
2:1:2:3
0%
3:2:2:1

Explanation

Hint: At the same temperature, pressure, and volume, each of the given gas will have an equal number of moles.

Explanation:

$1\ mole\ H_2$ molecule contains $\rightarrow$ $2$ moles of $H$ atoms.

$1\ mole\ He$ molecule contains $\rightarrow$ $1$ mole of $He$ atoms.

$1\ mole\ O_2$ molecule contains $\rightarrow$ $2$ moles of $O$ atoms.

$1\ mole\ O_3$ molecule contains $\rightarrow$ $3$ moles of $O$ atoms.

Since, number of moles of atoms is directly proportional to the number of atoms, ratio of number of atoms is $2:1:2:3$

Final Answer:

Therefore the ratio of total number of atoms of all the gases is $2:1:2:3$

Hence, the correct answer is option $C.$

Which kind of isomerism is exhibited by octahedral

$[Co(NH_3)_4Br_2]Cl$ ?

0%
Geometrical and ionization
0%
Geometrical and optical
0%
Optical and ionization
0%
Geometrical only

Explanation

The octahedral complex

$[Co(NH_3)_4Br_2]Cl$ exhibits geometrical as well as ionization isomerism.
The geometrical isomers are cis and trans isomers.
The ionization isomers are

$[Co(NH_3)_4Br_2]Cl and [Co(NH_3)_4Cl]Br_2$ .

The stability of ferric ion is due to:

0%
$\text{Half field f-orbital}$
0%
$\text{Half field d-orbital}$
0%
$\text{Complete field f-orbital}$
0%
$\text{Complete field d-orbital}$

Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect?

0%
Carboxypeptidase-A is an enzyme and contains zinc
0%
Haemoglobin is the red pigment of blood and contains iron
0%
Cyanocobalamin is $B^{12}$ and contains cobalt
0%
Chlorophylls are green pigments in plants and contain calcium

Complex species that exhibits isomerism is :

0%
$[Ag(NH_{3})_{2}]^{+}$
0%
$[Co(NO_{2})(NH_{3})_{5}]^{2+}$
0%
$[PtCl_{2}(en)]$
0%
$[CoCl(NH_{3})_{5}]^{2+}$

Explanation

None of the given complexes can exhibit either geometrical or optical isomerism. But the complex

$[Co(NO_2)(NH_3)_5]^{2+}$ shows linkage isomerism as it contains nitro group which is ambidentate ligand and can be attached through either N atom or O atom.

The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt (III) chloride represent :

0%
Linkage isomerism
0%
Ionization isomerism
0%
Coordination isomerism
0%
No isomerism

Explanation

Pentaamminesulphatocobalt (III) bromide is

$[Co(SO_4)(NH_3)_4]Br$ and pentaamminesulphatocobalt (III) chloride is

$[Co(SO_4)(NH_3)_5]Cl$ .
The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt (III) chloride represent no isomerism.

Coordination compounds are mainly known for transition metals.

0%
True
0%
False

True structure is predicted by :

0%
valence bond approach
0%
Sidgwick approach
0%
hybrid orbital formation
0%
None of the above

S in

$\displaystyle SF_{6}$ is ______________ hybridized.

0%
$\displaystyle sp^{3}d^{3}$
0%
$\displaystyle sp^{3}d^{4}$
0%
$\displaystyle sp^{3}d^{2}$
0%
None of these

Explanation

$SF_6$ , sulphur has six valence electrons and there are six fluorine atoms that are monovalent. So, by applying the hybridization formula answer comes for 6 hybrid orbitals, which is

$sp^3d^2$ hybridisation.
1297346_147061_ans_c3d52ff7feeb4135abfc9fc081c2de60.PNG

Consider following complexes :

$[Fe(H_2O)_6]^{2+}$
II :

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

$[Ni(CO)_4]$
IV :

$[Ni(H_2O)_4]^{2+}$
V :

$[Ni(CN)_4]^{2-}$

Out of this, select the complexes with octahedral geometry :

0%
$[Fe(H_2O)_6]^{2+}$
0%
$[Fe(CN)_6]^{4-}$
0%
$[Ni(H_2O)_4]^{2+}$
0%
$[Ni(CN)_4]^{2-}$

Explanation

The complexes,

$[Fe(H_2O)_6]^{2+}$ and

$[Fe(CN)_6]^{4-}$ have octahedral geometry.

The central iron atom undergoes

$sp^3d^2$ hybridization in

$[Fe(H_2O)_6]^{2+}$ . It results in outer orbital complex or high spin complex.

The central iron atom undergoes

$d^2sp^3$ hybridization in

$[Fe(CN)_6]^{4-}$ . It results in outer orbital complex or high spin complex.

Therefore, option A and B both are correct.

If the metal ion of

$[Fe(CN)_6]^{3-}$ has

${ d }^{ x }$ electronic configuration then value of

$x$ is:

0%
2
0%
4
0%
5
0%
3

Explanation

$[Fe(CN)_6]^{3-}$

$x+(6x+1)+3=0$

$x=3$

$Fe^{3+}=d^5$

$x=5$

Complex compound

$[Co(SCN)_{2}(NH_{3})_{4}]Cl$ exhibits.

0%
Ionization isomerism
0%
Geometrical isomerism
0%
Optical isomerism
0%
Linkage isomerism

Explanation

The complex

$[Co(SCN)_2(NH_3)_4]Cl$ exhibits ionization isomerism, geometrical isomerism as well as linkage isomerism.
Linkage isomerism is possible as SCN ligand is ambidentate ligand and has two donor atoms N and S.
Ionization isomerism is possible due to exchange of ions inside and outside the coordination sphere.
Two geometrical isomers cis and trans are possible. In the cis isomer, the two SCN ligands are adjacent to each other and in the trans isomer, they are opposite to each other.

Identify the type of isomerism shown by

$[Co(NH_3)_4Cl_2]^+$ :

0%
$geometrical$ $cis$ , $trans$
0%
$geometrical$ $fac-$ , $mer-$
0%
$optical$
0%
None of the above

Explanation

$[Co(NH_3)_4Cl_2]^+$ will show cis, trans geometrical isomerism.

State whether the given statement is true or false:

Use of

$cis$ -platin in cancer treatment also causes severe kidney damage.

0%
True
0%
False

State whether the given statement is true or false:

The

$d_{xy}, d_{yz}$ and

$d_{xz}$ orbitals lie between the axes and collectively called

$t_{2g}$ orbitals.

0%
True
0%
False

Explanation

The

$d_{xy}, d_{yz}$ and

$d_{xz}$ orbitals lie between the axes and collectively called

$t_{2g}$ orbitals.

This is true statement.

Chrome green is _____

0%
$PbCrO_4 + Fe^{III} [Fe^{II} (CN)_6]^-$
0%
$Cr_2O_5$
0%
$PbCrO_4$
0%
$FeCrO_4$

Explanation

Chrome green is

$PbCrO_4 + Fe^{III} [Fe^{II} (CN)_6]^-$ .

It is a composite pigment consisting of a combination of chrome yellow and prussian blue.

Hence, option A is correct.

The colour of a complex compound is due to :

0%
promotion of $3d$ -electrons of the central atom/ion to $4s$ -orbital.
0%
promotion of $3d$ -electrons of the central atom/ion to $4p$ -orbitals.
0%
promotion of $3d$ -electrons of the central atom/ion to $d$ -orbitals.
0%
promotion of $4s$ -electrons of the central atom/ion to $4p$ -orbitals.

Explanation

Option (C) is correct.
The colour of a complex compound is due to promotion of

$3d$ -electrons of the central atom/ion to

$d$ -orbitals.
The colour of the complex compound is due to

$d-d$ electronic transition.The

$d$ -orbitals of a free transition metal atom or ion are degenerate (all have the same energy.) When transition metals form coordination complexes, the d-orbitals of the metal interact with the electron cloud of the ligands in such a manner that the d-orbitals become non-degenerate (not all having the same energy).

Turnbull's blue is

$Fe_3{[Fe{(CN)}_6]}_2$ .

0%
True
0%
False

Explanation

$[Fe(CN)_6]^{4-}+ Fe^{3+}$ ---- (Prussian blue)

$[Fe(CN)_6]^{3-}+ Fe^{2+} \rightarrow Fe_3{[Fe{(CN)}_6]}_2$ ---- (Turnbull's Blue)

Hence, the given statement is true.

State whether the given statement is true or false:

Orange is the complimentary colour of blue spectral colour.

0%
True
0%
False

When

$KCN$ is added to

$CuS{ O }_{ 4 }$ solution, there is formation of the stable water soluble complex. This complex is :

0%
${ K }_{ 4 }\left[ Cu{ \left( CN \right) }_{ 6 } \right]$
0%
${ K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right]$
0%
${ K }_{ 2 }\left[ Cu{ \left( CN \right) }_{ 4 } \right]$
0%
${ K }\left[ Cu{ \left( CN \right) }_{ 3 } \right]$

Explanation

When

$KCN$ is added to

$CuS{ O }_{ 4 }$ solution, the following reaction occurs.

$4KCN+CuSO_4 \rightarrow { K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right]$

Hence, the complex formed is

${ K }_{ 3 }\left[ Cu{ \left( CN \right) }_{ 4 } \right]$ .

True structure is predicted by:

0%
Valence-bond approach
0%
Sidgwick approach
0%
Hybrid orbital
0%
None of the above

Turnbull's blue is :

0%
$Fe_3[Fe(CN)_6]_2$
0%
$K_4Fe(CN)_6$
0%
$K_3Fe(CN)_6$
0%
$Na_4Fe(CN)_6$

Explanation

Turnbull's blue is

$Fe_3[Fe(CN)_6]_2$ .
The structure is as follows:

In the complex

$CoCl_3\cdot 5NH_3$ , one

$Cl^-$ ion satisfies primary and secondary valencies of cobalt(III).

0%
True
0%
False

Explanation

Complex

$CoCl_3\cdot 5NH_3$ has octahedral structure so atleast one

$Cl^-$ ion satisfies primary valency of cobalt(III) and remaining chloride ions satisfies secondary valency.

Select the correct statements.

0%
van der Waals' radii is always larger than the covalent radii
0%
The bond length of a particular bond depends on the state of hybridization of the involved atoms
0%
When %s - character increases, then bond length increases
0%
All are incorrect

The complex used as an anticancer agent is :

0%
$trans$ - $[Co(NH_3)_3Cl_3]$
0%
$cis$ - $[PtCl_2(NH_3)_2]$
0%
$cis$ - $K_2[PtCl_2Br_2]$
0%
$Na_2CO_3$

Which of the following is first formed inert gas compound ?

0%
$Xe[Pt F_6]$
0%
$Kr[Pt F_6]$
0%
$XeF_2$
0%
$XeO_3$

Explanation

The ionisation energy of Xe is lower than that of Kr. Hence upon reaction with

$\\ { [PtF }_{ 6 }]\\$ ,

$\\ { Xe[PtF }_{ 6 }]\\$ should get formed first.

Linear combination of two hybridzed orbitals, belonging to two atoms and each having one electron leads to

0%
Sigma - bond
0%
Double bond
0%
Co-ordinate covalent bond
0%
Pi - bond

If a transition-metal compound absorbs violet-indigo radiation in the visible region. Its color would be :

0%
green.
0%
yellow.
0%
orange.
0%
blue.

Both geometrical and optical isomerisms are shown by :

0%
$[Co(en)_2Cl_2]^+$
0%
$[Co(NH_3)_5Cl_2]^{2+}$
0%
$[Co(NH_3)_4Cl_2]^+$
0%
$[Cr(OX)_3]^{3-}$

Explanation

$[Co(en)_2Cl_2]^+$ shows cis/trans isomerism.
also it have 2 optical isomers as

Transition metal compounds are usually colored. This is due to the electronic transition :

0%
from $p$ -orbital to $s$ -orbital.
0%
from $d$ -orbital to $s$ -orbital.
0%
from $p$ -orbital to $p$ -orbital.
0%
within the $d$ -orbitals.

10 ml of

$KMnO_4$ Solution is required it completely oxidise acidic solution of 30 ml of 1.5 Volume strength

$H_2O_2$ .

Calculate normality of

$KMnO_4$ Solution :

0%
0.4
0%
0.65
0%
0.8
0%
0.19

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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