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CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 14 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
The D-And F-Block Elements
Quiz 14
Transition metals show paramagnetic :
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due to characteristic configuration
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high lattice energy
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due to variable oxidation states
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due to unpaired electrons
Match List-I with List -II and select the correct answer using codes given ahead in the lists:
List-I
Metal ions
List-II
Magnetic moments (B.M)
$$Cr^{3+}$$
$$\sqrt{35}$$
$$Fe^{2+}$$
$$\sqrt{30}$$
$$Ni^{2+}$$
$$\sqrt {24}$$
$$Mn^{2+}$$
$$\sqrt{15}$$
$$\sqrt 8$$
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$$A-1, B-3, C-5, D-4$$
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$$A-2, B-3, C-5, D-1$$
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$$A-4, B-3, C-5, D-1$$
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$$A-4, B-5, C-3, D-1$$
Wilkinson's catalyst contains :
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rhodium
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iron
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aluminium
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cobalt
The catalysic activity of transition elements and their compounds is described to:
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their chemical reactivity
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their magnetic behaviour
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their unfilled $$d-$$orbital
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their ability to adopt multiple oxidation states and their complexing ability.
Explanation
Option D) is correct
The catalytic activity of transition elements and their compounds is ascribed to their ability to adopt multiple oxidation states and their complexing ability.
Catalysis
at a solid surface
involves the formation of bonds between reactant molecules and atoms
of the surface of the catalyst (first-row transition metals utilize
3d
and
4s
electrons for bonding).
This has the effect of increasing the concentration
of the reactants at the catalyst surface and also the weakening
of the bonds in the reacting molecules (the activation energy is lowering).
Also because the transition metal ions can change their oxidation states,
they become more effective as catalysts. For example, iron(III) catalyzes
the reaction between iodide and persulphate ions.
Select the correct statements :
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Diamond is hard while graphite is soft and slippery
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The degree of co valency in the following compounds runs as $$HgX_2 > CdX_2 > ZnX_2$$
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$$ZnS$$ is white, $$CdS$$ is yellow, while $$HgS$$ is black
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Diamond is soft while graphite is hard
Explanation
Diamond is hard
because the carbon atoms in
diamond
are bonded in a stronger tetrahedron pattern
but graphite is soft and slippery
because the carbon atoms in
graphite
are bonded in layers with only weak van der Waal force holding the layers together.
In general,
the degree of co-valency increases
as we go
down
a
group. This is
because the orbitals are more diffused and shielding by core electrons is greater. Thus, t
he degree of co-valency follows the order: HgX
₂ > CdX₂ > ZnX₂.
The colour in transition metal sulphides comes from absorption of light that leads to excitation of an electron from a filled valence band to an empty conduction band. As the size of the cation increases, upon moving down in the periodic table, the overlap between the orbitals decreases and the band gap shifts to lower energy. Since, the band gap decreases from ZnS to HgS, the colour of ZnS is white, CdS is yellow and HgS is black.
Hence, options A, B and C are correct.
Which of the following sulphides is yellow in colour?
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$$CuS$$
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$$CdS$$
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$$ZnS$$
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$$CoS$$
Explanation
Alkali and alkaline earth metal sulphides are colourless.
Heavy metal sulphides are unusually deeply coloured.
Both $$CuS$$ and $$CoS$$ are black coloured, $$ZnS$$ is bluish white.
$$CdS$$ is yellow in colour.
Hence the correct option is $$(B)$$.
The statement that is not correct for the periodic classification of elements is:
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The properties of elements are the periodic function of their atomic numbers.
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Non - metallic elements are lesser in number than metallic elements.
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The first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic number.
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For transition elements the d subshells are filled with electrons monotonically with increase in atomic number.
Explanation
The d- sub shells are not filled with electrons monotonically with increase in atomic number. There are some exceptions like Cr, Cu, etc.
Option D is correct.
Which of the following compounds is (are) coloured due to charge transfer spectra and not due to $$d-d$$ transition?
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$$KMnO_4$$
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$$K_2CrO_4$$
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$$CrO_3$$
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All of these
On heating chromite $$(FeCr_2O_4)$$ with $$Na_2CO_3$$ in air, the following product is obtained:
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$$Na_2CrO_4$$
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$$Na_2Cr_2O_7$$
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$$FeO$$
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$$Fe_3O_4$$
Number of moles of $$K_2Cr_2O_7$$ radius by $$1$$ mole of $$Sn^{2+}$$ is:
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$$1/6$$
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$$2/3$$
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$$1/3$$
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$$1$$
Explanation
The balanced chemical reactions (ionic reactions) for reduction of $$K_2Cr_2O_7$$ by $$Sn^{2+}$$ are;
$$Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O$$
$$3(Sn^{2+}\rightarrow Sn^{4+}+2e^-)$$
Balances net equation: $$3Sn^{2+}+Cr_2O_7^{2-}+14H^+\rightarrow 3Sn^{4+}+2Cr^{3+}+7H_2O$$
Thus,
According to the balanced reaction, $$1$$ mole of $$Cr_2O_7^{2-}$$ will be reduced by $$3$$ moles of $$Sn^{2+}$$
Thus, $$1$$ mole of $$Sn^{2+}$$ will be reduced $$=\dfrac 13$$ moles of $$Cr_2O_7^{2-}$$
Hence the correct answer is option C.
Which of the following are paramagnetic?
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$$[Fe(CN)_6]^{4-}$$
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$$Ni(CO)_4$$
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$$[NiCl_4]^{2-}$$
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$$[CoF_6]^{3-}$$
Predict the magnetic nature of $$A$$ and $$B$$:
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Both are diamagnetic
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Both are paramagnetic
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$$A$$ is paramagnetic and $$B$$ is diamagnetic
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$$A$$ is diamagnetic and $$B$$ is paramagnetic
Which of the following sulphides is yellow?
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$$ZnS$$
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$$CdS$$
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$$NiS$$
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$$PbS$$
Explanation
$$ZnS\rightarrow{White}$$
$$CdS\rightarrow{Yellow}$$ {Option (B) is correct.}
$$NiS\rightarrow{Brown}$$
$$PbS\rightarrow{Black}$$
Consider the following statements and arrange in the order of true/false as given in the codes.
$$S_{1}$$ : Both $$\left [ Co\left ( ox \right )_{3} \right ]^{3-}$$ and $$\left [ CoF_{6} \right ]^{3-}$$ are paramegnetic.
$$S_{2}:CoCl_{3}3NH_{3}$$ complex is non-conducing.
$$S_{3}$$ : The number of possible isomers for the complex $$\left [ Pt\left ( NO_{2} \right )\left ( py \right )\left ( OH \right )\left ( NH_{3} \right ) \right ]$$ is six.
$$S_{4}$$ : The oxidation state of iron in brown ring complex $$\left [ Fe\left ( H_{2}O \right )_{5}NO^{+} \right ]SO_{4}$$ is +II where NO is $$NO^{+}$$
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F T T F
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T T T T
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F T T F
0%
T T T F
Explanation
$$S_1:[CoF_6]^{3−}\text{ is paramagnetic with 5 unpaired electrons but }[Co(ox)_3]^{3−}\text{ is diamagnetic as all electrons}$$ $$\text{are paired.}$$
$$S_2:\text{ It exists as }[Co(NH_3)_3Cl_3],\text{ neutral molecule.}$$
$$S_3:\text{ It is a correct statement. It shows linkage and geometrical isomerism.}$$
$$S_4: [Fe(+I)(H_2O)_5NO^+]SO^{2−}_4$$
Option C is correct.
Which of the following complexes is / are paramagnetic?
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$$\left [ Mn\left ( CN_{6} \right ) \right ]^{3-}$$
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$$\left [ Cr\left ( NH_{3} \right )_{6} \right ]^{3+}$$
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$$\left [ Fe\left ( CN \right )_{6} \right ]^{4-}$$
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$$\left [ Co\left ( CN \right )_{6} \right ]^{3-}$$
Which of the following statements is correct ? (I) Anions of both (B) and (C) are diamagnetic and have tetrahedral geometry (II) Anions of both (B) and (C) are paramagnetic and have tetrahedral geometry (III) Anions of (B) is paramagnetic and that of (C) is diamagnetic but both have tetrahedral geometries (IV) Green coloured compound (B) in a neutral or acidic medium disproportionates to give (C) and (D)
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I and II only
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II and III only
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II and IV only
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III and IV only
Which of the following types of metal make the most efficient catalysts?
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Alkali metal
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Transition metal
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Alkali-earth metals
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Radioactive metals
Explanation
Transition metals are the most efficient catalysts due to partially or half-filled d-orbitals.
Transition metals can form unstable intermediate products with suitable reactants. These intermediate products lower the activation energy of the reaction which makes the reaction faster.
So, transition elements are the most efficient catalysts.
Which of the following pair of compounds is expected to exhibit same colour in aqueous solution ?
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$$FeCl_{2},CuCl_{2}$$
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$$VOCl_{2},CuCl_{2}$$
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$$VOCl_{2},FeCl_{2}$$
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$$FeCl_{2},MnCl_{2}$$
Titanium shows magnetic moment of 1.73 B M in its compound . What is the oxidation state of titanium in the compound ?
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+2
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+1
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+3
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+4
Consider the following statements and arrange in the order of true/false as given in the codes.
$$S_{1}$$ : The IUPAC name of the compound $$\left [ Cr\left ( NH_{3} \right )_{5} \left ( NCS \right )\right ]\left [ ZnCl_{4} \right ]$$ is pentaamminethiocyanato-N-chromate (III) tetraachlorozincate (II).
$$S_{2}: CoSO_{4}$$ is paramagnetic and coloured.
$$S_{3}:\left [ Co\left ( NH_{3} \right )_{4}\left ( NO_{2} \right )_{2} \right ]NO_{3}$$ shows ionisation, linkage and geometrical isomerism.
$$S_{4}$$ : The C-O bond length in the complex, $$\left [ Fe\left ( CO \right )_{5} \right ]$$ is same as that found in carbon monoxide itself.
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F T T F
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T T T T
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F T T F
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T T T F
Fe, Co and Ni have valuable catalytic properties in process involving
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Organic compound
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Oxidation
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Hydrogenation
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Compounds of hydrogen
Explanation
A Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. A catalyst accelerates the rate of a reaction by lowering the activation energy.
Fe, Co and Ni have valuable catalytic properties in the process involving hydrogenation due to large surface area and variable valency. Thus Ni is used for the hydrogenation of vegetable oils to vegetable fats.
Which has the maximum ferromagnetic character?
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Fe
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Co
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Ni
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Pt
Explanation
Iron has a maximum number of unpaired electrons so it has the maximum ferromagnetic character.
Which of the following may be colourless?
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$$ Cr^{3+} $$
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$$ Cu^{+} $$
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$$ Fe^{3+} $$
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$$ Cu^{2+} $$
Explanation
$$ Cu^{+} $$ may be colourless due to completely filled d-orbitals.
Which of the following species is expected to show the highest magnetic moment? (At. No.: Cr = 24, Mn = 25, Co = 27, Ni = 28, Cu =29)
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$$ Cr^{2+} $$
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$$ Mn^{2+} $$
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$$ Cu^{2+} $$
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$$ Co^{2+} $$
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$$ Ni^{2+} $$
Explanation
Highest magnetic moment depends upon number of unpaired electron since.
$$ Cr^{2+} = [Ar]3d^4 4s^0 , \quad Mn^{2+} = [Ar]3d^5 4s^0 $$
$$ Cu^{2+} = [Ar]3d^9 4s^0,\quad Co^{2+} = [Ar]3d^7 4s^0 ,\quad Ni^{2+} = [Ar]3d^8 4s^0 $$
So $$ Mn^{2+} $$ contain maximum number of unpaired electron i.e. 5.
The 3d metal ions form coloured compounds because the energy corresponding to the following lies in the visible range of electromagnetic spectrum
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free energy change of complex formation by 3d metal ions
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d-d transition of 3d electrons
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heat of hydration of 3d metal ions
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ionisation energy of 3d metal ions
Explanation
The 3d metal ions form coloured compounds because the energy corresponding to the d-d transition of 3d electrons lies in the visible range of the electromagnetic spectrum
Which one of the following characteristics of the transition metals is associated with their catalytic activity?
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Variable oxidation states
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High enthalpy of atomization
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Paramagnetic behaviour
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Colour of hydrated ions
Explanation
The transition metals form the reaction intermediates due to the presence of vacant orbitals or their tendency to form variable oxidation states.
The magnetic moment of a metal ion of first transition series is 2.83 BM. Therefore it will have unpaired electrons
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6
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4
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3
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2
Explanation
$$ \mu = \sqrt { n(n+2) } \quad ( \mu $$ = magnetic moment)
where, n. no. of unpaired electron.
$$ 2.83 = \sqrt { n(n+2) } $$
$$ n(n+2) = 8 $$
$$ n^2 +2n - 8 = 0 $$
$$ n = 2 $$
Which of the following trivalent ion has the largest atomic radii in the lanthanide series?
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$$ La $$
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$$ Ce $$
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$$ Pm $$
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$$ Lu $$
Explanation
Lanthanum is the first and repesentative element of Lanthanide series and size decreases with increase in atomic number. So, $$La$$ has the largest atomic radii.
A copper salt is isomorphic with $$ ZnSO_4 $$, the salt will be
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paramagnetic
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diamagnetic
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ferromagnetic
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none
Explanation
$$ ZnSO_4 \rightarrow Zn^{2+} + SO^{2-}_4 $$
Isomorphic with $$ Zn^{2+} \approx Cu^{2+} $$
$$ Cu^{2+} \rightarrow [Ar]3d^9 - 1 $$ unpaired electron.
$$ \therefore $$ Paramagnetic in nature
Effective magnetic moment of $$ Sc^{3+} $$ ion is
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1.73
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0
0%
5.92
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2.83
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3.87
Explanation
Magnetic moment, $$\mu=\sqrt{n(n+2)}$$
Where $$n =$$ number of unpaired electrons
For $$ Sc^{3+} = [Ar]3d^0 , \quad n = 0 , \quad \therefore \mu = 0 $$
Which one of the following ions is colourless?
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$$ Cu^+ $$
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$$ Co^{2+} $$
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$$ Ni^{2+} $$
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$$ Fe^{3+} $$
Explanation
$$ Cu^+ $$ ion is colourless as it has no unpaired electrons.
$$ Cu^+ - [Ar]3d^{10} 4s^0 $$
Which of the following ions has the highest magnetic moment?
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$$ Ti^{3+} $$
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$$ Sc^{3+} $$
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$$ Mn^{2+} $$
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$$ Zn^{2+} $$
Explanation
$$ Ti^{3+} \rightarrow [Ar]3d^1 4s^0 ; \ Sc^{3+} \rightarrow [Ar]3d^0 $$
$$ Mn^{2+} \rightarrow [Ar]3d^5 4s^0; \ Zn^{2+} \rightarrow [Ar]3d^{10} 4s^0 $$
In $$ Mn^{2+} \rightarrow $$ number of unpaired electrons $$= 5. $$
So it has maximum magnetic moment according to the formula $$ \mu = \sqrt { n( n+2)} $$
Chloride of which of the following element is coloured?
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Ag
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Hg
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Zn
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Co
Explanation
Chloride of Co element is coloured.
$$ CoCl_3 - Co^{3+} - [Ar]3d^6 4s^0 \to$$ 4 unpaired electrons. So, it will be coloured.
In nitroprusside ion, the iron and $$NO$$ exist as $$ Fe^{II} $$ and $$ NO^+ $$ rather than $$ Fe^{III } $$ and $$ NO $$. These forms can be differentiated by
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Estimating the concentration of iron
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Measuring the concentration of $$ CN^- $$
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Measuring the solid state magnetic moment
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Thermally decomposing the compound
Explanation
The existence of $$ Fe^{2+} $$ and $$ NO^+ $$ in Nitroprusside ion $$ [Fe(CN)_5NO]^{2-} $$ can be established by measuring the magnetic moment of the solid compound which should correspond to $$ (Fe^{2+} = 3d^6 ) $$ four unpaired electrons.
On adding excess of $$ NH_3 $$ solution to $$ CuSO_4 $$ solution, the dark blue colour is due to
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$$ [ Cu(NH_3)_4] ^{2+} $$
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$$ [ Cu(NH_3)_2]^{2+} $$
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$$ [ Cu(NH_3)]^+ $$
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None of the above
Explanation
On adding excess of $$ NH_3 $$ solution to $$ CuSO_4 $$ solution, the dark blue colour is due to formation of $$ [ Cu(NH_3)_4] ^{2+} $$ ion.
$$ CuSO_4 + 4NH_3 \xrightarrow{aqueous} [Cu(NH_3)_4]^{2+}(aq)+ SO^{2-}_4(aq) $$
Which of the following pair will have effective magnetic moment equal?
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$$ Cr^{3+} $$ and $$ Mn^{2+} $$
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$$ Cr^{2+} $$ and $$ Fe^{2+} $$
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$$ V^{2+} $$ and $$ Sc^{3+} $$
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$$ Ti^{2+} $$ and $$ V^{2+} $$
Explanation
$$ Cr^{2+} $$ and $$ Fe^{2+ }$$ will have an effective magnetic moment equal since they have the same number of unpaired electrons.
$$ Cr^{2+} - [Ar]3d^4 \to$$ 4 unpaired electrons
$$ Fe^{2+} - [Ar]3d^6\to $$ 4 unpaired electrons
On heating $$ Mn(OH)_2 $$ with $$ PbO_2 $$ and conc. $$ HNO_3 $$ pink colour is obtained due to the formation of
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$$ KMnO_4 $$
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$$ HMnO_4 $$
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$$ Pb(MnO_4 )_2 $$
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$$ PbMnO_4 $$
Explanation
Reaction of $$Mn(OH)_2$$ with $$PbO_2$$ and Conc. $$HNO_3$$ :
2Mn(NO
3
)
2
+ 5PbO
2
+ 6HNO
3
→ 2HMnO
4
+ 5Pb(NO
3
)
2
+ 2H
2
O
The Pink colour is obtained due to the formation of $$HMnO_4$$.
Hence, Option "B" is the correct answer.
Which of the following weighs less when weighed in magnetic field?
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$$ VCl_3 $$
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$$ ScCl_3 $$
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$$ TiCl_3 $$
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$$ FeCl_3 $$
Explanation
$$ ScCl_3 \rightarrow Sc^{3+} + 3Cl^- $$
$$ScCl_3$$ has no unpaired electron so will show diamagnetic character and will be repelled, so will weigh less.
Which of the following is more paramagnetic?
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$$ Fe^{2+} $$
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$$ Fe^{3+} $$
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$$ Cr^{3+} $$
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$$ Mn^{3+} $$
Explanation
$$ Fe^{3+} \to [Ar]3d^5\to $$ having 5 unpaired electrons have the highest number of unpaired electrons so it will be more paramagnetic.
Which of the following reagents can be used for the given reaction ?
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$$i. D_2O / D_3O^\bigoplus \,ii. Cl_2 + AcOH \,iii. NaBH_4 / H_2O$$
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$$i. Cl_2 + AcOH \,ii. D_2O \,iii. LAH / H_2O$$
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$$ii. D_2O \,iii. HgCl_2 / CdCO_3 / H_2O$$
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$$i. Base (PhLi), \,ii. HgCl_2 / CdCO_3-H_2O \, iii. D_2O$$
Explanation
The $$(C=O)$$ group is protected by converting into cyclic thioacetal and then an organometallic base like $$PhLi$$ is used to generate carbanion followed by treatment with $$D_2O$$ to incorporate $$D$$. Finally, the aldehyde is regenerated.
Hence, Option "C" is the correct answer.
Pick out wrong combination
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$$Fe^{+2}\rightarrow Haemoglobin$$
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$$Mg^{2+}\rightarrow Photosynthesis$$
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$$Se^{2+}\rightarrow Kreb \ Cycle$$
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$$CO^{+2}\rightarrow Vitamin \ B-12$$
Explanation
Iron atom in the Haemoglobin of blood is in the +2 Oxidation state. ie., Iron is present as $$Fe^{2+}$$.
$$\text{Mg2+ is the central atom of the chlorophyll molecule,}$$ and fluctuations in its levels in the chloroplast regulate the activity of key photosynthetic enzymes. Relatively little is known of the proteins mediating Mg2+ uptake and transport in plants.
Cobalt is also an essential trace element for humans, and is found at the centre of vitamin B12 and a range of other co-enzymes called cobalamins.
and, in there is no involvement of $$Se^{2+}$$ in kreb cycle. Thus, wrong combination is C.
Which of the following has maximum magnetic moment?
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$${V}^{3+}$$
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$${Cr}^{3+}$$
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$${Fe}^{3+}$$
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$${C}^{3+}$$
Which of the following is diamagnetic?
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$${Cu}^{2+}$$
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$${Zn}^{2+}$$
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$${Cr}^{2+}$$
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$${Ti}^{2+}$$
Transition metals are not characterized by :
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fixed valency
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coloured compound
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high melting and boiling points
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tendency to form complexes
Explanation
They do not have fixed valencies since they also have vacant orbitals due to which they show exciting property and when they loose/radiate their energy they impart color in the visible region and also due to availability of vacant orbitals it show wide oxidation numbers and hence make many complex compounds.
Hence option A is correct
The calculated spin only magnetic moment of $$Cr^{2+}$$ ion is
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$$4.90 BM$$
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$$5.92 BM$$
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$$2.84 BM$$
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$$3.87 BM$$
Explanation
Electronic configuration of the ion:
$$Cr^{2+} \Rightarrow [Ar] 4s^0 3d^4$$
$$n=4$$
Spin only magnetic moment $$\Rightarrow \sqrt {n(n+2)} = \sqrt {4 \times (4+2)}=4.9 BM$$
Option A is correct.
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