Coordination Compounds
338 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
Spin only magnetic moment of an octahedral complex of Fe2+ in the presence of a strong field ligand in BM is :
A.
4.89
B.
2.82
C.
0
D.
3.46
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value)?
A.
O$_2^ + $
B.
CuI
C.
[Cu(NH3)4]Cl2
D.
O$_2^ - $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
According to the valence bond theory the hybridization of central metal atom is dsp2 for which one of the following compounds?
A.
$NiC{l_2}.6{H_2}O$
B.
${K_2}[Ni{(CN)_4}]$
C.
$[Ni{(CO)_4}]$
D.
$N{a_2}[NiC{l_4}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
The correct order of intensity of colors of the compounds is :
A.
${[Ni{(CN)_4}]^{2 - }} > {[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}$
B.
${[Ni{({H_2}O)_6}]^{2 + }} > {[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }}$
C.
${[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }} > {[Ni{(CN)_4}]^{2 - }}$
D.
${[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO4 . 5H2O, respectively, are :
A.
5 and 1
B.
4 and 1
C.
6 and 5
D.
6 and 4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
The correct structures of trans-[NiBr2(PPh3)2] and meridonial-[Co(NH3)3(NO2)3], respectively, are
A.
B.
C.
D.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Match List - I with List - II :
Choose the correct answer from the options given below :
| List - I |
List - II |
||
|---|---|---|---|
| (a) | $[Co{(N{H_3})_6}][Cr{(CN)_6}]$ | (i) | Linkage isomerism |
| (b) | $[Co{(N{H_3})_3}{(N{O_2})_3}]$ | (ii) | Solvate isomerism |
| (c) | $[Cr{({H_2}O)_6}C{l_3}$ | (iii) | Co-ordination isomerism |
| (d) | $cis - {[CrC{l_2}{(ox)_2}]^{3 - }}$ | (iv) | Optical isomerism |
Choose the correct answer from the options given below :
A.
(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B.
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
C.
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
D.
(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
(i) [FeF6]3$-$
(ii) [Co(NH3)6]3+
(iii) [NiCl4]2$-$
(iv) [Cu(NH3)4]2+
(i) [FeF6]3$-$
(ii) [Co(NH3)6]3+
(iii) [NiCl4]2$-$
(iv) [Cu(NH3)4]2+
A.
(iii) > (iv) > (ii) > (i)
B.
(ii) > (iii) > (i) > (iv)
C.
(i) > (iii) > (iv) > (ii)
D.
(ii) > (i) > (iii) > (iv)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The hybridization and magnetic nature of ${[Mn{(CN)_6}]^{4 - }}$ and ${[Fe{(CN)_6}]^{3 - }}$, respectively are :
A.
sp3d2 and diamagnetic
B.
d2sp3 and paramagnetic
C.
sp3d2 and paramagnetic
D.
d2sp3 and diamagnetic
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
The calculated magnetic moments (spin only value) for species ${[FeC{l_4}]^{2 - }}$, ${[Co{({C_2}{O_4})_3}]^{3 - }}$ and $MnO_4^{2 - }$ respectively are :
A.
5.92, 4.90 and 0 BM
B.
4.90, 0 and 1.73 BM
C.
5.82, 0 and 0 BM
D.
4.90, 0 and 2.83 BM
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
The sum of oxidation states of two silver ions in [Ag(NH3)2] [Ag(CN)2] complex is _____________.
Correct Answer: 2
Explanation:
[Ag(NH3)2][Ag(CN)2] complex dissociates into [Ag(NH3)2]+ and [Ag(CN)2].
Oxidation of Ag in [Ag(NH3)2]+
Ag + 0 $\times$ 2 = + 1
Ag = + 1
Oxidation state of Ag in [Ag(CN)2]$-$
Ag + ($-$1) $\times$ 2 = $-$ 1
Ag $-$ 2 = $-$ 1
$\Rightarrow$ Ag = + 1
$\therefore$ Sum of oxidation states of two silver ions in [Ag(NH3)2][Ag(CN)2] complex is 2.
Oxidation of Ag in [Ag(NH3)2]+
Ag + 0 $\times$ 2 = + 1
Ag = + 1
Oxidation state of Ag in [Ag(CN)2]$-$
Ag + ($-$1) $\times$ 2 = $-$ 1
Ag $-$ 2 = $-$ 1
$\Rightarrow$ Ag = + 1
$\therefore$ Sum of oxidation states of two silver ions in [Ag(NH3)2][Ag(CN)2] complex is 2.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
The number of optical isomers possible for [Cr(C2O4)3]3$-$ is ____________.
Correct Answer: 2
Explanation:
The number of optical isomers for [Cr(C2O4)3]3$-$ is two.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
1 mol of an octahedral metal complex with formula MCl3 . 2L on reaction with excess of AgNO3 gives 1 mol of AgCl. The denticity of Ligand L is ____________. (Integer answer)
Correct Answer: 2
Explanation:
MCl3 . 2L octahedral
$\mathop {MC{l_3}.2L}\limits_{1\,mole} \buildrel {Ex.\,AgN{O_3}} \over \longrightarrow $ 1 mole of AgCl
Its means that one Cl$-$ ion present in ionization sphere.
$\therefore$ formula = [MCl2L2]Cl
For octahedral complex coordination no. is 6
$\therefore$ L act as bidentate ligand
$\mathop {MC{l_3}.2L}\limits_{1\,mole} \buildrel {Ex.\,AgN{O_3}} \over \longrightarrow $ 1 mole of AgCl
Its means that one Cl$-$ ion present in ionization sphere.
$\therefore$ formula = [MCl2L2]Cl
For octahedral complex coordination no. is 6
$\therefore$ L act as bidentate ligand
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
The overall stability constant of the complex ion [Cu(NH3)4]2+ is 2.1 $\times$ 1013. The overall dissociations constant is y $\times$ 10$-$14. Then y is __________. (Nearest integer)
Correct Answer: 5
Explanation:
Given ks = 2.1 $\times$ 1013
Kd = ${1 \over {{k_s}}}$ = 4.7 $\times$ 10$-$14
$\therefore$ y = 4.7 $ \approx $ 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
The ratio of number of water molecules in Mohr's salt and potash alum is ____________ $\times$ 10$-$1. (Integer answer)
Correct Answer: 5
Explanation:
Mohr's salt : (NH4)2 Fe(SO4)2 . 6H2O
The number of water molecules in Mohr's salt = 6
Potash alum : KAl(SO4)2 . 12H2O
The number of water molecules in potash alum = 12
So ratio of number of water molecules in Mohr's salt and potash alum
$ = {6 \over {12}}$
$ = {1 \over 2}$
= 0.5
= 5 $\times$ 10$-$1
The number of water molecules in Mohr's salt = 6
Potash alum : KAl(SO4)2 . 12H2O
The number of water molecules in potash alum = 12
So ratio of number of water molecules in Mohr's salt and potash alum
$ = {6 \over {12}}$
$ = {1 \over 2}$
= 0.5
= 5 $\times$ 10$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
3 moles of metal complex with formula Co(en)2Cl3 gives 3 moles of silver chloride on treatment with excess of silver nitrate. The secondary valency of Co in the complex is ___________.
(Round off to the nearest integer)
(Round off to the nearest integer)
Correct Answer: 6
Explanation:
$3\left[ {Co{{(en)}_2}C{l_2}} \right]Cl + \mathop {AgN{o_3}}\limits_{(excess)} \to \mathop {3AgCl}\limits_{(white\,ppt.)} $
Secondary valency of Co = 6
(C. N.)
Secondary valency of Co = 6
(C. N.)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X + Y is _______________.
Correct Answer: 2
Explanation:
Triamminetrinitrocobalt (III) $\to$ [Co(NO2)3(NH3)3]
trioxalatochromate (III) ion $\to$ [Cr(C2O4)3]3$-$[Co(NO2)3(NH3)3]
X + Y = 2 + 0 = 2.0
trioxalatochromate (III) ion $\to$ [Cr(C2O4)3]3$-$[Co(NO2)3(NH3)3]
X + Y = 2 + 0 = 2.0
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl3.3NH3.3H2O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______________.
Correct Answer: 0
Explanation:
Mole of AgCl precipitated is equal to mole of Cl- present in ionization sphere.
Since none of Cl- is present in the co-ordination sphere. Therefore answer is zero.
Since none of Cl- is present in the co-ordination sphere. Therefore answer is zero.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
The total number of unpaired electrons present in [Co(NH3)6]Cl2 and [Co(NH3)6]Cl3 is :
Correct Answer: 3
Explanation:
[Co(NH3)6]Cl2
Co2+ : [Ar]3d74s04p0
For this complex $\Delta$0 < P.E., so pairing of electron does not take place.
sp3d2 hybridisation
Total 3 unpaired electrons are present.
[Co(NH3)6]Cl3
Co3+ : [Ar]3d6 4s0 4p0
d2sp3 hybridistion
NH3 acts as SFL because $\Delta$0 > P.E.
So, here all electrons becomes paired.
Co2+ : [Ar]3d74s04p0
For this complex $\Delta$0 < P.E., so pairing of electron does not take place.
sp3d2 hybridisation
Total 3 unpaired electrons are present.
[Co(NH3)6]Cl3
Co3+ : [Ar]3d6 4s0 4p0
d2sp3 hybridistion
NH3 acts as SFL because $\Delta$0 > P.E.
So, here all electrons becomes paired.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form [Ni(CN)6]2$-$. The total change in number of unpaired electrons on metal centre is _______________.
Correct Answer: 2
Explanation:
[Ni(CN)6]2-
Ni+4 $\to$ d6, CN- strong field ligand. So pairing will happen.
Here zero unpaired electron
NiCl2 $\to$ Ni2+ $\to$ d8
$ \therefore $ Change = 2.
Ni+4 $\to$ d6, CN- strong field ligand. So pairing will happen.
Here zero unpaired electron
NiCl2 $\to$ Ni2+ $\to$ d8
$ \therefore $ Change = 2.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
The spin-only magnetic moment value for the complex [Co(CN)6]4$-$ is __________ BM.
[At. no. of Co = 27]
[At. no. of Co = 27]
Correct Answer: 2
Explanation:
[Co(CN)6]4$-$
x + 6 $\times$ ($-$1) = $-$4
where, x, 6, $-$1 and $-$4 are the oxidation number of Co, number of CN ligands, charge on one CN and charge on complex.
x = +2, i.e. Co2+
Electronic configuration of Co2+ : [Ar]3d7 and CN$-$ is a strong field ligand which can pair electron of central atom.
It has one unpaired electron (n) in 4d-subshell.
So, spin only magnetic moment ($\mu$) = $\sqrt {n(n + 2)} $ BM = $\sqrt {1(1 + 2)} $ BM = $\sqrt 3 $ BM
where, n = number of unpaired electrons
$\mu$ = $\sqrt 3 $ BM = 1.73 BM
Nearest integer = 2
x + 6 $\times$ ($-$1) = $-$4
where, x, 6, $-$1 and $-$4 are the oxidation number of Co, number of CN ligands, charge on one CN and charge on complex.
x = +2, i.e. Co2+
Electronic configuration of Co2+ : [Ar]3d7 and CN$-$ is a strong field ligand which can pair electron of central atom.
It has one unpaired electron (n) in 4d-subshell.
So, spin only magnetic moment ($\mu$) = $\sqrt {n(n + 2)} $ BM = $\sqrt {1(1 + 2)} $ BM = $\sqrt 3 $ BM
where, n = number of unpaired electrons
$\mu$ = $\sqrt 3 $ BM = 1.73 BM
Nearest integer = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is _____________.
Correct Answer: 3
Explanation:
In ${K_3}\left[ {Cr\left( {{C_2}{O_4}} \right)} \right]$, oxidation number of Cr :
$3( + 1) + x + 3( - 2) = 0$
$ \Rightarrow x = + 3$
$ \therefore $ ${}_{24}C{r^{ + 3}} = \left[ {Ar} \right]3{d^3}$
$ \therefore $ Number of unpaired electrons = 3
$3( + 1) + x + 3( - 2) = 0$
$ \Rightarrow x = + 3$
$ \therefore $ ${}_{24}C{r^{ + 3}} = \left[ {Ar} \right]3{d^3}$
$ \therefore $ Number of unpaired electrons = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
On complete reaction of FeCl3 with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is __________. (Round off to the Nearest Integer).
Correct Answer: 6
Explanation:
FeCl3 + 3H2C2O4 + 6KOH $ \to $ K3[Fe(C2O4)3] + 3KCl + 6H2O
Coordination number = 6
Secondary valency is 6
Coordination number = 6
Secondary valency is 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
[Ti(H2O)6]3+ absorbs light of wavelength 498 nm during a d $-$ d transition. The octahedral splitting energy for the above complex is ____________ $\times$ 10$-$19 J. (Round off to the Nearest Integer). h = 6.626 $\times$ 10$-$34 Js; c = 3 $\times$ 108 ms$-$1
Correct Answer: 4
Explanation:
Octahedral splitting energy = ${{hc} \over \lambda }$
$ = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {498 \times {{10}^{ - 9}}}}$
$ = 3.99 \times {10^{ - 19}}J$
$ \approx 4 \times {10^{ - 19}}$
$ = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {498 \times {{10}^{ - 9}}}}$
$ = 3.99 \times {10^{ - 19}}J$
$ \approx 4 \times {10^{ - 19}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl3 . 4NH3 is _________. (Round off to the Nearest Integer).
Correct Answer: 2
Explanation:
$[Co{(N{H_3}]_4}C{l_2}]Cl + 2en \to [Co{(en)_2}C{l_2}] + 4N{H_3}$
NH3 is the neutral monodentate ligand. Ethylene diamine is a neutral didentate ligand.
${H_2}\mathop N\limits^{ \bullet \,\, \bullet } - C{H_2} - C{H_2} - \mathop N\limits^{ \bullet \,\, \bullet } {H_2}(en)$
So, two ethylene diamine are equivalent to four 'NH3' ligand.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2$-$ is ___________. [ox = oxalate]
Correct Answer: 3
Explanation:
The coordination compound [Co(ox)2(Br)(NH3)]2$-$ of general formula [M(A - A)2BC] can show both geometrical and optical isomerism (stereoisomerism).
The cis-form produces non-superimposable mirror images, i.e. enantiomeric pairs (optically active)
The trans-form is optically inactive.
So, total number of stereoisomers possible
= cis( $ \pm $ ) + trans = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
Number of bridging CO ligands in [Mn2(CO)10] is __________.
Correct Answer: 0
Explanation:
$ \therefore $ Number of bridging CO ligands = 0.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is _________ BM.
Correct Answer: 2
Explanation:
Z = 29 [Cu] $\buildrel { - 2{e^ - }} \over \longrightarrow $ Cu2+ = [Ar] 3d9
Number of unpaired electron, n = 1
$\therefore$ Spin only magnetic moment,
$\mu = \sqrt {n(n + 2)} BM = \sqrt {1(1 + 2)} BM = \sqrt 3 BM$
= 1.73 BM $ \simeq $ 2 BM
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
For a d4 metal ion in an octahedral field, the
correct electronic configuration is :
A.
$t_{2g}^4e_g^0$ when $\Delta $0 < P
B.
$t_{2g}^3e_g^1$ when $\Delta $0 > P
C.
$e_g^2t_{2g}^2$ when $\Delta $0 < P
D.
$t_{2g}^3e_g^1$ when $\Delta $0 < P
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
The species that has a spin-only magnetic
moment of 5.9 BM, is :
(Td = tetrahedral)
(Td = tetrahedral)
A.
[MnBr4]2– (Td)
B.
[NiCl4]2– (Td)
C.
Ni(CO)4 (Td)
D.
[Ni(CN)4]2– (square planar)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Consider the complex ions,
trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B).
The correct statement regarding them is :
trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B).
The correct statement regarding them is :
A.
both (A) and (B) can be optically active.
B.
both (A) and (B) can not be optically active.
C.
(A) can not be optically active, but (B) can
be optically active.
D.
(A) can be optically active, but (B) can not
be optically active.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
The values of the crystal field stabilization
energies for a high spin d6 metal ion in
octahedral and tetrahedral fields, respectively,
are :
A.
–0.4$\Delta $0
and –0.27$\Delta $t
B.
–1.6$\Delta $0
and –0.4$\Delta $t
C.
–0.4$\Delta $0
and –0.6$\Delta $t
D.
–2.4$\Delta $0
and –0.27$\Delta $t
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The one that can exhibit highest paramagnetic
behaviour among the following is :
gly = glycinato; bpy = 2, 2'-bipyridine
gly = glycinato; bpy = 2, 2'-bipyridine
A.
[Fe(en)(bpy)(NH3)2]2+
B.
[Pd(gly)2]
C.
[Co(OX)2(OH)2]– ($\Delta $0 > P)
D.
[Ti(NH3)6]3+
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The Crystal Field Stabilization Energy
(CFSE) of [CoF3(H2O)3] ($\Delta $0 < P) is :
(CFSE) of [CoF3(H2O)3] ($\Delta $0 < P) is :
A.
-0.8 $\Delta $0
B.
-0.4 $\Delta $0
C.
-0.8 $\Delta $0 + 2P
D.
-0.4 $\Delta $0 + P
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The molecule in which hybrid MOs involve only
one d-orbital of the central atom is :
A.
XeF4
B.
[Ni(CN)4]2–
C.
[CrF6]3–
D.
BrF5
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
The pair in which both the species have the
same magnetic moment (spin only) is :
A.
[Cr(H2O)6]2+ and [CoCl4]2–
B.
[Co(OH)4]2– and [Fe(NH3)6]2+
C.
[Mn(H2O)6]2+ and [Cr(H2O)]2+
D.
[Cr(H2O)6]2+ and [Fe(H2O)6]2+
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
The number of isomers possible for
[Pt(en)(NO2)2] is :
[Pt(en)(NO2)2] is :
A.
3
B.
1
C.
4
D.
2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
The d-electron configuration of [Ru(en)3
]Cl2
and [Fe(H2O)6]Cl2 , respectively are :
and [Fe(H2O)6]Cl2 , respectively are :
A.
$t_{2g}^4e_g^2$ and $t_{2g}^6e_g^0$
B.
$t_{2g}^6e_g^0$ and $t_{2g}^6e_g^0$
C.
$t_{2g}^6e_g^0$ and $t_{2g}^4e_g^2$
D.
$t_{2g}^4e_g^2$ and $t_{2g}^4e_g^2$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc.H2SO4
loses
13.5% of its original mass, the correct molecular formula of A is :
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
A.
[Cr(H2O)5Cl]Cl2.H2O
B.
[Cr(H2O)4Cl2]Cl.2H2O
C.
[Cr(H2O)3Cl3].3H2O
D.
[Cr(H2O)6]Cl3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The complex that can show optical activity is :
A.
cis-[Fe(NH3)2(CN)4]–
B.
trans-[Cr(Cl2)(ox)2]3–
C.
trans-[Fe(NH3)2(CN)4]–
D.
cis-[CrCl2(ox)2]3– (ox = oxalate)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1
.
The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol-1, is :
A.
83.7
B.
242.5
C.
145.5
D.
97
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
The one that is not expected to show isomerism
is :
A.
[Pt(NH3)2Cl2]
B.
[Ni(NH3)4(H2O)2]2+
C.
[Ni(en)3]2+
D.
[Ni(NH3)2Cl2]
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
Simplified absorption spectra of three
complexes ((i), (ii) and (iii)) of Mn+ ion are
provided below; their $\lambda $max values are marked
as A, B and C respectively. The correct match
between the complexes and their $\lambda $max values is
(i) [M(NCS)6](–6 + n)
(ii) [MF6](–6 + n)
(iii) [M(NH3)6]n+
(i) [M(NCS)6](–6 + n)
(ii) [MF6](–6 + n)
(iii) [M(NH3)6]n+
A.
A-(i), B-(ii), C-(iii)
B.
A-(ii), B-(iii), C-(i)
C.
A-(ii), B-(i), C-(iii)
D.
A-(iii), B-(i), C-(ii)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
For octahedral Mn(II) and tetrahedral Ni(II)
complexes, consider the following statements:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV)aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are :
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV)aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are :
A.
(I), (III) and (IV) only
B.
(I) and (II) only
C.
(II), (III) and (IV) only
D.
(I), (II) and (III) only
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Consider that a d6 metal ion (M2+) forms a
complex with aqua ligands, and the spin only
magnetic moment of the complex is 4.90 BM.
The geometry and the crystal field stabilization
energy of the complex is
A.
tetrahedral and – 1.6 $\Delta $t
+ 1P
B.
octahedral and –2.4 $\Delta $0 + 2P
C.
tetrahedral and –0.6 $\Delta $t
D.
octahedral and –1.6 $\Delta $0
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
The correct order of the spin-only magnetic
moments of the following complexes is :
(I) [Cr(H2O)6]Br2
(II) Na4[Fe(CN)6]
(III) Na3[Fe(C2O4)3] ($\Delta $0 $>$ P)
(IV) (Et4N)2[CoCl4]
(I) [Cr(H2O)6]Br2
(II) Na4[Fe(CN)6]
(III) Na3[Fe(C2O4)3] ($\Delta $0 $>$ P)
(IV) (Et4N)2[CoCl4]
A.
(III) > (I) > (II) > (IV)
B.
(II) $ \approx $ (I) > (IV) > (III)
C.
(III) > (I) > (IV) > (II)
D.
(I) > (IV) > (III) > (II)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
The isomer(s) of [Co(NH3)4Cl2] that has/have
a Cl–Co–Cl angle of 90°, is/are :
A.
cis only
B.
cis and trans
C.
meridional and trans
D.
trans only
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Complex X of composition Cr(H2O)6Cln
has a spin only magnetic moment of 3.83
BM. It reacts with AgNO3 and shows
geometrical isomerism. The IUPAC
nomenclature of X is :
A.
Hexaaqua chromium (III) chloride
B.
Tetraaquadichlorido chromium(IV)
chloride dihydrate
C.
Tetraaquadichlorido chromium (III)
chloride dihydrate
D.
Dichloridotetraaqua chromium (IV)
chloride dihydrate
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
[Pd(F)(Cl)(Br)(I)]2– has n number of
geometrical isomers. Then, the spin-only
magnetic moment and crystal field stabilisation
energy [CFSE] of [Fe(CN)6]n–6, respectively,
are:
[Note : Ignore the pairing energy]
[Note : Ignore the pairing energy]
A.
1.73 BM and –2.0 $\Delta $0
B.
5.92 BM and 0
C.
2.84 BM and –1.6 $\Delta $0
D.
0 BM and –2.4 $\Delta $0
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
The correct order of the calculated spin-only
magnetic moments of complexs (A) to (D) is:
(A) Ni(CO)4
(B) [Ni(H2O)6]Cl2
(C) Na2[Ni(CN)4]
(D) PdCl2(PPh3)2
(A) Ni(CO)4
(B) [Ni(H2O)6]Cl2
(C) Na2[Ni(CN)4]
(D) PdCl2(PPh3)2
A.
(C) < (D) < (B) < (A)
B.
(C) $ \approx $ (D) < (B) < (A)
C.
(A) $ \approx $ (C) $ \approx $ (D) < (B)
D.
(A) $ \approx $ (C) < (B) $ \approx $ (D)