Basic Terms Questions in English

(C) Secondary valency corresponds to the coordination number of the central metal atom.
The formula of the given complex is $[Pt(NH_{3})_{4}Cl_{2}]Cl_{2}$.
In this complex,the central metal atom $Pt$ is bonded to $4$ ammonia ligands and $2$ chloride ligands.
Therefore,the coordination number of $Pt = 4 + 2 = 6$.
$\therefore$ The secondary valency of $Pt$ is $6$.

(C) ligand is a chemical species that is capable of donating one or more electron pairs to a central metal ion to form a coordinate bond.
According to the Lewis theory of acids and bases,a substance that acts as an electron pair donor is defined as a Lewis base.
Therefore,a ligand is considered a Lewis base.

(C) The coordination number of a central metal ion in a complex is defined as the number of ligand donor atoms to which the metal is directly bonded. For unidentate ligands,the coordination number is equal to the number of ligands attached.

(D) The chemical formula of hexamineplatinum$(IV)$ chloride is $[Pt(NH_3)_6]Cl_4$.
Upon ionization in an aqueous solution,it dissociates as follows:
$[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$.
This results in one complex cation $[Pt(NH_3)_6]^{4+}$ and four chloride anions $4Cl^-$.
Total number of ions = $1 + 4 = 5$ ions.

(C) Given,Molarity of the complex $= 0.1 \ M$.
Volume of the complex $= 100 \ mL$.
Mass of $AgCl$ obtained $= 2.86 \ g$.
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.
Number of moles of complex $= \frac{0.1 \times 100}{1000} = 0.01 \ mol$.
Number of moles of $AgCl$ precipitated $= \frac{2.86 \ g}{143.5 \ g/mol} \approx 0.02 \ mol$.
Since $1 \ mol$ of complex gives $2 \ mol$ of $AgCl$,there must be $2$ ionizable $Cl^-$ ions outside the coordination sphere.
Therefore,the complex is $[Cr(H_2 O)_5 Cl] Cl_2 \cdot H_2 O$.

(A) The reaction of $1$ mole of $PdCl_2 \cdot 4 NH_3$ with excess $AgNO_3$ produces $2$ moles of $AgCl$ precipitate,which indicates that $2$ chloride ions $(Cl^-)$ are present outside the coordination sphere.
Thus,the complex can be formulated as $[Pd(NH_3)_4]Cl_2$.
In aqueous solution,it dissociates as: $[Pd(NH_3)_4]Cl_2 \rightarrow [Pd(NH_3)_4]^{2+} + 2Cl^-$.
This produces $1$ cation and $2$ anions,making it a $1:2$ electrolyte.

(A) The complex is $[M(en)_{2}(C_{2}O_{4})]NO_{2}$.
In this complex,the counter ion is $NO_{2}^{-}$,which has a charge of $-1$. Therefore,the complex ion $[M(en)_{2}(C_{2}O_{4})]$ must have a charge of $+1$.
Let the oxidation state of $M$ be $x$.
$en$ (ethane-$1,2$-diamine) is a neutral bidentate ligand (charge $= 0$).
$C_{2}O_{4}^{2-}$ (oxalate) is a bidentate ligand with a charge of $-2$.
Setting up the equation for the charge of the complex ion:
$x + (2 \times 0) + (1 \times -2) = +1$
$x - 2 = +1$
$x = +3$
Thus,the oxidation state of $M$ is $+3$.
Coordination number calculation:
$en$ is bidentate ($2$ donor sites each,$2 \times 2 = 4$ sites).
$C_{2}O_{4}^{2-}$ is bidentate ($2$ donor sites).
Total coordination number $= 4 + 2 = 6$.
Therefore,the coordination number is $6$ and the oxidation state is $3$.

(A) The formula of the complex is $[Cr(H_2O)_5Cl]Cl_2$.
Each mole of the complex releases $2 \ mol$ of ionizable $Cl^-$ ions.
Number of moles of complex = $Molarity \times Volume (L) = 0.01 \ M \times 0.1 \ L = 0.001 \ mol$.
Number of moles of $Cl^-$ ions = $2 \times 0.001 \ mol = 0.002 \ mol$.
Reaction: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$.
Moles of $AgCl$ formed = $0.002 \ mol$.
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.
Mass of $AgCl = 0.002 \ mol \times 143.5 \ g/mol = 0.287 \ g = 287 \times 10^{-3} \ g$.

(B) $CrCl_3 \cdot 6 H_2 O$ exists as $[Cr(H_2 O)_6]Cl_3$,where $6$ water molecules are coordinated to the $Cr^{3+}$ ion.
$BaCl_2 \cdot 2 H_2 O$ is a hydrated salt where water molecules are present in the crystal lattice but not coordinated to the metal ion.
$CuSO_4 \cdot 5 H_2 O$ exists as $[Cu(H_2 O)_4]SO_4 \cdot H_2 O$,where $4$ water molecules are coordinated to the $Cu^{2+}$ ion.
Therefore,compounds $I$ and $III$ contain coordinated water.

(D) Element '$X$' has atomic number $13$,which corresponds to Aluminum $(Al)$.
In the complex $[XCl(H_2O)_5]^{2+}$,the central metal atom $X$ is bonded to $1$ chloride ion and $5$ water molecules.
Since both $Cl^-$ and $H_2O$ are unidentate ligands,the total number of coordinate bonds formed by $X$ is $1 + 5 = 6$. Thus,the covalency is $6$.
To find the oxidation state $(x)$: $x + (-1) + 5(0) = +2$.
$x - 1 = +2$.
$x = +3$.
Therefore,the covalency is $6$ and the oxidation state is $+3$.

(B) The given compound is Prussian blue,which is $Fe_4[Fe(CN)_6]_3$.
Comparing this with the general formula $Fe_x[Fe_y(CN)_6]_3$,we can identify the values of $x$ and $y$.
Here,$x = 4$ and $y = 1$.
Thus,the values of $x$ and $y$ are $4$ and $1$ respectively.

(C) The coordination number is defined as the number of ligand donor atoms to which the metal is directly bonded.
In the complex $[Cr(H_2O)_2(C_2O_4)_2]^-$,the ligands are:
$1$. Two $H_2O$ molecules,which are monodentate ligands (each donates $1$ pair of electrons).
$2$. Two $C_2O_4^{2-}$ (oxalate) ions,which are bidentate ligands (each donates $2$ pairs of electrons).
Calculation: $(2 \times 1) + (2 \times 2) = 2 + 4 = 6$.
Therefore,the coordination number of $Cr$ is $6$.

(B) An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms. Examples include $NO_2^{-}$,$CN^{-}$,and $SCN^{-}$.
In option $A$,$NO_2^{-}$ and $CN^{-}$ are ambidentate.
In option $B$,$C_2O_4^{2-}$ (oxalate) is a bidentate ligand,$H_2O$ is a monodentate ligand,and $SO_4^{2-}$ is a bidentate ligand. None of these are ambidentate.
In option $C$,$SCN^{-}$ is ambidentate.
In option $D$,$CN^{-}$ and $SCN^{-}$ are ambidentate.
Therefore,the set containing no ambidentate ligands is $C_2O_4^{2-}, H_2O, SO_4^{2-}$.

(C) Homoleptic complexes are those in which the central metal atom or ion is bonded to only one type of donor atom or ligand.
In the complex $[Co(NH_3)_6]^{3\oplus}$,all six ligands are ammonia $(NH_3)$ molecules,making it a homoleptic complex.
In the other options,the central metal ion is bonded to different types of ligands (e.g.,$NH_3$ and $Br^-$,$C_2O_4^{2-}$ and $NH_3$,or $CN^-$ and $Cl^-$),making them heteroleptic complexes.

(B) Ligands that possess two or more donor atoms but can coordinate through only one donor atom at a time are called ambidentate ligands.
$NO_2^{\ominus}$ can coordinate through $N$ (nitro) or $O$ (nitrito).
$CN^{\ominus}$ can coordinate through $C$ (cyano) or $N$ (isocyano).
$SCN^{\ominus}$ can coordinate through $S$ (thiocyanato) or $N$ (isothiocyanato).
Therefore,the set containing only ambidentate ligands is $NO_2^{\ominus}, CN^{\ominus}, SCN^{\ominus}$.

(D) The coordination number is the total number of coordinate bonds formed between the central metal ion and the ligands attached to it.
In the complex $[Co(en)_2Cl_2]$,the ligands are ethylenediamine $(en)$ and chloride ions $(Cl^-)$.
Ethylenediamine $(en)$ is a bidentate ligand,meaning each $en$ molecule forms $2$ coordinate bonds. Thus,$2$ $en$ molecules form $2 \times 2 = 4$ bonds.
Chloride $(Cl^-)$ is a monodentate ligand,meaning each $Cl^-$ ion forms $1$ coordinate bond. Thus,$2$ $Cl^-$ ions form $2 \times 1 = 2$ bonds.
The total coordination number is $4 + 2 = 6$.

(A) To find the oxidation state of $Al$ in $[AlCl(H_2O)_5]^{2+}$,let the oxidation state be $x$.
The charge on $Cl$ is $-1$ and $H_2O$ is a neutral ligand $(0)$.
$x + (-1) + 5(0) = +2$
$x - 1 = +2$
$x = +3$.
The covalency is defined as the total number of coordinate bonds formed by the central metal atom with the ligands.
Here,$Al$ is bonded to $5$ $H_2O$ molecules and $1$ $Cl^-$ ion.
Total covalency = $5 + 1 = 6$.
Therefore,the oxidation state is $+3$ and the covalency is $6$.

(D) In the complex $[AlCl(H_2O)_5]^{2+}$,let the oxidation state of $Al$ be $x$.
The oxidation state of $Cl$ is $-1$ and $H_2O$ is $0$.
So,$x + (-1) + 5(0) = +2$.
$x - 1 = +2$,which gives $x = +3$.
Thus,the oxidation state of $Al$ is $+3$.
The covalency (coordination number) is the total number of coordinate bonds formed by the central metal atom with the ligands.
Here,$1$ $Cl^-$ ion and $5$ $H_2O$ molecules are attached to $Al$,so the coordination number is $1 + 5 = 6$.

(B) The reaction of $AgNO_3$ with a coordination compound precipitates the chloride ions present outside the coordination sphere as $AgCl$.
Since $1 \ mole$ of $AgCl$ is formed from $1 \ mole$ of the complex,there must be only $1 \ Cl^-$ ion outside the coordination sphere.
The formula of the complex can be written as $[Co(NH_3)_X Cl_2]Cl$.
For $Co(III)$,the coordination number is $6$.
Therefore,$X + 2 = 6$,which gives $X = 4$.

(D) The oxidation states of the transition metals in the given compounds are calculated as follows:

Compound	Calculation of Oxidation State
$K_2Cr_2O_7$	$2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$
$KMnO_4$	$(+1) + x + 4(-2) = 0 \implies 1 + x - 8 = 0 \implies x = +7$
$CrO_5$	$x + 5(-2) = 0 \implies x = +10$ (Note: Due to peroxide linkages,the actual oxidation state is $+6$,but mathematically $x=+10$ based on standard rules).
$Fe(CO)_5$	$x + 5(0) = 0 \implies x = 0$ (Since $CO$ is a neutral ligand).

Therefore,the transition metal in $Fe(CO)_5$ has a zero oxidation state.

(B) The reaction of potassium cyanide with sodium hydroxide gives the cyanide ion. After that,the cyanide ion reacts with thiosulphate ions. Upon cooling and acidifying with hydrochloric acid,it forms the thiocyanate ion.
$KCN + NaOH \rightarrow NaCN + KOH$
$S_2O_3^{2-} + CN^{-} \xrightarrow{HCl} SCN^{-} + SO_3^{2-}$
The thiocyanate ion reacts with iron $(III)$ chloride to form the complex $[Fe(SCN)]^{2+}$,which has a blood-red colour.
$Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$ (blood-red colour)
Therefore,the solution produces a blood-red colour solution.

(C) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$CN^{-}$ can coordinate through $C$ or $N$.
$SCN^{-}$ can coordinate through $S$ or $N$.
$NO_2^{-}$ can coordinate through $N$ or $O$.
$SO_4^{2-}$ is a bidentate ligand that coordinates through two oxygen atoms simultaneously,but it does not have two different donor atoms to act as an ambidentate ligand.
Therefore,$SO_4^{2-}$ is not an ambidentate ligand.

(A) The reaction of the complex with $AgNO_3$ precipitates $Cl^-$ ions present outside the coordination sphere as $AgCl$.
Since $1$ mole of the complex yields $3$ moles of $AgCl$,there must be $3$ chloride ions outside the coordination sphere.
The formula $[Co(H_2O)_6]Cl_3$ dissociates as:
$[Co(H_2O)_6]Cl_3(aq) \rightarrow [Co(H_2O)_6]^{3+}(aq) + 3Cl^-(aq)$
The $3Cl^-$ ions react with $AgNO_3$ to form $3$ moles of $AgCl$:
$3Ag^+(aq) + 3Cl^-(aq) \rightarrow 3AgCl(s)$
Thus,the correct formula is $[Co(H_2O)_6]Cl_3$.

(D) The secondary valence of a metal complex is equal to its coordination number ($C$.$N$.),which is the total number of ligand donor atoms directly bonded to the central metal ion.
$1$. For $(I) \ CoCl_3 \cdot 6 H_2O$:
Since $3$ moles of $AgCl$ are precipitated,all $3$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Co(H_2O)_6]Cl_3$. The coordination number is $6$ (six $H_2O$ ligands).
$2$. For $(II) \ NiCl_3 \cdot 6 H_2O$:
Since $2$ moles of $AgCl$ are precipitated,$2$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Ni(H_2O)_5Cl]Cl_2 \cdot H_2O$. The coordination number is $6$ (five $H_2O$ and one $Cl^-$ ligand).
$3$. For $(III) \ Co(SO_4)Br \cdot 5 NH_3$:
Since $1$ mole of $AgBr$ is precipitated,the $Br^-$ ion is outside the coordination sphere. The formula is $[Co(NH_3)_5(SO_4)]Br$. The coordination number is $6$ (five $NH_3$ and one $SO_4^{2-}$ ligand).
Thus,the secondary valences for $(I), (II),$ and $(III)$ are $6, 6, 6$ respectively.

(C) The coordination number is the total number of coordinate bonds formed by the ligands with the central metal atom.
In the complex $K_3[Cr(C_2O_4)_3]$,the ligand is oxalate ion $(C_2O_4^{2-})$,which is a bidentate ligand.
Since there are $3$ oxalate ligands,the coordination number $= 3 \times 2 = 6$.
To find the oxidation state of $Cr$,let it be $x$.
The oxidation state of $K$ is $+1$ and the oxidation state of $C_2O_4^{2-}$ is $-2$.
For the neutral complex $K_3[Cr(C_2O_4)_3]$:
$3(+1) + x + 3(-2) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$.
Therefore,the coordination number is $6$ and the oxidation state is $+3$. The correct option is $(C)$.

(B) The coordination number of a central metal ion is defined as the number of ligand donor atoms directly bonded to it.
$1$. In $[Fe(CN)_6]^{4-}$,there are $6$ $CN^-$ ligands,each being monodentate. Thus,the coordination number is $6$.
$2$. In $[Fe(CN)_6]^{3-}$,there are $6$ $CN^-$ ligands,each being monodentate. Thus,the coordination number is $6$.
$3$. In $[FeCl_4]^-$,there are $4$ $Cl^-$ ligands,each being monodentate. Thus,the coordination number is $4$.
Therefore,the coordination numbers are $6, 6, 4$.

(C) In the complex $[Ni(CO)_4]$,$CO$ is a neutral ligand.
Let the oxidation state of $Ni$ be $x$.
$x + 4(0) = 0$
$x = 0$
Thus,in metal carbonyls,the metal is present in its zero-valent state.

(D) The given compound is a coordination compound.
In the complex ion $[AlCl(H_2O)_5]^{2+}$,the central metal atom is $Al$.
The ligands attached to the central metal atom are $1$ $Cl^-$ ion and $5$ $H_2O$ molecules.
The total number of coordinate bonds formed by the central metal atom with the ligands is $1 + 5 = 6$.
Thus,the coordination number (covalency) of $Al$ is $6$.

(B) Ethylenediamine is an organic compound with the formula $NH_2-CH_2-CH_2-NH_2$.
It acts as a bidentate ligand because it has two nitrogen atoms,each possessing a lone pair of electrons that can coordinate with a central metal ion.
Therefore,it can occupy two coordination positions in the coordination polyhedron.

(B) For the complex $[AlCl(H_2O)_5]^{2+}$:
Let the oxidation state of $Al$ be $x$.
$x + (-1) + 5(0) = +2$
$x = +3$
The covalency (coordination number) is the total number of ligands attached to the central metal atom.
Here,$1$ $Cl^-$ ion and $5$ $H_2O$ molecules are attached.
Total covalency $= 1 + 5 = 6$
Therefore,the oxidation state is $+3$ and the covalency is $6$.

(B) When the complex is dissolved in water,it dissociates to give $3$ ions. This indicates that there are $2$ $Cl^{-}$ ions present outside the coordination sphere as counter ions.
Therefore,the formula of the complex must be $[Co(NH_3)_5Cl]Cl_2$.
The dissociation reaction is:
$[Co(NH_3)_5Cl]Cl_2 \xrightarrow{H_2O} [Co(NH_3)_5Cl]^{2+} + 2Cl^{-}$

(B) The central metal ion is $Co^{3+}$ and the coordination number of $Co^{3+}$ is typically $6$. The molecular formula is $CoCl_3 \cdot yNH_3$ because $Co$ is in the $+3$ oxidation state.
When the complex dissolves in water,it produces $3$ ions. This implies the ionization sphere contains $2$ chloride ions,represented as $[CoCl(NH_3)_5]Cl_2$.
The dissociation is: $[CoCl(NH_3)_5]Cl_2 \rightleftharpoons [CoCl(NH_3)_5]^{2+} + 2Cl^-$.
In this structure,the $Cl^-$ ion inside the coordination sphere acts as a ligand (satisfying secondary valency) and also balances the charge of the metal (satisfying primary valency).
Thus,only $1$ $Cl^-$ ion satisfies both primary and secondary valencies.

(C) According to Werner's theory,the secondary valency of a central metal atom or ion in a coordination complex represents the number of ligands (neutral or negatively charged groups) directly bonded to it.
This is also known as the coordination number.
The primary valency corresponds to the oxidation state of the metal atom or ion.

(C) When aluminium chloride $(AlCl_3)$ is dissolved in water,it undergoes hydration to form the octahedral complex ion $[Al(H_2O)_6]^{3+}$ and releases chloride ions $(Cl^-)$.
The reaction is: $AlCl_3 + 6H_2O \rightarrow [Al(H_2O)_6]^{3+} + 3Cl^-$.
In this complex,the $Al^{3+}$ ion is coordinated to six water molecules,and the hybridization of $Al$ is $sp^3d^2$.

(A) In the brown ring complex $[Fe(H_2O)_5(NO)]SO_4$,the iron is in the $+1$ oxidation state $(Fe^{+})$.
To balance the charge of the complex,the nitric oxide ligand $(NO)$ acts as a nitrosonium ion $(NO^{+})$.
Thus,the oxidation state of $Fe$ is $+1$ and $NO$ is $+1$.

(B) Sodium nitroprusside is a coordination compound with the chemical formula $Na_2[Fe(CN)_5 NO]$.
It is commonly used in analytical chemistry as a reagent for the detection of sulfide ions $(S^{2-})$ and as a vasodilator in medicine.

(C) Mohr's salt is a double salt with the formula $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.
Upon complete dissociation in water,it ionizes as follows:
$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O \rightarrow Fe^{2+} + 2NH_4^{+} + 2SO_4^{2-} + 6H_2O$.
Counting the ions produced: $1$ mole of $Fe^{2+}$,$2$ moles of $NH_4^{+}$,and $2$ moles of $SO_4^{2-}$.
Total moles of ions = $1 + 2 + 2 = 5$ moles.

(A) The chemical formula for potassium trioxalatoaluminate $(III)$ is $K_3[Al(C_2O_4)_3]$.
$1$. The primary valency corresponds to the oxidation state of the central metal atom $(Al)$. In this complex,$Al + 3(-2) = -3$,which gives $Al = +3$. Thus,the primary valency is $3$.
$2$. The secondary valency corresponds to the coordination number of the central metal atom. The oxalate ion $(C_2O_4^{2-})$ is a bidentate ligand. Since there are $3$ oxalate ligands,the coordination number is $3 \times 2 = 6$. Thus,the secondary valency is $6$.
Therefore,the primary and secondary valencies are $3$ and $6$ respectively.

(A) The chemical formula of tetraaquadichloridochromium$(III)$ chloride is $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$.
Upon ionization in water,the complex dissociates as: $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^- + 2H_2O$.
Only the chloride ion outside the coordination sphere reacts with $AgNO_3$ to form $AgCl$ precipitate.
Number of moles of the complex = $\text{Molarity} \times \text{Volume (in L)} = 0.05 \ M \times 0.1 \ L = 0.005 \ \text{moles}$.
Since $1 \ \text{mole}$ of the complex releases $1 \ \text{mole}$ of $Cl^-$,the moles of $AgCl$ precipitated = $0.005 \ \text{moles}$.
Converting to the required format: $0.005 = 5 \times 10^{-3} \ \text{moles}$.
Thus,the answer is $5$.

(C) An ambidentate ligand is a ligand that can coordinate through two different atoms.
Thiocyanate $(SCN^{-})$ is an ambidentate ligand because it can coordinate through either the sulphur atom (thiocyanato-$S$) or the nitrogen atom (isothiocyanato-$N$).
Oxalate,ethane$-1,2-$diamine,and ethylenediaminetetraacetate ion are polydentate ligands but are not ambidentate.