MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 6

Which of the following is an EAN of iron in potassium ferricyanide?

0%
$28$
0%
$36$
0%
$35$
0%
$39$

Explanation

$EAN$ = Atomic number of central atom

$-$ Oxidation state+ (Number of ligands

$\times 2$ )

$EAN= 26-3+(6 \times 2)= 26+12-3=38-3$

Hence,

$EAN=35$ .

Primary and secondary valency of Pt in

$[Pt{ (en) }_{ 2 }{ Cl }_{ 2 }]{ Cl }_{ 2 }$ are:

0%
4,4
0%
2,6
0%
6,6
0%
None of these

Which of the following ligands is not a chelating agent?

0%
EDTA
0%
En
0%
Oxalate
0%
Pyridine

Which of the following complex will give white precipitate with BaCl

$_2$ (aq)?

0%
[Co(NH $_2)_4SO_4]NO_2$
0%
[Cr(NH $_3)_5SO_4$ ]Cl
0%
[Cr(NH $_3)_5Cl]SO_4$
0%
Both $B$ and $C$

Explanation

Since the reagent is

$BaCl_2(aq)$ , displacement reaction will take place. The resulting solution, if insoluble in water, will form a white precipitate. Both

$BaNO_2$ and

$BaCl_2$ are soluble in water and hence will not form a precipitate.

$BaSO_4$ , however, is insoluble in water and hence forms a white precipitate. Therefore C is the correct answer

Which of the molecules is not tetrahedral?

0%
$\left[ Pt(en)_{ 2 } \right] ^{ 2+ }$
0%
$\left[ Ni(Co)_{ 4 } \right]$
0%
$\left[ Zn(NH_{ 3 })_{ 4 } \right]$
0%
$\left[ { NiCl }_{ 4 } \right] ^{ 2- }$

Explanation

The Complex

$[Pt(en)_2]^{2+}$ is not tetrahedral because in this complex the metal ion is Pt which is a present in 5d series and the 5d series element have more diffused orbitals than 3d series elements and hence the ligand have the more closest approach towards metal ion and therefore the complex formed is inner orbital square planar. Hence the correct option is A.

Which of the following name is impossible ?

0%
Potassium tetrafluoridooxidochromate( $VI$ )
0%
Barium tetrafluoridobromate( $III$ )
0%
Dichlorobis(urea)copper( $II$ )
0%
All are impossible

Which of the following statements is true about hybridization ?

0%
Hybrid orbitals frequently undergo linear overlaps making sigma bonds.
0%
Hybrid orbitals frequently undergo lateral overlaps maing $\pi$ -bonds. In other words, there are several compounds in which $\pi$ -bonds are formed using hybrid orbitals.
0%
Hybrid orbitals are molecular orbitals.
0%
A hybrid orbital bigger in size makes shorter bond.

Explanation

$(A)$

Linear overlaps of bybridabitak from sigma bonds

$True$

$(B)$

HyBrid orbitals Do Not Form

$\pi$ -bonds.

$\pi$ -bonds are formed by arbybridized orbitals of the molecule by lateral overlap.

Statement is

$False$

$(C)$

Hybrid are not same as molecules orbital hence

$False$ statement

While of same atom, molecular orbitals are formed by mixing orbitals of different atoms

$False$

$(D)$

Only first Statement is

$True$

1364817_1178386_ans_c3ecf24b34804b6281759722f712588b.png

In 2sp hybridization, 2s-orbital can be mixed with

0%
Only $2p_x$
0%
Only $2p_y$
0%
Only $2p_z$
0%
Any one of $2p_x,2p_y$ and $2p_z$

Explanation

$S$ orbital is spherically symmetric and hence can be mixed with

$p$ orbital from any axis

So, answer should be

$d$

1364935_1178415_ans_97706cfe9181462b89cbaed0ee093e16.png

Hybridization of sulphur in

$SF$

$_{4}$ molecules are :

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

The bond angle and hybridization in ether

$(CH_3OCH_3)$ is ______.

0%
$106.51^0 , sp^{2}$
0%
$110^0$ , $sp^3$
0%
$109.28^0, sp^3$
0%
None of these.

The complex

$[Pt(NH_{3})_{4}]^{2+}$ has ......structure :

0%
square planer
0%
tetrahedral
0%
pyramidal
0%
pentagonal

Explanation

As refer to the above image $[Pt(NH_3)_4]^{2+}$ illustrates square planar geometry as there are only $4$ sigma bond around platinum with no lone pairs.

1100220_1224590_ans_1946a1b1e6994cc9a3d0ff1d4504c05d.PNG

Choose the correct options:

0%
In $O_2F_2$ , the $O-F$ bond length is lesser than expected value.
0%
In trislyl amine $(SiH_3)_3$ N, the bond length is lesser than expected value
0%
The bond angle in $OF_2$ is lesser than in $Cl_2O$
0%
The Be atom in $BeCl_2(s)$ is $sp^3$ hybridised.

In which of the following molecule, the central atom is used

$sp^{3}d$ hybridized orbitals?

0%
$SF$ $_{4}$
0%
$NH$ $_{3}$
0%
$XeF$ $_{2}$
0%
$XeF$ $_{4}$

Explanation

A) The hybridization that is involved in

$SF_{4}$ is

$sp^{3}d$ type.

When the central metal atom is surrounded with five pairs of electrons the electronic geometry is trigonal bi pyramidal. If all the electron pairs are shared, the molecular geometry will also be trigonal bi pyramidal. In this molecule

$SF_{4}$ , there are five electron pairs which should lead to trigonal bi pyramidal structure.

B) In

$NH_{3}$ , the central atom

$NH_{3}$ is

$sp^3$ hybridized.

$\rightarrow$ The geometry of such a molecule is pyramidal.

$\rightarrow$ The bond angle is

$107^{0}$ .

C)The hybridization of

$XeF_{2}$ (Xenon Difluoride) is an

$sp^{3}d$ type.

$\rightarrow$ It involves

$Sp^{3}$ hybridization of xenon atoms.

$\rightarrow$ The molecule has trigonal bipyramidial structure with 3 position occupied by lone pair. So,

$XeF_{2}$ is linear.

$XeF_{4}$

$\rightarrow$ The central atom has 6 electron pairs out of which two are lone pair electrons. The hybridization in Xenon is

$sp^{3}d^{2}$ because there is a migration of two electrons of p to d orbital which results in the formation of sigma bond with F.

$\rightarrow$

$XeF_{4}$ is

$AB_{4}L_{2}$ in which the Xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).

So, the hybridisation is

$sp^{3}d^{2}$ .

Hence, The correct option is A .

IUPAC name of the Wiilkinson's catalyst

$\left[ RhCl({ P\quad Ph }_{ 3 })_{ 3 } \right]$

0%
Chlorotris (triphenylphosphine) rhodium (I)
0%
Chlorotris (triphenylphosphine) rhodium (IV)
0%
Chlorotris (triphenylphosphine) rhodium (0)
0%
Chlorotris (triphenylphosphine) rhodium (VI)

Explanation

The IUPAC name of the wilkinson's catalyst

$[RhCl(PPh_3)_3]$ is chlorotris(triphenylphosphine)rhodium(I) because the oxidation state of Rh metal in the complex is x + (-1 x 1) + 0 = 0, implies x = + 1. hence the correct option is A.

The structural formula of a compound is

$CH_3-CH=C=CH_2$ . The type pf hybridization at the four carbon atoms from left to right is?

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

Statement: Secondary valency of

$Ni^{+2}$ is

$6$ in

${[Ni{{(H_2O)}_6}]}^{2+}$ .

State whether the given statement is true or false.

0%
True
0%
False

The number of geometrical isomers of the complex

$\left[ {Co{{(N{O_2})}_3}{{(N{H_3})}_3}} \right]$ is

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

Explanation

$2$ isomers i.e facial and meridional isomer.

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C,

$38.71\%$ and H,

$9.67\%$ . The empirical formula of the compound would be?

0%
$CH_4O$
0%
$CH_3O$
0%
$CH_2O$
0%
$CHO$

Explanation

The empirical formula is to be calculated as:

Element	% Composition	Atomic Mass	Mole Ratio	Simple Ratio
C	38.71	12	$\dfrac{38.71}{12} = 3.22$	$\dfrac{3.22}{3.22} = 1$
H	9.67	1	$\dfrac{9.67}{1} = 9.67$	$\dfrac{9.67}{3.22} = 3$
O	100-(38.71+9.67) =51.62	16	$\dfrac{51.62}{16} = 3.22$	$\dfrac{3.22}{3.22} = 1$

Hence the empirical formula of the compound is

$CH_{3}O$

When

$K_{4}Fe(CN)_{6}$ is added to

$FeCl^{3+}$ complex compound formed is:

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

The effective atomic number rules is not obeyed by

0%
$[Ni(CO)_4]$
0%
$K_4[Fe(CN)_6]$
0%
$K_3[Fe(CN)_6]$
0%
$[Fe(CO)_5]$

Which of the following involves transfer of five electrons?

0%
$MnO_4^ - \to M{n^{2 + }}$
0%
$CrO_4^{2 - } \to C{r^{3 + }}$
0%
$MnO_4^{2 - } \to Mn{O_2}$
0%
$C{r_2}O_7^{2 - } \to 2C{r^{3 + }}$

Explanation

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

$x - 8 = - 1$

$x = + 7$

1381125_1255302_ans_4a6ff40bf35c4b4686ba563a0f74e34c.jpg

The hybrid orbital having only 20% d character

0%
$CC{l }_{ 4}$
0%
$S{F }_{ 6}$
0%
$S{F }_{ 4}$
0%
$C{l }_{ 2}O$

The hybridization of S atom in

$SO^{2-}_4$ is?

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

The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is:

0%
$5.92$
0%
$3.87$
0%
$6.93$
0%
$4.90$

Explanation

$\mu=\sqrt{n(n+2)}$ BM

$n=$ Number of unpaired electrons

$n=$ Maximum number of unpaired electron

$=5$

$\mu=5.92BM$

$EX: Mn^{2+}$ complex.

Hence option A is correct.

The secondary valency of chromium in

$[Cr(en)_3]Cl_3$ is:

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

Which of the following is not

$sp^3$ hybridise

0%
$BH_3$
0%
$BH^-_4$
0%
$NH_4^+$
0%
$NH_3$

Explanation

$BH_3 \to Sp^2$

$NH_4^+ \to Sp^3$

$BH_4^- \to Sp^3$

$NH_3 \to Sp^3$

The complex having the molecular composition

$[CO(NO_{2})(SCN)(en)_{2}]Br$ exhibits

0%
linkage isomerism
0%
ionisation isomerism
0%
cis-trans isomerism
0%
all of these

Number of possible isomers for the complex

$[Co(en)_{2}Cl_{2}]Cl$ will be _____________.

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

The primary valency of the central transition metal ion in the complex compound

$[Cr(NH_3)_4Cl_2]Cl$ is:

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

Oxidation number of

$Cr$ in

$[Cr[NH_3]_4 Cl_2]^+$ is ____.

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

Explanation

Oxidation number of

$Cr$ in

$[Cr[NH_3]_4 Cl_2]^+$ is

$+3$

$x + (4 \times 0) + (-1 \times 2) = +1$

$x - 2 = +1$

$x = +3$

The valency of carbon is four. On what principle it can be explained in a better way?

0%
Resonance
0%
Hybridization
0%
Electron transfer
0%
None of the above

The units of solubility product of silver chromate

$(Ag_2CrO_4)$ will be __________.

0%
$mol^2L^{-2}$
0%
$mol^3L^{-3}$
0%
$mol L^{-1}$
0%
$mol L^{-2}$

The hybridisation of

$S$ in

$SF_6$ molecule is

0%
$sp^3$
0%
$sp^3d^2$
0%
$sp^3d$
0%
None

Explanation

Sulphur forms

$6$ bond pairs. It thus needs

$6$ hybrid orbitals and therefore, hybridization is

$sp^3d^2$ this molecule has octahedral structure.

1921361_1370880_ans_151c4db47c7a457fb4595b7191d43a8f.png

What is the proportion of sigma

$(\sigma)$ , and pi

$(\pi)$ bond in

$1, 4- Benzoquinone$ ?

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

The total number of possible isomer for the complex comp. :-

$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{I_4}} \right]$

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

The co-ordination number of zinc in zinc blende (Zns) is ___

0%
6
0%
4
0%
8
0%
12

In which of the following octahedral complexes of Co (at no. 27), will the magnitude of

$\triangle _0$ be the highest?

0%
$[Co(C_2O_4)_3]^{3-}$
0%
$[Co(H_2O)_6]^{3+}$
0%
$[Co(NH_3)_6]^{3+}$
0%
$[Co(CN)_6]^{3-}$

The total number of possible isomers for the complex compound

$\left[ \mathrm { Cu } ^ { \mathrm { II } } \left( \mathrm { NH } _ { 3 } \right) _ { 4 } \right) \left[ \mathrm { Pt } ^ { \mathrm { II } } \mathrm { Cl } _ { 4 } \right]$

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

A six coordinate complex of formula

$CrCl_{3} . 6H_{2}O$ has green colour. A

$0.1\ M$ solution of the complex when treated with excess of

$AgNO_{3}$ gave

$28.7\ g$ of white precipitate. The formula of the complex would be

0%
$[Cr(H_{2}O)_{6})]Cl_{3}$
0%
$CrCl(H_{2}O)_{5}]Cl_{2}.H_{2}O$
0%
$[CrCl_{2}(H_{2}O)_{4}]Cl.2H_{2}O$
0%
$[Cr(H_{2}O)_{3}Cl_{3}]$

The compound used in the treatment of lead poisoning is :

0%
EDTA
0%
Cis-platin
0%
D-penicillamine
0%
desferrioxime B

The unpaired electrons in

$[Ni(CO)_4]$ are:

0%
zero
0%
one
0%
three
0%
four

Which of the following has maximum number of unpaired electron (atomic number of

$Fe\ 26$ )

0%
$Fe$
0%
$Fe (II)$
0%
$Fe(III)$
0%
$Fe(IV)$

The number of water molecule(s) not coordinated to copper ion directly in

$Cu{SO}_{4}.5{H}_{2}O$ is:

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

Co-ordination number of central metal atom in

$Na_3[Al(C_2O_4)_3]$ and

$[Co(en)_2Cl]Cl$ are:

0%
$6$ and $6$
0%
$6$ and $5$
0%
$5$ and $5$
0%
$6$ and $4$

Explanation

$Na_3[Al(C_2O_4)_3]$ , co-ordination no

$= 3 \times 2 = 6$
In

$[Co(en)_2]Cl$ , co-ordination no

$= 2 \times 2 + 1 = 5$

Which of the following is diamagnetic complex?

0%
$[Co(OX)_{3}]^{3-}, [Fe(CN)_{6}]^{3-}$
0%
$[Co(Ox)_{3}]^{3-}, [FeF_{6}]^{3-}$
0%
$[Fe(Ox)_{3}]^{3-}, [FeF_{6}]^{3-}$
0%
$[Fe(CN)_{6}]^{3-}, [CoF_{6}]^{3-}$

Explanation

Diamagnetic complex is are

$[Co(Ox)_{3}]^{3-}$ and

$[Fe(CN)_{6}]^{3-}$ .

The primary valency and secondary valency of central metal ion and the no. of total ions produced in aqueous solution for

$K[Co(OX)_{2}(NH_{3})_{2}]$ complex respectively is ________.

0%
$3, 4, 2$
0%
$4, 4, 2$
0%
$3, 6, 2$
0%
$3, 6, 1$

Explanation

$k[Co(OX)_2(NH_3)_2] \rightarrow k^+ + [Co(OX)_2(NH_3)_2]^-$

So the number of total ions produced are 2

The oxidation state of

$K$ is +1 and that of

$(OX)_2$ is -2

Let the oxidation state of

$Co$ be

$x$

$\therefore +1+x+2(-2)+2(0)=0$

$\therefore x=3$

The primary valency of the

$Co$ is 3

The secondary valency of the

$Co$ is 6

$Cr^+3$ in aquous medium form green coloured complex with

$NH_3$ ligand. How many ligand associated

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

The conductivity measurement of a coordination compound of Cobalt (III) shows that it dissociates into 3 ions in solution. The compound is:

0%
Hexaamminecobalt (III) chloride
0%
Pentaamminesulphatocobalt (III) chloride
0%
Pentaamminechloridocobalt (III) sulphate
0%
Pentaamminechloridocobalt (III) chloride

Explanation

Pentaaminechloridocobalt (III) chloride

$[Co(NH_3)_5 Cl] Cl_2 \rightarrow \, \underbrace { [Co(NH_ 3)_ 5Cl]^{ 2+} +2Cl^-] }_{ 3 \, ions }$

$[Co(C_{2}O_{4})_{3}]^{3-}$ is a

0%
Low spin complex
0%
Paramagnetic
0%
High spin
0%
$sp^{3}d^{2}$ hybridized

Explanation

$Co^{+3}$ has

$d^{2}sp^{3}$ hybridisation so all electron are paired therefore it is Low spin complex and diamagnetic.

After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.

$AgBr+2Na_2S_2O_3\rightarrow 4Na^++Br^-+[Ag(S_2O_3)_2]^{3-}$
What is the function of the thiosulfate ion?

0%
To make the silver ions soluble
0%
To oxidise the silver ions
0%
To reduce the bromine
0%
To reduce the silver ions

Explanation

$AgBr + 2 {Na}_{2}{S}_{2}{O}_{3} \longrightarrow 4 {Na}^{+} + {Br}^{-} + {\left[ Ag {\left( {S}_{2}{O}_{3} \right)}_{2} \right]}^{3-}$

In the above reaction, thiosulfate ion

$\left( {{S}_{2}{O}_{3}}^{2-} \right)$ is use to make the silver ion

$\left( {Ag}^{+} \right)$ soluble.

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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