$$ [Fe(NH_4)_2] (SO_4)_2. 6H_2 O $$

$$(CN)$$ are linked with primary valency

$$K$$ are linked with secondary valency

$$K$$ are linked with non-ionic valency

MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 7

Correct increasing order for the wavelength of absorption in the visible region for the complexes of

$Co^{3+}$ is?

0%
$[Co(CN)_6]^{3-}, [Co(NH_3)_6]^{3+}, [Co(NH_3)_5(H_2O)]^{3+}[Co(NH_3)_5Cl]^{+2}$
0%
$[Co(CN)_6]^{3-}, [Co(NH_3)_5(H_2O)]^{3+}[Co(NH_3)_5Cl]^{+2}[Co(NH_3)_6]^{3+}$
0%
$[Co(NH_3)_6]^{3+}, [Co(CN)_6]^{3-}, [Co(NH_3)_5(H_2O)]^{3+}[Co(NH_3)_5Cl]^{+2}$
0%
$[Co(NH_3)_5Cl]^{+2}, [Co(NH_3)_5(H_2O)]^{3+}, [Co(NH_3)_6]^{3+}, [Co(CN)_6]^{3-}$

Explanation

Strength of ligands attached with

${Co}^{3+}$ ion is in the order of

${CN}^{-} > N{H}_{3} > {H}_{2}O > {Cl}^{-}$ . So order of spliting (value of

$\Delta_{0}$ ) will be in the same order.

Therefore,

Wavelength

$\propto \cfrac{1}{{\Delta}_{0}}$

Thus the correct increasing order of wavelength is-

$[Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+}< [Co(NH_3)_5Cl]^{+2}$

Hence, the correct option is

$A$

The Hybridisation and magnetic behavior of complex

$[Ni(CO)_{4}]$ is

0%
$dsp^{2}$ , paramagnetic
0%
$dsp^{2}$ , diamagnetic
0%
$sp^{3}$ , paramagnetic
0%
$sp^{3}$ , diamagnetic

The hybrid states of carbon atoms in

${(\overset{B}CN)_4}\overset{A}{C_2}$ are A and B and number of

$\pi$ bonds in compound is C. Then?

0%
$A=sp, B=sp^2$ and $C=9$
0%
$A=sp^2, B=sp$ , and $C=9$
0%
$A=sp^3, B=sp$ and $C=9$
0%
$A=sp^2, B=sp^2$ and $C=9$

Explanation

From the above shown strucutre we can observe that-
The hybrid state B is

$sp$ and that of A is

$sp^2$ and the total no. of

$\pi$ bond,

$C=9$ .

Option B is correct.
1552776_1717885_ans_5650fd72ff15474fac995d33331ae4c5.jpg

The compound

$[CoCl_2(NH_3)_2(en)]$ can form:

0%
Linkage isomers
0%
Coordination isomers
0%
Optical isomers
0%
Linkage as well as optical isomers

Explanation

The compound

$[CoCl_{2}(NH_{3})_{2}(en)]$ has two geometrical isomers. One is the cis-isomer (with respect to

$NH_{3}$ ) and the other is the trans- isomer (with respect to

$NH_{3}$ ).

The cis-isomer has two optically active forms,

$d-$ cis isomer and

$l-$ cis isomer. So, the compound

$[CoCl_{2}(NH_{3})_{2}(en)]$ can form geometrical isomers as well as optical isomers.

Hence, the correct answer is option (C).

1885381_1721749_ans_2b239aea794a469ca5bf2321fc1a0bdc.png

$d_{x^2-y^2}$ orbital is involved in which of the following hybridization?

0%
$sp^3d$ (sq. pyramidal)
0%
$dsp^2$
0%
$sp^3d^2$
0%
$sp^3d^3$

Explanation

$sp^3d\Rightarrow sp_xp_yp_zd_{x^2-y^2}$

$dsp^3\Rightarrow d_{x^2-y^2}sp_xp_yp_z$

$sp^3d^2\Rightarrow sp_xp_yp_zd_{x^2-y^2}d_{z^2}$

$sp^3d^3\Rightarrow sp_xp_yp_zd_{x^2-y^2}d_{z^2}d_{xy}$ .

In the hydrocarbon,

$H_3\underset{6}{C}-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}H_2-\underset{2}{C}\equiv \underset{1}{C}H$

The state of hybridization of carbons

$1, 3, 5$ are in the following sequence is:

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

Explanation

$H_3\underset{6}{\overset{sp^3}C}-\underset{5}{\overset{sp^2}C}H=\underset{4}{\overset{sp^2}C}H-\underset{3}{\overset{sp^3}C}H_2-\underset{2}{\overset{sp}C}\equiv \underset{1}{\overset{sp}C}H$

The state of hybridisation of carbon in

$1, 3$ and

$5$ position are

$sp, sp^3, sp^2$

Complexes A and B have similarity in which type of isomerism?

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

Which of the above is a pair of geometrical isomers?

0%
$I,II$
0%
$I,III$
0%
$II,IV$
0%
$III,IV$

Explanation

Geometric isomers are two or more coordination compounds which contain the same number and types of atoms, and bonds (i.e., the connectivity between atoms is the same),

$\text{but which have different spatial arrangements}$

$\text{of the}$

$\text{atoms.}$

II and IV are geometrical isomers.

Option C is correct.

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

$MAXBL$ (where A, B, X, and L are unidentate ligands)

$\text{shows two cis and}$

$\text{one trans isomer as shown in the image.}$
1828285_1718425_ans_28466cb75c72432880d2df5a6336e28b.png

The hybrid state and oxidation state of two oxygen atoms in

$H_2O_2$ are respectively:

0%
$sp^2, -1$
0%
$sp^2, +1$
0%
$sp^3, -1$
0%
$sp^3, +1$

Which of the following hybridization results in non-planar orbitals?

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

The hybridisation of oxygen atom in

$H_2O_2$ is:

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

Explanation

each oxygen atom in peroxide is

$sp^3$ hybridized. There is a covalent

$sp^3-sp^3$ bond between the two oxygens. Each oxygen atom is covalently linked to its respective hydrogen via.

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$

On hybridization of one s- and one p-orbital we get :

0%
Two mutually perpendicular orbitals
0%
Two orbitals at $180^0$
0%
Four orbitals directed tetrahedrally
0%
Three orbitals in a plane

On hybridization of one s-and one p-orbitals, we get:

0%
Two mutually perpendicular orbitals
0%
Two orbitals at $180^o$
0%
four orbitals directed tetrahedrally
0%
Three orbitals in the plane

Explanation

On hybridization of one s-and one p-orbital, we get two orbitals at

$180^o$ . The number of hybrid orbitals formed is equal to the number of participating atomic orbitals, which in present case is 2. The angle between these two is equal to

$360^o$ divided by the number of orbitals.

$\dfrac {360^o}{2} =180^o$

Hybridisation of the _____atom change sin:

0%
$AIH_3 to AlH_4^-$
0%
$H_2O to H_3 O^+$
0%
$NH_2^-$
0%
$NF_3$

Explanation

hybridization of central atom changes as the product is

in this change

$AlH_3$ to

$AlH_4^-$

The hybridization of I in ICl

$_2^+$ is :

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

In the following case(s) , hybridization of the underlined atom is affected:

0%
$PCl_5$ (solid) dissociates into $PCl_4^+ and PCl_6^-$
0%
$LiH$ reacts with $AlH_3$ forming $LiAlH_4$
0%
$NH_3$ is protonated
0%
$H_3PO_2$ is heated forming $PH_3$ and $H_3PO_3$

Explanation

$AlH_3$ has 3 bond pair, which means it is

$sp^2$ hybridized while in

${AlH_4}^{-}$ it has 4 bond pair which means it is

$sp^3$ hybridized.

When

$1$ mol

$CrCl_3\cdot 6H_2O$ is treated with excess of

$AgNO_3, 3$ mol of

$AgCl$ are obtained. The formula of the complex is?

0%
$[CrCl_3(H_2O)_3]\cdot 3H_2O$
0%
$[CrCl_2(H_2O)_4]Cl\cdot 2H_2O$
0%
$[CrCl(H_2O)_5]Cl_2\cdot H_2O$
0%
$[Cr(H_2O)_6]Cl_3$

Explanation

One mol of

$AgCL$ is precipitated by one mole of

$CL^-$ ion, therefore three moles of

$AgCL$ would get precipitared by three moles of chloride ions,

$CL^-$ and, in this case

$3 CL^-$ are present in ionization sphere ( ie. out side the coordination sphere) in complex at

$(iv)$ , therefore when

$1$ mol of it is treated with excess of this complex

$3$ mols of

$AgCL$ is precipitated. This satisfied the formula of the complex at

$(iv)$

Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type

$[Pd(C_6H_5)_2(SCN)_2]$ and

$[Pd(C_6H_5)_2(NCS)_2]$ are?

0%
Linkage isomers
0%
Coordination isomers
0%
Ionization isomers
0%
Geometrical isomers

Explanation

Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is by complexes containing the thiocyanate ligand,

$NCS-,$ which may bind through the nitrogen to give

$M-NCS$ or through sulphur to give

$M-SCN .$

$Cu$ -ammonia complex, the state of hybridisation of

$Cu^{2+}$ is:

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

Which of the following two are isostructral?

0%
$XeF_2, IF_2^-$
0%
$NH_3, BF_3$
0%
$CO_3^{2-}, SO_3^{2-}$
0%
$PCl_5, ICl_5$

Explanation

$XeF_2$ is a linear molecule (Xe is

$sp^3d$ hybridized with three equatorial positions occupied by lone pairs)

$IF_2^-$ is a linear molecule (I is

$sp^3d$ hybridized, with three positions occupied by lone pairs).

Option A is correct.

1623665_1729174_ans_07aa06cb24244f08a2998a3a5abd3c6c.webp

What is the formula of choclate colored ppt.?

0%
$Fe_4[Fe(CN)_6]$
0%
$Cu_2[Fe(CN)_6]$
0%
$Cu_2[Fe(CN)_4]$
0%
$Cu(CN)_2$

$[Co(NH_3)_5Br]SO_4$ and

$[Co(NH_3)_5SO_4]Br$ are related to each other as :

0%
ionization isomers
0%
linkage isomers
0%
coordination isomers
0%
geometrical isomers

The formulae of

$(A)$ and

$(B)$ with colours are:

0%
$Na_2CrO_4$ , yellow and $PbCrO_4$ , yellow
0%
$Na_2Cr_2O_7$ , orange and $PbCrO_4$ , yellow
0%
$Cr(OH)_3$ , green and $(CH_3COO)_3Cr$ , green
0%
$CrO_2Cl_2$ , red and $(CH_3COO)_3 Cr$ , green

The loss or reduction of chlorophyll in the leaves is termed:

0%
necrosis
0%
chlorosis
0%
epinasty
0%
lichen

Which of the following is the odd one out?

0%
Potassium ferrocyanide
0%
Tetramminezinc(II) chloride
0%
Ferrous ammonium sulphate
0%
Potassium ferricyanide

Explanation

All are the complexes other than ferrous ammonium sulphate are complexes with a single metal salt. But

$\text{ferrous}$

$\text{ammonium sulphate is a double salt with two different cations.}$ Note that double salts are different from coordination compounds. All others are coordination compounds, whereas ferrous ammonium sulphate is not a coordination compound and it ionizes into ferrous, ammonium and sulphate ions.

Option C is correct.

One mole of the complex compound

$Co\left(NH_{3}\right)_{5}Cl_{3}$ , gives

$3$ moles of ions on dissolution in water. One mole of the same complex reacts with two moles of

$AgNO_{3}$ solution to yield two moles of

$AgCl\left(s\right)$ . The structure of the complex is

0%
$[Co(NH_{3})_{3}Cl_{3}].2NH_{3}$
0%
$\left[Co\left(NH_{3}\right)_{4}Cl_{2}\right]Cl.NH_{3}$
0%
$\left[Co\left(NH_{3}\right)_{4}Cl\right]Cl_{2}.NH_{3}$
0%
$\left[Co\left(NH_{3}\right)_{5}Cl\right]Cl_{2}$

Explanation

As complex gives 2 moles of AgCl on reaction with water. This means that there are 2 moles of Chloride ions out of 3 moles of total ions.

$\left[Co\left(NH_{3}\right)_{5}Cl\right]Cl_{2}$ yields three moles of ions out of which two mole of ions are Chloride ions.

Hence, Option "D" is the correct answer.

Which one of the following compounds helps in achieving equilibrium between

$O_2$ and

$CO_2$ in the atmosphere?

0%
Chlorophyll
0%
Vitamin - B $_{12}$
0%
Porphyrin
0%
Acetyl salicylic acid

Explanation

Chlorophyll is a green substance in producers that traps light energy from the sun, which is then used to combine carbon dioxide and water into sugars in the process of photosynthesis. Chlorophyll is vital for photosynthesis, which helps plants get energy from light.

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (A)}$

$MgSO_{4}$ on reaction with

$NH_{4}OH$ and

$Na_{2}HPO_{4}$ forms a white crystalline precipitate. What is its formula?

0%
$Mg(NH_{4})PO_{4}$
0%
$Mg_{3}(PO_{4})_{2}$
0%
$MgCl_{2}-MgSO_{4}$
0%
$MgSO_{4}$

Explanation

Magnesium is Group VI radical.

The preliminary test of Mg involves following reaction,

$MgSO_4$ reacts with

$NH_4OH$ and

$Na_2HPO_4$ forming a white crystalline precipitate of Magnesium ammonium phosphate.

$MgSO_4+NH_4OH+Na_2HPO_4\rightarrow Mg(NH_4)PO_4\downarrow +Na_2SO_4+H_2O$

white ppt

Hence the correct option is

$(A)$ .

Which of the following species is/are isostructural with

$IF_4^-$ ?

0%
$[Cu(NH_3)_4]^{2+}$
0%
$XeF_4$
0%
$BF_4^-$
0%
$[Ni(CN)_4]^{2-}$

Explanation

	Hybridization	Structure
$IF_{4}^{-}$	$sp^{3}d^{2}$	square planar
$[Cu(NH_{3})_{4}]^{2+}$	$dsp^{2}$	square planar
$XeF_{4}$	$sp^{3}d^{2}$	square planar
$BF_{4}^{-}$	$sp^{3}$	Tetrahedral
$[Ni(CN)_{4}]^{2-}$	$dsp^{2}$	square planar

Ferrous Oxide is

0%
$FeO$
0%
$Fe_2O_3$
0%
$Fe_3O_4$
0%
None of these.

Explanation

Ferrous Oxide is

$FeO$ .

Hence, Option "A" is the correct answer.

Correct formula of Potassium Ferricyanide is

0%
$K_4Fe(CN)_6$
0%
$K_3[Fe(CN)_6]$
0%
$K_3[Fe(CN)_3]$
0%
$K_3[Fe(CN)_4]$

Explanation

Potassium Ferricyanide is

$K_3[Fe(CN)_6]$ .

Hence, Option "B" is the correct answer.

Mohr's salt is

0%
$FeSO_4. 7H_2O$
0%
$Fe(NH_4)SO_4 . 6H_2 O$
0%
$(NH_4)_2 SO_4 . FeSO_4 . 6H_2 O$
0%
$[Fe(NH_4)_2] (SO_4)_2. 6H_2 O$

Explanation

Mohr's salt is

$(NH_4)_2 SO_4 . FeSO_4 . 6H_2 O$ .

Hence, Option "C" is the correct answer.

In which one of the following changes there are transfer of five electrons

0%
$Mn{O}_{4}^{-}\rightarrow {Mn}^{2+}$
0%
$Cr{O}_{4}^{2-}{Cr}^{3+}$
0%
$Mn{O}_{4}^{2-}Mn{O}_{2}$
0%
${Cr}_{2}{O}_{7}^{2-} 2{Cr}^{3+}$

Explanation

$\overset{+7}{Mn}{O}_{4}^{-}\rightarrow {Mn}^{2+}+5{e}^{-}$

Option A.

The correct formula of the complex is

0%
$\left [ CoClBr(en)_{2} \right ]H_{2}O$
0%
$\left [ CoCl(en)_{2}(H_{2}O) \right ]BrCl$
0%
$\left [ CoBr(en)_{2} (H_{2}O)\right ] Cl_{2}$
0%
$\left [ CoBrCl(en)_{2} \right ]Cl.H_{2}O$

Explanation

Solution:

As per given conditions: Complex suffer loss in weight with conc. $H_2SO_4$ as it is dehydrating agent so ionization sphere must contain $H_2O$ molecules and on reaction with $AgNO_3$ solution it gives a white $ppt$ the ionization sphere should contain $Cl^-$ ion. So formula may be $[CoBrCl(en)_2]Cl.H_2O$

Option D is correct

Formula of ferric sulphate is

0%
$FeSO_4$
0%
$Fe(SO_4)_2$
0%
$Fe_2SO_4$
0%
$Fe_2(SO_4)_3$

Explanation

Ferric Sulphate is Iron (III) Sulphate i.e

$Fe_2(SO_4)_3$ .

Hence, Option "D" is the correct answer.

The primary valence of the metal ion in the co-ordination compound

$K_2[Ni(CN)_4]$ is

0%
four
0%
zero
0%
two
0%
six

Explanation

Primary valencies are also known as the oxidation state of the central metal atom.

$K_2[Ni(CN)_4]$ . Let

$x$ be the oxidation state of Ni.

Now,

$2\times1+x+4\times(-1)=0;\quad \Rightarrow x=+2$

$K_4 Fe(CN)_6$

0%
$(CN)$ are linked with primary valency
0%
$(CN)$ are linked with secondary valency
0%
$K$ are linked with secondary valency
0%
$K$ are linked with non-ionic valency

Explanation

$(CN)$ are linked with secondary valency to the central Fe metal ion. Ligand (CN) is linked to the central atom via a dative bond.

$K_4[Fe(CN)_6]\to 4K^++[Fe(CN)_6]^{4-}$

Haemoglobin is

0%
An enzyme
0%
A globular protein
0%
A vitamin
0%
A carbohydrate

Chlorophyll contains

0%
Fe
0%
Na
0%
Mg
0%
Zn

Explanation

$\text{Chlorophyll has magnesium as its central metal ion,}$ and the large organic molecule to which it bonds is known as a

$porphyrin.$

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

0%
Geometrical isomerism
0%
Optical isomerism
0%
Bonding isomerism
0%
Ionisation isomerism

Explanation

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

$Ma_4b_2$ type complex which show geometrical isomerism.
1668735_1760612_ans_b98985e0f3024fde8cc69131a8d3f085.png

The complex compounds which result from the coordination of carbon monoxides are known as

0%
electronic
0%
carbonyls
0%
carbonates
0%
carbon permono

Explanation

The complex compounds which result from the coordination of carbon monoxides are known as carbonyl compound.

Example:

$Fe(CO)_5, Ni(CO)_4,$ etc.

The valency of cuprammonium ion is

0%
$+4$
0%
$+2$
0%
$-2$
0%
$-4$

Explanation

Cuprammonium ion is

$[Cu(NH_3)_4]^{2+}$ . So, copper has

$+2$ valency.

$[Co(NH_3)_4Cl_2]NO_2$ and

$[Co(NH_3)_4Cl.NO_2]Cl$ are ___________ isomers.

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

Explanation

Both produce different ions in solution state so they are ionisation isomer.

$[Co(NH_3)_4Cl_2]NO_2 \rightleftharpoons [Co(NH_3)_4Cl_2]^+ +NO_2^-$

$[Co(NH_3)_4Cl. NO_2]Cl \rightleftharpoons [Co(NH_3)_4Cl. NO_2]^+ +Cl^-$

Number of unhybridised orbitals in vinyl acetylene are

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

Explanation

Number of unhybridised orbitals in vinyl acetylene is 6 as unhybridised orbitals are used in

$\pi-$ bonding.
1677688_1761240_ans_d29e66c47d93434ab6c5e45056959368.png

The number of neutral molecules or negative groups attached to the central metal atom in a complex ion is called

0%
atomic number
0%
effective atomic number
0%
coordination number
0%
primary valency

Explanation

The number of neutral molecules or negative groups attached to the central metal atom in a complex ion is called coordination number.

Example: In

$[Ni(CO)_4],$ coordination number is 4.

The compounds

$[Co(So_4)(NH_3)_5]Br$ and

$[Co(SO_4)(NH_3)_5]Cl$ represent

0%
Linkage isomerism
0%
Ionisation isomersim
0%
coordination isomerism
0%
no isomerism

Which of the following statement is correct ?

0%
In $BrF_3$ , maximum three halogen atoms can lie in same plane
0%
In $CH_2SF_2(CH_3)_2$ molecule all hydrogen atoms which bonded to $s - sp^2$ overlapping, lie in equatorial plane
0%
In $OSCl_4$ , Cl - S - Cl equatorial bond angle is greater than $120^0$
0%
Molecules $IOF_5$ and $XeO_2F_4$ have similar shape but have different number of lone pairs in whole molecule

Which of the following isomerism, exhibited by

$[CrCl_{2}(OH)_{2}(NH_{3})_{2}]^{-}$ ?

0%
Ionization
0%
Geometrical
0%
Hydrate
0%
Linkage

Explanation

$[CrCl_2(OH)_2(NH_3)_2]−$ exhibits geometrical isomerism. The geometrical isomers are cis and trans isomers. Cis isomer is optically active.

1770961_1817503_ans_15b936bb908f40d799e4b3648dbd5457.png

In which of the following combination hybridisation of central atom (*) does not change ?

0%
$H_{2}O + CO_{2}$
0%
$H_{3}BO_{3} + OH^{-}$
0%
$BF_{3} + NH_{3}$
0%
None of these

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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