MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 9

$2Cr_4^{2-} \underset{Y}{\overset{ X }{\rightleftharpoons}}Cr_2O_7^{2-}$ ,

$X$ and

$Y$ are respectively:

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$X = OH^-, \, Y = H^+$
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$X = H^+, \, Y = OH^-$
0%
$X = OH^-, \, Y = H_2O_2$
0%
$X = H_2O_2, \, Y = OH^-$

Explanation

$\textbf{Part 1: Explanation for X}$

When $CrO_{4}^{2-}$ is put in acidic medium, it changes its colour from yellow to orange by forming $Cr_{2}O_{7}^{2-}$ .

$2CrO_{4}^{2-}+H^+\rightleftharpoons Cr_{2}O_{7}^{2-}$

So, X is acidic medium $(H^+)$ .

$\textbf{Part 2: Explanation for Y}$

When $Cr_{2}O_{7}^{2-}$ is put in basic medium, it changes its colour from orange to yellow by forming $CrO_{4}^{2-}$ .

$Cr_{2}O_{7}^{2-}+OH^-\rightleftharpoons 2CrO_{4}^{2-}$

So, Y is basic medium

$(OH^-)$ .

The spin only magnetic moment of

$Mn^{4+}$ ion is nearly:

0%
$3\ BM$
0%
$6\ BM$
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$4\ BM$
0%
$5\ BM$

Magnetic moment of

${X^{3 + }}$ ion od

$3d$ series is

$\sqrt {35} \,BM$ . What is atomic number of

${X^{3 + }}$ ?

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$25$
0%
$26$
0%
$27$
0%
$28$

Which has the maximum magnetic moment?

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$Mn^{2+}$
0%
$Fe^{2+}$
0%
$Ti^{2+}$
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$Cr^{2+}$

Magnetic moment of

${}_{{\text{25}}}{\text{M}}{{\text{n}}^{{\text{x + }}}}{\text{ is }}\sqrt {{\text{15}}}$ B.M then the value of x is :

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1
0%
2
0%
3
0%
4

Which one of the following ion(s) is/are paramagnetic?

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$Al^{3+}$
0%
$Fe^{2+}$
0%
$K^+$
0%
$Mg^{2+}$

Magnetic moment

$2.83\ BM$ is given by which of the following ions?
(At. nos.

$Ti=22,Cr=24,Mn=25,Ni=28$ )

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${Ti}^{3+}$
0%
${Ni}^{2+}$
0%
${Cr}^{3+}$
0%
${Mn}^{2+}$

Which of the following becomes paramagnetic on heating?

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$MnO$
0%
$Fe_3O_4$
0%
$ZnFe_2O_4$
0%
$B$ and $C$ both

Explanation

Ferrimagnetic change their nature on heating and become paramagnetic

eg.

$Fe_3O_4, ZnFe_2O_4, MgFe_2O_4$

Actinoids belongs to following type of elements:

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s -block element
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d -block element
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f -block element
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p -block element

A complex

${\left[ {M{F_6}} \right]^{ - x}}$ has magnetic moment of 4.91 B.M. whereas

${\left[ {M{{(CN)}_6}} \right]^{ - x}}$ has zero magnetic moment, metal ion can be :

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$C{o^{ + 2}}$
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$M{n^{ + 2}}$
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$F{e^{ + 2}}$
0%
$F{e^{ + 3}}$

Explanation

$\begin{array}{l} According\, to\, question............ \\ \, \, \, \, \, \, \, from\, data\, can\, \, be\, oncluded\, that\, \, first\, complez\, comtains\, 4\, unpaired\, electron\, as\, its\, magnetic\, moment\, is\, \\ \, 4.91\, BM\, \, magnetic\, moment\, (\, \mu )=\sqrt { n } (n+2),where\, \, n\, is\, no.of\, unpaired\, electron: \\ In\, high\, spin\, complex\, ,\, metal\, \, \, ion\, has\, 4\, unpaired\, electron\, and\, \, in\, low\, spin\, complex\, it\, \, has\, zero\, unpaired\, electron.... \\ so, \\ \, \, \, \, Ion\, \, is\, F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, is\, high\, spin,\, 4\, unpaired\, \, electron. \\ F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, low\, spin\, ,zero\, unpaired\, electron. \\ \, \, \, \, \, \, \, \, \, \therefore \, \, \, ion\, is\, \, \, F{ e^{ +2 } }. \\ so\, the\, \, correct\, ption\, \, is\, C. \end{array}$

Which of the following is diamagnatie?

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$Cr^{3+}$
0%
$V^{2+}$
0%
$Sc^{3+}$
0%
none

Figure

0%
Ethane
0%
Propane
0%
n- Butane
0%
n-pentane

Compare magnetic moment of

$Mn,{ Mn }^{ + },{ Mn }^{ 2+ },{ Mn }^{ 3+ }:$

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${ Mn }^{ 3+ }>Mn>{ Mn }^{ 2+ }>{ Mn }^{ + }$
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${ Mn }^{ + }>Mn^{ 2+ }>{ Mn }^{ 3+ }>{ Mn }$
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$Mn={ Mn }^{ 2+ }>Mn^{ 3+ }>{ Mn }^{ + }$
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${ Mn }^{ + }>Mn=Mn^{ 2+ }>{ Mn }^{ 3+ }$

The magnetic moment for

${ Ti }^{ 2+ }$ ,

${ V }^{ 3+ }$ and

${ Ni }^{ 2+ }$ is ________ .

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4.90
0%
3.87
0%
2.83
0%
1.73

Which of the following compounds are coloured due to charge transfer spectra?

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${K}_{2}{Cr}_{2}{O}_{7}$
0%
$KMn{O}_{4}$
0%
$Cu{SO}_{4}.5{H}_{2}O$
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Both (A) and (C)

Uranium is a radioactive element but it is also a typical metal. What is not a property of uranium?

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It can be hammered into shape
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It conducts heat
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It is used as a source of energy
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It forms covalent compounds

Explanation

$\text{Uranium is a radioactive element which decay continuously. So it cannot form a covalent compound.}$

$Sc^{3+},Ti^{2+},Ti^{3+},V^{2+}$ , increasing order of spin only magnetic moment is:

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$Sc^{3+} < Ti^{2+} < Ti^{3+}< V^{2+}$
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$Sc^{3+} < Ti^{3+} < Ti^{2+}< V^{2+}$
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$Ti^{2+} < Sc^{3+} < Ti^{3+}< V^{2+}$
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$Sc^{3+} < Ti^{2+} < V^{2+}< Ti^{3+}$

Explanation

Solution:- (B)

${Sc}^{3+} < {Ti}^{+3} < {Ti}^{+2} < {V}^{2+}$

$\mu=\sqrt{n(n+2)}$ B.M

For magnetic moment to be higher, number of unpaired electron must be higher

$Sc^{3+}\Rightarrow 3d^0$

$\mu=0$

$Ti^{2+}\Rightarrow 3d^2$

$\sqrt8$ B.M

$Ti^{3+}\Rightarrow 3d^1$

$\sqrt3$ B.M

$V^{2+}\Rightarrow 3d^3$

$\sqrt{15}$ B.M

Transition metal compounds are usually colored. This is due to the electronic transition

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from p-orbital to s-orbital
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from d-orbital to s-orbital
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from d-orbital to p-orbital
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within the d-orbitals

Explanation

Colour in transition series metal compounds arises due to two types of transitions. they are:

(a) Charge transfer transitions [Ligand-to-metal charge transfer (LMCT), Metal-to-ligand charge transfer]

(b)

$d-d$ transition

In transition metal complexes, all the

$d-$ orbitals do not possess same energy. So, electrons can jump from a lower energy

$d-$ orbital to a higher energy

$d-$ orbital, by absorbing energy. When it returns to the ground state, excess energy is released, and a corresponding wavelength is found in the visible region.

Therefore, transition metals are colored due to the electronic transition within the

$d-$ orbitals.

So, option D is the correct answer.

Which of the following statements is correct for the complex,

$K_4[Fe(CN)_5O_2]$ having

$t^6_{2g}, e^0_g$ electronic configuration?

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$d^2sp^3$ hybridized and diamagnetic
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$sp^3d^2$ hybridized and paramagnetic
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$sp^3d^2$ hybridized and diamagnetic
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$d^2sp^3$ hybridized and paramagnetic

Choose the true statement regarding compounds X and Y.

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Both compounds X and Y have different magnetic moment(spin only)
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Both complexes X and Y have same magnetic moment(spin only)
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Magnetic moment of both X and Y is calculated by assuming they are low-spin complex
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Both compounds X and Y have different primary valency

Among the following the lowest degree of paramagnetism per mole of the compound at

$298$ K will be shown by?

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$MnSO_4\cdot 4H_2O$
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$FeSO_4\cdot H_2O$
0%
$CuSO_4\cdot 5H_2O$
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$NiSO_4\cdot 6H_2O$

Explanation

The degree of paramagnetism will depend on the number of unpaired electrons because spin magnetic moment =

$\sqrt{n(n+2)}$ B.M

Water is a weak field ligand, so there will be no pairing of electrons as the interaction of the central metal with water molecules is very weak. All the metal ions are in

$+2$ oxidation state. The valence shell electronic configuration of

$Cu^{+2}$ is

$3d^{9}4s^{0}$ . So,

$Cu^{2+}$ has only one unpaired electron.

$Mn^{2+}$ has four unpaired electrons.

$Fe^{2+}$ has four unpaired electrons.

$Ni^{+2}$ has two unpaired electrons.

Therefore, the lowest degree of paramagnetism per mole of the compound is shown by

$CuSO_4.5H_2O$ .

Hence, the correct option is (C).

1892914_1722260_ans_57b945c6e1e744908a168f2b57fe5a27.png

Which of the following statements is correct?

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With $d^2sp^3$ hybridization, $[FeCl(CN)_4(O_2)]^{4-}$ complex is diamagnetic.
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$[NiCl_4]^{2-}$ complex is more stable than $[Ni(dmg)_2]$ due to higher C.F.S.E. value.
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$[V(CO)_6]$ is not very stable and easily reduces to $[V(CO)_6]^-$ .
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Ligands such as $CO, CN^-, NO^+$ are $\pi e^-$ donor due to the presence of filled $\pi$ -molecular orbital.

Explanation

The complex

$[Fe(CN)_{4}(O_{2})]^{-4}$ is an octahedral complex. So, it may have the hybridization of

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

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

$CN^{-}$ is a strong field ligand, the electronic configuration of

$Fe^{+3}$ in the complex will be

$t_{2g}^{6}e_{g}^{0}$ due to the pairing of electrons.

So, it will be an inner-orbital complex involving the inner

$3d$ orbitals (

$d^{2}sp^{3}$ hybridisation). It is also a diamagnetic complex as all the

$t_{2g}$ electrons are paired up.

So, option (A) is correct.

An aqueous solution of titanium chloride, when subjected to magnetic measurement, measured zero magnetic moment. Assuming the complex as octahedral in aqueous solution, the formula of the complex is?

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$[Ti(H_2O)_6]Cl_2$
0%
$[Ti(H_2O)_6]Cl_4$
0%
$[TiCl_6]^{3-}$
0%
$[Ti(H_2O)_4Cl_2]$

Explanation

Magnetic moment of any complex arises due to the presence of unpaired electron(s) in the central metal ion or atom. An octahedral complex is a complex surrounded by six monodentate ligands inside the coordination sphere. In the compound

$[Ti(H_{2}O)_{6}]Cl_{4}$ ,

$Ti$ is in

$+4$ oxidation state. It has an electronic configuration of

$[Ar]3d^{0}$ . Hence, it has zero magnetic moment. So, the formula

$[Ti(H_2O)_6]Cl_4$ fits for the conditions mentioned in the question.

Therefore, the correct option is (B).

Atomic numbers of

$Cr$ and

$Fe$ are, respectively,

$24$ and

$26$ . Which of the following is paramagnetic with spin of electron?

0%
$Cr(CO)_6$
0%
$[Fe(CO)_5]$
0%
$[Fe(CN)_6]^{4-}$
0%
$[Cr(NH_3)_6]^{3+}$

Explanation

The increasing order of field strength of the ligands is

$NH_{3}<CN^{-}<CO$ . So, only the complex

$[Cr(NH_{3})_{6}]^{+3}$ will be a comparatively high-spin complex.

$Fe^{+2}$ has

$24$ electrons and

$Cr^{+3}$ has

$21$ electrons. The unpaired electrons of

$Fe$ and

$Cr$ can pair up in the complexes containing

$CO$ and

$CN^{-}$ ligands because they are strong field ligands. But in the complex

$[Cr(NH_{3})_{6}]^{+3}$ , there is a single unpaired electron present as

$NH_3$ is a weak field ligand. So, the complex

$[Cr(NH_3)_6]^{3+}$ will be paramagnetic.

Therefore, the correct option is (D).

The d-electron configurations of

$Mn^{2+}, Fe^{2+}, Co^{3+}$ and

$Ni^{2+}$ are

$3d^5, 3d^6, 3d^6, 3d^8$ , respectively. Which of the following aqua complexes will exhibit the minimum paramagnetic behavior?

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$[Fe(H_2O)_6]^{2+}$
0%
$[Co(H_2O)_6]^{3+}$
0%
$[Ni(H_2O)_6]^{2+}$
0%
$[Mn(H_2O)_6]^{2+}$

The color of

$Cu^+$ compounds is?

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White
0%
Blue
0%
Orange
0%
Yellow

Explanation

$Cu^{+}$ compounds are generally colourless (white). The electronic configuration is

$3d^{10}4s^{0}$ . So there is no unpaired electron present in the ion. So the electronic transition between two energy levels

$t_{2g}$ and

$e_{g}$ (responsible for the colour of transition metals) is not possible. So the correct option is A.

The complex

$[Fe(H_3O)_5NO]^{2+}$ is formed in the brown ring test for nitrates when freshly prepared

$FeSO_4$ solution is added to an aqueous solution of

$NO^-_3$ followed by addition of conc.

$H_2SO_4$ . Select correct statements about this complex:

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Color change is due to charge transfer
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It has iron in $+1$ oxidation state and nitrosyl as $NO^+$
0%
It has magnetic moment of $3.87$ B.M. confirming three unpaired electrons in Fe
0%
In complex Fe has $d^2sp^3$ hybridization

$K_2 [HgI_4]$ detects the ion/group:

0%
$NH^-_2$
0%
$NO$
0%
$NH_4^+$
0%
$AlCl_3$

Arrange the following ions in order of their magnetic moment:
(i)

$V^{4+}$
(ii)

$Mn^{4+}$
(iii)

$Fe^{3+}$
(iv)

$Ni^{2+}$
(Atomic number of V = 23, Mn = 25, Fe = 26, Ni = 28)

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(ii) > (iii) > (i) > (iv)
0%
(iii) > (ii) > (iv) > (i)
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(i) > (iv) > (iii) > (ii)
0%
(iv) > (iii) > (i) > (ii)

The colorless spccits is:

0%
$VCI_3$
0%
$VOSO_4$
0%
$Na_3 VO_4$
0%
$[V(H_2O]SO_4] H_2O$

Yellow color solution of

$FeCl_{3}$ changew to light green when

0%
$SnCl_{2}$ is added
0%
Zn is added
0%
$H_{2}S$ gas is passed
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All are true

For Ni and pt dofferent L.P. in KJ

$mol^{-1}$ are given below.

$\frac { (IP)_ 1+(IP)_ 2 }{ \overset { Ni }{ Pt } \quad \quad \quad \quad \overset { 2.49 }{ 2.60 } } \frac { (IP)_ 3+(IP)_ 4 }{ \overset { 8.80 }{ 6.70 } }$

0%
Nickel (II) compounds tend to be thermodynami -cally more stable than platinum (II)
0%
Platinum (IV) compounds lend to be more stable than nickel (IV)
0%
Both are correct
0%
None is correct

$Cr_2O_7^{-2}$ every Cr is linked to :

0%
Two O-atoms
0%
Three O-atoms
0%
Four O-atoms
0%
Five O-atoms

Elements which generally exhibit multiple oxidation states and whose ions are usually coloured are

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Metalloids
0%
Transition elements
0%
Non-metals
0%
Gases

$Ti^{2+}$ is purple while

$Ti^{4+}$ is colorless,because:

0%
There is no crystal field effect in $Ti^{4+}$
0%
$Ti^{4+}$ has $3d^2$ configuration
0%
$Ti^{4+}$ has $3d^0$ configuration
0%
$Ti^{4+}$ is a very small cation when compared to $Ti^{2+}$ and hence does not absorb any radiation

Explanation

Electronic configuration of

$Ti=[Ar]3d^24s^2$

So,

$Ti^{2+}=[Ar]3d^2$

$Ti^{4+}=[Ar]3d^0$

$Ti^{2+}$ ion has two unpaired electrons due to which d-d transition is possible by absorption of light in visible region. Because of which

$Ti^{2+}$ ion is purple and

$Ti^{4+}$ is colorless due to absence of unpaired electrons.

Hence, the correct option is (B).

Increasing basic properties of

$TiO_2$ ,

$ZrO_2$ and

$HfO_2$ are in order.

0%
$TiO_2 < ZrO_2 < HfO_2$
0%
$HfO_2 < ZrO_2 < TiO_2$
0%
$HfO_2 < TiO_2 < ZrO_2$
0%
$ZrO_2 < TiO_2 < HfO_2$

In the dichoromate dianion:-

0%
Four Cr-O bonds are equivalent
0%
Six Cr-O bonds are equivalent
0%
All Cr-O bonds are equivalent
0%
All Cr-O bonds are non-equivalent

One heating ammonium dichromate, the gas evolved is:

0%
Oxygen
0%
Ammonia
0%
Nitrous oxide
0%
Nitrogen

Explanation

$(NH_4) Cr_2 O_7 \xrightarrow{\triangle} \ N_2 \uparrow +Cr_2O_3(s)+4H_2O (l)$

One heating ammonium dichromate, the gas evolved is nitrogen gas.

Out of

$[Fe(CN)_6]^{4-} , [Ni(CN)_4]^{2-} ,$ and

$[Ni(CO)_4]$ select the incorrect statements.

0%
All have identical geometry
0%
All are paramagnetic
0%
All are diamagnetic
0%
$[Fe(CN)_6]^{4-}$ is diamagnetic but $[Ni(CN)_4]^{2-}$ and $[Ni(CO)_4]$ are paramagnetic

Which of the following ions show higher spin only magnetic moment value ?

0%
$Ti^{3+}$
0%
$Mn^{2+}$
0%
$Fe^{3+}$
0%
$Co^{3+}$

$[Fe(CN)_5(NO)]^{2-}, Fe$ has +2 state. It cannot be decided by.

0%
Magnetic measurement
0%
Colligative property
0%
Colour
0%
Hybridization

In the dichromate dianoin:

0%
Four Cr-O bonds are equivalent
0%
Six Cr-O bonds equivalent
0%
All Cr-O bonds are equivalent
0%
All Cr-O bonds are non-equivalent

In context of the lanthanoids , which of the following statements is not correct ?

0%
There is gradual decrease in the raddi of the members with increasing atomic number in the series
0%
All the members exhibit $+3$ oxidation state
0%
Because of similar properties, the separation of lanthanoids is not easy
0%
Availability of $4f$ electrons results in the formation of compounds in $+4$ state for all the members of the series.

The correct statement about the magnetic properties of

$[Fe(CN)_6]^{3-}$ and

$[FeF_6]^{3-}$ is:

$(Z=26)$ .

0%
$[Fe(CN)_6]^{3-}$ is paramagnetic, $[FeF_6]^{3-}$ is diamagnetic
0%
Both are diamagnetic
0%
$[Fe(CN)_6]^{3-}$ is diamagnetic, $[FeF_6]^{3-}$ is paramagnetic
0%
Both are paramagnetic

Hg is found in nature as HgS because :

0%
$Hg^{2+}$ is soft acid and $S^{2-}$ soft base
0%
Lattice energy of HgS is high
0%
Hg and S are close in periodic table
0%
Ionisation energy of Hg is very high

Explanation

Mercuric ion (

$Hg^{2+}$ ) is a soft acid and sulfide ion (

$S^{2-}$ ) is a soft base. Hence

$Hg^{2+}$ reacts more strongly with

$S^{2-}$ to form mercuric sulfide

$HgS$ .

Hence, mercury is available in the form of cinnabar,

$HgS$ , whereas carbonate, oxide and silicate ions are hard bases. Therefore, mercuric ion (

$Hg^{2+}$ ) does not prefer to combine with these (carbonate, oxide and silicate) ions.

Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour?

0%
$[Co(CN)_6]^{3-}$
0%
$[Fe(CN)_6]^{3-}$
0%
$[Mn(CN)_6]^{3-}$
0%
$[Cr(CN)_6]^{3-}$

The electronic configuration of actinoids can not be assigned with degree of certanity because of :

0%
overlapping of inner orbitals
0%
free movement of electron over all the orbitals
0%
small energy difference between $5f$ and $6d$ levels
0%
none of above

Lanthanoid contraction implies :

0%
Decrease in density
0%
Decreases in mass
0%
Decrease in ionic radii
0%
Decrease in radioactivity

Most of the man-made synthetic elements occur:

0%
in actinide series
0%
in lanthanide series
0%
in coinage metals
0%
in alkaline earth metals

Explanation

The first actinides to be discovered were Uranium by Klaproth in 1789 and Thorium by Berezelius in 1829, but most of the Actinides were

$man-made$ products of the 20th century. Actinium and Protactinium are found in small portions in nature, as decay products of 253-Uranium and 238-Uranium.

Hence (A) is the answer.

Arrange

$Ce^{3+} , La^{3+} ,Pm^{3+}$ and

$Yb^{3+}$ in increasing order of their ionic radii:

0%
$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$
0%
$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$
0%
$La^{3+} < Ce^{3+} < Yb^{3+} < Pm^{3+}$
0%
$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 9 - MCQExams.com

Lanthanoid contraction effect results from poor shielding of nuclear charge (nuclear attractive force on electrons) by 4f electrons; the 6s electrons are drawn towards the nucleus, thus resulting in a smaller atomic radius.

Hence, option C is correct.

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 9 - MCQExams.com

Lanthanoid contraction effect results from poor shielding of nuclear charge (nuclear attractive force on electrons) by 4f electrons; the 6s electrons are drawn towards the nucleus, thus resulting in a smaller atomic radius.

Hence, option C is correct.

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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