EAN number Questions in English

(D) The $EAN$ (Effective Atomic Number) is calculated using the formula:
$EAN = (\text{Atomic number} - \text{Oxidation state} + 2 \times \text{Coordination number})$
For $Ni(CO)_4$:
Atomic number of $Ni = 28$
Oxidation state of $Ni = 0$ (since $CO$ is a neutral ligand)
Coordination number of $Ni = 4$
$EAN = 28 - 0 + 2 \times 4 = 36$

(C) The $E.A.N.$ (Effective Atomic Number) is calculated as:
$E.A.N. = \text{Atomic number} - \text{Oxidation state} + 2 \times \text{Coordination number}$.
For $K_4[Fe(CN)_6]$:
$1.$ The oxidation state of $Fe$ is $x + 6(-1) = -4$,so $x = +2$.
$2.$ The atomic number of $Fe$ is $26$.
$3.$ The coordination number is $6$.
$E.A.N. = 26 - 2 + 2(6) = 24 + 12 = 36$.

(A) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $[PtCl_6]^{2-}$,the atomic number of $Pt$ $(Z)$ is $78$.
The oxidation state of $Pt$ is $x + 6(-1) = -2$,so $x = +4$.
The coordination number is $6$.
$EAN = 78 - 4 + 2(6) = 74 + 12 = 86$.
Since the given value in option $A$ is $84$,it is incorrectly matched.

(A) The complex is $[Ni(NH_3)_4]SO_4$.
First,determine the oxidation state of $Ni$: Let the oxidation state be $x$.
$x + 4(0) + (-2) = 0$,so $x = +2$.
The atomic number of $Ni$ is $28$.
The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = (\text{Atomic number}) - (\text{Oxidation state}) + (2 \times \text{Coordination number})$.
$E.A.N. = 28 - 2 + (2 \times 4) = 26 + 8 = 34$.

(A) The formula for $EAN$ is: $EAN = Z - x + 2n$,where $Z$ is the atomic number,$x$ is the oxidation state,and $n$ is the coordination number.
For the complex $[Co(NH_3)_6]^{3+}$:
Atomic number of $Co$ $(Z)$ = $27$.
Oxidation state of $Co$ $(x)$: $x + 6(0) = +3 \implies x = +3$.
Coordination number $(n)$ = $6$.
$EAN = 27 - 3 + 2(6) = 24 + 12 = 36$.

(C) The formula for $EAN$ is: $EAN = Z - \text{oxidation state} + (2 \times \text{coordination number})$.
In potassium ferricyanide,$K_3[Fe(CN)_6]$,the iron is in the $+3$ oxidation state.
The atomic number $(Z)$ of iron $(Fe)$ is $26$.
The coordination number of $Fe$ in $[Fe(CN)_6]^{3-}$ is $6$.
Therefore,$EAN = 26 - 3 + (2 \times 6) = 23 + 12 = 35$.

(A) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom in a complex is equal to the atomic number of the next noble gas.
For $K_4[Fe(CN)_6]$:
Atomic number of $Fe = 26$.
Oxidation state of $Fe = +2$.
Number of electrons lost $= 2$.
Number of electrons gained from $6$ ligands $(CN^-)$ $= 6 \times 2 = 12$.
$EAN = (26 - 2) + 12 = 36$.
Since $36$ is the atomic number of $Krypton$ $(Kr)$,it follows the $EAN$ rule strictly.

(B) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion in a complex should be equal to the atomic number of the next noble gas.
$1$. Potassium ferrocyanide: $K_4[Fe(CN)_6]$. $Fe$ is in $+2$ state. $EAN = 26 - 2 + (6 \times 2) = 36$ (Follows).
$2$. Potassium ferricyanide: $K_3[Fe(CN)_6]$. $Fe$ is in $+3$ state. $EAN = 26 - 3 + (6 \times 2) = 35$ (Does not follow).
$3$. Tetracarbonyl nickel: $[Ni(CO)_4]$. $Ni$ is in $0$ state. $EAN = 28 - 0 + (4 \times 2) = 36$ (Follows).
$4$. Cobalt $(III)$ hexaammine chloride: $[Co(NH_3)_6]Cl_3$. $Co$ is in $+3$ state. $EAN = 27 - 3 + (6 \times 2) = 36$ (Follows).
Therefore,potassium ferricyanide does not follow the $EAN$ rule.

(D) The formula for Effective Atomic Number $(EAN)$ is: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $Ni(CO)_4$:
$Z$ (atomic number of $Ni$) = $28$.
Oxidation state of $Ni$ = $0$ (since $CO$ is a neutral ligand).
Coordination number = $4$.
$EAN = 28 - 0 + 2 \times 4 = 28 + 8 = 36$.

(B) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - \text{oxidation state} + (2 \times \text{coordination number})$.
For $[Ni(CO)_4]$: $Z=28$,$O.S.=0$,$C.N.=4$. $EAN = 28 - 0 + (2 \times 4) = 36$.
For $[Fe(CN)_6]^{4-}$: $Z=26$,$O.S.=+2$,$C.N.=6$. $EAN = 26 - 2 + (2 \times 6) = 36$.
Since both complexes have an $EAN$ of $36$,option $B$ is the correct pair.

(C) The formula for $EAN$ is: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
In potassium ferricyanide,$K_3[Fe(CN)_6]$,the iron is in the $+3$ oxidation state.
The atomic number $(Z)$ of $Fe$ is $26$.
The coordination number is $6$.
$EAN = 26 - 3 + (2 \times 6) = 23 + 12 = 35$.

(C) The effective atomic number $(EAN)$ is calculated using the formula: $EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number})$.
In $[A(NH_3)_6]Cl_3$,the oxidation state of $A$ is $+3$ and the coordination number is $6$.
Given $EAN = 33$,we have: $33 = Z - 3 + 2(6)$.
$33 = Z - 3 + 12$.
$33 = Z + 9$.
$Z = 33 - 9 = 24$.

(C) The formula for Effective Atomic Number $(EAN)$ is given by:
$EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$
For $[Co(NH_3)_6]^{3+}$:
$1$. The atomic number $(Z)$ of $Co$ is $27$.
$2$. The oxidation state of $Co$ is $x + 6(0) = +3$,so $x = +3$.
$3$. The coordination number is $6$ (since $NH_3$ is a monodentate ligand).
$EAN = 27 - 3 + 2(6) = 24 + 12 = 36$.
Thus,the $EAN$ of $Co$ is $36$.

(C) The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = Z - O.S. + 2 \times C.N.$,where $Z$ is the atomic number,$O.S.$ is the oxidation state,and $C.N.$ is the coordination number.
For $K_4[Fe(CN)_6]$:
$Z$ for $Fe = 26$.
$O.S.$ of $Fe = +2$.
$C.N.$ of $Fe = 6$.
$E.A.N. = 26 - 2 + 2(6) = 24 + 12 = 36$.
Since $36$ is the atomic number of the inert gas Krypton $(Kr)$,this complex satisfies the condition.

(D) The complex is $K_4[Fe(CN)_6]$.
The atomic number $(Z)$ of $Fe$ is $26$.
Calculate the oxidation state of $Fe$: $4(+1) + x + 6(-1) = 0 \implies 4 + x - 6 = 0 \implies x = +2$.
The coordination number $(CN)$ of $Fe$ is $6$ (since $CN^-$ is a monodentate ligand).
The formula for $EAN$ is: $EAN = Z - (\text{Oxidation state}) + 2 \times (CN)$.
$EAN = 26 - 2 + 2(6) = 24 + 12 = 36$.

(C) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom in a complex should be equal to the atomic number of the next noble gas.
$1$. For $Fe(CO)_5$: $EAN = 26 - 0 + 2 \times 5 = 36$ (Follows $EAN$ rule).
$2$. For $[Mn(CO)_5(C_2H_4)]^{\oplus}$: $Mn$ is in $+1$ oxidation state. $EAN = 25 - 1 + 2 \times 5 + 2 = 36$ (Follows $EAN$ rule).
$3$. For $[Ag(NH_3)_2]^{\oplus}$: $Ag$ is in $+1$ oxidation state. $EAN = 47 - 1 + 2 \times 2 = 50$ (Does not follow $EAN$ rule,as $50 \neq 54$).
$4$. For $[PtCl_2(NH_3)_4]^{+2}$: $Pt$ is in $+4$ oxidation state. $EAN = 78 - 4 + 2 \times 6 = 86$ (Follows $EAN$ rule).

(B) The oxidation state of $Fe$ in the complex $[(\eta^5 - C_5H_5) Fe(CO)_2Cl]$ is calculated as follows:
Let the oxidation state of $Fe$ be $x$.
$x + (-1) + 2(0) + (-1) = 0 \implies x = +2$.
Electrons in $Fe^{2+}$ ion = $26 - 2 = 24$.
Electrons donated by $\eta^5-C_5H_5^-$ (cyclopentadienyl anion) = $6$.
Electrons donated by $2(CO)$ ligands = $2 \times 2 = 4$.
Electrons donated by $Cl^-$ ligand = $2$.
Total $EAN = 24 + 6 + 4 + 2 = 36$.

(C) The complex $Fe(CO)_x$ follows the $18$-electron rule (or Sidgwick's $EAN$ rule) for stability.
For a stable metal carbonyl,the total number of valence electrons around the central metal atom should be $36$ (for $3d$ series metals).
The formula for $EAN$ is $Z - O.S. + 2(x) = 36$,where $Z$ is the atomic number of $Fe$ $(26)$ and $O.S.$ is the oxidation state of $Fe$ ($0$ in metal carbonyls).
Substituting the values: $26 - 0 + 2x = 36$.
$2x = 36 - 26 = 10$.
$x = 5$.
Therefore,the complex is $Fe(CO)_5$.

(C) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number})$.
$1$. For $K_2[Ni(CN)_4]$: $Ni$ is in $+2$ oxidation state. $Z = 28$. $EAN = 28 - 2 + 2(4) = 26 + 8 = 34$.
$2$. For $[Cu(NH_3)_4]SO_4$: $Cu$ is in $+2$ oxidation state. $Z = 29$. $EAN = 29 - 2 + 2(4) = 27 + 8 = 35$.
$3$. For $K_2[PtCl_6]$: $Pt$ is in $+4$ oxidation state. $Z = 78$. $EAN = 78 - 4 + 2(6) = 74 + 12 = 86$.
Thus,the values are $34, 35, 86$.

(D) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For option $A$:
$[Cu(NH_3)_4]SO_4$: $Cu$ is $+2$,$CN=4$. $EAN = 29 - 2 + (2 \times 4) = 35$.
$K_3[Fe(CN)_6]$: $Fe$ is $+3$,$CN=6$. $EAN = 26 - 3 + (2 \times 6) = 35$.
Both have $EAN = 35$.
For option $B$:
$K_4[Fe(CN)_6]$: $Fe$ is $+2$,$CN=6$. $EAN = 26 - 2 + (2 \times 6) = 36$.
$[Co(NH_3)_6]Cl_3$: $Co$ is $+3$,$CN=6$. $EAN = 27 - 3 + (2 \times 6) = 36$.
Both have $EAN = 36$.
For option $C$:
$K_3[Cr(C_2O_4)_3]$: $Cr$ is $+3$,$CN=6$ (bidentate ligand). $EAN = 24 - 3 + (2 \times 6) = 33$.
$[Cr(NH_3)_6]Cl(NO_2)_2$: $Cr$ is $+3$,$CN=6$. $EAN = 24 - 3 + (2 \times 6) = 33$.
Both have $EAN = 33$.
Since all pairs have the same $EAN$ for their respective central metal ions,the correct answer is $D$.

(C) The $EAN$ (Effective Atomic Number) rule states that the metal ion in a complex tends to achieve the electronic configuration of the next noble gas ($EAN = 36$ for $3d$ series metals).
Calculation of $EAN$ for the given complexes:
$1$. $K_4[Fe(CN)_6]: EAN = 26 - 2 + (6 \times 2) = 36$
$2$. $[Co(NH_3)_6]Cl_3: EAN = 27 - 3 + (6 \times 2) = 36$
$3$. $[Cu(NH_3)_4]SO_4: EAN = 29 - 2 + (4 \times 2) = 35$
$4$. $K_3[Co(Cl)_2(C_2O_4)_2]: EAN = 27 - 3 + (2 \times 2) + (2 \times 4) = 36$
Since the $EAN$ of $[Cu(NH_3)_4]SO_4$ is $35$,it violates the $EAN$ rule.

(C) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the number of electrons donated by ligands.
For $Fe(NO)_2(CO)_2$: The oxidation state of $Fe$ is $0$ (assuming $NO$ as $NO^+$ and $CO$ as neutral,or $NO$ as neutral and $CO$ as neutral depending on convention; in standard $EAN$ calculations for these complexes,$Fe$ is $0$). $Fe$ $(Z=26)$,$2 \times CO$ ($2 \times 2 = 4$ electrons),$2 \times NO$ ($2 \times 3 = 6$ electrons). $EAN = 26 - 0 + 10 = 36$.
For $Co_2(CO)_8$: Each $Co$ $(Z=27)$ has one $Co-Co$ bond ($1$ electron) and $4$ $CO$ ligands ($8$ electrons). Oxidation state is $0$. $EAN = 27 - 0 + 9 = 36$.

(C) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom should be equal to the atomic number of the next noble gas.
For $Co$ $(Z = 27)$,the next noble gas is $Kr$ $(Z = 36)$.
In $Co_2(CO)_8$,there is a $Co-Co$ bond.
Each $Co$ atom contributes $9$ electrons (valence electrons) + $8$ electrons from $4$ $CO$ ligands $(4 \times 2 = 8)$ + $1$ electron from the $Co-Co$ bond.
Total electrons per $Co$ atom = $9 + 8 + 1 = 18$ electrons in terms of valence shell,or $27 + 8 + 1 = 36$ as $EAN$.
Thus,$Co_2(CO)_8$ follows the $EAN$ rule.

(C) The $EAN$ (Effective Atomic Number) rule states that a complex is stable if the total number of electrons around the central metal atom equals the atomic number of the next noble gas.
For $[Fe(CO)_5]$:
Atomic number of $Fe = 26$.
Oxidation state of $Fe = 0$.
Number of electrons donated by $5$ $CO$ ligands $= 5 \times 2 = 10$.
$EAN = 26 - 0 + 10 = 36$.
Since $36$ is the atomic number of $Krypton$ $(Kr)$,$[Fe(CO)_5]$ obeys the $EAN$ rule.

(B) The formula for Effective Atomic Number $(EAN)$ is:
$EAN = Z - \text{Oxidation State} + 2 \times \text{Coordination Number}$
For $[Fe(CO)_5]$: $EAN = 26 - 0 + 2(5) = 36$
For $[Cr(CN)_6]^{-3}$: $EAN = 24 - 3 + 2(6) = 33$
For $[Mn_2(CO)_{10}]$: $EAN = 25 - 0 + 2(5) + 1 = 36$ (where $1$ is the contribution from the $Mn-Mn$ bond)
For $[Fe(CN)_6]^{-4}$: $EAN = 26 - 2 + 2(6) = 36$
Thus,$[Cr(CN)_6]^{-3}$ is the odd one out as its $EAN$ is $33$,while the others have an $EAN$ of $36$ (following the $EAN$ rule).

(B) The formula of potassium hexachloroplatinate $(IV)$ is $K_2[PtCl_6]$.
In this complex,the oxidation state of $Pt$ is $+4$.
The atomic number of $Pt$ is $78$.
The coordination number is $6$,so the number of electrons donated by $6$ ligands is $6 \times 2 = 12$.
$EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number})$
$EAN = 78 - 4 + 12 = 86$.

(C) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
Given $EAN = 36$ and $Z = 27$ for $Co$.
Let the charge on $O_2$ be $x$. The total charge on the complex is $0$.
$Co + 2(NH_3) + x + (en) + Cl = 0$
$Co + 0 + x + 0 - 1 = 0 \Rightarrow Co = 1 - x$.
Substituting into $EAN$ formula: $36 = 27 - (1 - x) + 2 \times 6 = 27 - 1 + x + 12 = 38 + x$.
$36 = 38 + x \Rightarrow x = -2$.
Since the charge on $O_2$ is $-2$,it is a peroxide ion $(O_2^{2-})$.

(C) The oxidation state of $Mg$ in $[Mg(EDTA)]^{2-}$ is calculated as follows:
$x + 4(-1) = -2 \Rightarrow x = +2$.
The atomic number $(Z)$ of $Mg$ is $12$.
The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - (\text{oxidation state}) + 2 \times (\text{coordination number})$.
For $EDTA^{4-}$,the denticity is $6$,so the coordination number is $6$.
$EAN = 12 - 2 + 2 \times 6 = 10 + 12 = 22$.

(B) According to Sidgwick's $EAN$ rule,the Effective Atomic Number $(EAN)$ of the central metal atom/ion should be equal to the atomic number of the nearest noble gas (which is $36$ for $3d$ series metals like $Cr$,$Z = 24$).
Formula: $EAN = Z - O.S. + 2 \times (C.N.)$
For $[M(CO)_6]$:
Atomic number $(Z)$ = $24$
Oxidation state $(O.S.)$ = $0$ (since $CO$ is a neutral ligand)
Coordination number $(C.N.)$ = $6$
$EAN = 24 - 0 + 2 \times 6 = 24 + 12 = 36$
Since $36$ is the atomic number of Krypton $(Kr)$,$[M(CO)_6]$ follows the $EAN$ rule.

(B) In $Fe_2(CO)_9$,each $Fe$ atom is bonded to $3$ terminal $CO$ ligands,$3$ bridging $CO$ ligands,and one $Fe-Fe$ bond.
The atomic number of $Fe$ is $26$.
The oxidation state of $Fe$ in $Fe_2(CO)_9$ is $0$.
The number of electrons donated by $3$ terminal $CO$ ligands is $3 \times 2 = 6$.
The number of electrons donated by $3$ bridging $CO$ ligands is $3 \times 1 = 3$.
The number of electrons from the $Fe-Fe$ bond is $1$.
$EAN = Z - \text{oxidation state} + \text{electrons from ligands} + \text{electrons from metal-metal bond}$
$EAN = 26 - 0 + 6 + 3 + 1 = 36$.

(A) The brown ring complex is $[Fe(H_2O)_5(NO)]SO_4$.
Case $1$: When $NO$ is considered as $NO^+$ ($2e^-$ donor):
$Fe$ is in $+1$ oxidation state.
$EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number}) = 26 - 1 + 2 \times 6 = 26 - 1 + 12 = 37$.
Case $2$: When $NO$ is considered as a neutral ligand ($NO$,$3e^-$ donor):
$Fe$ is in $0$ oxidation state.
$EAN = Z - \text{oxidation state} + 2 \times (\text{number of } H_2O) + 3 \times (\text{number of } NO) = 26 - 0 + 2 \times 5 + 3 \times 1 = 26 + 10 + 3 = 39$.
Wait,re-evaluating the standard convention: In the brown ring complex $[Fe(H_2O)_5(NO)]^{+2}$,$Fe$ is $+1$ and $NO$ is $NO^+$. If we treat $NO$ as neutral,the oxidation state of $Fe$ becomes $0$.
$EAN = 26 - 0 + 10 + 3 = 39$.
Comparing $37$ and $39$,the $EAN$ increases.

(C) $E.A.N. = \text{Atomic number of central metal ion} - \text{Oxidation number of central metal ion} + 2 \times \text{Coordination number of central metal ion.}$
Atomic number of $Fe = 26$.
Oxidation number of $Fe$ in $[Fe(C_2O_4)_3]^{3-}$ is calculated as: $x + 3(-2) = -3 \implies x = +3$.
Coordination number of $Fe$ with bidentate ligand $C_2O_4^{2-}$ is $3 \times 2 = 6$.
$E.A.N. = 26 - 3 + 2 \times 6 = 23 + 12 = 35$.
Hence,option $C$ is correct.

(C) $E.A.N. = \text{Atomic number of central metal ion} - \text{Oxidation number of central metal ion} + 2 \times \text{Coordination number of central metal ion}$.
Atomic number of $Fe = 26$.
Oxidation number of $Fe$ in the given compound $[Fe(C_2O_4)_3]^{3-}$ is calculated as: $x + 3(-2) = -3 \implies x = +3$.
Coordination number of $Fe$ with bidentate ligand $C_2O_4^{2-}$ is $3 \times 2 = 6$.
$E.A.N. = 26 - 3 + 2 \times 6 = 23 + 12 = 35$.

(D) The $EAN$ of $Co(CO)_4$ is calculated as: $27 + (4 \times 2) = 35$.
Since $35 \neq 36$,it does not follow the Sidgwick $EAN$ rule.
It can attain stability by gaining one electron (reduction of $Co$) to form $[Co(CO)_4]^-$,or by dimerization to form $Co_2(CO)_8$,where each $Co$ atom effectively shares an electron,resulting in an $EAN$ of $36$ for each $Co$ atom.

(D) The $EAN$ (Effective Atomic Number) is calculated as: $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the number of electrons donated by ligands.
For option $D$:
For $[Ni(CO)_4]$: $Z=28, X=0, Y=4 \times 2 = 8$. $EAN = 28 - 0 + 8 = 36$.
For $[Ni(CN)_4]^{2-}$: $Z=28, X=2, Y=4 \times 2 = 8$. $EAN = 28 - 2 + 8 = 34$.
Since $36 \neq 34$,the $EAN$ values are not the same.

(C) The $E.A.N.$ (Effective Atomic Number) is calculated using the formula: $E.A.N. = Z - \text{oxidation state} + 2 \times (\text{coordination number})$.
Given: $E.A.N. = 36$,$Z = 26$,and oxidation state of $M^{2+} = 2$.
Let $n$ be the number of monodentate ligands,which is equal to the coordination number.
Substituting the values: $36 = 26 - 2 + 2n$.
$36 = 24 + 2n$.
$12 = 2n$.
$n = 6$.
Thus,the number of monodentate ligands is $6$.

(C) $I. [(Ph_3P)_2PdCl_2PdCl_2]$: The $Pd$ atom is in $+2$ oxidation state. $EAN = 46 - 2 + (4 \times 2) = 52 \neq 54$ (Atomic number of $Xe$). Thus,it is $NF$.
$II. [NiBrCl(en)]$: The $Ni$ atom is in $+2$ oxidation state. $EAN = 28 - 2 + (4 \times 2) = 34 \neq 36$ (Atomic number of $Kr$). Thus,it is $NF$.
$III. Na_4[Fe(CN)_5NOS]$: The $Fe$ atom is in $+2$ oxidation state. $EAN = 26 - 2 + (5 \times 2) + 2 = 36 = [Kr]$. Thus,it is $F$.
$IV. Cr(CO)_3(NO)_2$: The $Cr$ atom is in $0$ oxidation state. $EAN = 24 - 0 + (3 \times 2) + (3 \times 2) = 36 = [Kr]$. Thus,it is $F$.
Therefore,the sequence is $NF-NF-F-F$.

(C) The stability of metal carbonyls is often governed by the $18$-electron rule or the $EAN$ (Effective Atomic Number) rule,which states that the complex is most stable when the total number of electrons equals the nearest noble gas configuration ($36$ for $Mn$ series).
$1$. For $[Mn(CO)_6]^+$,the $EAN$ is $25 - 1 + (6 \times 2) = 36$,which is stable.
$2$. For $[Mn(CO)_6]$,the $EAN$ is $25 + 12 = 37$. This species has one extra electron compared to the stable $36$-electron configuration.
$3$. Therefore,$[Mn(CO)_6]$ tends to lose an electron to achieve the stable $36$-electron configuration: $[Mn(CO)_6] \rightarrow [Mn(CO)_6]^+ + e^-$.
$4$. Since it loses an electron,it acts as a reducing agent.

(D) The $EAN$ is calculated using the formula: $EAN = Z - O.S. + 2 \times C.N.$,where $Z$ is the atomic number,$O.S.$ is the oxidation state,and $C.N.$ is the coordination number.
In $Co_2(CO)_8$,the complex contains a $Co-Co$ bond.
Each $Co$ atom is bonded to $3$ terminal $CO$ groups,$1$ bridging $CO$ group,and $1$ $Co$ atom.
Thus,the coordination number $(C.N.)$ of each $Co$ is $3 + 1 + 1 = 5$.
The oxidation state $(O.S.)$ of $Co$ in $Co_2(CO)_8$ is $0$.
The atomic number $(Z)$ of $Co$ is $27$.
$EAN = 27 - 0 + 2 \times 5 = 27 + 10 = 37$.

(D) The $EAN$ of $Co$ in $Co(CO)_4$ is $27 + 8 - 0 = 35$.
To achieve the stable $36$ electron configuration (Krypton core),it can either gain an electron to form $[Co(CO)_4]^-$ (reduction) or undergo dimerization to form $Co_2(CO)_8$ where each $Co$ atom shares an electron,effectively reaching $36$ electrons per $Co$ atom.
Thus,both reduction and dimerization are pathways to stability.

(C) The brown ring complex is $[Fe(H_2O)_5(NO)]SO_4$,which contains the complex ion $[Fe(H_2O)_5(NO)]^{2+}$.
In this complex,the oxidation state of $Fe$ is $+1$ (since $H_2O$ is neutral and $NO$ acts as $NO^+$).
The atomic number of $Fe$ is $26$.
The number of electrons in $Fe^+$ is $26 - 1 = 25$.
The complex has $5$ $H_2O$ ligands ($5 \times 2 = 10$ electrons) and $1$ $NO^+$ ligand ($2$ electrons).
$EAN = 25 + 10 + 2 = 37$.

(A) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - ON + 2 \times CN$,where $Z$ is the atomic number,$ON$ is the oxidation number,and $CN$ is the coordination number.
For $[Ni(CO)_4]$:
$Z$ of $Ni = 28$.
$ON$ of $Ni = 0$ (since $CO$ is a neutral ligand).
$CN = 4$ (there are $4$ $CO$ ligands).
$EAN = 28 - 0 + 2 \times 4 = 28 + 8 = 36$.

(B) The chemical formula for potassium hexachloroplatinate $(IV)$ is $K_2[PtCl_6]$.
In this complex,the central metal ion is $Pt^{4+}$.
The atomic number of $Pt$ is $78$.
The number of electrons in $Pt^{4+}$ is $78 - 4 = 74$.
The ligand $Cl^-$ is a monodentate ligand,and there are $6$ such ligands.
Each $Cl^-$ ligand donates $2$ electrons to the central metal ion.
Total electrons donated by $6$ $Cl^-$ ligands = $6 \times 2 = 12$.
$EAN = (\text{Number of electrons in metal ion}) + (\text{Number of electrons donated by ligands})$.
$EAN = 74 + 12 = 86$.

(A) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom in a complex should be equal to the atomic number of the next noble gas.
$1$. For $[Cu(NH_3)_4]^{2+}$: Atomic number of $Cu = 29$. Oxidation state $= +2$. Electrons in $Cu^{2+} = 27$. Electrons from $4 \ NH_3$ ligands $= 4 \times 2 = 8$. $EAN = 27 + 8 = 35$. This does not equal $36$ (Krypton).
$2$. For $[Zn(OH)_4]^{2-}$: Atomic number of $Zn = 30$. Oxidation state $= +2$. Electrons in $Zn^{2+} = 28$. Electrons from $4 \ OH^-$ ligands $= 4 \times 2 = 8$. $EAN = 28 + 8 = 36$.
$3$. For $[HgI_4]^{2-}$: Atomic number of $Hg = 80$. Oxidation state $= +2$. Electrons in $Hg^{2+} = 78$. Electrons from $4 \ I^-$ ligands $= 4 \times 2 = 8$. $EAN = 78 + 8 = 86$ (Radon).
$4$. For $[Fe(CO)_5]$: Atomic number of $Fe = 26$. Oxidation state $= 0$. Electrons in $Fe = 26$. Electrons from $5 \ CO$ ligands $= 5 \times 2 = 10$. $EAN = 26 + 10 = 36$ (Krypton).
Thus,$[Cu(NH_3)_4]^{2+}$ does not obey the $EAN$ rule.

(D) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom in a complex is equal to the atomic number of the next noble gas.
For $[Fe(CO)_5]$: $Z = 26$,$O.S. = 0$,$CN = 5$. $EAN = 26 - 0 + (5 \times 2) = 36$ (Krypton).
For $[Ni(CO)_4]$: $Z = 28$,$O.S. = 0$,$CN = 4$. $EAN = 28 - 0 + (4 \times 2) = 36$ (Krypton).
For $K_4[Fe(CN)_6]$: $Z = 26$,$O.S. = +2$,$CN = 6$. $EAN = 26 - 2 + (6 \times 2) = 36$ (Krypton).
Since all complexes follow the $EAN$ rule,the correct option is $D$.

(B) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - x + 2n$,where $Z$ is the atomic number of the central metal atom,$x$ is the oxidation state,and $n$ is the coordination number.
For $[Cr(SCN)_6]^{3-}$:
$Z$ for $Cr = 24$.
Let the oxidation state of $Cr$ be $x$. Since $SCN^-$ has a charge of $-1$,$x + 6(-1) = -3$,so $x = +3$.
The coordination number $n = 6$.
$EAN = 24 - 3 + 2(6) = 21 + 12 = 33$.

(A) . $i$. $Cu$. In $Cu$,the second electron is removed from a $3d^{10}$ configuration,which is highly stable.
$ii$. $Zn$. The third electron is removed from a $d^{10}$ configuration,which is highly stable.
$iii$. $Zn$. Zinc has all electrons paired and hence it cannot form strong metallic bonds.
$b$. $i$. $Fe(CO)_5$. According to the Effective Atomic Number $(EAN)$ rule,$EAN = Z - \text{electrons lost} + \text{electrons gained}$.
$EAN = 26 - 0 + (5 \times 2) = 36$.
$ii$. $MnO_3F$. $Mn$ shows a $(+7)$ oxidation state.

(C) To be isoelectronic with $Ni(CO)_4$,the species must have the same Effective Atomic Number $(EAN)$.
$EAN$ of $Ni(CO)_4 = 28 + 4 \times 2 = 36$.
Calculating $EAN$ for the given species:
$1. V(CO)_6 = 23 + 6 \times 2 = 35$
$2. Cr(CO)_5 = 24 + 5 \times 2 = 34$
$3. Cu(CO)_3 = 29 + 3 \times 2 = 35$
$4. Mn(CO)_5 = 25 + 5 \times 2 = 35$
$5. Fe(CO)_5 = 26 + 5 \times 2 = 36$ (Matches)
$6. [Co(CO)_3]^{3-} = 27 + 3 + 3 \times 2 = 36$ (Matches)
$7. [Cr(CO)_4]^{4-} = 24 + 4 + 4 \times 2 = 36$ (Matches)
$8. Ir(CO)_3 = 77 + 3 \times 2 = 83$
The species with $EAN = 36$ are $Fe(CO)_5$,$[Co(CO)_3]^{3-}$,and $[Cr(CO)_4]^{4-}$.
Thus,the total number of species isoelectronic with $Ni(CO)_4$ is $3$.

(C) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - \text{Oxidation State} + 2 \times (\text{Coordination Number})$.
For $[Co(NH_3)_x]Cl_3$,the oxidation state of $Co$ is $+3$ (since $Cl$ is $-1$ and $NH_3$ is neutral).
The atomic number $(Z)$ of $Co$ is $27$.
The coordination number is $x$.
Given $EAN = 36$,we have: $36 = 27 - 3 + 2x$.
$36 = 24 + 2x$.
$2x = 12$.
$x = 6$.