(Atomic numbers: Mn = 25; Fe = 26, Co =27)
The number of isoelectronic species among $Sc^{3+}, Cr^{2+}, Mn^{3+}, Co^{3+}$ and $Fe^{3+}$ is ‘n’. If ‘n’ moles of AgCl is formed during the reaction of complex with formula $CoCl_3(en)_2NH_3$ with excess of $AgNO_3$ solution, then the number of electrons present in the $t_{2g}$ orbital of the complex is ________.
Explanation:
$ \begin{array}{|l|l|} \hline \mathrm{Sc}^{+3} & 18 \\ \hline \mathrm{Cr}^{+2} & 22 \\ \hline \mathrm{Mn}^{+3} & 22 \\ \hline \mathrm{Co}^{+3} & 24 \\ \hline \mathrm{Fe}^{+3} & 23 \\ \hline \end{array} $
In the table, the number written for each ion is its total number of electrons.
$\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ both have $22$ electrons, so they are isoelectronic (same number of electrons).
So, the number of isoelectronic species is
$ \mathrm{n}=2 $
Now, for the complex $CoCl_3(en)_2NH_3$:
$en$ and $NH_3$ are neutral ligands, so they do not change the charge.
In the given formula, total $Cl$ are $3$. If $n=2$ moles of $AgCl$ are formed with excess $AgNO_3$, it means $2$ chloride ions are outside the coordination sphere (they are ionisable and precipitate as $AgCl$).
So the complex must be:
Complex is : $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{NH}_3 \mathrm{Cl}\right] \mathrm{Cl}_2$
Here, $2$ $Cl^-$ are outside, so the complex ion has charge $+2$.
Let oxidation state of Co be $x$.
Inside the bracket: $en$ and $NH_3$ contribute $0$, and one coordinated $Cl^-$ contributes $-1$.
So, $x-1=+2 \Rightarrow x=+3$.
Thus cobalt is $Co^{3+}$, and its $d$-electron configuration is:
$ \Rightarrow \mathrm{Co}^{3+} \quad 3 \mathrm{~d}^6 \quad \mathrm{t}_{2 \mathrm{~g}}^{2,2,2} \quad \mathrm{e}_{\mathrm{g}}^{0,0} $
So, total electrons in $t_{2g}$ orbitals $=2+2+2=6$.
X is the number of geometrical isomers exhibited by $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{BrCl}\right]$.
Y is the number of optically inactive isomer(s) exhibited by $\left[\mathrm{CrCl}_2(\mathrm{ox})_2\right]^{3-}$
Z is the number of geometrical isomers exhibited by $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3\right]$.
The value of $\mathrm{X}+\mathrm{Y}+\mathrm{Z}$ is $\_\_\_\_$ .
Explanation:
1) Find $X$ for $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)\left(\mathrm{H}_2 \mathrm{O}\right)\mathrm{BrCl}\right]$
Pt in such complexes is usually $\mathrm{Pt(II)}$, which (as per NCERT) forms square planar complexes.
The complex is of type $[\mathrm{MABCD}]$ (four different ligands).
For square planar $[\mathrm{MABCD}]$, the number of geometrical isomers is 3 (based on which ligand is trans to a chosen ligand; distinct trans-pairings give 3 arrangements).
So, $X = 3.$
2) Find $Y$ for $\left[\mathrm{CrCl}_2(\mathrm{ox})_2\right]^{3-}$
Here $\mathrm{ox}$ (oxalate) is a bidentate ligand.
Two oxalate ligands occupy $4$ coordination positions, and $2$ positions are occupied by $\mathrm{Cl^-}$.
So the complex is octahedral of type $[\mathrm{M(AA)}_2\mathrm{B}_2]$.
Such complexes show cis–trans isomerism:
cis form is optically active (exists as $\Delta$ and $\Lambda$ enantiomers)
trans form is optically inactive
So optically inactive isomers = only 1 (trans).
Thus, $Y = 1.$
3) Find $Z$ for $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3\right]$
This is an octahedral complex of type $[\mathrm{MA}_3\mathrm{B}_3]$.
It shows fac–mer isomerism:
fac
mer
So, $Z = 2.$
Final Answer
$X+Y+Z = 3+1+2 = 6.$
$ \boxed{6} $
A chromium complex with a formula $\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}$ has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to $1: 2$ electrolyte. 2.75 g of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of $\mathrm{AgNO}_3$. The amount of AgCl formed in the above process is $\_\_\_\_$ g. (Nearest integer)
[Given: Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{Cr}: 52 ; \mathrm{Cl}: 35.5, \mathrm{Ag}: 108, \mathrm{O}: 16, \mathrm{H}: 1$ ]
Explanation:
$ \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}+\mathrm{AgNO}_3 \rightarrow 2 \mathrm{AgCl} $
$\,\,\,\,\,\,\,\ 2.75 / 266.5 \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\,-$
$ =0.0103 \text { moles }\,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\ \quad 0.02063 \text { mole } $
$ \text { Mass of } \mathrm{AgCl}=0.02063 \times 143.5=2.96 \mathrm{gm} $
Total number of unpaired electrons present in the central metal atoms/ions of
$\left[\mathrm{Ni}(\mathrm{CO})_4\right],\left[\mathrm{NiCl}_4\right]^{2-},\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right],\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ and $\left[\mathrm{Pt}(\mathrm{CN})_4\right]^{2-}$ is $\_\_\_\_$。
Explanation:
$\mathrm{In}\left[\mathrm{Ni}(\mathrm{CO})_4\right], \mathrm{Ni}^0: 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
Hybridisation state: $\mathrm{sp}^3$
Unpaired electron $=0$
In $\left[\mathrm{NiCl}_4\right]^{2-}, \mathrm{Ni}^{2+}: 3 \mathrm{~d}^8$
Hybridisation state: $\mathrm{sp}^3$
Unpaired electron $=2$
In $\left[\mathrm{PtCl}_4\right]^{2-}, \mathrm{Pt}^{2+}: 5 \mathrm{~d}^8$
Hybridisation state : $\mathrm{dsp}^2$
Unpaired electron $=0$
In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}, \mathrm{Ni}^{2+}: 3 \mathrm{~d}^8$
Hybridisation state: $\mathrm{dsp}^2$
Unpaired electron $=0$
In $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right], \mathrm{Pt}^{2+}: 5 \mathrm{~d}^8$
Hybridisation state: $\mathrm{dsp}^2$
Unpaired electron $=0$
The crystal field splitting energy of $\left[\mathrm{Co}(\text { oxalate })_3\right]^{3-}$ complex is ' $n^{\prime}$ times that of the $\left[\mathrm{Cr}(\text { oxalate })_3\right]^{3-}$ complex. Here ' $n$ ' is $\_\_\_\_$ . (Assume $\Delta_0 \gg P$ )
Explanation:
Since $\Delta_0 \gg P$, we ignore pairing energy and decide the electron arrangement mainly using $\Delta_0$.
For ${\left[\operatorname{Co}(\mathrm{ox})_3\right]^{3-}}$:
$\mathrm{Co}^{+3}$ has configuration $\mathrm{d}^6$.
In an octahedral field (given $\Delta_0 \gg P$), all 6 electrons occupy $t_{2g}$ first, so the distribution is $\mathrm{t}_2 \mathrm{~g}^{2,2,2}\ \mathrm{eg}^{0,0}$.
Each electron in $t_{2g}$ gives CFSE of $\left(-0.4 \Delta_0\right)$, so
$\operatorname{CFSE}=6 \times\left(-0.4 \Delta_0\right)=-2.4 \Delta_0$
For ${\left[\operatorname{Cr}\left(\mathrm{ox}_3\right)\right]^{3-}}$:
$\mathrm{Cr}^{+3}$ has configuration $\mathrm{d}^3$.
So the electrons occupy $t_{2g}$ as $\mathrm{t}_2 \mathrm{~g}^{1,1,1}\ \mathrm{eg}^{0,0}$.
Hence,
$\operatorname{CFSE}=3 \times\left(-0.4 \Delta_0\right)=-1.2 \Delta_0$
Now,
$\frac{(\mathrm{CFSE})_{\mathrm{Co}^{+3}}}{(\mathrm{CFSE})_{\mathrm{Cr}^{+3}}}= \frac{-2.4\Delta_0}{-1.2\Delta_0}=2$
So, $n = 2$.
Identify the metal ions among $\mathrm{Co^{2+}}$, $\mathrm{Ni^{2+}}$, $\mathrm{Fe^{2+}}$, $\mathrm{V^{3+}}$ and $\mathrm{Ti^{2+}}$ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is _________ .
Explanation:
$\begin{aligned} & \mathrm{V}^{3+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^2 \\ & \mathrm{Ti}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^2 \\ & \mathrm{Ni}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^8\end{aligned}$
$ \begin{aligned} & \mathrm{Fe}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^6 \\ & \mathrm{Co}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^7 \end{aligned} $
Only for $\mathrm{Fe}^{2+}$ and $\mathrm{Co}^{2+} \mu$ is more than 3.0 B.M.
$\therefore $ Number of unpaired electrons $=4+3=7$
Explanation:
To determine which metal complexes are paramagnetic and have the same number of unpaired electrons, let's analyze each one:
$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$:
Cobalt ion: $\mathrm{Co}^{3+}$
Electron configuration: $3d^6$
Orbital occupancy: $\mathrm{t}_{2g}^{2,2,2}$ $\mathrm{e}_{g}^{0,0}$
Magnetic property: Diamagnetic (unpaired electrons = 0)
$[\mathrm{Co}(\mathrm{C}_2 \mathrm{O}_4)_3]^{3-}$:
Cobalt ion: $\mathrm{Co}^{3+}$
Electron configuration: $3d^6$
Orbital occupancy: $\mathrm{t}_{2g}^{2,2,2}$ $\mathrm{e}_{g}^{0,0}$
Magnetic property: Diamagnetic (unpaired electrons = 0)
$[\mathrm{MnCl}_6]^{3-}$:
Manganese ion: $\mathrm{Mn}^{3+}$
Electron configuration: $3d^4$
Orbital occupancy: $\mathrm{t}_{2g}^{1,1,1}$ $\mathrm{e}_{g}^{1,0}$
Magnetic property: Paramagnetic (unpaired electrons = 4)
$[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$:
Manganese ion: $\mathrm{Mn}^{3+}$
Electron configuration: $3d^4$
Orbital occupancy: $\mathrm{t}_{2g}^{2,1,1}$ $\mathrm{e}_{g}^{0,0}$
Magnetic property: Paramagnetic (unpaired electrons = 2)
$[\mathrm{CoF}_6]^{3-}$:
Cobalt ion: $\mathrm{Co}^{3+}$
Electron configuration: $3d^6$
Orbital occupancy: $\mathrm{t}_{2g}^{2,1,1}$ $\mathrm{e}_{g}^{1,1}$
Magnetic property: Paramagnetic (unpaired electrons = 4)
$[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$:
Iron ion: $\mathrm{Fe}^{3+}$
Electron configuration: $3d^5$
Orbital occupancy: $\mathrm{t}_{2g}^{2,2,1}$ $\mathrm{e}_{g}^{0,0}$
Magnetic property: Paramagnetic (unpaired electrons = 1)
$[\mathrm{FeF}_6]^{3-}$:
Iron ion: $\mathrm{Fe}^{3+}$
Electron configuration: $3d^5$
Orbital occupancy: $\mathrm{t}_{2g}^{1,1,1}$ $\mathrm{e}_{g}^{1,1}$
Magnetic property: Paramagnetic (unpaired electrons = 5)
From this analysis, the paramagnetic complexes having the same number of unpaired electrons (4 unpaired electrons) are:
$[\mathrm{MnCl}_6]^{3-}$
$[\mathrm{CoF}_6]^{3-}$
These two complexes show the same paramagnetic property, each with 4 unpaired electrons.
The number of paramagnetic complexes among $\left[\mathrm{FeF}_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$, $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-},\left[\mathrm{MnCl}_6\right]^{3-}$, and $\left[\mathrm{CoF}_6\right]^{3-}$, which involved $\mathrm{d}^2 \mathrm{sp}^3$ hybridization is _________.
Explanation:
$\begin{array}{lll} {\left[\mathrm{FeF}_6\right]^{3-}} & \text { Paramagnetic } & \mathrm{sp}^3 \mathrm{~d}^2 \\ {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}} & \text { Paramagnetic } & \mathrm{d}^2 \mathrm{sp}^3 \\ {\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}} & \text { Paramagnetic } & \mathrm{d}^2 \mathrm{sp}^3 \\ {\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}} & \text { Diamagnetic } & \mathrm{d}^2 \mathrm{sp}^3 \\ {\left[\mathrm{MnCl}_6\right]^{3-}} & \text { Paramagnetic } & \mathrm{sp}^3 \mathrm{~d}^2 \\ {\left[\mathrm{CoF}_6\right]^{3-}} & \text { Paramagnetic } & \mathrm{sp}^3 \mathrm{~d}^2 \end{array}$
Only $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ are paramagnetic and $\mathrm{d}^2 \mathrm{sp}^3$ hybridisation of metal.
A metal complex with a formula $\mathrm{MCl}_4 \cdot 3 \mathrm{NH}_3$ is involved in $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation. It upon reaction with excess of $\mathrm{AgNO}_3$ solution gives ' $x$ ' moles of AgCl . Consider ' $x$ ' is equal to the number of lone pairs of electron present in central atom of $\mathrm{BrF}_5$. Then the number of geometrical isomers exhibited by the complex is _________.
Explanation:
It shows 2 geometrical isomers $\left(\mathrm{Ma}_3 \mathrm{b}_3\right.$ type$)$
facial (fac) & meridional (Mer)
The number of optical isomers exhibited by the iron complex $(\mathrm{A})$ obtained from the following reaction is___________.
$ \mathrm{FeCl}_3+\mathrm{KOH}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{~A} $
Explanation:
$\Rightarrow\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ is $\left[\mathrm{M}(\mathrm{AA})_3\right]$ type complex.
So total optical isomers $=2$
The spin-only magnetic moment value of $\mathrm{M}^{\mathrm{n}+}$ ion formed among $\mathrm{Ni}, \mathrm{Zn}, \mathrm{Mn}$ and Cu that has the least enthalpy of atomisation is_________ . (in nearest integer) Here n is equal to the number of diamagnetic complexes among $\mathrm{K}_2\left[\mathrm{NiCl}_4\right],\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$, $\mathrm{K}_3\left[\mathrm{Mn}(\mathrm{CN})_6\right]$ and $\left[\mathrm{Cu}\left(\mathrm{PPh}_3\right)_3 \mathrm{I}\right]$
Explanation:
$\mathrm{K}_2\left[\mathrm{NiCl}_4\right] \Rightarrow \mathrm{sp}^3$, Paramagnetic
$\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2 \Rightarrow \mathrm{sp}^3 \mathrm{~d}^2$, Diamagnetic
$\mathrm{K}_3\left[\mathrm{Mn}(\mathrm{CN})_6\right] \Rightarrow \mathrm{d}^2 \mathrm{sp}^3$, Paramagnetic
$\left[\mathrm{Cu}\left(\mathrm{PPh}_3\right)_3 \mathrm{I}\right] \Rightarrow \mathrm{sp}^3$, Diamagnetic
Hence the value of $n$ is 2
Least value of enthalpy of atomisation among Ni,
$\mathrm{Zn}, \mathrm{Mn}$ and Cu is of Zn
$\begin{aligned} & \mathrm{Zn}^{+2}:-[\mathrm{Ar}] 3 \mathrm{~d}^{10} \\ & \mu=0 \end{aligned}$
A transition metal (M) among $\mathrm{Mn}, \mathrm{Cr}, \mathrm{Co}$ and Fe has the highest standard electrode potential $\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)$. It forms a metal complex of the type $\left[\mathrm{M}(\mathrm{CN})_6\right]^{4-}$. The number of electrons present in the $\mathrm{e}_{\mathrm{g}}$ orbital of the complex is ___________.
Explanation:
Co has highest standard electrode potential $\left(\mathrm{M}^{+3} / \mathrm{M}^{+2}\right)$ among $\mathrm{Mn}, \mathrm{Cr}, \mathrm{Co}, \mathrm{Fe}$
$\therefore$ Complex is $\left[\mathrm{Co}(\mathrm{CN})_6\right]^4$ and its splitting is as follows.
$\therefore$ electron in $\mathrm{e}_{\mathrm{g}}$ orbital is one.
Consider the following low-spin complexes
$ \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right], \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] \text { and } \mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] $
The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer in nearest integer)
Explanation:
Given low-spin complexes:
${K_3}\left[ {Co{{(N{O_2})}_6}} \right] - $ Yellow colour
${K_4}\left[ {Fe{{(CN)}_6}} \right] - $ light yellow colour
${K_3}\left[ {Fe{{(CN)}_6}} \right] - $ bright red colour not yellow.
$C{u_2}\left[ {Fe{{(CN)}_6}} \right] - $ reddish brown color not yellow
$Z{n_2}\left[ {Fe{{(N{O_2})}_6}} \right] - $ bluish-white colour not yellow
${K_3}\left[ {Co{{(N{O_2})}_6}} \right]$
Oxidation state of $Co: + 3$
Configuration:
$Co\,{d^7}\,{s^2}$
$C{o^{3 + }}\,{d^6}$
$3( + 1) + x + 6( - 1) = 0$
$3 + x - 6 = 0$
$x = 6 - 3 = + 3$
For low-spin d$^6$ configuration, no unpaired electrons are present : 
spin-only magnetic moment, $\mu = \sqrt {n(n + 2)} $
$n = 0$, $\mu = 0$
This complex shows yellow colour (bright yellow)
${K_4}\left[ {Fe{{(CN)}_6}} \right]$
Oxidation state of $Fe: + 2$
$4( + 1) + x + 6( - 1) = 0$
$4 + x - 6 = 0$
$x = 6 - 4 = + 2$
Configuration:
$Fe\,{d^6}\,{s^2}$
$F{e^{ + 2}}\,{d^6}$
$\mu = 0$
For low-spin d$^6$ configuration, no unpaired electrons are present : 
This complex shows yellow colour (light yellow)
${K_3}\left[ {Fe{{(CN)}_6}} \right]$
Oxidation state of $Fe: + 3$
$3( + 1) + x + 6( - 1) = 0$
$3 + x - 6 = 0$
$x = 6 - 3 = + 3$
Configuration:
$Fe\,{d^6}\,{s^2}$
$F{e^{ + 2}}\,{d^5}$
low spin d$^5$ 
One unpaired electron is present.
Spin-only magnetic moment, $\mu = \sqrt {n(n + 2)} $
n (number of unpaired electrons) = 1
So, $\mu = \sqrt {n(n + 2)} $
$ = \sqrt {1 \times 3} $
$ = \sqrt 3 $
= 1.732 BM
Colour of the complex is not yellow.
$C{u_2}\left[ {Fe{{(CN)}_6}} \right],\,C{u^{2 + }}$ and ${\left[ {Fe{{(CN)}_6}} \right]^{4 - }}$
$x + 6( - 1) = - 4$
$x = - 4 + 6$
$ = + 2$
Oxidation state of $Fe: + 2$
Configuration:
$Fe\,{d^6}\,{s^2}$
$F{e^{2 + }}\,{d^6}$
For low-spin d$^6$ configuration, no unpaired electrons are present 
$n = 0,\mu = \sqrt {n(n + 2)} = 0$
Colour of the complex is not yellow.
$Z{n_2}\left[ {Fe{{(CN)}_6}} \right]$
${\left[ {Fe{{(CN)}_6}} \right]^{4 - }}$ and $Z{n^{2 + }}$
$x + 6( - 1) = - 4$
$x = - 4 + 6 = + 2$
Oxidation state of $Fe: + 2$
Configuration:
$Fe\,{d^6}\,{s^2}$
$F{e^{2 + }}\,{d^6}$
For low-spin d$^6$ configuration, no unpaired electrons are present.
$n = 0,\,\mu = \sqrt {n(n + 2)} = 0$
Colour of the complex is not yellow
The complexes having yellow colour are ${K_3}\left[ {Co{{(N{O_2})}_6}} \right]$ and ${K_4}\left[ {Fe{{(CN)}_6}} \right]$
The sum of the spin-only magnetic moment values of complexes having yellow colour is 0 (zero).
Explanation:
The compound with the strongest oxidizing power among Mn₂O₃, TiO, and VO is Mn₂O₃. This conclusion is based on the standard electrode potential ($ E^\circ $) value:
$ E_{\text{Mn}^{+3} / \text{Mn}^{+2}}^\circ = +1.57 \, \text{V} $
For Mn³⁺, the electronic configuration is $ d^4 $. The spin-only magnetic moment ($ \mu $) is calculated as follows:
$ \mu = \sqrt{n(n+2)} \, \text{BM} $
Where $ n $ is the number of unpaired electrons. Here, $ n = 4 $:
$ \mu = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM} $
This calculates to approximately 4.89 BM, which rounds to an integer value of 5 BM.
$\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{NO}, \mathrm{NO}_2, \mathrm{CO}, \mathrm{K}_2\left[\mathrm{NiCl}_4\right],\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3, \mathrm{~K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$
Explanation:
To determine which molecules or species are paramagnetic, we need to identify those with unpaired electrons. Here's the analysis:
$\mathrm{O}_2$: This molecule has 2 unpaired electrons as predicted by the Molecular Orbital Theory (MOT), making it paramagnetic.
$\mathrm{O}_2^{+}$: This ion has 1 unpaired electron according to MOT, which also makes it paramagnetic.
$\mathrm{O}_2^{-}$: This ion contains 1 unpaired electron as determined by MOT, indicating it is paramagnetic.
$\mathrm{NO}$: As a species with an odd number of electrons, $\mathrm{NO}$ has unpaired electrons, making it paramagnetic.
$\mathrm{NO}_2$: This molecule is also an odd-electron species, which means it has unpaired electrons and is paramagnetic.
$\mathrm{K}_2\left[\mathrm{NiCl}_4\right]$: In this compound, $\mathrm{Ni}^{2+}$ (nickel with a +2 oxidation state) has an electron configuration of $3d^8$. Given the presence of weak field ligands and a coordination number of 4, this forms a tetrahedral complex. As a result, it is paramagnetic with 2 unpaired electrons.
Thus, the paramagnetic species in the provided list are $\mathrm{O}_2$, $\mathrm{O}_2^{+}$, $\mathrm{O}_2^{-}$, $\mathrm{NO}$, $\mathrm{NO}_2$, and $\mathrm{K}_2\left[\mathrm{NiCl}_4\right]$.
The complex of $\mathrm{Ni}^{2+}$ ion and dimethyl glyoxime contains __________ number of Hydrogen (H) atoms.
Explanation:
$\mathrm{Ni}^{2+}$ with $(\mathrm{dmg})$ forms $\left[\mathrm{Ni}(\mathrm{dmg})_2\right]^{2+}$ having 2 H -Bonds as shown :
The no. of H atoms $=14$
Consider the following test for a group-IV cation.
$\mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{A} \text { (Black precipitate)+ byproduct }$
$\mathrm{A}+\text { aqua regia } \rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}$
$\mathrm{B}+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{C}+\text { byproduct }$
The spin-only magnetic moment value of the metal complex $\mathrm{C}$ is _________ $\mathrm{BM}$ (Nearest integer)
Explanation:
$\begin{aligned} & \text { In } \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right], \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 4 \mathrm{~s}^0 \\ & \mathrm{Co}^{3+}: \mathrm{d}^2 \mathrm{sp}^3 \text { Hybridisation } \\ & \text { Number of unpaired } \mathrm{e}^{-}=0 \\ & \text { Magnetic moment }=\sqrt{n(n+2)}=0 \text { B.M } \\ & \end{aligned}$
Number of ambidentate ligands among the following is _________.
$\mathrm{NO}_2^{-}, \mathrm{SCN}^{-}, \mathrm{C}_2 \mathrm{O}_4^{2-}, \mathrm{NH}_3, \mathrm{CN}^{-}, \mathrm{SO}_4^{2-}, \mathrm{H}_2 \mathrm{O} \text {. }$
Explanation:
Ambidentate ligands are ligands that can attach to a central metal atom through two different atoms. Let’s analyze the given ligands one by one:
- $\mathrm{NO}_2^{-}$: This ligand can bind through either the nitrogen (N) or the oxygen (O). Therefore, $\mathrm{NO}_2^{-}$ is an ambidentate ligand.
- $\mathrm{SCN}^{-}$: This ligand can bind through either the sulfur (S) or the nitrogen (N). Therefore, $\mathrm{SCN}^{-}$ is an ambidentate ligand.
- $\mathrm{C}_2\mathrm{O}_4^{2-}$ (oxalate): This is a bidentate ligand, but it coordinates through both oxygen atoms only. Therefore, it is not an ambidentate ligand.
- $\mathrm{NH}_3$: This ligand can only bind through nitrogen (N). Therefore, it is not an ambidentate ligand.
- $\mathrm{CN}^{-}$: This ligand can bind through either carbon (C) or nitrogen (N). Therefore, $\mathrm{CN}^{-}$ is an ambidentate ligand.
- $\mathrm{SO}_4^{2-}$ (sulfate): This ligand typically coordinates through oxygen atoms. It is not considered an ambidentate ligand.
- $\mathrm{H}_2\mathrm{O}$: This ligand can only bind through oxygen (O). Therefore, it is not an ambidentate ligand.
Based on this analysis, the ambidentate ligands are $\mathrm{NO}_2^{-}$, $\mathrm{SCN}^{-}$, and $\mathrm{CN}^{-}$.
Therefore, the number of ambidentate ligands among the given options is 3.
Total number of unpaired electrons in the complex ions $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ and $[\mathrm{NiCl}_4]^{2-}$ is ________.
Explanation:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}: \mathrm{Co}^{3+}: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^0: \mathrm{n}=0$
$\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{2+}: t_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^2: \mathrm{n}=2$
The 'spin only' magnetic moment value of $\mathrm{MO}_4{ }^{2-}$ is ________ BM. (Where M is a metal having least metallic radii. among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Zn}$ ).
(Given atomic number: $\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25$ and $\mathrm{Zn}=30$)
Explanation:
To determine the 'spin only' magnetic moment of the complex $\mathrm{MO}_4^{2-}$, we first need to identify the metal $\mathrm{M}$.
Step 1: Identify the metal M based on metallic radii.
The problem states that $\mathrm{M}$ has the least metallic radii among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$, and $\mathrm{Zn}$.
The trend for metallic radii across the first transition series (3d series) generally decreases from $\mathrm{Sc}$ to $\mathrm{Cr/Mn}$, stays relatively constant for $\mathrm{Fe, Co, Ni}$, and then increases for $\mathrm{Cu}$ and $\mathrm{Zn}$. However, $\mathrm{Mn}$ has a complex crystal structure leading to a metallic radius that is anomalously high compared to its neighbors. The actual values (in pm) are approximately:
$\mathrm{Sc}$: 164
$\mathrm{Ti}$: 147
$\mathrm{V}$: 135
$\mathrm{Cr}$: 129
$\mathrm{Mn}$: 137 (anomalously high)
$\mathrm{Zn}$: 137 (larger due to full d-subshell shielding)
Among the options provided ($\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}, \mathrm{Zn}$), Chromium ($\mathrm{Cr}$) has the smallest metallic radius.
Therefore, $\mathrm{M} = \mathrm{Cr}$.
Step 2: Determine the oxidation state of Cr in the complex.
The complex is $\mathrm{CrO}_4^{2-}$ (Chromate ion).
Let the oxidation state of $\mathrm{Cr}$ be $x$.
The oxidation state of Oxygen ($\mathrm{O}$) is generally $-2$.
The sum of oxidation states equals the charge on the ion:
$x + 4(-2) = -2$
$x - 8 = -2$
$x = +6$
So, Chromium is in the $+6$ oxidation state ($\mathrm{Cr}^{6+}$).
Step 3: Determine the electronic configuration.
The atomic number of Chromium ($\mathrm{Cr}$) is 24.
Ground state configuration of $\mathrm{Cr}$: $[\mathrm{Ar}] 3d^5 4s^1$
To form $\mathrm{Cr}^{6+}$, we remove the outer 6 electrons (one from $4s$ and five from $3d$).
Configuration of $\mathrm{Cr}^{6+}$: $[\mathrm{Ar}] 3d^0 4s^0$
Step 4: Calculate the number of unpaired electrons.
In the $3d^0$ configuration, there are no electrons in the d-orbitals.
Number of unpaired electrons ($n$) = $0$.
Step 5: Calculate the spin-only magnetic moment.
The formula for spin-only magnetic moment ($\mu$) is:
$\mu = \sqrt{n(n+2)} \text{ BM}$
Substituting $n = 0$:
$\mu = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ BM}$
Conclusion:
The 'spin only' magnetic moment value is 0.
Answer: 0
The difference in the 'spin-only' magnetic moment values of $\mathrm{KMnO}_4$ and the manganese product formed during titration of $\mathrm{KMnO}_4$ against oxalic acid in acidic medium is ________ $\mathrm{BM}$. (nearest integer)
Explanation:
The 'spin-only' magnetic moment for $\mathrm{Mn}^{7+}$ is $0$ BM. During the titration of $\mathrm{KMnO}_4$ with oxalic acid in an acidic medium, manganese is reduced to $\mathrm{Mn}^{2+}$, which has a 'spin-only' magnetic moment of $5.91$ BM.
Thus, the difference in magnetic moment values is calculated as follows:
$0 \, \text{BM}$ (for $\mathrm{Mn}^{7+}$) subtracted from $5.91 \, \text{BM}$ (for $\mathrm{Mn}^{2+}$), which results in $5.91 \, \text{BM}$.
Rounding this to the nearest integer, we get a difference of approximately $6$ BM.
The spin-only magnetic moment value of the ion among $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Co}^{3+}$ and $\mathrm{Cr}^{2+}$, that acts as strong oxidising agent in aqueous solution is _________ BM (Near integer).
(Given atomic numbers : $\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Co}: 27$)
Explanation:
The ion which acts as strong oxidising agent in aqueous solution is $\mathrm{Cr}^{2+}:[\mathrm{Ar}] 4 s^{\circ} 3 d^4$
$\mu=\sqrt{4(4+2)}=4.89 \Rightarrow 5$
The 'Spin only' Magnetic moment for $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ is _________ $\times 10^{-1} \mathrm{~BM}$. (given $=$ Atomic number of $\mathrm{Ni}: 28$)
Explanation:
$\mathrm{NH}_3$ act as WFL with $\mathrm{Ni}^{2+}$
$\mathrm{Ni}^{2+}=3 \mathrm{~d}^8$
No. of unpaired electron $=2$
$\begin{aligned} \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} & =\sqrt{8}=2.82 \mathrm{~BM} \\ & =28.2 \times 10^{-1} \mathrm{~BM} \\ \mathrm{x} & =28 \end{aligned}$
Number of complexes which show optical isomerism among the following is ________.
$\text { cis- }\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-},\left[\mathrm{Co}(\text {en})_3\right]^{3+}, \text { cis- }\left[\mathrm{Pt}(\text {en})_2 \mathrm{Cl}_2\right]^{2+}, \text { cis- }\left[\mathrm{Co}(\text {en})_2 \mathrm{Cl}_2\right]^{+}, \text {trans- }\left[\mathrm{Pt}(\text {en})_2 \mathrm{Cl}_2\right]^{2+}, \text { trans- }\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-}$
Explanation:
cis $-\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-} \rightarrow$ can show optical isomerism (no POS & COS)
$\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+} \rightarrow \text { can show (no POS & COS) }$
$\text { cis }-\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+} \rightarrow$ can show (no POS & COS)
$\text { cis }-\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+} \rightarrow$ can show (no POS & COS)
$\operatorname{trans}-\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+} \rightarrow$ can't show (contains POS & COS)
$\text { trans }-\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-} \rightarrow$ can't show (contains POS & COS)
The Spin only magnetic moment value of square planar complex $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}$ is _________ B.M. (Nearest integer)
(Given atomic number for $\mathrm{Pt}=78$)
Explanation:
$\mathrm{Pt^2+ (d^8)}$
$\mathrm{Pt}^{2+} \rightarrow \mathrm{dsp}^2$ hybridization and have no unpaired $\mathrm{e}^{-} \mathrm{s}$.
$\therefore$ Magnetic moment $=0$
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 $=7$
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 $=9$