Coordination Compounds
354 Questions
Start JEE Mains Test
2022
Q201
JEE Mains
Numerical
14 Mar 2026
Reaction of [Co(H2O)6]2+ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t2g-orbitals of the product is ___________.
Correct Answer: 6
Explanation:
$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+2} \underset{\mathrm{O}_{2}}{\stackrel{\mathrm{NH}_{3}}{\longrightarrow}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}+e^{-}$
$ \mathrm{Co}^{+3} \longrightarrow 3 \mathrm{~d}^{6} $
$\mathrm{NH}_{3}$ is a strong field ligand.
$ 3 \mathrm{~d}^{6} \longrightarrow \mathrm{t}_{2 \mathrm{~g}}^{6} ~\mathrm{eg}^{\circ} $
$ \mathrm{Co}^{+3} \longrightarrow 3 \mathrm{~d}^{6} $
$\mathrm{NH}_{3}$ is a strong field ligand.
$ 3 \mathrm{~d}^{6} \longrightarrow \mathrm{t}_{2 \mathrm{~g}}^{6} ~\mathrm{eg}^{\circ} $
2022
Q202
JEE Mains
Numerical
14 Mar 2026
The spin-only magnetic moment value of an octahedral complex among CoCl3.4NH3, NiCl2.6H2O and PtCl4.2HCl, which upon reaction with excess of AgNO3 gives 2 moles of AgCl is ___________ B.M. (Nearest integer)
Correct Answer: 3
Explanation:
From the given information, we are looking for a complex which, upon reaction with excess of AgNO₃, gives 2 moles of AgCl. This implies that the complex has 2 chloride ions involved.
Considering the complexes :
- CoCl₃.4NH₃ : This complex has 3 chloride ions, so it's not the one we are looking for.
- NiCl₂.6H₂O : This complex has 2 chloride ions, so it's a potential candidate.
- PtCl₄.2HCl : This complex has 4 chloride ions, so it's not the one we are looking for.
Therefore, the complex we are interested in is NiCl₂.6H₂O.
$\mathrm{CoCl}_{3} \cdot 4 \mathrm{NH}_{3} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \cdot \mathrm{Cl}_{2}\right]+\mathrm{AgCl}$$ \left.\left.\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\right[ \mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+2 \mathrm{AgCl} $
$\mathrm{PtCl}_{4} \cdot 2 \mathrm{HCl} \longrightarrow\left[\mathrm{PtCl}_{6}\right]^{4-}+\mathrm{No} ~\mathrm{AgCl} ~\mathrm{ppt}$
The next step is to find the oxidation state of the nickel ion. The nickel ion must have a charge of +2 to balance the -2 charge from the two chloride ions, thus it is Ni2+.
In the case of Ni²⁺, the electron configuration is [Ar]3d8. For an octahedral complex, the d-orbitals split into two sets under the influence of ligands: the $e_g$ set which includes d(x²-y²) and d(z²) orbitals, and the $t_{2g}$ set which includes the d(xy), d(xz), and d(yz) orbitals. Electrons will occupy the lower energy $t_{2g}$ orbitals first.
The 3d8 electron configuration implies there are 8 electrons in the 3d orbitals. The first six electrons pair up in the three $t_{2g}$ orbitals, and the next two electrons will go into the two $e_g$ orbitals, with each one having one unpaired electron.
The spin-only magnetic moment (μ) can be calculated using the formula :
$ \mu = \sqrt{n(n+2)} \, \text{B.M.} $
where n is the number of unpaired electrons. In this case, n = 2, so
$ \mu = \sqrt{2 \times (2+2)} = \sqrt{8} \, \text{B.M.} $
Rounding to the nearest integer, the spin-only magnetic moment is approximately 3 B.M.
2022
Q203
JEE Mains
Numerical
14 Mar 2026
Amongst FeCl3.3H2O, K3[Fe(CN)6] and [Co(NH3)6]Cl3, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is ____________ B.M. [nearest integer]
Correct Answer: 2
Explanation:
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \rightarrow$ Outer-orbital complex
$ \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow \text { Inner-orbital complex } $
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3} \rightarrow$ Inner-orbital complex
Since $\mathrm{CN}^{-}$is a strong field ligand than $\mathrm{NH}_{3}$. Hence $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is the inner-orbital complex that absorbs light at shortest wavelength.
$\mathrm{Fe}(\text{III}) \rightarrow$ valence shell configuration $3 \mathrm{~d}^{5}$
Since $\mathrm{CN}^{-}$will do pairing, so unpaired electron $=1$
$ \mu=\sqrt{1(1+2)}=\sqrt{3} \mathrm{BM} \simeq 2 \mathrm{BM} $
$ \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow \text { Inner-orbital complex } $
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3} \rightarrow$ Inner-orbital complex
Since $\mathrm{CN}^{-}$is a strong field ligand than $\mathrm{NH}_{3}$. Hence $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is the inner-orbital complex that absorbs light at shortest wavelength.
$\mathrm{Fe}(\text{III}) \rightarrow$ valence shell configuration $3 \mathrm{~d}^{5}$
Since $\mathrm{CN}^{-}$will do pairing, so unpaired electron $=1$
$ \mu=\sqrt{1(1+2)}=\sqrt{3} \mathrm{BM} \simeq 2 \mathrm{BM} $
2022
Q204
JEE Mains
Numerical
14 Mar 2026
If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be ____________ $\times$ 10$-$21 J. [Nearest Integer]
(Given : h = 6.63 $\times$ 10$-$34 Js and c = 3.08 $\times$ 108 ms$-$1)
Correct Answer: 766
Explanation:
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}$ is tetrahedral
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is octahedral
$ \because \Delta_{\mathrm{t}}=\frac{4}{9} \times \Delta_{0} $
$ \begin{aligned} &\Delta_{\mathrm{t}}=\frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \\\\ &\Delta_{0}=\frac{9}{4} \times \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \approx 765.7 \times 10^{-21} \mathrm{~J} \end{aligned} $
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is octahedral
$ \because \Delta_{\mathrm{t}}=\frac{4}{9} \times \Delta_{0} $
$ \begin{aligned} &\Delta_{\mathrm{t}}=\frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \\\\ &\Delta_{0}=\frac{9}{4} \times \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \approx 765.7 \times 10^{-21} \mathrm{~J} \end{aligned} $
2022
Q205
JEE Mains
Numerical
14 Mar 2026
In the cobalt-carbonyl complex : [Co2(CO)8], number of Co-Co bonds is "X" and terminal CO ligands is "Y". X + Y = ___________.
Correct Answer: 7
Explanation:
x = 1
y = 6
$\therefore$ x + y = 7
2021
Q206
JEE Mains
MCQ
14 Mar 2026
The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are $-$0.8 $\Delta$0 and 3.87 BM, respectively. Identify (Mz+) :
A.
V3+
B.
Cr3+
C.
Mn4+
D.
Co2+
2021
Q207
JEE Mains
MCQ
14 Mar 2026
The potassium ferrocyanide solution gives a Prussian blue colour, when added to :
A.
CoCl3
B.
FeCl2
C.
CoCl2
D.
FeCl3
2021
Q208
JEE Mains
MCQ
14 Mar 2026
Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is :
A.
5.92
B.
0
C.
1
D.
1.73
2021
Q209
JEE Mains
MCQ
14 Mar 2026
The denticity of an organic ligand, biuret is :
A.
2
B.
4
C.
3
D.
6
2021
Q210
JEE Mains
MCQ
14 Mar 2026
Indicate the complex/complex ion which did not show any geometrical isomerism :
A.
[CoCl2(en)2]
B.
[Co(CN)5(NC)]3$-$
C.
[Co(NH3)3(NO2)3]
D.
[Co(NH3)4Cl2]+
2021
Q211
JEE Mains
MCQ
14 Mar 2026
Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value.
Complexes : $\mathop {{{[Co{F_6}]}^{3 - }}}\limits_A ,\mathop {{{[Co{{({H_2}O)}_6}]}^{2 + }}}\limits_B ,\mathop {{{[Co{{(N{H_3})}_6}]}^{3 + }}}\limits_C and \mathop {{{[Co{{({en})}_3}]}^{3 + }}}\limits_D $
Choose the correct option :
Complexes : $\mathop {{{[Co{F_6}]}^{3 - }}}\limits_A ,\mathop {{{[Co{{({H_2}O)}_6}]}^{2 + }}}\limits_B ,\mathop {{{[Co{{(N{H_3})}_6}]}^{3 + }}}\limits_C and \mathop {{{[Co{{({en})}_3}]}^{3 + }}}\limits_D $
Choose the correct option :
A.
A < B < C < D
B.
B < A < C < D
C.
B < C < D < A
D.
C < D < B < A
2021
Q212
JEE Mains
MCQ
14 Mar 2026
Given below are two statements :
Statement I : ${[Mn{(CN)_6}]^{3 - }}$, ${[Fe{(CN)_6}]^{3 - }}$ and ${[Co{({C_2}{O_4})_3}]^{3 - }}$ are d2sp3 hybridised.
Statement II : ${[MnCl)_6}{]^{3 - }}$ and ${[Fe{F_6}]^{3 - }}$ are paramagnetic and have 4 and 5 unpaired electrons, respectively.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : ${[Mn{(CN)_6}]^{3 - }}$, ${[Fe{(CN)_6}]^{3 - }}$ and ${[Co{({C_2}{O_4})_3}]^{3 - }}$ are d2sp3 hybridised.
Statement II : ${[MnCl)_6}{]^{3 - }}$ and ${[Fe{F_6}]^{3 - }}$ are paramagnetic and have 4 and 5 unpaired electrons, respectively.
In the light of the above statements, choose the correct answer from the options given below :
A.
Statement I is correct but statement II is false
B.
Both statement I and Statement II are false
C.
Statement I is incorrect but statement II is true
D.
Both statement I and statement II are true
2021
Q213
JEE Mains
MCQ
14 Mar 2026
The type of hybridisation and magnetic property of the complex [MnCl6]3$-$, respectively, are :
A.
sp3d2 and diamagnetic
B.
d2sp3 and diamagnetic
C.
d2sp3 and paramagnetic
D.
sp3d2 and paramagnetic
2021
Q214
JEE Mains
MCQ
14 Mar 2026
The number of geometrical isomers found in the metal
complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+ respectively, are :
complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+ respectively, are :
A.
1, 1, 1, 1
B.
2, 1, 2, 2
C.
2, 0, 2, 2
D.
2, 1, 2, 1
2021
Q215
JEE Mains
MCQ
14 Mar 2026
Which one of the following metal complexes is most stable?
A.
[Co(en)(NH3)4]Cl2
B.
[Co(en)3]Cl2
C.
[Co(en)2(NH3)2]Cl2
D.
[Co(NH3)6]Cl2
2021
Q216
JEE Mains
MCQ
14 Mar 2026
Which one of the following species responds to an external magnetic field?
A.
[Fe(H2O)6]3+
B.
[Ni(CN)4]2$-$
C.
[Co(CN)6]3$-$
D.
[Ni(CO)4]
2021
Q217
JEE Mains
MCQ
14 Mar 2026
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
Q218
JEE Mains
MCQ
14 Mar 2026
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
Q219
JEE Mains
MCQ
14 Mar 2026
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
Q220
JEE Mains
MCQ
14 Mar 2026
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
Q221
JEE Mains
MCQ
14 Mar 2026
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
Q222
JEE Mains
MCQ
14 Mar 2026
The correct structures of trans-[NiBr2(PPh3)2] and meridonial-[Co(NH3)3(NO2)3], respectively, are
A.
B.
C.
D.
2021
Q223
JEE Mains
MCQ
14 Mar 2026
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
Q224
JEE Mains
MCQ
14 Mar 2026
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
Q225
JEE Mains
MCQ
14 Mar 2026
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
Q226
JEE Mains
MCQ
14 Mar 2026
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
Q227
JEE Mains
Numerical
14 Mar 2026
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
Q228
JEE Mains
Numerical
14 Mar 2026
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
Q229
JEE Mains
Numerical
14 Mar 2026
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
Q230
JEE Mains
Numerical
14 Mar 2026
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
Q231
JEE Mains
Numerical
14 Mar 2026
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
Q232
JEE Mains
Numerical
14 Mar 2026
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
Q233
JEE Mains
Numerical
14 Mar 2026
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
Q234
JEE Mains
Numerical
14 Mar 2026
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
Q235
JEE Mains
Numerical
14 Mar 2026
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$0 > P.E.
So, here all electrons becomes paired.
2021
Q236
JEE Mains
Numerical
14 Mar 2026
An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form [Ni(CN)6]2$-$. 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
Q237
JEE Mains
Numerical
14 Mar 2026
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
Q238
JEE Mains
Numerical
14 Mar 2026
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
Q239
JEE Mains
Numerical
14 Mar 2026
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
Q240
JEE Mains
Numerical
14 Mar 2026
[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
Q241
JEE Mains
Numerical
14 Mar 2026
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
Q242
JEE Mains
Numerical
14 Mar 2026
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
Q243
JEE Mains
Numerical
14 Mar 2026
Number of bridging CO ligands in [Mn2(CO)10] is __________.
Correct Answer: 0
Explanation:
$ \therefore $ Number of bridging CO ligands = 0.
2021
Q244
JEE Mains
Numerical
14 Mar 2026
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
Q245
JEE Mains
MCQ
14 Mar 2026
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 $0 < P
2020
Q246
JEE Mains
MCQ
14 Mar 2026
The species that has a spin-only magnetic
moment of 5.9 BM, is :
(Td = tetrahedral)
(Td = tetrahedral)
A.
[MnBr4]2– (Td)
B.
[NiCl4]2– (Td)
C.
Ni(CO)4 (Td)
D.
[Ni(CN)4]2– (square planar)
2020
Q247
JEE Mains
MCQ
14 Mar 2026
Consider the complex ions,
trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B).
The correct statement regarding them is :
trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B).
The correct statement regarding them is :
A.
both (A) and (B) can be optically active.
B.
both (A) and (B) can not be optically active.
C.
(A) can not be optically active, but (B) can
be optically active.
D.
(A) can be optically active, but (B) can not
be optically active.
2020
Q248
JEE Mains
MCQ
14 Mar 2026
The values of the crystal field stabilization
energies for a high spin d6 metal ion in
octahedral and tetrahedral fields, respectively,
are :
A.
–0.4$\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 $t
2020
Q249
JEE Mains
MCQ
14 Mar 2026
The one that can exhibit highest paramagnetic
behaviour among the following is :
gly = glycinato; bpy = 2, 2'-bipyridine
gly = glycinato; bpy = 2, 2'-bipyridine
A.
[Fe(en)(bpy)(NH3)2]2+
B.
[Pd(gly)2]
C.
[Co(OX)2(OH)2]– ($\Delta $0 > P)
D.
[Ti(NH3)6]3+
2020
Q250
JEE Mains
MCQ
14 Mar 2026
The Crystal Field Stabilization Energy
(CFSE) of [CoF3(H2O)3] ($\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
