MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 1

On treatment of

$100\, mL$ of

$0.1\, M$ solution of

$CoCl_{3}.6H_{2}O$ with excess

$AgNO_{3};1.2\times 10^{22}$ ions are precipitated. The complex is :

0%
$\left [ Co\left ( H_{2}O \right )_{3}Cl_{3} \right ].2H_{2}O$
0%
$\left [ Co\left ( H_{2}O \right )_{6} \right ]Cl_{3}$
0%
$\left [ Co\left ( H_{2}O \right )_{5}Cl \right ]Cl_{2}.H_{2}O$
0%
$\left [ Co\left ( H_{2}O \right )_{4}Cl \right ]Cl_{2}.H_{2}O$

Explanation

Avogadro's number is

$\displaystyle 6.023\times 10^{23}$

$\displaystyle 1.2\times 10^{22}$ ions corresponds to

$\displaystyle \dfrac {1.2\times 10^{22}}{6.023\times 10^{23}}=0.02$ moles.

$100$ mL of 0.1M solution of

$\displaystyle CoCl_{3}.6H_{2}O$ corresponds to

$\displaystyle \dfrac {100 \: mL}{1000 \: mL/L} \times 0.1 \: M = 0.01 \: moles$ .

$0.01$ moles of

$\displaystyle CoCl_{3}.6H_{2}O$ reacts with excess

$\displaystyle AgNO_3$ to form

$0.02$ moles of

$\displaystyle AgCl$ precipitate, which means

$0.01$ moles of

$Ag^+$ and

$0.01$ moles of

$Cl^-$ ions.

Thus, from one molecule of

$\displaystyle CoCl_{3}.6H_{2}O$ , one chlorine atom is displaced. Hence, the complex is

$\displaystyle \left [ Co\left ( H_{2}O \right )_{5}Cl\right ]Cl_2.H_{2}O$ as number of chloride ion outside co-ordination sphere should be one.

Hence, the option (C) is the correct answer.

Which of the following complexes is used to be as an anticancer agent?

0%
$mer-[Co(NH_3)_3Cl]$
0%
$cis-[PtCl_2NH_3)_2]$
0%
$cis-K_2[Pt Cl_2Br_2]$
0%
$Na_2CoCl_4$

Explanation

$Cis-[PtCl_2NH_3)_2]$ known as cis platin is used as an anticancer agent.

Iron carbonyl,

$Fe(CO)_{5}$ is __________.

0%
trinuclear
0%
tetranuclear
0%
dinuclear
0%
mononuclear

Explanation

Iron carbonyl,

$Fe(CO)_{5}$ is mononuclear. In

$Fe(CO)_{5}$ , one Fe atom is surrounded by 5 CO ligands. Mononuclear complexes are those complexes in which one metal atom/ion is surrounded by ligands.

The hypothetical complex chloro-diaquatriamminecobalt(III) chloride can be represented as:

0%
$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 } \right] { Cl }_{ 2 }$
0%
$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) Cl }_{ 3 } \right]$
0%
$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 }Cl \right]$
0%
$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 3 }{ Cl }_{ 3 } \right]$

Explanation

The hypothetical complex chloro-diaquatriamminecobalt(III) chloride can be represented as

$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 } \right] { Cl }_{ 2 }$ .

The complex

$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) Cl }_{ 3 } \right]$ is tetra

$chloro-aquatriamminecobalt(III)$ .

The complex

$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 2 }Cl \right]$ is di

$chloro-diaquatriamminecobalt(III)$ .

The complex

$\left[ CoCl{ \left( { NH }_{ 3 } \right) }_{ 3 }{ \left( { H }_{ 2 }O \right) }_{ 3 }{ Cl }_{ 3 } \right]$ is tetra

$chloro-triaquatriamminecobalt(III)$ .

The colour of blood is red due to the presence of:

0%
magnesium
0%
cytoglobin
0%
haemoglobin
0%
all of these

The properties of a compound are:

0%
Similar to the properties of its constituents
0%
Similar to the properties of one of the constituents
0%
Different from the properties of its constituents
0%
None of these

The brown complex obtained in the detection of nitrate radical is formulated as

$\displaystyle \left[ { Fe\left( { H }_{ 2 }O \right) }_{ 5 }NO \right] { SO }_{ 4 }$ . What is the oxidation number of

$\displaystyle Fe$ in this complex?

0%
+1
0%
+2
0%
+3
0%
0

Explanation

The brown colour complex obtained in the detection of nitrate radical is formulated as

$\displaystyle \left[ { Fe\left( { H }_{ 2 }O \right) }_{ 5 }NO \right] { SO }_{ 4 }$ . In this complex, oxidation state of iron is +1. The univalent character of iron in the complex is justified by the presence of a coordinated

$\displaystyle { NO }^{ + }$ group.

Which of the following compounds will exhibit the highest magnetic moment?

0%
(a) $[Ti{(NH_3)}_6]^{3+}$
0%
(b) $[Cr{(NH_3)}_6]^{3+}$
0%
(c) $[Co{(NH_3)}_6]^{3+}$
0%
(d) $[Zn{(NH_3)}_6]^{2+}$

The hybridization of Ag in a

$[Ag(N{ H }_{ 3 }{ ) }_{ 2 }{ ] }^{ (+) }$ is:-

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

The cobalt hexammine cation has an atom of cobalt in the centre, surrounded by six molecules of ammonia that are attached to it. The shape of this ion is?

0%
Tetrahedral
0%
Square
0%
Linear
0%
Octahedral

Explanation

REF.Image.

$CO^{3+} \Rightarrow [Ar]d^{6} \Rightarrow$

$\Rightarrow SP^{3}d^{2}$

$\Rightarrow [Co(NH_{3})_{6}]^{3P}$

octagonal

$\Rightarrow PP^{3}d^{2}\rightarrow$ hydrogenation.

1122378_930469_ans_b9667a10b8444cd990066d21a79c228e.png

Pair of compounds which is planar:

0%
${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 }$
0%
${ \left[ Ni{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }\quad$
0%
${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Rh{ \left( CO \right) }_{ 2 }{ \left( P{ Ph }_{ 3 } \right) }_{ 2 } \right] }^{ + }$
0%
${ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }$

Which of the following statement is correct?

0%
Energy difference should be more between orbitals which undergo hybridization.
0%
Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.
0%
Hybrid orbitals arrange around the center of the atom unsymmetrically.
0%
Hybrid orbitals can form $\pi$ bonds.

Explanation

Number of hybrid orbitals formed is equal to number of atomic orbitals involved in hybridisation.

$\pi$ bonds will be formed always by pure orbitals (p or d orbitals). Hybrid orbitals arrange around the centre of atom gymmetrically depending on number of bond pairs and lone pairs.

Transition metals are coloured due to the following electronic transition

0%
d - s
0%
d - d
0%
s - p
0%
f - s

Which of the following metal ion is present in Ziegler-Natta catalyst?

0%
$Fe^{2+}$
0%
$Zr^{4+}$
0%
$Rh^{+}$
0%
$Ti^{4+}$

Explanation

Ziegler-Natta catalyst is triethylaluminium

$Al(C_2H_5)_3$ along with

$TiCl_4$ .

Hence, option D is correct.

The correct IUPAC name of

$Mn_{3}(CO)_{12}$ is:

0%
mangesedodecylcarbonyl(0)
0%
dodecacarbonylmanganate(0)
0%
dodecacarbonylmanganic(II)
0%
dodecacarbonyltrimanganese(0)

Explanation

The correct IUPAC name of

$Mn_{3}(CO)_{12}$ is

$dodecacarbonyltrimanganese(0)$ . The central metal atom is a cobalt atom in zero oxidation state. The oxidation state is written in parenthesis. Three such

$Mn$ atoms are present. Hence, the prefix tri is used.

$12$ carbonyl groups are present. Hence, the prefix

$dodeca$ is used.

The number of isomers exhibited by

$\left [ Cr\left ( NH_{3} \right )_{3}Cl_{3} \right ]$ is:

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

Explanation

The complex

$\left [ Cr\left ( NH_{3} \right )_{3}Cl_{3} \right ]$ is of the type

$Ma_3b_3$ , has two isomers cis and trans or facial and meridional.

Which of the following will give

$Fe^{3+}$ ions in the solution?

0%
$[Fe(CN)_{6}]^{3-}$
0%
$[Fe(CN)_{6}]^{4-}$
0%
$NH_{4}(SO_{4})_2.FeSO_{4}.6H_{2}O$
0%
$Fe_{2}(SO_{4})_{3}$

Explanation

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

$[Fe(CN)]_{6}^{4-}$ are complex ions.

They retain their identity in the solution.

Hence, they will not give

$Fe^{3+}$ ions in the solution.

$(NH_{4})_{2}SO_{4}.FeSO_{4}.6H_{2}O$ is a double salt.

In the solution, it breaks into individual ions, but it will give

$Fe^{2+}$ ions in the solution.

Similarly,

$Fe_{2}(SO_{4})_{3}$ also breaks into individual ions.

Hence, they will give

$Fe^{3+}$ ions in the solution.

Thus, option D is the correct answer.

Hybridisation involves:

0%
addition of an electron pair
0%
combination and redistribution of atomic orbitals
0%
removal of an electron pair
0%
separation of orbitals

Explanation

$\bf{Hint:-}$ Mixing of different energy orbitals to form new equal energy orbitals known as hybridization.

$\bf{Explanation-}$

$\bullet$ Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals.

$\bullet$ Atomic orbitals with similar energy combine to give degenerated hybrid orbitals.

$\bullet$ The number of combining atomic orbitals is equal to the hybridised orbitals.

$\bullet$ They involve redistribution of electrons of combined atomic orbital to hybrid orbitals.

$\bf{Conclusion-}$ Hence, option

$B$ is correct.

Co-ordination compounds are mostly formed by:

0%
s-block elements
0%
p-block elements
0%
d-block elements
0%
f-block elements

The chemical formula of iron(III) hexacyanoferrate(II) is:

0%
$Fe\left [ Fe\left ( CN \right )_{6} \right ]$
0%
$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]$
0%
$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]_{4}$
0%
$Fe_{4}\left [ Fe\left ( CN \right )_{6} \right ]_{3}$

Explanation

We proceed from left to right, iron (III) means

$Fe^{3+}$ , hexacyano means

$6CN^{-}$ ions are present and ferrate (II) means

$Fe^{2+}$ is the central metal atom.

So, the complex ion is

$[Fe(CN)_{6}]^{4-}$ and the counter ion is

$Fe^{3+}$ . So, after balancing the charge, the formula becomes

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

Option D is correct.

IUPAC name of

$Na_{3}\left [ Co\left ( ONO \right )_{6} \right ]$ is:

0%
Sodium cobaltinitrite
0%
Sodium hexanitritocobaltate (III)
0%
Sodium hexanitrocobalt (III)
0%
Sodium hexanitritocobaltate (II)

Explanation

The above complex

${ Na }_{ 3 }[Co(ONO)_{ 6 }]$ has two ions:

${ Na }^{ + }$ and a complex ion which is negatively charged

$[Co(ONO)_{ 6 }]^{3 - }$

According to the rules, name of the positive ion comes first (i.e., sodium), do not mention the number of sodium as it is not present in a square bracket.

After naming the positive ion, next will be the negative ion, Now in this example, the negative ion is a complex compound, which contain's 'metal' and 'ligand'. While naming a complex, name of ligand comes first, also mention the number of ligand by using prefix (di, tri, tetra. etc..) and if a bidentate ligand is present then mention the atom which is attached to central metal (i.e., hexanitrito

${ - }O{ - })$ followed by name of metal with the ending 'ate' (i.e., cobaltate, always use the word 'ate' if the complex is negatively charge) and than oxidation number of metal.

To show the oxidation state, we use Roman numerals inside parenthesis.

Option B is correct.

Example for a coordination compound is ____________.

0%
$KCl.MgCl_{2}.6H_{2}O$
0%
$K_{2}SO_{4}Al_{2}(SO_{3})24H_{2}O$
0%
$CoCl_{3}.6NH_{3}$
0%
$FeSO_{4}.(NH_{4})_{2}SO_{4}.6H_{2}O$

Explanation

Carnallite

$KCl.MgCl_{2}.6H_{2}O$
Alum

$K_{2}SO_{4}Al_{2}(SO_{4})_3.24H_{2}O$
Ferrous ammonium sulphate

$(NH_4)_2Fe(SO_4)_26H_2O$ .
All are double salts.
Only

$[Co(NH_3)_6].Cl_3$ is complex compound.

Which is a coordination compund ?

0%
Ferrous ammonium sulphate
0%
Carnallite
0%
Potassium ferrocyanide
0%
Gypsum

Explanation

Only 'Potassium ferrocyanide' (i.e.

$[Fe(CN)_{ 5 }]$ ) shows properties of coordination compound.

When it gets dissolve they do not form simple ion like

$Fe^{ +2 }$ or

$CN^{ - }$ , but instead their complex ions remain intact.

whereas carnallite (i.e.

$KMg{ Cl }_{ 3 }$ ), Gypsum (i.e.

$Ca{ SO }_{ 4 }$ ) and Ferrous ammonium sulphate (i.e.

$({ NH }_{ 4 })_ { 2 }Fe (SO_4)_2$ ) they dissovle and give simple ions.

Option C is correct.

The primary valence of the metal ion is satisfied by :

0%
neutral molecules
0%
positive ions
0%
negative ions
0%
all

Which of the following has the highest molar conductivity?

0%
$[Co(NH_{3})_{6}]Cl_{3}$
0%
$[Co(NH_{3})_{5}Cl]Cl_{2}$
0%
$[Co(NH_{3})_{4}Cl_{2}]Cl_{2}$
0%
$[Co(NH_{3})_{3}Cl_{3}]$

Explanation

Hint: Molar conductivity is dependent on the number of ions produced in the solution.

Explanation:

The dissociation of $[Co{(N{H_3})_6}]C{l_3}$ is as follows;

$[Co{(N{H_3})_6}]C{l_3} \rightleftharpoons {[Co{(N{H_3})_6}]^{3 + }} + 3C{l^ - }$

The number of ions produced here is $4$ . So the molar conductivity is the highest.

Final Answer: The molar conductivity of is the highest. Hence the correct answer is option $A$ .

Transition elements form complexes readily because :

0%
Small size of cation
0%
Large ionic Charge
0%
Vacant d orbitals
0%
All the above

According to Werner’s theory of valency, transition metals possess:

0%
only one type of valency
0%
two types of valencies
0%
three types of valencies
0%
four types of valencies

Explanation

According to Werner’s theory of valence, transition metals has two valencies are primary valencies and secondary valency.

The primary valency relates to the oxidation state and the secondary valency relates to the coordination number or it is the number of ligands attached to metal ions.

Hence, the correct option is

$B$

During the splitting of degenerate d-orbitals under the influence of ligand the average d-orbital energy

0%
remains same
0%
increases
0%
decreases
0%
may increase or decrease

The ionizable valency of Ni in Ni(CO) $_{4}$ is __________.

0%
2
0%
4
0%
0
0%
1

The following ion shows colour not due to d-d transistion.

0%
$Cr_{2}O{_{7}}^{2-}$
0%
$MnO{_{4}}^{-}$
0%
$CrO{_{4}}^{2-}$
0%
All

The splitting of degenerated d-orbitals takes place into which of the following two sets :

0%
$d_{xy},d_{z^{2}},d_{xz}$ and $d_{yz},d_{x^{2}-y^{2}}$
0%
$d_{xy},d_{yz},d_{zx}$ and $d_{x^{2}-y^{2}},d_{z^{2}}$
0%
$d_{xy},d_{x^{2}-y^{2}},d_{z^{2}}$ and $d_{yz},d_{xz}$
0%
$d_{xy},d_{x^{2}-y^{2}},d_{xz}$ and $d_{yz},d_{z^{2}}$

Explanation

The d-orbitals split into two sets with an energy difference, the crystal-field splitting parameter where the

$d_{xy}, d_{xz} \ and \ d_{yz}$ orbitals will be lower in energy than the

$d_{z^2} \ and \ d_{x^2-y^2}$ , which will have higher energy, because the former group is farther from the ligands than the latter and therefore experience less repulsion. The three lower-energy orbitals are collectively referred to as

$t_{2g}$ , and the two higher-energy orbitals as

$e_g$ .

Number of $Cl^{-}$ ions satisifying both primary and secondary valency are in $CoCl_{3}.5NH_{3}$ .

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

Explanation

There is one

$Cl^{-}$ ion which satisfies both primary and secondary valency in

$CoCl_{3}.5NH_{3}$ .

Option A is correct.

In the presence of strong eletrical field, the following set of orbitals are not degenerate :

0%
$3d_{xy} and 3d_{z}2$
0%
$3d_{xy} and 3d_{yz}$
0%
$3d_{xy}, 3d_{yz} and 3d_{zx}$
0%
all the above

Explanation

Due to their different axis orientation; in presence of strong electrical field,

$3d_{xy}$ and

$3d_{z^2}$ orbitals are not degenerated.

1 mole of amino cobalt chloride on treating with an excess of $AgNO$ $_{3}$ solution gives $2$ moles of $AgCl$ precipitate. The number of $Cl^-$ ions which satisfies both primary and secondary valency of cobalt in the complex is:

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

Explanation

1 mole of amino cobalt chloride gives 2 moles of

$AgCl$ so number of

$Cl^-$ ions which satisfies both primary and secondary valency of cobalt in the complex is 1.

When mole of

$CoCl_{3}.6NH_{3}$ is treated with excess of AgNO

$_{3}$ solution. The number of moles of AgCl precipitated is:

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

Explanation

One mole of

$CoCl_{3}.6NH_{3}$ gives 3 moles of

$Cl$ ion so when mole of

$CoCl_{3}.6NH_{3}$ is treated with excess of

$AgNO_{3}$ solution, it gives 3 moles of

$AgCl$ .

$CoCl_3.6NH_3(aq) + excess AgNO_3(aq) \rightarrow 3AgCl + Co(NO_3)_3.6NH_3(aq)$

The primary and secondary valency of

$Co$ in the octahedral complex

$CoCl_{3}.5NH_{3}$ are __________.

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

Explanation

The primary valency of

$Co$ is 3 and the secondary valency of

$Co$ is 6.
The primary valency represents the number of ions in the ionization sphere and the secondary valency represents the number of ligands.
In this example, one chlorine atom serves dual role.

IUPAC name for the complex $[Cu(NH_{3})_{4}]SO_{4}$ is:

0%
cupramonium sulphate
0%
copper sulphate tetraamonia
0%
tetrammine copper (II) sulphate
0%
copper amonium (IV) sulphate

The molecular interactions responsible for hydrogen bonding in HF:

0%
$\text{ion induced dipole}$
0%
$\text{dipole dipole}$
0%
$\text{dipole induced dipole}$
0%
$\text{none}$

An excess of AgNO $_{3}$ is added to 100 ml of a 0.1 M solution of dichlorotetraaquachromium (III) chloride. The number of mol of AgCl precipitate would be:

0%
0.001
0%
0.002
0%
0.003
0%
0.01

Explanation

dichlorotetra aquachromium (III) cholride is $[Cr(H_2O)_4Cl_2]Cl$ , so one mole of it will give one mole of $Cl^-$ ion so 100 ml of 0.1M will give 0.01 mole of $Cl^-$ ion and 0.01 mole of $AgCl$ formed.

The correct order of ligands for writing the formula of complex compounds is _________.

0%
neutral, anionic, cationic
0%
anionic, neutral, cationic
0%
anionic, cationic, neutral
0%
cationic, neutral, anionic

$[Co(NH_{3})_{4}(NO_{2})_{2}]Cl$ exhibits:

0%
linkage, ionization and geometrical isomerism
0%
ionisation, geometrical and optical isomerism
0%
linkage, geometrical and optical isomerism
0%
linkage, ionization and optical isomerism

Explanation

By changing

$Cl$ and

$NO_2$ , it will show ionization isomerism
Due to

$-NO_2$ and

$-ONO$ , it will show linkage isomerism.
Also, it show cis/trans isomerism.
But due to symmetry it will not show optical isomerism.

Which of the following is considered to be ananticancer species :

Explanation

Cisplatin i.e (

$cis-diamminedichloroplatinum$ ) is considered as anticancer species. Hence the option B is correct

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

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

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

0%
Chlorophylls are green pigments in plants and contain calcium
0%
Haemoglobin is the red pigments of blood and contains iron
0%
Cyanocobalamin is vitamin B12 and contains cobalt
0%
Carboxypeptidase-A is an enzyme and contains zinc

Explanation

Chlorophyll is the green pigment in plants that helps in photosynthesis. The metal inside the ring structure of chlorophyll is magnesium

$(Mg^{2+})$ . The four nitrogen atoms in the ring structure of chlorophyll are bonded to the magnesium ion.

Compounds having equal number of hybrid orbitals among the following are:

0%
0%
0%
0%
none of these

Explanation

Compound A & C has equal no. of hybrid orbitals (as each carbon is

$sp^3$ hybridized and each compound has

$4$ carbon so total no. of

$sp^3$ hybridized orbitals are

$16$ ).

Option A and C are correct.

In which of the compound given below contains more than one kind of hybridization (

$sp,sp^{2}$ ,

$sp^{3}$ ) for carbon:

0%
$CH_{3}CH_{2}CH_{2}CH_{3}$
0%
$CH_{3}-CH=CH-CH_{3}$
0%
$CH_{2}=CH-CH=CH_{2}$
0%
$H-C\equiv C-H$

Explanation

(i)

$\overset{sp^3}{CH_3}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_3}$

Only

$sp^3$ hybridized carbon.

(ii)

$\overset{sp^3}{CH_3}-\overset{sp^2}{CH}=\overset{sp^2}{CH}-\overset{sp^3}{CH_3}$

Both

$sp^2$ and

$sp^3$ hybridized carbon.

(iii)

$\overset{sp^2}{CH_2}=\overset{sp^2}{CH}-\overset{sp^2}{CH}=\overset{sp^2}{CH_2}$

Only

$sp^2$ hybridized carbon.

(iv)

$H- \underset{sp}{C}\equiv \underset{sp}{C}-H$

Only

$sp$ hybridized carbon.

Option B is correct.

Total numbers of hybrid orbitals present in

$CH_{3}-C\equiv CH$ are:

0%
$8$
0%
$6$
0%
$4$
0%
$2$

Explanation

Propyne has

$8$ hybrid orbitals.

Statement I: The geometrical isomer of the complex $[M(NH_{3})_{4}Cl_{2}]$ are optically inactive.

Statement II: Both geometrical isomers of the complex $[M(NH_{3})_{4}Cl_{2}]$ possess axis of symmetry

Select the correct option regarding given statements:

0%
Statement I is true, Statement II is true,Statement II is a correct explanation of Statement I.
0%
Statement I is true, Statement II is true,Statement II is not the correct explanation of statement I
0%
Statement I is true, Statement II is false.
0%
Statement I is false, Statement II is true.

The neutral complex, diamine dibromo dichloro platinum (IV) is best represented as:

0%
$[Pt(NH_{3})_{2}Br_{2}Cl_{2}]$
0%
$[PtCl_{2}Br_{2}(NH_{3})_{2}]$
0%
$[PtBr_{2}Cl_{2}(NH_{3})_{2}]$
0%
$[Pt(NH_{3})_{2}Cl_{2}Br_{2}]$

$CuSO_{4}$ when reacts with

$KCN$ forms

$CuCN$ which is insoluble in water. It is soluble in excess of

$KCN$ , due to formation of the following complex:

0%
$K_{2}[Cu(CN)_{4}]$
0%
$K_{3}[Cu(CN)_{4}]$
0%
$CuCN_{2}$
0%
Cu $[KCu(CN)_{4}]$

Explanation

Here,

$CuCN$ is insoluble in water. This means

$Cu$ in

$CuCN$ is in

$+1$ state. Because

$Cu$ is unstable in water only when it is in

$+1$ state.

Co-ordination sphere exactly contains

$4 \ CN^-$ ions. So, its structure would be

$[Cu(CN)_4]^{x}$

Calculating

$x$ ,

$Cu$ is

$+1$ state

$1 + 4(-1)=x$

$1-4=x$

$x=-3$

So,

$[Cu(CN)_4]^{-3}$

It combines with

$3K^+$ ions to form

$K_3[Cu(CN)_4]$

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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