Basic Terms Questions in English

(A) . Metal carbonyls,such as $Ni(CO)_4$ or $Fe(CO)_5$,contain the neutral carbonyl ligand $(CO)$.
Since the ligand $CO$ is neutral,the oxidation state of the central metal atom in these complexes is $0$.

(B) The complex ion is given as $[Ag(CN)_2]^-$.
The superscript outside the square brackets represents the net charge on the coordination entity.
Therefore,the charge on the $[Ag(CN)_2]^-$ complex is $-1$.

(B) The cuprammonium ion is represented as $[Cu(NH_3)_4]^{2+}$.
In this complex ion,the central metal atom is copper $(Cu)$ and the ligand is ammonia $(NH_3)$.
Since $NH_3$ is a neutral ligand,the charge on the complex ion is equal to the oxidation state of the central metal ion.
Therefore,the valency (oxidation state) of copper in the cuprammonium ion is $+2$.

(C) To determine the oxidation state of the central metal atom,we set the sum of oxidation states equal to the charge of the complex:
$(A)$ For $[Cu(NH_3)_4]SO_4$:
$x + 4(0) - 2 = 0 \implies x = +2$
$(B)$ For $[Pt(NH_3)_2Cl_2]$:
$x + 2(0) + 2(-1) = 0 \implies x = +2$
$(C)$ For $[Ni(CO)_4]$:
$x + 4(0) = 0 \implies x = 0$
$(D)$ For $K_3[Fe(CN)_6]$:
$3(+1) + x + 6(-1) = 0 \implies 3 + x - 6 = 0 \implies x = +3$
Since the oxidation state of $Ni$ in $[Ni(CO)_4]$ is $0$,it is a zero-valent metal complex. Therefore,option $C$ is correct.

(B) The complex compound $[Co(NH_3)_6]Cl_3$ dissociates in an aqueous solution as follows:
$[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$
From the dissociation,we get $1$ complex cation $[Co(NH_3)_6]^{3+}$ and $3$ chloride anions $Cl^-$.
Total number of ions produced = $1 + 3 = 4$ ions.

(C) The chemical formula for pentamminechlorocobalt$(III)$ chloride is $[Co(NH_3)_5Cl]Cl_2$.
Upon dissolution in water,it dissociates as follows:
$[Co(NH_3)_5Cl]Cl_2 \rightleftharpoons [Co(NH_3)_5Cl]^{2+} + 2Cl^{-}$.
This results in one complex cation $[Co(NH_3)_5Cl]^{2+}$ and two chloride anions $2Cl^{-}$,totaling $3$ ions.

(C) white precipitate with $BaCl_2$ is formed due to the presence of free sulfate ions $(SO_4^{2-})$ in the solution.
In the complex $[Cr(NH_3)_5Cl]SO_4$,the $SO_4^{2-}$ ion is present outside the coordination sphere as a counter-ion.
When dissolved in water,it dissociates as follows: $[Cr(NH_3)_5Cl]SO_4 \to [Cr(NH_3)_5Cl]^{2+} + SO_4^{2-}$.
The free $SO_4^{2-}$ ions react with $Ba^{2+}$ ions from $BaCl_2$ to form a white precipitate of $BaSO_4$: $Ba^{2+} + SO_4^{2-} \to BaSO_4 \downarrow$ (white precipitate).

(D) The given coordination compound is $[Pt(NH_3)Cl_2Br]Cl$.
In an aqueous solution,the complex dissociates as follows:
$[Pt(NH_3)Cl_2Br]Cl \rightleftharpoons [Pt(NH_3)Cl_2Br]^+ + Cl^-$.
Only the $Cl^-$ ion present outside the coordination sphere is ionizable and can be precipitated by adding silver nitrate $(AgNO_3)$.
Therefore,the number of precipitable halide ions is $1$.

(C) The complex $[Co(NH_3)_5Cl]Cl_2$ ionizes in an aqueous solution as follows:
$[Co(NH_3)_5Cl]Cl_2(aq) \rightarrow [Co(NH_3)_5Cl]^{2+}(aq) + 2Cl^{-}(aq)$
From the dissociation,we get $1$ complex cation $[Co(NH_3)_5Cl]^{2+}$ and $2$ chloride anions $Cl^{-}$.
Therefore,the total number of ions produced per mole of the complex is $1 + 2 = 3$ ions.

(A) The complex $Co(NH_3)_5Cl_3$ reacts with $2$ moles of $AgNO_3$ to form $2$ moles of $AgCl_{(s)}$,which indicates that there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.
Therefore,the formula is $[Co(NH_3)_5Cl]Cl_2$.
Upon dissolution in water,it dissociates as: $[Co(NH_3)_5Cl]Cl_2 \rightleftharpoons [Co(NH_3)_5Cl]^{2+} + 2Cl^-$.
This produces $1$ complex cation and $2$ chloride anions,totaling $3$ moles of ions per mole of the complex.

(D) In the complex $K_4[Fe(CN)_6]$,the following types of bonds are present:
$1$. Ionic bond: Exists between the $K^+$ ions and the $[Fe(CN)_6]^{4-}$ complex ion.
$2$. Covalent bond: Exists between the carbon and nitrogen atoms within the cyanide $(CN^-)$ ligand ($C \equiv N$ bond).
$3$. Co-ordinate covalent bond (dative bond): Exists between the lone pair donor carbon atom of the $CN^-$ ligand and the central metal ion $Fe^{2+}$,forming the coordination bond.

(C) Ammine $(NH_3)$ is a neutral ligand because it has no net electrical charge.

(B) Ligands that possess two or more donor atoms and can bind to the central metal ion through more than one atom simultaneously are known as $Polydentate$ ligands.
If a ligand has only one donor atom,it is called a $monodentate$ ligand.
$Ambident$ ligands are those that have two donor atoms but can only bind through one at a time.

(B) $CO$ (Carbon monoxide) is a $\pi$-acid ligand because it possesses empty $\pi^*$ antibonding molecular orbitals that can accept electron density from the filled $d$-orbitals of the metal atom,forming a $M \to L$ $\pi$-backbond.

(A) The molecular formula is $Co(NH_3)_5(NO_2)Cl_2$.
When this complex is dissolved in water,it dissociates to produce $3$ moles of ions.
The reaction with excess $AgNO_3$ confirms the presence of $2$ chloride ions outside the coordination sphere.
The dissociation is: $[Co(NH_3)_5(NO_2)]Cl_2(aq) \to [Co(NH_3)_5(NO_2)]^{2+}(aq) + 2Cl^-(aq)$.
Thus,the correct formula is $[Co(NH_3)_5(NO_2)]Cl_2$.

(C) The complex ion $[Co^{III}(NH_3)_5Cl]^X$ has a charge $X$. Given the oxidation state of $Co$ is $+3$,$NH_3$ is $0$,and $Cl$ is $-1$,the charge $X$ is calculated as: $X = (+3) + 5(0) + (-1) = +2$.
To form a neutral complex salt,the complex cation $[Co(NH_3)_5Cl]^{2+}$ must combine with two chloride ions $(2Cl^-)$.
Therefore,the reaction is: $[Co(NH_3)_5Cl]^{2+} + 2Cl^- \rightarrow [Co(NH_3)_5Cl]Cl_2$.

(B) The complex is $K_3[Cr(C_2O_4)_3]$.
$1$. Coordination Number: The ligand oxalate $(C_2O_4)^{2-}$ is a bidentate ligand. Since there are $3$ such ligands,the coordination number is $3 \times 2 = 6$.
$2$. Oxidation State: Let the oxidation state of $Cr$ be $x$. The charge on the oxalate ion is $-2$ and the charge on the potassium ion is $+1$. The total charge of the complex is $0$.
$3(+1) + x + 3(-2) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$
Therefore,the coordination number is $6$ and the oxidation state is $+3$.

(B) The coordination number is defined as the total number of ligand donor atoms directly bonded to the central metal ion.
In the complex $[Fe(CN)_6]^{4-}$,there are $6$ $CN^-$ ligands,so the coordination number is $6$.
In the complex $[Fe(CN)_6]^{3-}$,there are $6$ $CN^-$ ligands,so the coordination number is $6$.
In the complex $[FeCl_4]^{-}$,there are $4$ $Cl^-$ ligands,so the coordination number is $4$.
Therefore,the coordination numbers are $6, 6, 4$ respectively.

(C) The correct answer is $(C)$.
$DMG$ (Dimethylglyoxime) is a polydentate ligand that coordinates to a central metal ion (like $Ni^{2+}$) through two donor atoms,forming a stable five-membered ring structure known as a chelate.
$H_2O$,$OH^{-}$,and $Cl^{-}$ are monodentate ligands and cannot form chelates.

(D) Potassium ferrocyanide,$K_4[Fe(CN)_6]$,is a coordination compound that dissociates in an aqueous solution as follows:
$K_4[Fe(CN)_6] \to 4K^{+} + [Fe(CN)_6]^{4-}$
From the dissociation equation,we can see that one formula unit of $K_4[Fe(CN)_6]$ produces $4$ potassium ions $(K^{+})$ and $1$ ferrocyanide complex ion $([Fe(CN)_6]^{4-})$.
Therefore,the total number of ions produced is $4 + 1 = 5$ ions.

(D) $[EDTA]^{4-}$ stands for Ethylenediamine tetraacetate ion.
It contains six donor atoms: two nitrogen atoms and four oxygen atoms from the carboxylate groups.
Therefore,it acts as a hexadentate ligand,capable of forming six coordinate bonds with a central metal ion.

(A) According to Werner's theory,negative ligands present in the coordination sphere can satisfy both primary and secondary valencies.
In ${K_3}[Fe(CN)_6]$,the $CN^-$ ligands satisfy both the primary valency (oxidation state of $Fe$ is $+3$) and the secondary valency (coordination number is $6$).
Therefore,statement $(a)$ is incorrect because the ligands satisfy both valencies.
Statement $(c)$ is correct because $NH_3$ is a neutral ligand and only satisfies the secondary valency.
Since statement $(a)$ is incorrect,the correct choice is $(d)$.

(C) Tartar emetic is a chemical compound known as potassium antimony tartrate.
Its chemical formula is $K_2Sb_2(C_4H_2O_6)_2 \cdot 3H_2O$ or $KSbOC_4H_4O_6 \cdot 0.5H_2O$.
Among the given options,$C$ represents the correct chemical formula for tartar emetic.

(A) Ferric ferrocyanide is commonly known as Prussian blue.
In this compound,the iron in the coordination sphere is in the $+2$ oxidation state (ferrocyanide,$[Fe(CN)_6]^{4-}$),and the iron outside the coordination sphere is in the $+3$ oxidation state (ferric,$Fe^{3+}$).
To balance the charges,we need four $Fe^{3+}$ ions for every three $[Fe(CN)_6]^{4-}$ ions.
Thus,the formula is $Fe_4[Fe(CN)_6]_3$.

(C) In ${K_4}[Fe(CN)_6]$,there is an ionic bond between $4K^+$ ions and the $[Fe(CN)_6]^{4-}$ complex ion.
Within the complex ion,there is a coordinate covalent bond between the central metal ion $Fe^{2+}$ and the $6CN^-$ ligands.
Within the cyanide ligand $(CN^-)$,there is a covalent bond between the carbon and nitrogen atoms $(C \equiv N^-)$.

(B) ligand is defined as an atom,ion,or molecule that donates a pair of electrons to a central metal atom or ion to form a coordinate bond.
According to the Lewis theory of acids and bases,a species that donates an electron pair is a Lewis base.
Therefore,all ligands act as Lewis bases.

(A) complex gives a white precipitate with $BaCl_2$ if it contains ionizable sulfate ions $(SO_4^{2-})$ in the coordination sphere.
In the complex $[Cr(NH_3)_5Cl]SO_4$,the $SO_4^{2-}$ ion is present outside the coordination sphere.
When this complex reacts with $BaCl_2$,it undergoes the following reaction:
$[Cr(NH_3)_5Cl]SO_4 + BaCl_2 \rightarrow [Cr(NH_3)_5Cl]Cl_2 + BaSO_4 \downarrow$ (white precipitate).
Therefore,option $A$ is correct.

(B) According to the Lewis acid-base theory,a ligand acts as a Lewis base because it donates a pair of electrons to the central metal atom or ion to form a coordinate bond.

(D) The total number of chlorine atoms in the complex is $4$.
Since $1/4$ of the chlorine is precipitated by $AgNO_3$,it means $4 \times (1/4) = 1$ chlorine atom is present as a free chloride ion $(Cl^-)$ outside the coordination sphere.
The formula must be of the form $[Pt(H_2O)_xCl_3]Cl$.
Among the given options,$PtCl_4 \cdot 3H_2O$ can be written as $[Pt(H_2O)_3Cl_3]Cl$,which provides $1$ mole of $Cl^-$ ions per mole of the complex.
Therefore,the correct formula is $PtCl_4 \cdot 3H_2O$.

(C) neutral ligand is a ligand that carries no net electrical charge. Among the given options,$NH_3$ (Ammine) is a neutral molecule. The other options,$Cl^-$ (Chloro),$OH^-$ (Hydroxo),and $CN^-$ (Cyano),are anionic ligands.

(D) The chemical formula for ferric hexacyanoferrate $(II)$ is $Fe_4[Fe(CN)_6]_3$.
When this compound dissociates in an aqueous solution,it produces $4$ ferric ions $(Fe^{3+})$ and $3$ hexacyanoferrate $(II)$ ions $([Fe(CN)_6]^{4-})$.
Total number of ions $= 4 + 3 = 7$.

(C) The coordination number of a central metal ion is defined as the number of ligand donor atoms to which the metal is directly bonded.
In the complex $[Cr(H_2O)_6]^{3+}$,the central metal ion is $Cr^{3+}$.
It is bonded to $6$ water $(H_2O)$ molecules.
Since $H_2O$ is a monodentate ligand,each molecule donates one pair of electrons to the central metal.
Therefore,the coordination number of $Cr$ is $6 \times 1 = 6$.

(C) bidentate ligand is a ligand that can donate two pairs of electrons to the central metal atom.
$1$. Cyano $(CN^-)$ is a monodentate ligand.
$2$. Acetato $(CH_3COO^-)$ is a monodentate ligand.
$3$. Dimethylglyoximato $(dmg^-)$ is a bidentate ligand with a negative charge.
$4$. Ethylenediamine $(en)$ is a bidentate ligand but it is neutral.
Therefore,the correct answer is Dimethylglyoximato.

(A) The given complex is $[Cr(H_2O)_4Br_2]Cl$.
When dissolved in water,it ionizes as:
$[Cr(H_2O)_4Br_2]Cl \rightarrow [Cr(H_2O)_4Br_2]^+ + Cl^-$.
The $Cl^-$ ion is present in the ionization sphere (outside the coordination sphere),so it is free to react and can be detected by a test (e.g.,with $AgNO_3$).
The $Br^-$ ions are inside the coordination sphere,bonded to the central metal atom,and do not ionize in the aqueous solution.
Therefore,only the $Cl^-$ ion test is given.

(B) Sodium nitroprusside is a coordination compound with the chemical formula $Na_2[Fe(CN)_5NO]$.
In this complex,the central metal ion is iron $(Fe)$ in the $+2$ oxidation state,coordinated with five cyanide $(CN^-)$ ligands and one nitrosyl $(NO^+)$ ligand.

(B) The coordination compound $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates in an aqueous solution as follows:
$[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$.
From the dissociation,we get one complex cation $[Pt(NH_3)_4Cl_2]^{2+}$ and two chloride anions $2Cl^-$.
The total number of ions produced is $1 + 2 = 3$.

(B) In the complex $[E(en)_2(C_2O_4)]NO_2$,both $en$ (ethylenediamine) and $C_2O_4^{2-}$ (oxalate) are bidentate ligands.
Coordination Number $= (2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
To find the oxidation state $(x)$ of $E$:
$x + 2(0) + (-2) + (-1) = 0$
$x - 3 = 0$
$x = +3$.
Therefore,the coordination number is $6$ and the oxidation state is $3$.

(A) Triphenylphosphine $(PPh_3)$ has a phosphorus atom with a lone pair of electrons.
It does not carry any net charge,so it is a neutral ligand.
It donates only one lone pair to the central metal atom,making it a monodentate ligand.

(B) An ambidentate ligand is a ligand which has two donor atoms but can only use one of them to form a coordinate bond with the central metal atom at a given time. Examples include $NO_2^-$ (which can bond through $N$ or $O$) and $SCN^-$ (which can bond through $S$ or $N$).

(A) The total volume of the mixture is $2 \ L$,containing $0.02 \ mol$ of each complex. Thus,$1 \ L$ of the mixture contains $0.01 \ mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01 \ mol$ of $[Co(NH_3)_5Br]SO_4$.
When $1 \ L$ of the mixture reacts with $AgNO_3$,only the ionizable $Br^-$ reacts: $[Co(NH_3)_5SO_4]Br + AgNO_3 \to [Co(NH_3)_5SO_4]NO_3 + AgBr(s)$. The moles of $AgBr$ $(Y)$ formed = $0.01 \ mol$.
When $1 \ L$ of the mixture reacts with $BaCl_2$,only the ionizable $SO_4^{2-}$ reacts: $[Co(NH_3)_5Br]SO_4 + BaCl_2 \to [Co(NH_3)_5Br]Cl_2 + BaSO_4(s)$. The moles of $BaSO_4$ $(Z)$ formed = $0.01 \ mol$.

(C) The coordination number of a metal is defined as the number of ligand donor atoms to which the metal is directly attached through $\sigma$-bonds.

(B) The coordination number of a central metal atom is the number of ligand donor atoms to which it is directly bonded.
In the complex $[Co(en)_2Br_2]Cl_2$,the ligand $en$ (ethylenediamine) is a bidentate ligand,and $Br^-$ is a monodentate ligand.
Coordination number $= (2 \times \text{denticity of } en) + (2 \times \text{denticity of } Br^-)$
Coordination number $= (2 \times 2) + (2 \times 1) = 4 + 2 = 6$.

(B) The coordination number is defined as the total number of ligand donor atoms directly bonded to the central metal ion.
In $[Zn(CN)_4]^{2-}$,the coordination number is $4$.
In $[Cr(H_2O)_6]^{3+}$,there are $6$ water ligands,so the coordination number is $6$.
In $[Cu(CN)_4]^{2-}$,the coordination number is $4$.
In $[Ni(NH_3)_4]^{2+}$,the coordination number is $4$.
Therefore,the correct option is $B$.

(D) The reaction with $AgNO_3$ indicates that there are $2 \ mol$ of ionizable $Cl^-$ ions outside the coordination sphere.
This is represented as $[Co(NH_3)_5Cl]Cl_2$.
Upon dissociation,it gives $[Co(NH_3)_5Cl]^{2+} + 2Cl^-$,resulting in a total of $3 \ mol$ of ions ($1 \ mol$ of complex cation + $2 \ mol$ of chloride anions).
Therefore,the correct formula is $[Co(NH_3)_5Cl]Cl_2$.

(A) $1$. The complex is $[E(en)_2(C_2O_4)]NO_2$.
$2$. Let the oxidation state of $E$ be $x$.
$3$. The ligand $(en)$ (ethylenediamine) is neutral $(0)$,and the oxalate ion $(C_2O_4^{2-})$ has a charge of $-2$. The counter ion $NO_2^-$ has a charge of $-1$.
$4$. The total charge of the complex is $0$.
$5$. Equation: $x + 2(0) + 1(-2) + 1(-1) = 0 \implies x - 2 - 1 = 0 \implies x = +3$.
$6$. The coordination number is calculated by the number of donor atoms attached to the central metal ion.
$7$. $(en)$ is a bidentate ligand ($2$ donor atoms each,$2 \times 2 = 4$) and $(C_2O_4^{2-})$ is a bidentate ligand ($2$ donor atoms).
$8$. Total coordination number = $4 + 2 = 6$.
$9$. Thus,the oxidation state is $3$ and the coordination number is $6$.

(D) tridentate ligand is a ligand that has three donor atoms capable of binding to a central metal atom simultaneously.
Propyltriamine,also known as $1,2,3$-triaminopropane or $NH_2-CH_2-CH(NH_2)-CH_2-NH_2$,contains three nitrogen donor atoms.
$NO_2^-$ is a monodentate ligand.
Oxalate ion $(C_2O_4^{2-})$ is a bidentate ligand.
$EDTA$ (Ethylenediaminetetraacetate) is a hexadentate ligand.
Therefore,the correct option is $D$.

(A) The coordination number is defined as the total number of sigma bonds formed by the central metal ion with the donor atoms of the ligands.
In the complex $[Cu(H_2O)_4]^{2+}$,there are $4$ water molecules acting as monodentate ligands.
Therefore,the coordination number of $Cu$ is $4 \times 1 = 4$.

(A) $1$. The coordination number is determined by the number of donor atoms attached to the central metal ion. The ligand $en$ (ethylenediamine) is a bidentate ligand,and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
$2$. Coordination number = $(2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
$3$. To find the oxidation state of $E$,let it be $x$. The charge on $en$ is $0$,the charge on $C_2O_4$ is $-2$,and the charge on the counter ion $NO_2$ is $-1$.
$4$. The sum of charges in the complex is $0$: $x + 2(0) + 1(-2) + 1(-1) = 0$.
$5$. $x - 2 - 1 = 0 \implies x = +3$.

(C) The complex $[Co(NH_3)_5Cl]Cl_2$ dissociates in water as follows:
$[Co(NH_3)_5Cl]Cl_2(aq) \to [Co(NH_3)_5Cl]^{2+}(aq) + 2Cl^-(aq)$.
This gives a total of $3 \ mol$ of ions ($1 \ mol$ of complex cation and $2 \ mol$ of chloride ions).
When treated with $AgNO_3$,the $2 \ mol$ of free chloride ions react to form $2 \ mol$ of $AgCl_{(s)}$:
$2Cl^-(aq) + 2Ag^+(aq) \to 2AgCl(s)$.

(A) The complex $CoCl_3 \cdot 5NH_3 \cdot H_2O$ is formulated as $[Co(NH_3)_5(H_2O)]Cl_3$.
This coordination compound is known as pentaamminaaquacobalt$(III)$ chloride.
Experimental observations confirm that this specific complex exhibits an orange-yellow color.