Coordination Compounds
(a) CoCl3.4NH3, (b) CoCl3.5NH3, (c) CoCl3.6NH3 and (d) CoCl(NO3)2.5NH3.
Number of complex(es) which will exist in cis-trans form is/are _______________.
Explanation:
$\mathrm{CoCl}_{3} \cdot 5 \mathrm{NH}_{3} \Rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
$\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3} \Rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$
Only $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]$ can show geometrical isomerism. Hence can exist in cis-trans form.
Number of complexes which will exhibit synergic bonding amongst, $[Cr{(CO)_6}]$, $[Mn{(CO)_5}]$ and $[M{n_2}{(CO)_{10}}]$ is ___________.
Explanation:
Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ____________ B.M. (Nearest integer)
Explanation:
$\mathrm{Mn}^{+2}$ has 5 unpaired electrons
$\therefore $ Spin only magnetic moment $=\sqrt{5(5+2)}$
$ \begin{aligned} &=\sqrt{5 \times 7} \\\\ &=\sqrt{35} \\\\ &\simeq 5.92 \text { B.M. } \\\\ &\simeq 6 \text { B.M. } \end{aligned} $
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 ___________.
Explanation:
$ \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} $
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)
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.
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]
Explanation:
$ \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} $
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)
Explanation:
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ 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} $
In the cobalt-carbonyl complex : [Co2(CO)8], number of Co-Co bonds is "X" and terminal CO ligands is "Y". X + Y = ___________.
Explanation:
x = 1
y = 6
$\therefore$ x + y = 7
LIST-I contains metal species and LIST-II contains their properties.
| List-I | List-II |
|---|---|
| (I) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}$ |
(P) $t_{2 \mathrm{g}}$ orbitals contain 4 electrons |
| (II) $\left[\mathrm{RuCl}_{6}\right]^{2-}$ | (Q) $\mu$ (spin-only $)=4.9 \mathrm{BM}$ |
| (III) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ |
(R) low spin complex ion |
| (IV) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ |
(S) metal ion in $4+$ oxidation state |
| (T) $d^{4}$ species |
[Given: Atomic number of $\mathrm{Cr}=24, \mathrm{Ru}=44, \mathrm{Fe}=26$ ]
Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option
$ \text { Match the following } $
| Column -I (Reaction) |
Column - II (Colour of the product or nature) | ||
|---|---|---|---|
| (A) | $\mathrm{FeCl}_3(\mathrm{aq})+\mathrm{NH}_3(\mathrm{aq}) \longrightarrow$ | (l) | Green ppt. |
| (B) | $\mathrm{AgCl}(\mathrm{aq})+\mathrm{NH}_3(\mathrm{aq}) \longrightarrow$ | (II) | Deep blue |
| (C) | $\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{NH}_3(\mathrm{aq}) \longrightarrow$ | (III) | Brown ppt. |
| (IV) | Colourless | ||
$ \text { The correct match } $
| A | B | C |
|---|---|---|
| I | II | III |
| A | B | C |
|---|---|---|
| I | III | IV |
| A | B | C |
|---|---|---|
| III | IV | II |
| A | B | C |
|---|---|---|
| III | I | IV |
Secondary valences of the following complexes based on their reactions with excess $\mathrm{AgNO}_3$ are
$ \begin{array}{llc} \hline & \begin{array}{l} \text { Formula of the } \\ \text { complexes } \end{array} & \begin{array}{c} \text { Moles of } \mathrm{AgCl} \text { precipitated } \\ \text { per mole of complex } \end{array} \\ \hline \text { (I) } & \mathrm{CoCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O} & 3 \\ \hline \text { (II) } & \mathrm{NiCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O} & 2 \\ \hline \text { (III) } & \mathrm{Co}\left(\mathrm{SO}_4\right) \mathrm{Br} \cdot 5 \mathrm{NH}_3 & 1 \\ \hline \end{array} $| I | II | III |
|---|---|---|
| 4 | 6 | 6 |
| I | II | III |
|---|---|---|
| 6 | 4 | 4 |
| I | II | III |
|---|---|---|
| 6 | 4 | 6 |
| I | II | III |
|---|---|---|
| 6 | 6 | 6 |
The pair in which both the species have same magnetic moment (spin only) is
$\left[\mathrm{CoCl}_4\right]^{2-},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2-}$
$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{CoCl}_4\right]^{2-}$
A metal complex absorbed orange light. The colour in which it appears is
yellow
yellow- green
red
green - blue
Among the given complexes the possess " $\mathrm{CO}^{\prime \prime}$ as a bridged ligands are
I. $\left[\mathrm{Co}_2(\mathrm{CO})_8\right]$
II. $\left[\mathrm{Fe}_3(\mathrm{CO})_{12}\right]$
III. $\left[\mathrm{Mn}_2(\mathrm{CO})_{10}\right]$
IV. $\left[\mathrm{Fe}_2(\mathrm{CO})_9\right]$
I, II and III
II, III and IV
I, II and IV
I, III and IV
The correct order of decreasing field strength of the below given ligands is
I $>$ II $>$ IV $>$ III
III $>$ II $>$ IV $>$ I
III $>$ I $>$ IV $>$ II
III $>$ IV $>$ I $>$ II
An element '$X$' with the atomic number 13 forms a complex of the type $[\mathrm{XCl}(\mathrm{H}_2 \mathrm{O})_5]^{2+}$. The covalency and oxidation state of $X$ in it are respectively
Identify the species, which does not exist?
The homoleptic complex in the following is
The crystal field theory is successful in explaining which of the following?
I. Ligands as point charges.
II. Formation and structures of complexes.
III. Colour.
IV. Magnetic properties.
V. Covalent character of metal-ligand bonding.
Which of the following is correct related to the colours of $\mathrm{TiCl}_3(X)$ and $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3(Y)$ ?
What is the hybridisation of [CrF6]3$-$ ?
Select the correct statement about the complex [Co(NH3)5SO4] Br.
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 :
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 :
complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+ respectively, are :
| 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 :
(i) [FeF6]3$-$
(ii) [Co(NH3)6]3+
(iii) [NiCl4]2$-$
(iv) [Cu(NH3)4]2+
Explanation:
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.
Explanation:
Explanation:
$\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
Explanation:
Given ks = 2.1 $\times$ 1013
Kd = ${1 \over {{k_s}}}$ = 4.7 $\times$ 10$-$14
$\therefore$ y = 4.7 $ \approx $ 5
Explanation:
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
(Round off to the nearest integer)
Explanation:
Secondary valency of Co = 6
(C. N.)
Explanation:
trioxalatochromate (III) ion $\to$ [Cr(C2O4)3]3$-$[Co(NO2)3(NH3)3]
X + Y = 2 + 0 = 2.0