$${ \left[ { Sc\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +3 }$$

$${ \left[ { Fe\left( CN \right) }_{ 6 } \right] }^{ -3 }$$

$${ \left[ { Ti\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +3 }$$

$${ \left[ { Cu\left( NH_{ 3 } \right) }_{ 4 } \right] }^{ +2 }$$

$${ FeSO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$$

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

When a brown compound of Mn(A) is treated with

$HCl$ , it gives a gas B. The gas B taken in excess reacts with

$NH_3$ to give an explosive compound C. The compounds A, B and C are:

0%
$A=MnO_2, B=Cl_2, C=NCl_3$
0%
$A=MnO, B=Cl_2, C=NH_4Cl$
0%
$A=Mn_3O_4, B=Cl_2, C=NCl_3$
0%
$A=MnO_3, B=Cl_2, C=NCl_2$

Explanation

When brown compound

$\displaystyle Mn_3O_4$ is treated with

$HCl$ , it gives chlorine gas.

$\displaystyle Mn_3O_4 + 8HCl \rightarrow 3MnCl_2+4H_2O+ Cl_2 \uparrow$

Chlorine gas taken in excess reacts with ammonia to give an explosive compound

$\displaystyle NCl_3$ .

$\displaystyle NH_3 + 3Cl_2(excess) \rightarrow NCl_3+3HCl$

The compounds A, B and C are

$\displaystyle A=Mn_3O_4, B=Cl_2, C=NCl_3$

In a transition series, with the increase in atomic-number, the paramagnetism :

0%
increases gradually
0%
decreases gradually
0%
first increases to a maximum and then decreases
0%
first decreases to a minimum and then increases

Explanation

For transition elements, with increase in atomic number electrons enter the empty

$d$ -orbitals. First they are singly occupied which increases paramagnetism until the point where all the

$d$ -orbitals are half-filled (point of maximum paramagnetism) and then they get paired which decreases paramagnetism. Hence, paramagnetism first increases till maximum then decreases.

Which of the following complex compounds will exhibit the highest paramagnetic behaviour?

$[At. no. Ti = 22, Cr = 24, Co = 27, Zn = 30]$

0%
$[Ti(NH_{3})_{6}]^{3+}$
0%
$[Cr(NH_{3})_{6}]^{3+}$
0%
$[Co(NH_{3})_{6}]^{3+}$
0%
$[Zn(NH_{3})_{6}]^{2+}$

Explanation

(a) Electronic configuration of

$Ti^{3+}$ in

$[Ti(NH_{3})_{6}]^{3+}$

$Ti^{3+} = 3d^{1}$ [ref. image (a)]

(b) Electronic configuration of

$Cr^{3+}$ in

$[Cr(NH_{3})_{6}]^{3+}$

$Cr^{3+} = 3d^{3}$ ; [ref. image (b)]

$Co^{3+}$ in

$[Co(NH_{3})_{6}]^{3+}$ ;

$Co^{3+} = 3d^{6}$ .
In the presence of strong field ligand

$NH_{3}$ , pairing of electrons takes place and hence, octahedral complex,

$[Co(NH_{3})_{6}]^{3+}$ is diamagnetic.

$[Co(NH_{3})_{6}]^{3+} \begin{bmatrix}\text {inner orbital or}\\ \text {low spin complex} \end{bmatrix} (6NH_{3} molecles)$ [ref. image (c)]

(d) Electronic configuration of

$Zn^{2+}$ in

$[Zn(NH_{3})_{6}]^{2+}$

$\therefore [Zn(NH_{3})_{6}]^{2+}$ is an outer orbital complex and is diamagnetic. [ref. image (d)]

$[Cr(NH_{3})_{6}]^{3+}$ complex has the highest unpaired electron. Hence it will be highly paramagnetic.

Option B is correct.

The magnetic moment of

$\left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 3 }$ is:

0%
$1.73$
0%
$2.83$
0%
$6.6$
0%
Zero

Explanation

${ \left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] }^{ 3+ }$ , the oxidation state of

$Co$ is

$+3$ .

${ \left[ Co{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] }^{ 3+ }$ does not contain unpaired electron hence, its magnetic moment is zero.

Actinides exhibit a larger number of oxidation state than that of corresponding lanthanides. The reason behind this aspect is :

0%
lesser energy difference between 5f and 6d-orbitals than between 4f and 5d-orbitals
0%
larger atomic size of actinides than the lanthanides
0%
more energy difference between 5f and 6d orbitals than between 4f and 5d-orbitals
0%
greater reactive nature of the actinides than the lanthanides

Explanation

Because of lesser energy difference between

$5f$ and

$6d$ orbitals than that between

$4f$ and

$5d$ -orbitals actinides exhibit larger number of oxidation state than that of corresponding lanthanides. There is no large energy difference for pulling out the electrons in case of actinides.

If a transition-metal compound absorbs violet-indigo radiation in the visible region. Its color would be?

0%
Green
0%
Yellow
0%
Orange
0%
Blue

When dil.

${ H }_{ 2 }S{ O }_{ 4 }$ is added to aqueous solution of potassium chromate, yellow colour of solution turns to orange colour. It indicates :

0%
chromate ions are reduced
0%
chromate ions are oxidized
0%
mono-centric complex is converted into dicentric complex
0%
oxygen gets removed from chromate ions

Explanation

When dil.

${ H }_{ 2 }S{ O }_{ 4 }$ is added to the yellow solution of

${ K }_{ 2 }Cr{ O }_{ 4 }$ yellow colour changes to orange red due to the formation of

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ .

$2{ K }_{ 2 }Cr{ O }_{ 4 }+{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ K }_{ 2 }S{ O }_{ 4 }+{ H }_{ 2 }O$
Above reaction shows that the monocentric complex

$\left( { K }_{ 2 }Cr{ O }_{ 4 } \right)$ is converted into dicentric

$\left( { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 } \right)$ complex and it can be explained with the help of diagram.

In the dichromate dianion, the nature of bonds are:

0%
four equivalent $Cr - O$ bonds
0%
six equivalent $Cr - O$ bonds and one $O - O$ bond
0%
six equivalent $Cr - O$ bonds and one $Cr - Cr$ bond
0%
six non-equivalent $Cr - O$ bonds
0%
six equivalent $Cr - O$ bonds and one $Cr - O - Cr$ bond

Explanation

In the dichromate dianion structure, the negative charge can be on any

$O$ attached to

$Cr$ , therefore all

$Cr-O$ are equivalent. Also have

$Cr-O-Cr$ bond.

Thus, option D is correct.

The colour of

$Cu{ Cr }_{ 2 }{ O }_{ 7 }$ solution in water is green because__________.

0%
${ Cu }^{ 2+ }$ ion is green
0%
${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$
0%
both the ions are green
0%
${ Cu }^{ 2+ }$ ion is blue and ${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$ ion is yellow

Explanation

$CuCr_2O_7$ solution is green because $Cu^{2+}$ ions are blue and $Cr_2O_7^{-2}$ ions are yellow in aqueous medium. Both blue and yellow colours mix up to give a green colour.

Hence option D is correct.

The magnetic moment of a transition metal ion is

$\sqrt { 15 } BM$ . Therefore, the number of unpaired electrons present in it is:

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

Explanation

The magnetic moment of a transition metal ion is

$\sqrt { 15 } BM$ . Therefore, the number of unpaired electrons present in it is

$3$ .
If n is the number of unpaired electrons, then the magnetic moment is given by the expression:

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

$\displaystyle \sqrt { 15 } = \sqrt {n(n+2)}$

$\displaystyle { 15 } = {n(n+2)}$

$\displaystyle 15 = n^2 + 2n$

$\displaystyle n^2+2n- { 15 } = 0$

This is the quadratic equation with solution

$\displaystyle n = \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$

$\displaystyle n = \dfrac {-2 \pm \sqrt {2^2-4(1)(- { 15 } )}}{2(1)}$

$\displaystyle n = \dfrac {-2 \pm 8}{2(1)}$

$\displaystyle n=3$ or

$\displaystyle n=-5$ .

The negative value is discarded as the number of unpaired electrons cannot be negative.

Hence,

$\displaystyle n=3$ .

Which one of the following complex ions has the highest magnetic moment?

0%
$[Cr(NH_{3})_{6}]^{3+}$
0%
$[Fe(CN)_{6}]^{3-}$
0%
$[Fe(CN)_{6}]^{4-}$
0%
$[Zn(NH_{3})_{6}]^{2+}$

Explanation

Higher the number of unpaired electrons, the higher is the magnetic moment.

$[Cr(NH_{3})_{6}]^{3+}, Cr$ is present as

$Cr^{3+}$ . Thus,

$Cr$ in

$[Cr(NH_{3})_{6}]^{3+} = [Ar] 3d^{3} 4s^{0} 4p^{0}$ . So, it will have 3 unpaired electrons.

$[Fe(CN)_{6}]^{3-}$ , one unpaired electron is present and in

$[Fe(CN)_{6}]^{4-}$ and

$[Zn(NH_{3})_{6}]^{2+}$ , both, unpaired electrons are absent.

$CN^-$ is a strong field ligand, thus causes pairing.

Hence,

$[Cr(NH_{3})_{6}]^{3+}$ has the highest magnetic moment among the given complex ions.

$[\mu = \sqrt {3(3 + 2)} = 3.87\ BM]$

The magnetic moment of a divalent ion in aqueous solution with atomic number 25 is:

0%
5.47 B.M
0%
2.9 B.M
0%
6.9 B.M
0%
6.7 B.M

Explanation

electronic configuration of the element with atomic number 25(divalent ion) will be

$[Ar]3d^5$ ,having

$5$ unpaired electrons.

magnetic moment will be

$\sqrt{n(n+2)}$

where n is unpaired electrons

n=5

$\sqrt {5(5+2) }=\sqrt {35}=5.47$

A transition metal

$X$ has a configuration

$\left[Ar\right]3d^{5}$ in its

$+3$ oxidation state. Its atomic number is:-

0%
$22$
0%
$26$
0%
$28$
0%
$19$

Magnetic moments of

$V(Z = 23), Cr(Z = 24)$ and

$Cu(Z = 29)$ are x,y and z respectively, then which of the following is correct?

0%
$x = y = z$
0%
$y > x > z$
0%
$x > y > z$
0%
$z > y > x$

Explanation

$V\left( z=23 \right) \quad \left[ Ar \right] { 3d }^{ 3 }\rightarrow n=3$

$Cr\left( z=24 \right) \quad \left[ Ar \right] { 4s }^{ 1 }{ 3d }^{ 5 }\rightarrow n=6$

$Cu\left( z=29 \right) \quad \left[ Ar \right] { 4s }^{ 1 }{ 3d }^{ 10 }\rightarrow n=1$

where n= no. of unpaired electrons

So,

$y>x>z$

Hence, the correct option is

$B$

Which of the following ion will not form coloured aqueous solution?

0%
$Ni^{2+}$
0%
$Fe^{3+}$
0%
$Ti^{4+}$
0%
$Cu^{2+}$

Explanation

$\displaystyle Ni^{2+}$ ion with

$\displaystyle 3d^8$ outer electronic configuration has two unpaired electrons. It will form a green coloured aqueous solution.

$\displaystyle Fe^{3+}$ ion with

$\displaystyle 3d^5$ outer electronic configuration has five unpaired electrons. It will form yellow coloured aqueous solution.

$\displaystyle Ti^{4+}$ ion will not form coloured aqueous solutionaas it does not contain unpaired electrons.

$\displaystyle Ti^{4+}$ ion with

$\displaystyle 3d^0$ outer electronic configuration has zero unpaired electrons.

$\displaystyle Cu^{2+}$ ion with

$\displaystyle 3d^9$ outer electronic configuration has one unpaired electron. It will form blue coloured aqueous solution.

Hence, option

$C$ is correct.

(i)

$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+AcOH\rightarrow \underset{ppt}{M}$
(ii)

$\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+HCN\rightarrow \underset{ppt.}{N}$
Which is the correct option regarding the given reactions?

0%
M and N are the same and characterized as $BaCrO_4$
0%
M is $BaCrO_4$ and N stands for no precipitate
0%
M is $BaCrO_4$ and N is $BaCO_3$
0%
M stands for no precipitate and N is $BaCrO_4$

Explanation

(i) $\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+AcOH\rightarrow \underset{ppt} BaCrO_4+Na_2CO_3$

(ii) $\underset{aq. suspension}{BaCO_3}+Na_2CrO_4+HCN\rightarrow \underset{ppt.} BaCrO_4+Na_2CO_3$

So M and N are $BaCrO_4$ .

Hence option A is correct.

Identify the oxidation states of titanium

$(Z=22)$ and copper

$(Z=29)$ in their colourless compounds:

0%
$Ti^{3+}, Cu^{2+}$
0%
$Ti^{2+}, Cu^{2+}$
0%
$Ti^{4+}, Cu^{+}$
0%
$Ti^{4+}, Cu^{2+}$

Explanation

The oxidation states of titanium

$(Z=22)$ and copper

$(Z=29)$ in their colourless compounds are

$Ti^{4+}, Cu^{+}$ respectively.

The electronic configurations are given below.

$\displaystyle Ti : [Ar] 4s^23d^2$

$\displaystyle Ti^{4+} : [Ar] 4s^03d^0$

$\displaystyle Cu : [Ar] 4s^13d^{10}$

$\displaystyle Cu^+ : [Ar] 4s^03d^{10}$

The above electronic configurations of ions correspond to zero unpaired electrons.

Which one of the following shows maximum paramagnetic character?

0%
${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 3- }$
0%
${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 4- }$
0%
${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$
0%
${ \left[ Cu{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ }$

Explanation

(i)

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

Here, oxidation state of

$Fe = +3$

Electronic configuration of

$Fe$ (ground state)

$= [Ar]3d^64s^2$

Electronic configuration of

$Fe$ (excited state)

$= [Ar]3d^54s^0$

Number of unpaired electrons

$=5$

(ii)

$[Fe(CN)_6]^{4-}$

Here, oxidation state of

$Fe = +2$

Electronic configuration of

$Fe$ (ground state)

$= [Ar]3d^64s^2$

Electronic configuration of

$Fe$ (excited state)

$= [Ar]3d^64s^0$

Number of unpaired electrons

$=4$

(iii)

$[Cr(H_2O)_6]^{3+}$

Here, oxidation state of

$Cr = +3$

Electronic configuration of

$Cr^{3+}= [Ar]3d^34s^0$

Number of unpaired electrons

$=3$

(iv)

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

Here, oxidation state of

$Cu = +2$

Electronic configuration of

$Cu^{2+} = [Ar] 3d^94s^0$

Number of unpaired electrons

$=1$

A number of unpaired electrons present is directly proportional to a paramagnetic character.

So,

$[Fe(CN)_6]^{3-}$ has maximum paramagnetic character.

The incorrect statement is :

0%
$Cu^{2+}$ salts give red colour borax bead test in reducing flame.
0%
$Cu^{2+}$ and $Ni^{2+}$ ions give black colour precipitate with $H_2S$ in the presence of $HCl$ solution.
0%
Ferric ion gives blood red colour with potassium thiocyanate.
0%
$Cu^{2+}$ ion gives chocolate coloured precipitate with potassium ferrocyanide solution.

What would be the correct order if Zn,Fe,Al and Cu are arranged in increasing order of reactivity ?

0%
Cu, Fe, Zn, Al
0%
Al, Cu, Fe, Zn
0%
Zn, Al, Cu, Fe
0%
Fe, Zn, Al, Cu

The bonds generally formin compounds of Actinoids are:

0%
convalent
0%
ionic
0%
coordinate covalent
0%
metallic

Out of

$TiF_{6}^{2-},\ CoF_{6}^{3-},\ Cu_{2}Cl_{2}$ and

$NiCl_{4}^{2-}$ , the colourless sphere are.....

0%
$Cu_{2}Cl_{2},\ NiCl_{4}^{2-}$
0%
$TiF_{6}^{2-},\ Cu_{2}Cl_{2}$
0%
$CoF_{6}^{3-},\ NiCl_{4}^{2-}$
0%
$TiF_{6}^{2-},\ CiF_{6}^{3-}$

Which one of the following statement is incorrect about transition elements?

0%
The last electron enters into them in the d-orbital
0%
Their properties are in between those of s-block elements and p-block elements
0%
The transition elements with smallest atomic number is scandium
0%
Lanthanum is not a transition elements but is a lanthanoid

Select the incorrect match (reason of color):-

0%
$KMnO_{4},K_{2}Cr_{2}O_{7}\Rightarrow$ charge transfer spectra
0%
Halogen $\Rightarrow$ HOMO-LUMO transition
0%
$[Ti(NH_{3})_{6}]^{+3}\Rightarrow$ d-d transition
0%
$CrO_{2}Cl_{2},CrO_{4}^{-2}\Rightarrow$ d-d transition

$M^{x+}$ ion has magnetic moment 2.84 BM.Then 'x' is (Z of M=22)

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

Which of the following set of elements are Ferromagnetic in nature?

0%
Zn, Cd and Hg
0%
Cu,Ag and Au
0%
Fe, Co and Ni
0%
Sc, Ti and U

The spin only magnetic moment of

$[Cu(NH_3)_4]^{2+}$ complex is:-

0%
$\text{1.732 B.M.}$
0%
$\text{2.2 B.M.}$
0%
$\text{3.87 B.M.}$
0%
$\text{5.96 B.M.}$

Which of the following pair have almost similar size

0%
$Ti_{22}$ and $Zr_{40}$
0%
$Nb_{41}$ and $Ta_{73}$
0%
$Y_{39}$ and $La_{57}$
0%
$Ca_{20}$ and $Ir_{31}$

$K_2Cr_2O_7$ , every

$Cr$ is linked to:

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

Find the spin magnetic moment of the

$Fe^{3+}$ .

0%
$5.91$
0%
$4.89$
0%
$3.87$
0%
none of these

$CrO_4^{2-}$ (yellow)

$\xrightarrow[{pH=Y}]{pH=X}$

$Cr_2O_7^{-2}$ (orange)
The

$pH$ values

$X$ and

$Y$ are respectively:

0%
6, 8
0%
6, 6
0%
8, 5
0%
7, 7

The magnetic moment of

$V(Z= 23), Cr (Z= 34)$ and

$Mn (Z= 25)$ are

$x, y, z$ . Hence:

0%
$x = y = z$
0%
$z < y < x$
0%
$x < y < z$
0%
$x < z < y$

Which of the following ion has magnetic moment 3.87 BM?

0%
$Co^{2+}$
0%
$Mn^{3+}$
0%
$Cu^{+}$
0%
$Ni^{2+}$

The effect of lanthanoid contration in the lanthanid series of elements by and large means:

0%
decrease in atomic radii and increases in ionic radii
0%
decrease in both atomic and ionic radii
0%
increase in both atomic and ionic radii
0%
increases in atomic radii and decrease in ionic radii

A transition metal cation M

$^{2+}$ has a magnetic moment

$\sqrt35$ . what is the atomic number of the metal?

0%
$24$
0%
$25$
0%
$26$
0%
$27$

Orbital angular momentum of

$e^-$ (last filled ) in Cr is

0%
Zero
0%
$\sqrt { 2 } \frac { h }{ 2\pi }$
0%
$\sqrt { 6 } \frac { h }{ 2\pi }$
0%
$\sqrt { 12 } \frac { h }{ 2\pi }$

An iron

$_{25}Mn^{x+}$ has the magnetic moment of 4.9 BM. What is the value of x ?

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

Which of the following complex is anion ?

0%
Fluoro pentaammine cobalt (III)
0%
Trioxalato ferrate (III)
0%
Penta Carbonyl iron
0%
Dichloro diammine platinum(II)

The magnetic moment (spin only) of

$[NiCI_4]^{2-}$ is:

0%
5.46 BM
0%
2.83 BM
0%
1.41 BM
0%
1.82 BM

The metal ion which is not colored is:

0%
${ Fe }^{ 3+ }$
0%
${ V }^{ 2+ }$
0%
${ Zn }^{ 2+ }$
0%
${ Ti }^{ 3+ }$

Which of the following lanthanum ions is diamagnetic?

0%
${ Ce }^{ 4+ }$
0%
${ Sm }^{ 2+ }$
0%
$Eu^{ 2+ }$
0%
$Yb^{ 2+ }$

Magnetic moment 2.84 B.M is given by:-

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

Colourless ion is:

0%
${ \left[ { Fe\left( CN \right) }_{ 6 } \right] }^{ -3 }$
0%
${ \left[ { Ti\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +3 }$
0%
${ \left[ { Sc\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +3 }$
0%
${ \left[ { Cu\left( NH_{ 3 } \right) }_{ 4 } \right] }^{ +2 }$

The correct order of spin -only magnetic moments among the following is :
(Atomic number :

$Mn =25, Co + 27 , Ni = 28, Zn = 30)$

0%
$[ZnCl_4]^{2-} > [NiCl_4]^{2-} > [CoCl_4]^{2-} > [MnCl_4]^{2-}$
0%
$[CoCl_4]^{2-} > [MnCl_4]^{2- } > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$
0%
$[MnCl_{ 4 }]^{ 2- }>[CoCl_{ 4 }]^{ 2- }>[NiCl_{ 4 }]^{ 2- }>[ZnCl_{ 4 }]^{ 2- }$
0%
$[NiCl_{ 4 }]^{ 2- }>[CoCl_{ 4 }]^{ 2- }>[MnCl_{ 4 }]^{ 2- }>[ZnCl_{ 4 }]^{ 2- }$

An orange solid (A) on heating gave a green residue (B), a colourless gas (C) and water vapour. The gas (C) on passing over heated Mg gave white solid (D) which on reaction with water gave a gas (E). The gas (E) forms white fumes with HCl.

An orange solid A is

0%
Sodium dichromate
0%
Potassium dichromate
0%
Potassium chromate
0%
Ammonium dichromate

Fenton reagent is

0%
${ FeSO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$
0%
$Zn + HCl$
0%
$Sn + HCl$
0%
None of the above

The number of unpaired electron(s) in

$[Fe(H_2O)_6]^{2+}$ is

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

Copper sulphate disolves in excess of

$KCN$ to give:

0%
$Cu\left( CN \right) _{ 2 }$
0%
$CuCN$
0%
$[Cu\left( CN \right) _{ 4 }]^{ 3- }$
0%
$[Cu\left( CN \right) _{ 4 }]^{ 2- }$

Which of the following ions has maximum magnetic moment?

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

The d-electronic configuration of

${ Cr }^{ 2+ },{ Mn }^{ 2+ },{ Fe }^{ 2+ }$ and

${ Ni }^{ 2+ }$ are

${ 3d }^{ 4 },{ 3d }^{ 5 },{ 3d }^{ 6 }$ and

${ 3d }^{ 8 }$ respectively. Which one of the following aqua complexes will exhibit the maximum paramagnetic character?

0%
$\left[ \left( Fe\left( { H }_{ 2 }O \right) \right) _{ 6 } \right] ^{ 2+ }$
0%
$\left[ \left( Ni\left( { H }_{ 2 }O \right) \right) _{ 6 } \right] ^{ 2+ }$
0%
$\left[ \left( { C }{ r }\left( { H }_{ 2 }O \right) \right) _{ 6 } \right] ^{ 2+ }$
0%
$\left[ \left( { Mn }\left( { H }_{ 2 }O \right) \right) _{ 6 } \right] ^{ 2+ }$

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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