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CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 8 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
The D-And F-Block Elements
Quiz 8
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $$Cr^{3+}$$ ion is :
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2.87 B.M.
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3.87 B.M.
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3.47 B.M.
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3.57 B.M.
Explanation
$${ Cr }^{ +3 }$$ has an electronic configuration of $$\left[ Ar \right] 3{ d }^{ 3 }4{ s }^{ 0 }$$ ,having unpaired electrons as $$3$$.Hence magnetic moment of $${ Cr }^{ +3 }$$ is $$\sqrt { 3(3+2) }=3.87 BM $$
The second and third row elements of transition metals resemble each other much more than they resemble the first row because of:
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lanthanoid contraction which results in almost same radii of second and third row metals
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diagonal relationship between second and third row elements
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similar ionisation enthalpy of second and third row elements
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similar oxidation states of second and third row metals
Explanation
The second and third row of transition elements resemble each other more than they do with the first row of transition elements due to lanthanide contraction calmost same atomic radius because of increase in nuclear charge.
Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?
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$$Ag_2SO_4$$
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$$CuF_2$$
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$$ZnF_2$$
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$$Cu_2Cl_2$$
Explanation
The respective transition elements exists in
$$Ag^{+}-{4d}^{10}5s^0$$, (have completely filled d orbital)
$$Cu^{+2}-3d^{9}4s^0$$, (have incompletely filled d orbital)
$$Zn^{+2}-3d^{10}4s^0$$ , (have completely filled d orbital)
$$Cu^{+1}-3d^{10}4s^0$$ (have completely filled d orbital),
Since, $$Cu^{+2}$$ has unpaired electron it will show electron transitions.
Hence will be coloured.
Metallic radii of some transition elements are given below. Which of these elements will have the highest density?
Element
$$Fe$$
$$Co$$
$$Ni$$
$$Cu$$
Metallic radii/pm
126
125
125
128
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$$Fe$$
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$$Ni$$
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$$Co$$
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$$Cu$$
Explanation
On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal. Hence $$Cu$$ will have highest density.
Arrange the following in increasing value of magnetic moments.
(i) $$[Fe(CN)_6]^{4-}$$
(ii) $$Fe(CN)_6]^{3-}$$
(iii) $$Cr(NH_3)_6^{ 3+}$$
(iv) $$Ni(H_2O)^4]^{ 2+}$$
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$$(I) < (II) < (III) < (IV)$$
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$$(I) < (II) < (IV) < (III)$$
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$$(II) < (III) < (I) < (IV)$$
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$$(IV) < (III) < (II) < (I)$$
Explanation
1) [Fe(CN)$$_6]^{4-}$$
Here, Fe is present as Fe$$^{2+}$$
Electronic configuration of
Fe$$^{2+}$$ is
[Ar]
3d$$^6 4s^0$$
Since CN$$^-$$ is a strong field ligand so when it approaches Fe
$$^{2+}$$ it pair up the e$$^-$$ s and result in the formation of low spin complex.
No. of unpaired e$$^-$$ s =0
So, magnetic moment for n unpaired e$$^-$$s
= $$\sqrt{n(n+2)} BM$$
=0
2)
[Fe(CN)$$_6]^{3-}$$
Here, Fe is present as Fe$$^{3+}$$
Electronic configuration of
Fe$$^{3+}$$ is
[Ar]
3d$$^5 4s^0$$
Since , CN$$^-$$ is a strong field ligand so when it approaches Fe
$$^{2+}$$ it pair up the e$$^-$$s so,
No. of unpaired e$$^-$$ s =1
So, magnetic moment for n unpaired e$$^-$$s
= $$\sqrt{1\times (1+2)} $$
=
$$\sqrt{3} $$ BM.
3) [Cr(NH$$_3)_6]^{3+}$$
Electronic configuration of Cr
$$^{3+}$$ is
[Ar]
3d$$^3 4s^0$$
Since , NH$$_3$$ is a strong field ligand so it pairs up the electrons when it approaches when it approaches
Cr
$$^{3+}$$
No. of unpaired e$$^-$$ s =3
So, magnetic moment
= $$\sqrt{3(3+2)} $$
=$$\sqrt{1\times 5} $$
=$$\sqrt{15}$$ BM
4)[Ni(H$$_2O)_4]^{2+}$$
Electronic configuration of Ni
$$^{2+}$$ is
[Ar]
3d$$^8 4s^0$$
Since , H$$_2$$O is a weak field ligand. it does not pair up the e$$^-$$s and forms outer orbital complex.
No. of unpaired e$$^-$$ s =2
So, magnetic moment =
$$\sqrt{2(2+2)} $$
=
$$\sqrt{8}$$ BM
But because of orbital effect it will be little lesser. So, correct order is iii)>iv)> ii)> i)
Which of the following transition metal ions has highest magnetic moment?
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$$cu^{ 2+ }$$
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$$Ni^{ 2+ }$$
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$$Co^{2+}$$
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$$Fe^{2+}$$
Explanation
Magnetic moment depends on number of unpaired electrons,in $$Fe^{+2}$$ has $$4$$ unpaired electrons while $$Co^{+2}$$ ,$$Ni^{+2}$$ , $$Cu^{+2}$$has $$3,2,1$$ unpaired electrons respectively.
Select the correct option, among $$Sc(III),\ Ti(IV),\ Pd(II)$$ and $$Cu(II)$$ ions.
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All are paramagnetic
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All are diamagnetic
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$$Sc(III),\ Ti(IV)$$ are paramagnetic and $$Pd(III),\ Cu(II)$$ are diamagnetic
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$$Sc(III),\ Ti(IV)$$ are diamagnetic and $$Pd(II),\ Cu(II)$$ are paramagnetic
Explanation
$${ SC }^{ 3+ }\left( { [Ar]3d }^{ 0 } \right) $$ ] $$\rightarrow $$ since no unpaired electrons so diamagnetic
$${ Ti }^{ 4+ }\left([Ar] { 3d }^{ 0 } \right) $$ ] $$\rightarrow $$ since no unpaired electrons so diamagnetic
$${ Pd }^{ 2+ }\left( [Kr]{ 4d }^{ 8 } \right) $$ ] $$\rightarrow $$ since there is unpaired electrons so paramagnetic
$${ Cu }^{ 2+ }\left([Ar] { 3d }^{ 9 } \right) $$ ] $$\rightarrow $$ since there is unpaired electrons so paramagnetic.
So, the correct option is D.
$$CuSO_4$$ is paramagnetic while $$ZnSO_4$$ is diamagnetic because:
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$$Cu^{2+}$$ ion has $$3d^9$$ configuration while $$Zn^{2+}$$ ion has $$3d^{10}$$ configuration.
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$$Cu^{2+}$$ ion has $$3d^5$$ configuration while $$Zn^{2+}$$ ion has $$3d^6$$ configuration.
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$$Cu^{2+}$$ has half filled orbitals while $$Zn^{2+}$$ has fully filled orbitals.
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$$CuSO_4$$ is blue in colour while $$ZnSO_4$$ is white.
Explanation
$$CuSO_4$$ has $$Cu^{+2}$$ ion which has electronic configuration of $$[Ar]3d^94s^0$$ while in $$ZnSO_4$$ ,$$Zn^{+2}$$ has $$[Ar]3d^{10}4s^0$$ configuration.
The correct order of a number of unpaired electrons is :
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$$Cu^{2+} > Ni^{2+} > Cr^{3+} > Fe^{3+}$$
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$$Ni^{2+} > Cu^{2+} > Fe^{3+} > Cr^{3+}$$
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$$Fe^{3+} > Cr^{3+} > Ni^{2+} > Cu^{2+}$$
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$$Cr^{3+}> Fe^{3+} > Ni^{2+} > Cu^{2+}$$
Explanation
The no. of unpaired e$$^-$$ in the given ions is:
Cr$$^{3+}$$ - [Ar] 3d$$^34s^0$$
Fe
$$^{3+}$$
- [Ar] 3d$$^54s^0$$
Ni
$$^{2+}$$ - [Ar] 3d$$^84s^0$$
Cu
$$^{2+}$$ - [Ar] 3d$$^94s^0$$
$$Fe^{3+} - 3d^{5}$$ No. of unpaired electrons = $$5$$
$$Cr^{3+} - 3d^{3}$$ No. of unpaired electrons = $$3$$
$$Ni^{2+} - 3d^{8}$$ No. of unpaired electrons = $$2$$
$$Cu^{2+} - 3d^{9}$$ No. of unpaired electrons = $$1$$
The order is: Fe
$$^{3+}$$ > Cr
$$^{3+}$$> Ni
$$^{2+}$$ >Cu
$$^{2+}$$
In the dichromate anion $$Cr_2O_7$$ :
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all Cr- O bonds are equivalent
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6 Cr - O bonds are equivalent
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3 Cr - O bonds are equivalent
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no bonds in $$Cr_2O_7$$ are equivalent
Explanation
There are $$6$$ $$Cr-O$$ terminal bonds which are equivalent due to resonance.
(Refer to Image)
Which of the following statements for the reaction, is correct?
$$Na_2CrO_4 + H_2SO_4 \rightarrow$$
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It is a redox reaction in which green solution of $$[Cr(H_2O)_6]^{+3}$$ is produced.
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One of the product is reaction has trigonal planar structure.
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Dimeric bridged tetrahedral metal ion is produced.
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Dark blue color in obtained in reaction.
Explanation
$${ 2Na }_{ 2 }{ CrO }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { Na }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ Na }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }O$$
$$\Downarrow $$
dimeric bridged tetrahedral
The correct order of ionic radii of Ce, La, Pm, and Yb in +3 oxidation state is:
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$$La^{3+} \, < Pm^{3+} \, < Ce^{3+} \, < Yb^{3+}$$
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$$Yb^{3+} \, < Pm^{3+} \, < Ce^{3+} \, < La^{3+}$$
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$$La^{3+} \, < Ce^{3+} \, < Pm^{3+} \, < Yb^{3+}$$
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$$Yb^{3+} \, < Ce^{3+} \, < Pm^{3+} \, < La^{3+}$$
Explanation
$${ La }^{ 3+ }>{ Ce }^{ 3+ }>{ Pm }^{ 3+ }>{ Yb }^{ 3+ }$$
Due to lanthanoid contraction ionic radii decreases.
Ionic radii decreases from $$La^{3+} -Lu^{3+}$$ in +3 oxidation state.
Hence, the correct option is $$C$$.
Identify the correct Structure of dichromate ion.
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0%
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Explanation
Structure
of dichromate ion is as shown.
Compound that is both paramagnetic and coloured is:
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$$K_2Cr_2O_7$$
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$$(NH_4)_2[TiCI_6]$$
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$$VOSO_4$$
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$$K_3[Cu(CN)_4]$$
Explanation
In VOSO$$_4$$ we have no VO$$^+$$ ion in which oxidation state of V is +3.
So, the electronic configuration for $$V^{3+}$$ is [Ar] 3d$$^24s^0$$ because of presence of 2 unpaired e$$^-$$s it is paramagnetic and coloured as well.
K$$_2Cr_2O_7$$ - diamagnetic -orange
(NH$$_4)_2$$[TiCl$$_6$$]- diamagnetic - colourless
$$K_3[Cu(CN)_4$$] - diamagnetic- colourless.
Transition metals make the most efficient catalysts because of their ability to:
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adopt multiple oxidation states and to form complexes
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form coloured ions
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show paramagnetism due to unpaired electrons
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form a large number of oxides
Explanation
Transition metals have partially filled d- orbitals so they can easily withdraw the electrons from the reagents or give electrons to them depending on the nature of the reaction. They also have a tendency to show large no. of oxidation states and the ability to form complexes which makes them a good catalyst.
Most of the transition metals exhibit:
(i) paramagnetic behaviour
(ii) diamagnetic behaviour
(iii) variable oxidation states
(iv) formation of coloured ions
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(ii), (iii) and (iv)
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(i), (iii) and (iv)
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(i), (ii) and (iii)
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(i), (ii) and (iv)
Explanation
Most of the transition metals exhibit:
i) Paramagnetic Behaviour- Because of presence of unpaired e$$^-$$s in their orbitals.
iii) Variable oxidation states- Because of vacant d-orbitals
iv) Formation of coloured ions- Because of unpaired e$$^-$$s in their orbitals which undergoes d-d transition or charge transfer phenomenon.
Which of the following is paramagnetic?
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$$V(CO)_{6}$$
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$$Fe(CO)_{5}$$
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$$Fe_{2}(CO)_{9}$$
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$$Cr(CO)_{6}$$
Explanation
$${ V\left( CO \right) }_{ 6 }\Rightarrow V-23$$
$${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 2 }{ 3d }^{ 3 }$$
$$CO-$$weak ligand
$$\Rightarrow $$ para-magnetic in nature.
Consider the following statements
I. $$La(OH)_3$$ is least basic among hydroxides of lanthanides.
II. $$Zr^{4+}$$ and $$Hf^{4+}$$ possess almost the same ionic radii.
III. $$Ce^{4+}$$ can act as an oxidizing agent.
Which of the above is/are true?
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I and III
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II and III
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II only
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I and II
Explanation
The ionic radii of $$Zr^{4+}$$ and $$Hf^{4+}$$ are almost similar due to the lanthanide contraction which means the steady decrease in the size of the atoms with the increase from lanthanum through Lutetium due to the poor shielding of the $$4f$$ $$e^-$$, the $$e^-$$ pull towards the nucleus increases thereby decreasing the size.
$$Ce^{4+}$$ act as oxidizing agent as it oxidizes other elements by itself getting reduced as $$Ce^{3+}$$ is more stable than $$Ce^{4+}$$ oxidation state so in order to attain stable $$+3$$ oxidation it gets reduced.
The basic characteristics of lanthanoid hydroxides decrease down the group.
Thus $$La(OH)_3$$ is the most basic.
Thus only statement II and III are correct.
Which of the following statements concerning lanthanide elements is false?
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All lanthanides are highly dense metals.
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The characteristic oxidation state of lanthanide elements is +3.
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Lanthanides are separated from one another by ion exchange method.
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Ionic radii of trivalent lanthanides steadily increase with an increase in the atomic number.
Explanation
Ionic radii decreases with increasing atomic number in Lanthanide series. So, the ionic radii of trivalent Lanthanide will also follow the same property.
Which ion has magnetic moment $$5.90$$ B.M?
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$$Fe^{2+}$$
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$$Fe^{3+}$$
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$$Co^{2+}$$
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$$Mn^{4+}$$
Explanation
$$ \displaystyle \mu = \sqrt {n(n+2)}=5.9$$ B.M.
$$ \displaystyle {n(n+2)}=5.9 \times 5.9 = 34.8$$
$$ \displaystyle n^2+2n= 34.8$$
$$ \displaystyle n^2+2n- 34.8=0$$
$$ \displaystyle n= \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$$
$$ \displaystyle n= \dfrac {-2 \pm \sqrt {2^2-4(1)(- 34.8)}}{2(1)}$$
$$ \displaystyle n= \dfrac {-2 \pm \sqrt {4+139}}{2}$$
$$ \displaystyle n= \dfrac {-2 \pm \sqrt {143}}{2}$$
$$ \displaystyle n= \dfrac {-2 \pm 12}{2}$$
$$ \displaystyle n= \dfrac {10}{2}$$ or $$ \displaystyle n= \dfrac {-14}{2}$$
$$ \displaystyle n= 5$$ or $$ \displaystyle n= -7$$
The negative value is discarded as the number of electrons cannot be negative.
$$ \displaystyle \therefore n= 5$$
$$ \displaystyle Fe ^{3+}$$ ion (with valence shell electronic configuration $$ \displaystyle 3d^5 4s^0$$ has magnetic moment 5.90 B.M. It has 5 unpaired electrons.
Hence, option $$B$$ is correct.
Which of the following ion is paramagnetic?
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$$Cu^{+}$$
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$$Zn^{2+}$$
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$$O_{2}^{2-}$$
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$$Cr^{3+}$$
Explanation
Ions with unpaired electron is paramagnetic. Elelctronic configuration and unpaired electron in the given ions are:
$$Cu^+:3d^{10}$$: 0 unpaired electron
$$Zn^{2+}:3d^{10}$$: 0 unpaired electron
$$O_2^{2-}:\sigma2p^2\pi2p^4\pi^{*}2p^4$$: 0 unpaired electron
$$Cr^{3+}:3d^{3}$$: 3 unpaired electron
thus option D is correct.
Which of the following ion has the maximum theoretical magnetic moment?
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$$V^{3+}$$
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$$Cr^{3+}$$
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$$Ti^{3+}$$
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$$Co^{3+}$$
Explanation
Metal ion with maximum number of unpaired electrons will have highest magnetic momemt.
$$V^{3+}$$: $$3d^2$$: 2 unpaired electrons
$$Cr^{3+}$$: $$3d^3$$: 3 unpaired electrons
$$Ti^{3+}$$: $$3d^1$$: 1 unpaired electrons
$$Co^{3+}$$: $$3d^6$$: 4 unpaired electrons
Thus
$$Co^{3+}$$ will have maximum magnetic moment.
option D is correct
Which of the following is not an actinide?
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Uranium
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Curium
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Californium
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Erbium
Explanation
Erbium with atomic number $$68$$, electronic configuration $$[Xe]4f^{12}6s^2$$ belongs to lanthanides.
$$\therefore$$, Erbium is not an actinide.
Which of the following catalysts is not correctly matched with the reaction?
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Vanadium (V) oxide in contact process for oxidation of $$SO_2 \, to \, SO_3.$$
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Finely divided iron in Habers process in conversion of $$N_2 \, and \, H_2 \, to \, NH_3.$$
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$$PtCl_2$$ catalyzes the oxidation of ethyne to ethanal in the Wacker process.
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Ni in presence of hydrogen for conversion of vegetable oil to ghee.
Explanation
Wacker's process is the oxidation of ethylene to acetaldehyde in the presence of $$PdCl_2$$ as the catalyst.
All Cu(II) halides are known except the iodide. The reason for it is that:
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iodide is a bulky ion.
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$$Cu^{2+}$$ oxidizes iodide to iodine.
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$$Cu^{2+}_{(aq)}$$ has much more negative hydration enthalpy.
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$$Cu^{2+}$$ ion has smaller size.
Explanation
All $$Cu(II)$$ halides are known except the iodine because $${ Cu }^{ 2+ }$$ oxidizes iodine to iodine
$${ 2Cu }^{ 2+ }+{ 4I }^{ -1 }\rightarrow 2{ CuI }_{ \left( S \right) }+{ I }_{ 2 }$$.
Hence,
option B is correct.
Which of the following is paramagnetic as well as coloured ion?
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$$Cu^+$$
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$$Cu^{2+}$$
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$$Sc^{3+}$$
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$$Zn^{2+}$$
Explanation
$$_{27}Cu\longrightarrow 4s^23d^7$$
$$Cu^{2+}\longrightarrow 4s^0 3d^7$$
Due to the presence of unpaired electrons, the transition of $$e^-$$ is easier hence it shows color.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
(A)
Ni in the presence of hydrogen
(i)
Ziegler Natta catalyst
(B)
$$Cu_2Cl_2$$
(ii)
Contact process
(C)
$$V_2O_5$$
(iii)
Vegetable oil to ghee
(D)
Finely divided iron
(iv)
Sandmeyer reaction
(E)
$$TiCl_4 \, + \, Al (CH_3)_3$$
(v)
Haber's Process
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$$(A) \rightarrow (iv), (B) \rightarrow (ii), (C) \rightarrow (iii), (D) \rightarrow (i), (E) \rightarrow (v)$$
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$$(A) \rightarrow (ii), (B) \rightarrow (v), (C) \rightarrow (i), (D) \rightarrow (iii), (E) \rightarrow (iv)$$
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$$(A) \rightarrow (v), (B) \rightarrow (iii), (C) \rightarrow, (iv), (D) \rightarrow( ii), (E) \rightarrow (i)$$
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$$(A) \rightarrow (iii), (B) \rightarrow (iv), (C) \rightarrow (ii), (D) \rightarrow (v), (E) \rightarrow(i)$$
Explanation
A) $$Ni$$ in the presence of hydrogen is used in the reduction of vegetable oil to ghee.
B) $$Cu_2Cl_2$$ is used in Sandmeyer Reaction for the synthesis of aryl halides.
C) $$V_2O_5$$ is used in contact process for the oxidation of $$SO_2$$ to $$SO_3$$.
D) Finely divided iron is used as catalyst in Haber's Process.
E) $$TiCl_4+Al(CH_3)_3$$ is the Zieglar Natta Catalyst.
Which one of the following ions is coloured?
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$$Sc^{3+}$$
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$$Ti^{4+}$$
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$$Zn^{2+}$$
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$$V^{2+}$$
Explanation
$$_{23}V:4s^23d^3$$
$$V^{2+}:4s^03d^3;V^{2+}$$ has three unpaired $$e^-s$$ which are easily available for electronic transitions and hence exhibit colour.
Whereas, in $$Sc^{3+}\longrightarrow 4s^03d^0;Ti^{4+}\longrightarrow 4s^03d^0$$ and $$Zn^{2+}\longrightarrow 4s^03d^{10}$$; no unpaired $$e^-$$ present in d-orbital for transition. Hence they don't show colour.
Which of the following compounds is not coloured ?
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$$ Na_{2}[CuCl_{4}] $$
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$$ Na_{2}[CdCl_{2}] $$
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$$ [Cr(H_{2}O)_{6}]Cl_{3} $$
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all of the above
Explanation
The outer shell electronic configuration of $$Cu^{2+}$$ has $$3d^{9}$$ configuration and
contain 1 unpaired electron, thus it is coloured.
$$Na_2[CdCl_2]$$, compound is not coloured because of central atom $$Cd$$ which has $$4d^{10} 5s^2$$ electronic configuration so transition do not occur.
Hence option B
Smallest ionic radius is:
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$${La}^{3+}$$
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$${U}^{3+}$$
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$${Yb}^{3+}$$
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$${Cf}^{3+}$$
Explanation
Atomic and ionic radius of lanthanides decreases from La to Lu.
Due to the contraction of lanthanide
the decrease in ionic radius
of the element
in the lanthanide
series from lanthanum
to lutetium occurs.
The silver salts like $$AgBr$$ and $$AgI$$ are coloured because of:
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d-d transition of electrons
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charge transfer
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polirisation of halid$${Ag}^{+}$$ by
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both (b) and (c)
Explanation
The silver salts like$$AgBr$$ and
$$AgI$$
are
colored,
the
color of transition metal ions is due to the presence of unpaired electron . The electrons which absorbs radiations of color
that
from the visible or UV spectrum and undergo transition from ground state to excited state within d-
sub shells
,so it is d-d transitions of electrons.
The radius of $$La^{3+}$$(Atomic number of $$La=57$$) is $$1.06\mathring A$$. Which one of the following given values will be closest to the radius of $$Lu^{3+}$$ (Atomic number of $$Lu=71$$)?
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$$1.40\mathring{ A}$$
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$$1.06\mathring{ A}$$
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$$0.85\mathring{ A}$$
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$$1.60\mathring{ A}$$
Explanation
The ionic radii of $$Ln^{3+}$$ reduces from $$La^{3+}$$ to $$Lu^{3+}$$ due to Lanthanide Contraction. Therefore the lowest value here must be the one closest ionic radii of $$Lu^{3+}\implies 0.85 A.$$
Which of the following has highest value of magnetic moment?
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$$Fe^{2+}$$
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$$Co^{3+}$$
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$$V^{3+}$$
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$$Ti^{3+}$$
Explanation
$$ Magnetic\ moment = \sqrt {n({n+2})}$$
$$ n$$= no. of unpaired electrons,
$$ Fe^{+2}$$ has $$4$$ unpaired electrons
Other compound have less unpaired electrons.
More the number of electron, more the magnetic moment. Thus $$Fe^{2+}$$ has the highest value of the magnetic moment among the given elements.
Which one of the following compound is both colored and paramagnetic?
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$$ScCl_3$$
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$$TiCl_4$$
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$$CrCl_3$$
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$$CuCl$$
Explanation
The species which have unpaired electron in d-subshell is both paramagnetic and coloured.
$$Sc^{3+}\longrightarrow 4s^03d^0$$
$$Ti^{4+}\longrightarrow 4s^03d^0$$
$$Cr^{3+}\longrightarrow 4s^03d^3$$
$$Cu^{+}\longrightarrow 4s^03d^{10}$$
$$Cr^{3+}$$ is having $$3$$ unpaired $$e^-$$, thus paramagnetic and hence show color due to transition of these unpaired.
In the dichromate di-anion:
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4$$Cr-O$$ bonds are equivalent
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6$$Cr-O$$ bonds are equivalent
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all $$Cr-O$$ bonds are equivalent
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all $$Cr-O$$ are non equivalent
Explanation
The Dichromate di-anion consists of two tetrahedra sharing an oxygen at the common corner.The chromate ion has the tetahedral structure. The six terminal bond in the dichromate ion have same bond length due to resonance. Hence all six $$Cr-O$$ bonds are equal.
Arrange $$Ce^{3+}$$, $$La^{3+}$$, $$Pm^{3+}$$, and $$Yb^{3+}$$ in increasing order of their ionic radius:
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$$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$$
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$$Ce^{3+} > Yb^{3+} < Pm^{3+} < La^{3+}$$
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$$Yb^{3+} > Pm^{3+} < La^{3+} < Ce^{3+}$$
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none of these
Explanation
Atomic and ionic radii of Lanthanides decrease from $$La$$ to $$Lu$$.
Their order of ionic radii: $$Yb^{+3}<Pm^{3+}< Ce^{3+}< La^{+3}$$
Among $$V(Z=23)$$, $$Cr(Z=24)$$, $$Mn(Z=25)$$ and $$Fe(Z=26)$$, which will have the highest magnetic moment?
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$$V$$
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$$Cr$$
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$$Mn$$
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$$Fe$$
Explanation
$$V$$,$$Cr$$,$$Mn$$ and $$Fe$$ has three, six, five and four unpaired $$e^-$$ respectively.
So, $$M_V=\sqrt {3(3+2)}=\sqrt {15}=3.18$$ BM
$$M_{Cr}=\sqrt {6(6+2)}=\sqrt {48}=6.92$$ BM
$$M_{Mn}=\sqrt{5(5+2)}=\sqrt {35}=5.91$$ BM
$$M_{Fe}=\sqrt{4(4+2)}=\sqrt {24}=4.89$$ BM
So, $$Cr$$ is having highest magnetic moment. It can be concluded that more is the number of unpaired $$e^-s$$, higher is the magnetic moment.
Which of the following species have same magnetic moment?
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$$Cr^{+3}$$
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$$Fe^{+3}$$
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$$Co^{+2}$$
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$$Ni^{+2}$$
Explanation
$$Cr^{3+}:4s^03d^3$$ number of unpaired $$e^-=3$$
$$Fe^{3+}:4s^03d^5$$ number of unpaired $$e^-=5$$
$$Ca^{2+}:4s^03d^6$$ number of unpaired $$e^-=4$$
$$Ni^{2+}:4s^03d^7$$ number of unpaired $$e^-=3$$
$$Cr^{3+}$$ and $$Ni^{2+}$$ have same number of unpaired $$e^-$$.
$$\therefore$$ They have same magnetic moment.
The ratio of magnetic moments of $$Fe(III)$$ and $$Co(II)$$ is:
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$$7 : 3$$
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$$3 : 7$$
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$$\sqrt7 : \sqrt3$$
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$$\sqrt3 : \sqrt7$$
Explanation
Electronic configuration of $$Fe^{3+}:[Ar]3d^54s^0$$
So number of unpaired elctrons is $$n=5$$
Hence, $$\mu=\sqrt{n(n+2)}=\sqrt{35}$$
Now,
Electronic configuration of $$Co^{2+}:[Ar]3d^74s^0$$
So number of unpaired electros is $$n=3$$
Hence, $$\mu=\sqrt{n(n+2)}=\sqrt{15}$$
So, Ratio of Magnetic moments of $$Fe^{3+}$$ and $$Co^{2+}$$ is$$:\dfrac{\sqrt{7}}{\sqrt{3}}$$
Which of the following is diamagnetic ion?
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0%
$$Zn^{2+}$$
0%
$$Ni^{2+}$$
0%
$$Co^{2+}$$
0%
$$Cu^{2+}$$
Explanation
$${ Zn }^{ 2+ }\equiv \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 10 }$$
It is diamagnetic, so no unpaired $${ e }^{ - }s$$.
$$\mu = \sqrt{15}$$ is true for the pair:
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0%
$$Co^{+2}, Cr^{+3}$$
0%
$$Fe^{+2}, Cr^{+3}$$
0%
$$Fe^{+3}, Cr^{+2}$$
0%
$$Mn^{+2}, Fe^{+2}$$
Explanation
The electronic configuration of $$Co^{2+}$$ is;$$[Ar]4s^0 3d^7$$
Hence there are three unpaired electrons in $$Co^{2+}$$
Now the elctronic configuration of $$Cr^{3+}$$ is;$$[Ar]3d^3$$
Hence there are 3 unpaired electrons in $$Cr^{3+}$$
We know magnetic moment $$\mu=\sqrt{(n(n+2)}$$$$
where n is number of unpaired electrons.
So when n=3,
$$\mu=\sqrt{(3(3+2)}=\sqrt{15}$$
Hence option A is correct answer.
Aqueous solution of which of the following is green?
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0%
$$V^{2+}$$
0%
$$V^{+3}$$
0%
$$VO^{2+}$$
0%
$$VO_{2}^{+}$$
Explanation
$$V^{2+}\longrightarrow$$ Mauve/Lavender.
$$V^{3+}\longrightarrow$$ Green.
$$VO^{2+}\longrightarrow$$ Blue.
$$VO_2^{+}\longrightarrow$$ Yellow.
Maximum ferromagnetism is found in ?
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0%
$$Fe$$
0%
$$Ni$$
0%
$$Co$$
0%
$$None$$
Explanation
As Iron have highest value of saturation, Magnetization $$M_s$$ which is defined as the magnetic moment density when the material is subjected to a field strong enough to align all its moments.
$$Fe \ 1707 \ M_s$$
$$Co \ 1400 \ M_s$$
$$Ni \ 485 \ M_s$$
An aqueous solution containing $$V^{x+}$$ ion has magnetic moment equal to $$\sqrt {15} \ BM$$. Therefore, the x is:
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0%
$$5$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Let $$n$$ be the number of unpaired electrons.
Magnetic moment $$= \sqrt {n(n+2)}= \sqrt {15}$$
$$\Rightarrow n=3$$
Atomic number of Vanadium is $$23$$. So, it must loose all its $$4s$$ electrons to get $$3$$ unpaired d electrons.
$$\therefore$$ value of $$x$$ will be $$2$$
Which of the following species do not exist?
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0%
$$PbF_4$$
0%
$$PbCl_4$$
0%
$$PbO_2$$
0%
$$Pbl_4$$
Explanation
$$PbF_4$$ exists but $$PbI_4$$ does not exist as $$I_2$$ reduces $$Pb(II)$$ to $$Pb$$ and oxidizes $$I$$ to $$I_2$$. Since, $$I_2$$ is not strong reducing agent to reduce $$Pb(II)$$ to $$Pb$$. Hence, $$PbI_2$$ is formed and not $$PbI_4$$.
Which of the following complex compounds will exibit highest paramagnetic behavior?
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$$[Ti({ NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }$$
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$$[Cr({ NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }$$
0%
$$[Co({ NH }_{ 3 }{ ) }_{ 6 }{ ] }^{ 3+ }$$
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$$[{ Zn }(NH_{ 3 }{ ) }_{ 4 }{ ] }{ Cl }_{ 2 }$$
Explanation
$${ T }i^{ 3+ }-\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 1 }$$
$${ Zn }^{ 2+ }-\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 10 }\quad \} -$$ not much paramagnetic
$${ Cr }^{ 3+ }\rightarrow { d }_{ 3 }$$ configuration $$\rightarrow { t }_{ 2 g}$$ means paramagnetic
$${ Co }^{ 3+ }\rightarrow { d }_{ 6 }$$ configuration $$\rightarrow { NH }_{ 3 }$$ strong field ligand
So, it would be $${ d }_{ 6 }$$ (low spin) diamagnetic.
Which of the following ions have zero value of magnetic moment?
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0%
$$Sc^{3+}$$
0%
$$Ti^{4+}$$
0%
$$Zn^{2+}$$
0%
None of these
Explanation
It's because
$$Sc^{3+}$$
does not have any
electrons
left in its d orbital. Sc has the
electron configuration
$$[Ar](4s)^2(3d)^1$$ but in order to get its +3 oxidation state, it has to lose 3
electrons
, starting with the two 4s
electrons
and followed by the lone 3d
electron
.
Hence n=0 which makes magnetic moment zero.
The ratio of spin only magnetic moment of $$Fe^{3+}$$ and $$CO^{2+}$$ is:
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0%
$$\sqrt { 24 } : \sqrt { 15 }$$
0%
$$7 : 3$$
0%
$$\sqrt { 35 } : \sqrt { 15 } $$
0%
$$\sqrt { 5 } : \sqrt { 7 }$$
Number of identical Cr-O bonds in dichromate ion $$Cr_2O_7^{2-}$$ is :
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0%
4
0%
6
0%
7
0%
8
Explanation
The correct answer is - There are '6' identical Cr - O bonds in the dichromate ion $$(Cr_2O_7^{-2}).$$
In a dichromate ion
$$(Cr_2O_7^{-2}).$$
, there are 6 terminal Cr – O bonds. All these bonds are equivalent due to resonance.
The Dichromate ions
$$(Cr_2O_7^{-2}).$$
are oxoanions of chromium in the oxidation state +6.
Hence option B is correct answer.
The coloured species is ?
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0%
$${ VCl }_{ 3 }$$
0%
$${ VOSO }_{ 4 }$$
0%
$${ Na }_{ 3 }{ VO }_{ 4 }$$
0%
$$[{ V(H }_{ 2 }O)_{ 6 }]{ SO }_{ 4 }{ H }_{ 2 }O$$
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