MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 3

Hypo is used in photography because it is :

0%
A strong reducing agent
0%
A strong oxidizing agent
0%
A strong Complexing agent
0%
Photo sensitive Compound

Explanation

Correct Answer: Option C

Explanation:

Hypo forms the complex $[Ag(S_2O_3)_{2}]^{3-}$ from insoluble $Ag$ .

$[Ag(S_2O_3)_{2}]^{3-}$ is formed by treatment of $Na_2S_2O_3. 5H_2O$ with $AgBr$ .

Therefore, Hypo is used in photography because it is a strong Complexing agent.

Hence, the correct answer is option $C$ .

Select the correct I.U.P.A.C. name for [PtCl

$_6$ ]

$^{2-}$ :

0%
Hexachloridoplatinate (IV) ion
0%
Hexachloridoplatinum (II) ion
0%
Hexachloridoplatinate (2-) ion
0%
Hexachloridoplatinum (2-) ion

Explanation

The stoek system puts the calculated oxidation number of the metal ion as a Roman numeral in parentheses after the name of the metal This is the more common convention, although there are cases in which it is difficult to assign oxidation numbers.
The Ewing-Bassett system puts the charge on the coordination sphere in parentheses after the name of the metal. This convention is used by chemical abstracts and offers an unambiguous identification of the species.
[PtCl

$_6$ ]

$^{2-}$
Hexachloridoplatinate (2-) ion
Hexachloridoplatinate (IV) ion

Find the ratio of geometrical isomers in [M(AA)

$_2$ b

$_2$ ] and optical isomers of [M(AA)

$_3$ ].

0%
1
0%
2
0%
1.5
0%
2.5

Explanation

The complex [M(AA)

$_2$ b

$_2$ ] has 2 geometrical isomers cis and trans isomers.
The complex [M(AA)

$_3$ ] has two optical isomers d and l forms.

$\displaystyle{\frac{2}{2}}$ = 1
Thus, the ratio of geometrical isomers in [M(AA)

$_2$ b

$_2$ ] and optical isomers of [M(AA)

$_3$ ] is 1.

Silver halides are used in photography because they are -

0%
Photosensitive
0%
Soluble in hypo solution
0%
Soluble in $NH_4OH$
0%
Insoluble in acids

Which of the following inert gas has the smallest ionization energy ?

0%
He
0%
Ne
0%
Ar
0%
Xe

Mg is an important component of which biomolecule occurring extensively in living world ?

0%
Hemoglobin
0%
Chlorophyll
0%
Florigen
0%
ATP

Developed photography film is due to :

0%
$AgBr$
0%
$[Ag(S_2O_3)_2]^{3+}$
0%
$Ag$
0%
$Ag_2O$

Secondary valencies correspond to ____________ of the metal atom and are satisfied by ligands.

0%
coordination number
0%
valency
0%
element
0%
molecules

Total number of stereoisomers of

$[Co(acac)_{2}BrCl]^{\ominus}$ are.

0%
$4$
0%
$3$
0%
$6$
0%
$2$

Explanation

Option (B) is correct.

Total number of stereo isomers of

$[Co(acac)_{2}BrCl]^{-}$ are 3 i.e. cis,trans,optical.

Chlorobenzene is o, p -directing in electrophilic substitution reaction. The directing influence is explained by

0%
+M of Ph
0%
+I of Cl
0%
+M of Cl
0%
-I of Ph

Calculate the octahedral crystal field splitting energy in kJ/mol for

$[Fe(CN)_6]^{4-}$ , if the wavelength of the most intensely absorbed light is 305 nm.

0%
392 kJ/mol
0%
643 kJ/mol
0%
238 kJ/mol
0%
456 kJ/mol

Amongest

$N{ O }_{3}^{-}\,, AsO_{3}^{3-}\, CO_{3}^{2-},\, Cl{ O }_{3}^{-},\, SO_{3}^{2-}$ and

$BO_{3}^{3-}$ , the non - planar species are:

0%
$CO_{3}^{2-}, SO_{3}^{2-}$ and $BO_{3}^{3-}$
0%
$AsO_{3}^{3-},\, ClO_{3}^{-}$ and $SO_{3}^{2-}$
0%
$N{ O }_{3}^{-}\, CO_{3}^{2-},\, CO_{3}^{2-}$ and $BO_{3}^{3-}$
0%
$AsO_{3}^{3-},\, Cl{ O }_{3}^{-}$ and $BO_{3}^{3-}$

Explanation

$AsO_{3}^{3-},\, Cl{ O }_{3}^{-}$ and

$SO_{3}^{2-}$ are non - planar species.

In these species, the central atom has

$sp^3$ hybridization due to presence of

$3$ bonding domains and one lone pair of electrons. The molecular geometry is pyramidal and the electronic geometry is tetrahedral.

State True or False.

In hybridisation, only orbitals are involved and not electrons.

0%
True
0%
False

State True or False.

All the orbitals in a set of hybridised orbitals have the same shape and energy

0%
True
0%
False

A hybrid orbital formed from s-and p-orbital can contribute to:

0%
a $\sigma$ bond only
0%
$\pi$ bond only
0%
either $\sigma \;or\;\pi$ bond
0%
none of these

Which of the following statements is not correct ?

0%
Hybridization is the mixing of atomic orbitals of large energy difference.
0%
$sp^2$ -hybrid orbitals are formed from two p - atomic orbitals and ones - atomic orbitals.
0%
$dsp^2$ -hybtrid orbitals are all at 90 to one another.
0%
$d^2sp^3$ -hybrid orbitals are directed towards the corners of a regular octahedron.

In the electronic configurations of elements

$A\;and\;B$ are

$1s^2,\,2s^2\;2p^6,\,3s^1\;and\;1s^2,\,2s^2\;2p^6,\,3s^2\;3p^4$ respectively, then the formula of the compound formes by the combination of these elements will be

0%
$AB$
0%
$AB_3$
0%
$AB_2$
0%
$A_2B$

Explanation

Electronic configuration of

$A$ is

$1s^2\,2s^2\,2p^6\,3s^1$
Electronic configuration of

$B$ is

$1s^2\,2s^2\,2p^6\,3s^2\,3p^4$
Valency of

$A=1$
Valency of

$B=8-6=2$

$\therefore$ The compound formed by

$A\;\&\;B$ is
1313633_329861_ans_06f34f4359e6433f9e993bbab0147700.bmp

In the complex

$CoCl_3\cdot 5NH_3$ , the number of ions satisfying primary and secondary valencies of cobalt(III) are:

0%
3, 0
0%
2, 1
0%
1, 2
0%
0, 3

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.

Which element is not present in chlorophyll?

0%
Carbon
0%
Calcium
0%
Magnesium
0%
Hydrogen

In the electronic configurations of elements A and B are

$1s^2, 2s^2 2p^6, 3s^1$ and

$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$ respectively, then the formula of the compound formed by the combination of these elements will be

0%
$AB$
0%
$AB_3$
0%
$AB_2$
0%
$A_2B$

Explanation

Electronic configuration of A is

$1S^2 2s^22p^6 3s^1$
Electronic configuration of B is

$1S^2 2s^22p^6 3s^23p^4$
Valency of A =1
Valency of B

$= 8 -6 = 2$
$$\therefore The compound formed by A & B is

1313642_331879_ans_f219e4461f1d4958861d0056d6c234cf.bmp

The correct increasing order of trans-effect of the following species is:

0%
$Br^- > CN^- > NH_3 > {C_6 H_5}^-$
0%
$CN^- > Br^- > C_6H_5^- > NH_3$
0%
$NH_3 > CN^- > Br^- > {C_6 H_5}^-$
0%
$CN^- > {C_6 H_5}^- > Br^- > NH_3$

Explanation

Trans effect can be defined as the effect of a ligand towards substitution on other ligand which is trans to it.

(weak) $F^- < HO^- < H_2O < NH_3 < py < Cl^- < Br^- < I^-< SCN^-< NO_2^{-}< SC(NH_3)_2< Ph^-$ $< SO_3^{2-}< PR_3 < AsR_3< SR_2< H_3C^- < H^-< NO< CO< NC^-< C_2H_4$ (strong).

Trans effect is nothing but those ligands which are trans directing. That means, which prefer a trans structure. Mostly in Square planar complexes.

Hence option D is correct.

The formula for lithium iodotris(trifluorophosphine)nickelate(0) is:

0%
$Li[Ni(PF_3)_3]$
0%
$Li[Ni(PF_3)_3I_3]$
0%
$Li[(PF)_3I_3]$
0%
none of these

Explanation

Trifluorophosphine is

$PF_3$

The formula should be

$Li[Ni(PF_3)_3I]$ .

Use the figure below to answer the question that follows.
For which of the molecules in the figure does the central atom, or atoms, have

$sp^{2}$ hybridization?

0%
III only
0%
IV only
0%
III and IV only
0%
II and IV only

Explanation

$N$ in

$NO_2^-$ is having

$sp^2$ hybridization.

As hybridization =

$\cfrac{1}{2}$ [number of valence electrons in central atom - number of monovalent atom attached - cationic charge + anionic charge]

$\cfrac{1}{2}[5+1] = 3 = sp^2$

Select the correct IUPAC name for

$[Co(NH_3)_6]^{2+}$ .

0%
Hexammoniacobaltate(II) ion
0%
Hexaamminecobaltate(II) ion
0%
Hexammoniacobalt(II) ion
0%
Hexaamminecobalt(II) ion

Explanation

$[CO(NH_3)_6]^{2+}$ the oxidation state of CO is 2. So the correct IUPAC name is hexaamminecobalt(II) ion.

What is the correct name of

$I{F}_{7}$ ?

0%
Iron heptafluoride
0%
Iodine heptafluoride
0%
Iodine septafluoride
0%
Iron (IV) septafluoride

Explanation

Iodine heptafluoride, also known as iodine(VII) fluoride or iodine fluoride, is an interhalogen compound with the chemical formula $IF_7$ . It has an unusual pentagonal bipyramidal structure.

In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion

$(Ca^{2+})$ and fluoride ion

$(F^-)$ are:

0%
8 and 4
0%
4 and 8
0%
4 and 2
0%
6 and 6

Explanation

Calcium fluoride crystallizes in a Face-Centered Cubic unit cell (FCC) having an edge length of 5.463 Angstroms.

The cell is displayed here with calcium cations (in blue) defining FCC lattice sites, and fluoride anions (in green) occupying all tetrahedral sites.

Fluoride anions have 4 neighbors of opposite charge arranged at vertices of an tetrahedron. Calcium cations have EIGHT neighbors of opposite charge arrange at corners that outline a smaller cube. So the Ca: F coordination ratio is 8 : 4 or 2 : 1.

In $4CaF_2$ the the co ordination number of

$Ca^{+2}$ is 8

$F^{-}$ is 4
Hence option A is correct

Select the correct IUPAC name for

$[FeF_4(OH_2)_2]{^-}$ .

0%
Diaquatetrafluoroiron(III) ion
0%
Diaquatetrafluoroferrate(III) ion
0%
Diaquatetrafluoroiron(I) ion
0%
Diaquatetrafluoroferrate(I) ion

Explanation

$[FeF_4(OH_2)_2]^-$ the oxidation state of iron is

$+3$ so correct IUPAC name is diaquatetrafluoroferrate(III) ion.

In C-C bond is

$C_2H_6$ undergoes heterolytic fission, the hybridisation of two resulting carbon atoms is/are :

0%
$sp^2$ both
0%
$sp^3$ both
0%
$sp^2,sp^3$
0%
$sp,sp^2$

Explanation

${ H }_{ 3 }C-{ CH }_{ 3 }\quad \underrightarrow { heterolytic } \quad \underbrace { { CH }_{ 3 }^{ + } } \quad \underbrace { { CH }_{ 3 }^{ - } }$

${ SP }^{ 2 }$

${ SP }^{ 3 }$

Among the following coordination compounds, the one giving a white precipitate with

$BaCl_2$ solution is:

0%
$[Cr(H_2O)_5 Br]SO_4$
0%
$[Cr(H_2O)_5 SCN]$
0%
$[Co(NH_3)_5 SO_4]Br$
0%
$[Pt(NH_3)_6]Cl_4$

Explanation

Among the following coordination compounds, the one giving a white precipitate with

$\displaystyle BaCl_2$ solution is

$\displaystyle [Cr(H_2O)_5 Br]SO_4$ .

It dissociates in aqueous solution to give sulphate ions. Sulphate ions react with barium chloride to form a white precipitate of barium sulphate.

$\displaystyle [Cr(H_2O)_5 Br]SO_4 \rightarrow [Cr(H_2O)_5 Br]^{2+} + SO_4^{2-}$

$\displaystyle SO_4^{2-} + BaCl_2 \rightarrow BaSO_4 \downarrow + 2Cl^-$

Hence, option

$A$ is correct.

The shapes of

$IF_3$ and

$IF_7$ are respectively:

0%
square pyramidal and pentagonal bipyramidal
0%
octahedral and pyramidal
0%
trigonal bipyramidal and square antiprismatic
0%
distorted square planar and distorted octahedral

Explanation

As refer to IMAGE

$01$

$I\rightarrow 5s^25p^5$

at ground state.

For

$IF_3$ ;

$I$ at excited state.

Geometry is Trigonal bipyramidal as refer to IMAGE

$02$

As refer to IMAGE

$03$

For

$IF_7$ ,

$I$ at excited state is

$sp^3d^3$

Geometry

$=$ Pentagonal bipyramidal

Shape

$=$ Pentagonal bipyramidal as refer to IMAGE

$04$

There is change in the type of hybridisation when:

0%
$NH_3$ combines with $H^+$
0%
$AlH_3$ combines with $H{^-}$
0%
$NH_3$ forms $NH_2{^-}$
0%
$SiF_4$ forms $SiF_4^{2-}$

Explanation

Change in hybridization will take place only when there is change in no. of electrons around the central atom in a molecule.

$\Rightarrow$ Combination of

${ H }^{ + }$ to N of

$N{ H }_{ 3 }$ will bring no change in hybridization as

${ H }^{ + }$ has no electrons.

$\Rightarrow$ Combination of

${ H }^{ - }$ to Al of

$Al{ H }_{ 3 }$ will bring a change in hybridization as there will be addition of

$2$ electrons around Al by

${ H }^{ - }$

$\Rightarrow N{ H }_{ 2 }^{ - }$ formation from

$N{ H }_{ 3 }$ will take place when

${ H }^{ + }$ leaves

$N{ H }_{ 3 }$ molecule. This will not lead to removal of electrons from N atom. There will be no change in type of hybridization.

$\Rightarrow$ Formation of

$Si{ F }_{ 4 }^{ 2- }$ from

$Si{ F }_{ 4 }$ will lead to addition of one lone pair on Si atom.

There will be a change in type of hybridization.

The hybridisation and geometry of

$BrF_3$ molecules are:

0%
$sp^3d$ and $T$ shaped
0%
$sp^2d^2$ and tetragonal
0%
$sp^3d$ and bent
0%
none of these

Explanation

Steric no. of

$BrF_3=\cfrac{1}{2}[7+3]=5 \rightarrow sp^3d$

Molecular geometry is trigonal bipyramidal.

Shape is

$T-$ shaped (Refer IMAGE

$01$ )

$Br$ at ground state (Refer IMAGE

$02$ )

$Br- 4s^{2}4p^{5}$

$Br$ at excited state (Refer IMAGE

$03$ )

$Br- sp^{3}d$ with

$2lp$

The geometry of

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

0%
tetrahedral
0%
square planar
0%
triangular
0%
octahedral

Explanation

The geometry of

${ \left[ Ni{ (CN) }_{ 4 } \right] }^{ 2- }$ is square planar as shown in the image.

The IUPAC name of

$[Ni(CO)_4]$ is :

0%
Tetracarbonyl nickelete (0)
0%
Tetracarbonyl nickelete (II)
0%
Tetracarbonyl nickel (0)
0%
Tetracarbonyl nickel (II)

Explanation

The IUPAC name of

$Ni{ (CO) }_{ 4 }$ is Tetracarbonyl nickel (0) since the oxidation state of carbonyl ligand is 0, so

$Ni$ will also be 0 and four ligand are bonded to central metal atom.

Square planar complex of the type

$MAXBL$ (where A, B, X and L are unidentate ligands) shows following set of isomers.

0%
Two cis and one trans
0%
Two trans and one cis
0%
Two cis and two trans
0%
Three cis and one trans

Explanation

Square planar complex of the type

$M{AXBL}$ (where A, B, X, and L are unidentate ligands) shows two cis and one trans isomer as shown in the image.

What is the angle between any two bonds in methane molecule?

0%
$105^o54'$
0%
$109^o28'$
0%
$119^o28'$
0%
$190^o28'$

Explanation

The angle between any two bonds to the central carbon atom in the methane molecule is

$109^o28'$ . The molecule of methane has a tetrahedral geometry.
1316795_625830_ans_bf6db116459d4bf19413e3a1bcd411e3.png

Which of the following biomolecules contain non-transition metal ion?

0%
Vitamin $B_{12}$
0%
Chlorophyll
0%
Haemoglobin
0%
Insulin

Explanation

Chlorophyll contains a non-transition metal ion. Chlorophyll is magnesium porphyrin complex. It is a green pigment present in plant leaves and is used in photosynthesis.

Hence, option

$B$ is correct.

Which of the following

$0.1M$ complex compound solutions will have the minimum electrical conductivity?

0%
Hexammine platinum (IV) chloride
0%
Chloropenta ammine platinum (IV) chloride
0%
Dichloro tetrammine platinum (IV) chloride
0%
Trichloro tetrammine platinum (IV) chloride

Explanation

The formula of given complex are as follows:
(a) Hexammine platinum (IV) chloirde

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 4 }$
(b) Chloropentammine platinum (IV) chloirde

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 3 }$
(c) Dichlorotetrammine platinum (IV) chloirde

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] { Cl }_{ 2 }\quad$
(d) Trichlorotriammine platinum (IV) chloirde

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] Cl$
In aqueous solution the complexes ionise as

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 4 }\rightleftharpoons \underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] }^{ 4+ }+4{ Cl }^{ - }}_{5\, ions}$

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 4 }\rightleftharpoons\underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] }^{ 3+ }+3{ Cl }^{ - }}_{4\, ions}$

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] Cl\rightleftharpoons\underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] }^{ 2+ }+2{ Cl }^{ - }}_{3\, ions}$

$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] Cl\rightleftharpoons\underbrace{{ \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] }^{ + }+{ Cl }^{ - }}_{2\, ions}$

Trichlorotriammine platinum (IV) chloride gives the minimum number of ions in the solution. Hence, it has the minimum electrical conductivity

Which compound is formed when excess of

$KCN$ is added to an aqueous solution of copper sulphate?

0%
$Cu(CN)_2$
0%
$K_2[Cu(CN)_2]$
0%
$K[Cu(CN)_2]$
0%
$K_3[Cu(CN)_4]$

Explanation

When excess of

$KCN$ is added to an aqueous solution of

$CuSO_4$ then-

1. Initially, cupric cyanide is formed,

$Cu(CN)_2$ .

$CuSO_4+2KCN\rightarrow K_2CO_4+Cu(CN)_2$

2.Cupric Cyanide will decompose to produce cuprous cyanide,

$Cu_2(CN)_2$ .

$2Cu(CN)_2\rightarrow 2Cu_2(CN)_2 + 2(CN)$

3. Cuprous cyanide reacts with an excess of

$KCN$ to form a complex,

$K_3[Cu(CN)_4]$ , which is a stable complex.

$CuCN+3Ka\rightarrow K_3[Cu(CN)_4]$ .

Hence, option D is correct.

The isomerism exhibited by following compounds

$[Co(NH_3)_6][Cr(CN)_6]$ and

$[Cr(NH_3)_6][Co(CN)_6]$ is:

0%
ionization isomerism
0%
polymerization isomerism
0%
linkage isomerism
0%
coordination isomerism

Explanation

Coordination isomerism occurs in those complexes in which the total ratio of ligand to metal remains the same but the ligands attached to the specific metal ion varies.

Hence, the correct option is

$\text{D}$

An octahedral complex is formed when hybrid orbitals of the following type are involved:

0%
${sp}^{3}$
0%
$d{sp}^{2}$
0%
${d}^{2}{sp}^{3}$
0%
${sp}^{2}{d}^{2}$

Explanation

Hybridisation	Geometry of complex
$sp_3$	Tetrahedral
$dsp^2$	Square planar
$d^2sp^3$	Octahedral
$sp^3d^2$	Not possible

Hence option C is correct.

The number of isomers for the compound with the molecular formula

$\mathrm{C}_2\mathrm{Br Cl F I}$ is :

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

The coordination of

$Pt$ in the complex ion

$[Pt (en)_2Cl_2]^{2+}$ is:

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

Explanation

The coordination of Pt in the complex ion

$\displaystyle [Pt (en)_2Cl_2]^{2+}$ is

$6.$
en (ethylene diamine) is bidentate ligand.

So,

$2\ en$ ligands and

$2 \ Cl^-$ ligands corresponds to six donor atoms.

$Al_4C_3$ on hydrolysis yields

0%
nitrogen
0%
methane
0%
hydrogen
0%
all the above

Explanation

The hydrolysis reaction of $Al_4C_3$

$Al_4C_3 + 12 H_2O \rightarrow 4Al(OH)_3 + 3CH_4$

So, $Al_4C_3$ on hydrolysis gives methane

Hence option B is correct.

In which one of the following the central atom is

$sp^{3}$ hydridised?

0%
$NH_{4}^{+}$
0%
$BF_{3}$
0%
$SF_{6}$
0%
$PCl_{5}$
0%
$XeF_{4}$

$TeCl_4$ , the central atom tellurium involves:

0%
$sp^3$ hybridisation
0%
$sp^3d$ hybridisation
0%
$dsp^2$ hybridisation
0%
$sp^3d^2$ hybridisation

Explanation

Number of hybrid orbitals

$=\dfrac{1}{2}$ (no. of electrons in valence shell of atom + no. of monovalent atoms - charge no cation + charge on the anion

Number of hybride orbitals

$=\dfrac{1}{2}(6+4+0+0)=5$

Hence,

$TeCl_4$ shows

$sp^3d$ hybridisation.

Therefore, option B is correct.

Coordinate bond is formed in:

0%
$NaCl$
0%
${ NH }_{ 4 }Cl$
0%
${ Cl }_{ 2 }$
0%
$B{ Cl }_{ 3 }$

Explanation

$NH_4Cl$ exists as

$NH_4^+Cl^-$

$NH_4^+$

only coordinate bond is present.

1171431_711815_ans_4b099b2e52294721804bb1b3d34e9cfb.PNG

When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

0%
$[Co(NH_3)_6]Cl_3$
0%
$[Co(NH_3)_6Cl_3]$
0%
$[Co(NH_3)_5Cl]Cl_2$
0%
$[Co(NH_3)_4Cl_2]NO_3$

Explanation

$4.305 \, g \, AgCl =\dfrac {4.305}{143.5} mol= 0.03\, mol$

As 0.01 mole of the complex gives 0.03 mole of

$AgCI$ , this shows that there are three ionisable

$Cl$ .

Hence, formula is

$[Co(NH_3)_6]Cl_3$ .

The concept hybridisation of orbits of an atom was introduced by ______.

0%
Linus Pauling
0%
Moseley
0%
Lewis
0%
Kossel

In spectrochemical series, chlorine is above water i.e.,

$Cl> {H}_{2}O$ . This is due to:

0%
Good $\pi -$ acceptor properties of $Cl$
0%
Strong $\sigma -$ and good $\pi -$ acceptor properties of $Cl$
0%
Good $\pi -$ donor properties of $Cl$
0%
Larger size of $Cl$ than ${H}_{2}O$
0%
None of the option

Explanation

None of the given options is correct.

${H}_{2}O$ is a stronger ligand than

$Cl$ and the order of the ligands in spectrochemical series is determined experimentally.

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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