Coordination Compounds
Chiral complex from the following is :
Here en = ethylene diamine
Match List I with List II
| List I Coordination entity |
List II Wavelength of light absorbed in nm |
||
|---|---|---|---|
| A. | $\mathrm{[CoCl(NH_3)_5]^{2+}}$ | I. | 310 |
| B. | $\mathrm{[Co(NH_3)_6]^{3+}}$ | II. | 475 |
| C. | $\mathrm{[Co(CN)_6]^{3-}}$ | III. | 535 |
| D. | $\mathrm{[Cu(H_2O)_4]^{2+}}$ | IV. | 600 |
Choose the correct answer from the options given below :
The hybridization and magnetic behaviour of cobalt ion in $\mathrm{[Co(NH_3)_6]^{3+}}$ complex, respectively is :
Which of the following cannot be explained by crystal field theory?
The primary and secondary valencies of cobalt respectively in $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ are :
Explanation:
The $t_{2g}$ orbitals are lower in energy than the $e_g$ orbitals.
The $\mathrm{Co}^{2+}$ ion has a d$^7$ electron configuration, with three electrons in the $t_{2g}$ set and two electrons in the $e_g$ set. In an octahedral crystal field, the three $t_{2g}$ orbitals will be lower in energy than the two $e_g$ orbitals. Therefore, the three electrons in the $t_{2g}$ set will occupy all three $t_{2g}$ orbitals, and the two electrons in the $e_g$ set will occupy the higher energy $e_g$ orbitals. As a result, there is only $\boxed{1}$ unpaired electron in the $t_{2g}$ set of orbitals.
The ratio of spin-only magnetic moment values $\mu_{\text {eff }}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} / \mu_{\text {eff }}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is _________.
Explanation:
Despite the difference in ligands (CN- and H2O), both complexes have a low-spin configuration because CN- is a strong field ligand and H2O is a weak field ligand. This configuration is denoted as $\mathrm{t}_{2g}^3 \mathrm{e}_{\mathrm{g}}^0$, indicating that there are three unpaired electrons in the t2g orbital and no unpaired electrons in the eg orbital.
Using the spin-only magnetic moment formula, $\mu = \sqrt{n(n+2)}$, where n is the number of unpaired electrons, we find that :
For $[\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}$, the spin-only magnetic moment is $\mu_{1}=\sqrt{3(3+2)} = \sqrt{15}$ BM.
And for $[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$, the spin-only magnetic moment is also $\mu_{2}=\sqrt{3(3+2)} = \sqrt{15}$ BM.
Therefore, the ratio of the spin-only magnetic moments is $\frac{\mu_{1}}{\mu_{2}} = \frac{\sqrt{15}}{\sqrt{15}} = 1$.
So, the spin-only magnetic moment of $[\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}$ is equal to the spin-only magnetic moment of $[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$.
For a metal ion, the calculated magnetic moment is $4.90 ~\mathrm{BM}$. This metal ion has ___________ number of unpaired electrons.
Explanation:
Using the equation $\mu = \sqrt{n(n+2)}$, where $\mu$ is the magnetic moment and $n$ is the number of unpaired electrons, we can substitute $\mu = 4.90$:
$4.90 = \sqrt{n(n+2)}$
Squaring both sides of the equation:
$(4.90)^2 = n(n+2)$
$24 = n^2 + 2n$
Rearranging the equation:
$n^2 + 2n - 24 = 0$
Solving the quadratic equation, we find:
$n =4$
Therefore, the metal ion has 4 unpaired electrons.
In potassium ferrocyanide, there are ________ pairs of electrons in the $t_{2g}$ set of orbitals.
Explanation:
$ \begin{aligned} & \mathrm{Fe}^{+2}=[\mathrm{Ar}] 3 \mathrm{~d}^6 \\\\ & \mathrm{CN}^{-}=\mathrm{SFL} \end{aligned} $
$\mathrm{t}_{2 \mathrm{~g}}$ contain 6 electron so it become 3 pairs
The observed magnetic moment of the complex $\left.\left[\operatorname{Mn}(\underline{N} C S)_{6}\right)\right]^{x^{-}}$ is $6.06 ~\mathrm{BM}$. The numerical value of $x$ is __________.
Explanation:
The complex is given as $[\mathrm{Mn}(\mathrm{NCS})_{6}]^{x-}$. Here, $\mathrm{NCS}^{-}$ acts as a ligand. Each $\mathrm{NCS}^{-}$ has a charge of -1 and since there are six of them, they contribute a total charge of -6 to the complex.
Manganese (Mn) in this complex is in the +2 oxidation state. We know this because in its ground state, Mn has 5 electrons in its 3d orbitals and 2 electrons in its 4s orbital. But in the Mn²⁺ cation, the 2 electrons from the 4s orbital have been removed, leaving 5 unpaired electrons in the 3d orbitals. This matches the given magnetic moment of 6.06 BM, which corresponds to 5 unpaired electrons.
So, the overall charge of the complex is -4: the Mn²⁺ ion contributes a charge of +2 and the six $\mathrm{NCS}^{-}$ ligands contribute a total charge of -6. When you add these together, you get -4. Hence, $x = -4$.
In summary, the $[\mathrm{Mn}(\mathrm{NCS})_{6}]^{x-}$ complex has an overall charge of -4, meaning x = -4.
So the numerical value of x will be 4.
Number of ambidentate ligands in a representative metal complex $\left[\mathrm{M}(\mathrm{en})(\mathrm{SCN})_{4}\right]$ is ___________.
[en = ethylenediamine]
Explanation:
Ambidentate ligands are ligands that can bond to a metal atom through two different atoms. They can attach through one site or the other, but not both at the same time.
In the given complex $[\mathrm{M}(\mathrm{en})(\mathrm{SCN})_{4}]$:
$\mathrm{en}$, or ethylenediamine, is a bidentate ligand, meaning it can form two bonds with the metal ion, but it is not ambidentate because it always binds through the same two nitrogen atoms.
$\mathrm{SCN}^{-}$, on the other hand, is an example of an ambidentate ligand. It can bind to the metal either through the sulfur atom ($\mathrm{S}$) or the nitrogen atom ($\mathrm{N}$).
Since there are four $\mathrm{SCN}^{-}$ ligands in the given complex, the number of ambidentate ligands is 4.
The spin only magnetic moment of $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ complexes is _________ B.M. (Nearest integer)
(Given : Atomic no. of Mn is 25)
Explanation:
The spin only magnetic moment of the $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ complex can be calculated using the formula:
μ = $\sqrt {n\left( {n + 2} \right)} $ Bohr magnetons
Where n is the number of unpaired electrons in the complex. To find the number of unpaired electrons in $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$, we can use the electron configuration of the Mn ion:
For Mn2+ : 1s2 2s2 2p6 3s2 3p6 3d5
From the electron configuration, we can see that Mn2+ has 5 unpaired electrons. Thus, the spin only magnetic moment of $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is:
μ = $\sqrt {5\left( {5 + 2} \right)} $ = $\sqrt {35} $ = 5.9 Bohr Magnetons
The nearest integer to 5.9 is 6.
Hence, the spin only magnetic moment of $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is 6 Bohr magnetons.
Assume Planck's constant (h) $=6.4 \times 10^{-34} \mathrm{Js}$, Speed of light $(\mathrm{c})=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and Avogadro's
Constant $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} / \mathrm{mol}$
Explanation:
$ \begin{aligned} \frac{\mathrm{hc}}{\lambda} & =\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}} \\\\ \lambda & =\frac{0.4 \times 6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3 \times 10^8}{96 \times 10^3} \\\\ & =0.48 \times 10^{-6} \mathrm{~m} \\\\ & =480 \times 10^{-9} \mathrm{~m} \\\\ & =480 \mathrm{~nm} \end{aligned} $
The sum of bridging carbonyls in $\mathrm{W(CO)_6}$ and $\mathrm{Mn_2(CO)_{10}}$ is ____________.
Explanation:
Total number of moles of AgCl precipitated on addition of excess of AgNO$_3$ to one mole each of the following complexes $\mathrm{[Co(NH_3)_4Cl_2]Cl,[Ni(H_2O)_6]Cl_2,[Pt(NH_3)_2Cl_2]}$ and $\mathrm{[Pd(NH_3)_4]Cl_2}$ is ___________.
Explanation:
$ \begin{aligned} & {\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2} \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} 2 \mathrm{AgCl}} \\\\ & {\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2} \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} 2 \mathrm{AgCl}} \end{aligned} $
Total moles of $\mathrm{AgCl}$ precipitated $=5$
The number of paramagnetic species from the following is _____________.
$\mathrm{{[Ni{(CN)_4}]^{2 - }},[Ni{(CO)_4}],{[NiC{l_4}]^{2 - }}}$
$\mathrm{{[Fe{(CN)_6}]^{4 - }},{[Cu{(N{H_3})_4}]^{2 + }}}$
$\mathrm{{[Fe{(CN)_6}]^{3 - }}\,and\,{[Fe{({H_2}O)_6}]^{2 + }}}$
Explanation:
The d-electronic configuration of $\mathrm{[CoCl_4]^{2-}}$ in tetrahedral crystal field in ${e^mt_2^n}$. Sum of "m" and "number of unpaired electrons" is ___________
Explanation:
Configuration $\mathrm{e}^4 \mathrm{t}_2{ }^3: \mathrm{m}=4$
Number of unpaired electrons $=3$
So, answer $=7$
[Atomic Number: $\mathrm{Fe}=26, \mathrm{Mn}=25, \mathrm{Co}=27$ ]
| List - I | List - II |
|---|---|
| (P) $t_{2 g}^6 e_g^0$ | (1) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| (Q) $t_{2 g}^3 e_g^2$ | (2) $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| (R) $\mathrm{e}^2 \mathrm{t}_2^3$ | (3) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ |
| (S) $t_{2 g}^4 e_g^2$ | (4) $\left[\mathrm{FeCl}_4\right]^{-}$ |
| (5) $\left[\mathrm{CoCl}_4\right]^{2-}$ |
[en $=\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{NH}_2$ ]
The sum of oxidation state and co-ordination number of central metal atom is maximum with respect to which of the following complex?
$\mathrm{K}_3\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$
$\left[\mathrm{Cr}(\mathrm{CO})_6\right]$
$\mathrm{K}_2\left[\mathrm{PtCl}_6\right]$
$\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
$ \text { Match the following. } $
$ \begin{array}{llll} \hline & \begin{array}{l} \text { List-I } \\ (\text { Complex }) \end{array} & & \begin{array}{l} \text { List-II } \\ \text { (Spin only magnetic } \\ \text { moment) } \end{array} \\ \hline \text { A. } & {\left[\mathrm{CoF}_6\right]^{3-}} & \text { I. } & 0 \\ \hline \text { B. } & {\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}} & \text { II. } & \sqrt{24} \\ \hline \text { C. } & {\left[\mathrm{FeF}_6\right]^{3-}} & \text { III. } & \sqrt{8} \\ \hline \text { D. } & {\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}} & \text { IV. } & \sqrt{35} \\ \hline & & \text { V. } & \sqrt{15} \\ \hline \end{array} $
The correct answer is
A-V, B-II, C-IV, D-I
A-II, B-I, C-IV, D-III
A-II, B-I, C-V, D-III
A-III, B-II, C-I, D-V
Identify the correct set for $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ ion.
(Hybridisation of $\mathrm{Co}^{3+}$, type of complex, number of unpaired electrons in the complex ion respectively.)
$d^2 s p^3$, inner orbital complex, zero
$s p^3 d^2$, outer orbital complex, three
$d^2 s p^3$, inner orbital complex, two
$s p^3 d^2$, outer orbital complex, zero
According to Werner's theory, the number of groups bonded to the central metal atom/ion in a coordination complex represent.
oxidation state
primary valency
secondary valency
polyhedron
Identify the incorrect match from the following.
The correct answer is Table
The complex showing a spin-only magnetic moment of $2.82 \mathrm{~BM}$ is
The correct statement with respect to the complexes $\mathrm{Ni}(\mathrm{CO})_4$ and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ is
Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; $\mathrm{H}_{2} \mathrm{~N}_{-} \mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$)
Match List I with List II
| List - I (Complex) | List - II (Hybridization) | ||
|---|---|---|---|
| (A) | $Ni{(CO)_4}$ | (I) | $s{p^3}$ |
| (B) | ${[Ni{(CN)_4}]^{2 - }}$ | (II) | $s{p^3}{d^2}$ |
| (C) | ${[Co{(CN)_6}]^{3 - }}$ | (III) | ${d^2}s{p^3}$ |
| (D) | ${[Co{F_6}]^{3 - }}$ | (IV) | $ds{p^2}$ |
Choose the correct answer from the options given below :
Low oxidation state of metals in their complexes are common when ligands :
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of :
The metal complex that is diamagnetic is (Atomic number: $\mathrm{Fe}, 26 ; \mathrm{Cu}, 29)$
The correct order of energy of absorption for the following metal complexes is :
A : [Ni(en)3]2+ , B : [Ni(NH3)6]2+ , C : [Ni(H2O)6]2+
Correct formula of the compound which gives a white precipitate with BaCl2 solution, but not with AgNO3 solution, is :
Given below are two statements.
$\bullet$ Statement I : In CuSO4 . 5H2O, Cu-O bonds are present.
$\bullet$ Statement II : In CuSO4 . 5H2O, ligands coordinating with Cu(II) ion are O-and S-based ligands.
In the light of the above statements, choose the correct answer from the options given below.
Given below are two statements :
Statement I : [Ni(CN)4]2$-$ is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4] is tetrahedral, paramagnetic and with sp3-hybridication for Ni.
Statement II : [NiCl4]2$-$ and [Ni(CO)4] both have same d-electron configuration have same geometry and are paramagnetic.
In light the above statements, choose the correct answer from the options given below :
Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : Mn = 25; Fe = 26)
A. [FeF6]3$-$
B. [Fe(CN)6]3$-$
C. [MnCl6]3$-$ (high spin)
D. [Mn(CN)6]3$-$
Choose the correct answer from the options given below :
Which of the following will have maximum stabilization due to crystal field?
Which statement is not true with respect to nitrate ion test?
Transition metal complex with highest value of crystal field splitting ($\Delta$0) will be :
Match List - I with List - II :
| List - I | List -II | ||
|---|---|---|---|
| (A) | ${[PtC{l_4}]^{2 - }}$ | (I) | $s{p^3}d$ |
| (B) | $Br{F_5}$ | (II) | ${d^2}s{p^3}$ |
| (C) | $PC{l_5}$ | (III) | $ds{p^2}$ |
| (D) | ${[Co{(N{H_3})_6}]^{3 + }}$ | (IV) | $s{p^3}{d^2}$ |
Choose the most appropriate answer from the options given below :
Sum of oxidation state (magnitude) and coordination number of cobalt in $\mathrm{Na}\left[\mathrm{Co}(\mathrm{bpy}) \mathrm{Cl}_{4}\right]$ is _________.
Explanation:
Oxidation state of cobalt $=+3$
Coordination number of cobalt $=6$
[As bpy is bidentate]
So, sum $=9$
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is $(-)$ ____________ $\Delta_{0}$. (Nearest integer)
Explanation:
Configuration $\Rightarrow t_{2 g}^{5}$
CFSE $=5 \times \frac{-2}{5} \Delta_{0}=-2 \Delta_{0}$
Total number of relatively more stable isomer(s) possible for octahedral complex $\left[\mathrm{Cu}(\mathrm{en})_{2}(\mathrm{SCN})_{2}\right]$ will be _________.
Explanation:
The conductivity of a solution of complex with formula $\mathrm{CoCl}_{3}\left(\mathrm{NH}_{3}\right)_{4}$ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is __________.
Explanation:
$\therefore$ Possible compound is $\rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \mathrm{Cl}^{-}$
Oxidation state of central atom represents the total number of primary valency
$\therefore$ Primary valency will be 3 .
The difference between spin only magnetic moment values of $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\mathrm{Cl}_{2}$ and $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ is ___________.
Explanation:
$\mathrm{H}_{2} \mathrm{O}$ is weak field ligand.
$\mathrm{Co}^{+2} \rightarrow 3 d^{7}$
$ \begin{aligned} \mathrm{n}=3 \quad \mu_{1} &=\sqrt{n(\mathrm{n}+2)} \\ &=\sqrt{15} \text { B.M. } \end{aligned} $
$\mathrm{Cr} \rightarrow 4 s^{1} 3 d^{5}$
$\mathrm{Co}^{+3} \rightarrow 3 d^{3}$
$\mathrm{n}=3 \quad \mu_{2}=\sqrt{15}$ B.M.
$\mu_{1}-\mu_{2}=0$
Consider the following metal complexes :
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
$\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$
The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________ B. M. (Nearest integer)
Explanation:
The stronger the ligand, the higher the crystal field splitting.
So, the order of crystal field splitting is
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}>$ $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$
Shortest wavelength is shown by complex having maximum crystal field splitting.
Spin only magnetic moment $=\sqrt{0(0+2)}=0$ B.M
In the following brown complex, the oxidation state of iron is +_____________.
${[Fe{({H_2}O)_6}]^{2 + }} + NO \to \mathop {{{[Fe{{({H_2}O)}_5}(NO)]}^{2 + }}}\limits_{\text{Brown complex}} + {H_2}O$
Explanation:
To determine the oxidation state of iron in the brown complex ${[Fe({H_2}O)_5(NO)]^{2+}}$, we need to analyze the ligands and their charges, as well as the overall charge of the complex.
Step 1: Understanding the Complex
The complex is:
$ {[Fe({H_2}O)_5(NO)]^{2+}} $
Iron (Fe): The central metal atom whose oxidation state we need to find.
Ligands:
5 Water Molecules ($H_2O$): Neutral ligands (charge = 0).
1 Nitrosyl Group (NO): Can have different charges depending on its mode of bonding.
Step 2: Assigning Charges to Ligands
Water ($H_2O$): Neutral ligand, so it contributes 0 to the overall charge.
Nitrosyl (NO): Can act as:
NO$^+$ (nitrosonium ion) with a +1 charge.
NO (neutral molecule) with 0 charge.
NO$^-$ (nitroxide ion) with a –1 charge.
Step 3: Setting Up the Oxidation State Equation
Let $ x $ be the oxidation state of Fe.
The sum of the oxidation states of all components equals the overall charge of the complex:
$ x + (5 \times 0) + (\text{Charge of NO}) = +2 $
Simplifying:
$ x + (\text{Charge of NO}) = +2 $
Step 4: Determining the Charge of NO in the Complex
In the brown ring complex, experimental evidence shows that the nitrosyl ligand acts as NO$^+$. This is because:
The NO ligand forms a linear bond with Fe, characteristic of NO$^+$.
The complex is known to involve a reduction of the oxidation state of Fe to an unusual value.
Thus, Charge of NO = +1.
Step 5: Calculating the Oxidation State of Fe
Substitute the charge of NO into the equation:
$ x + (+1) = +2 $
Solving for $ x $:
$ x = +2 - (+1) = +1 $
Therefore, the oxidation state of Fe in the brown complex is $ +1 $.
Conclusion
The oxidation state of iron in ${[Fe({H_2}O)_5(NO)]^{2+}}$ is +1.
Spin only magnetic moment ($\mu$s) of ${K_3}[Fe{(CN)_6}]$ is ____________ B.M.
(Nearest integer)
Explanation:
$\mathrm{CN}^{-}$ have $-1$ oxidation State.
Let oxidation state of $\mathrm{Fe}$ in the complex is $x$.
So, $x-6=-3 \Rightarrow x=+3$
Now, $\mathrm{CN}^{-}$ion is a strong field ligand.
For $Fe^{3+}: -3d^54s^0$
Unpaired electron $=1$
$\mu_{\mathrm{B}}=\sqrt{n(n+2)}=\sqrt{1 \times(1+2)}=1.7$
$\mu_{\mathrm{B}}=2$ B.M
${[Fe{(CN)_6}]^{4 - }}$
${[Fe{(CN)_6}]^{3 - }}$
${[Ti{(CN)_6}]^{3 - }}$
${[Ni{(CN)_4}]^{2 - }}$
${[Co{(CN)_6}]^{3 - }}$
Among the given complexes, number of paramagnetic complexes is ____________.
Explanation:
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \quad$ Paramagnetic (1 unpaired electron)
$\left[\mathrm{Ti}(\mathrm{CN})_6\right]^{3-} \quad$ Paramagnetic (1 unpaired electron)
$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \quad$ Diamagnetic
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \quad$ Diamagnetic