EAN number Questions in English

(D) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal atom in a complex should equal the atomic number of the next noble gas.
For Vanadium $(V)$,the atomic number $(Z)$ is $23$.
The next noble gas is Krypton $(Kr)$ with an atomic number of $36$.
The complex is $[V(CO)_6]^{n-}$.
The number of electrons contributed by $6$ $CO$ ligands is $6 \times 2 = 12$ electrons.
The number of electrons in the metal ion $V^{n+}$ is $23 - n$.
According to the $EAN$ rule: $(23 - n) + 12 = 36$.
$35 - n = 36$.
$n = 35 - 36 = -1$.
Therefore,the value of $n$ is $-1$.

(A) The hexacyanoferrate$(II)$ ion is $[Fe(CN)_6]^{4-}$.
$1$. The central metal ion is $Fe^{2+}$.
$2$. The atomic number $(Z)$ of $Fe$ is $26$.
$3$. The number of electrons in $Fe^{2+}$ is $26 - 2 = 24$.
$4$. Each $CN^-$ ligand donates $2$ electrons. Since there are $6$ ligands,the total electrons donated by ligands = $6 \times 2 = 12$.
$5$. $EAN = (\text{Number of electrons in metal ion}) + (\text{Number of electrons donated by ligands})$.
$6$. $EAN = 24 + 12 = 36$.

(A) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $[Co(NH_3)_6]^{3+}$:
$1$. The atomic number $(Z)$ of $Co$ is $27$.
$2$. The oxidation state of $Co$ is calculated as $x + 6(0) = +3$,so $x = +3$.
$3$. The coordination number is $6$ (since $NH_3$ is a monodentate ligand).
$4$. $EAN = 27 - 3 + 2(6) = 24 + 12 = 36$.

(C) 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 $[Zn(NH_3)_4]^{2+}$:
$Z$ of $Zn = 30$.
$ON$ of $Zn$: $x + 4(0) = +2 \implies x = +2$.
$CN$ of $Zn = 4$.
$EAN = 30 - 2 + 2(4) = 28 + 8 = 36$.

(B) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $[Fe(CN)_6]^{3-}$:
$1$. The atomic number $(Z)$ of $Fe$ is $26$.
$2$. Let the oxidation state of $Fe$ be $x$. Then $x + 6(-1) = -3$,which gives $x = +3$.
$3$. The coordination number of $Fe$ is $6$.
$4$. $EAN = 26 - 3 + 2(6) = 23 + 12 = 35$.

(D) 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.
$1$. For $[Co(NH_3)_6]^{3+}$: $Co$ $(Z=27)$. $EAN = 27 - 3 + (6 \times 2) = 36$ (Obeys $EAN$ rule).
$2$. For $[Cr(CO)_6]$: $Cr$ $(Z=24)$. $EAN = 24 - 0 + (6 \times 2) = 36$ (Obeys $EAN$ rule).
$3$. For $[Zn(NH_3)_4]^{2+}$: $Zn$ $(Z=30)$. $EAN = 30 - 2 + (4 \times 2) = 36$ (Obeys $EAN$ rule).
$4$. For $[Cu(NH_3)_4]^{2+}$: $Cu$ $(Z=29)$. $EAN = 29 - 2 + (4 \times 2) = 35$ (Does $NOT$ obey $EAN$ rule).
Therefore,the correct option is $D$.

(D) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number of the metal,$X$ is the oxidation state,and $Y$ is the number of electrons donated by the ligands.
For $[Cu(NH_3)_4]^{2+}$:
$Z = 29$ (for $Cu$)
$X = +2$ (as $NH_3$ is neutral,$Cu + 4(0) = +2$)
$Y = 2 \times 4 = 8$ (each $NH_3$ donates $2$ electrons)
$EAN = 29 - 2 + 8 = 35$.

(D) The atomic number $(Z)$ of $Zn$ is $30$.
In $[Zn(NH_3)_4]^{2+}$,the oxidation state of $Zn$ is $+2$,so $X = 2$.
The number of electrons donated by $4$ $NH_3$ ligands is $Y = 4 \times 2 = 8$.
The $EAN$ is calculated as $EAN = Z - X + Y$.
$EAN = 30 - 2 + 8 = 36$.

(D) The formula for the effective atomic number $(EAN)$ is given by: $EAN = Z - \text{oxidation state} + \text{number of electrons donated by ligands}$.
In the complex $[Co(NH_3)_6]^{3+}$:
$1$. The atomic number $(Z)$ of $Co$ is $27$.
$2$. Let the oxidation state of $Co$ be $x$. Since $NH_3$ is a neutral ligand,$x + 6(0) = +3$,so $x = +3$.
$3$. There are $6$ $NH_3$ ligands,each donating $2$ electrons,so the total number of electrons donated is $6 \times 2 = 12$.
$4$. $EAN = 27 - 3 + 12 = 36$.

(D) The effective atomic number $(EAN)$ is the total number of electrons present with the central metal atom or ion in a coordination complex,including the electrons donated by the ligands.
The formula to calculate the $EAN$ is:
$EAN = Z - O.N. + 2 \times C.N.$
Where:
$Z = \text{Atomic number} = 27$
$O.N. = \text{Oxidation number of } Co \text{ in } [Co(NH_3)_6]^{3+} = +3$
$C.N. = \text{Coordination number of } Co = 6$
Substituting the values:
$EAN = 27 - 3 + 2 \times 6 = 24 + 12 = 36$.

(B) The effective atomic number $(EAN)$ is calculated using the formula: $EAN = Z - X + Y$,where:
$Z$ = atomic number of the metal = $78$
$X$ = number of electrons lost by the metal to form the ion = $2$ (since the oxidation state of $Pt$ in $[Pt(NH_3)_4]^{2+}$ is $+2$)
$Y$ = number of electrons donated by the ligands = $4 \times 2 = 8$ (since there are $4$ $NH_3$ ligands,each donating $2$ electrons)
$EAN = 78 - 2 + 8 = 84$

(D) The Effective Atomic Number $(EAN)$ 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.
Given: $EAN = 36$,$Z = 27$,and $C.N. = 6$.
Since each ligand donates $2$ electrons,$Y = 6 \times 2 = 12$.
Substituting the values into the formula:
$36 = 27 - x + 12$
$36 = 39 - x$
$x = 39 - 36$
$x = +3$
Therefore,the oxidation state of cobalt is $+3$.

(C) The effective atomic number $(EAN)$ is calculated using the formula: $EAN = Z - X + Y$,where $Z$ is the atomic number of the metal,$X$ is the oxidation state of the metal,and $Y$ is the number of electrons donated by the ligands.
For $[Fe(CN)_6]^{4-}$:
$Z = 26$ (for $Fe$).
Let the oxidation state of $Fe$ be $x$. Then $x + 6(-1) = -4$,which gives $x = +2$. So,$X = 2$.
Since $CN^-$ is a monodentate ligand,$6$ ligands donate $6 \times 2 = 12$ electrons. So,$Y = 12$.
$EAN = 26 - 2 + 12 = 36$.

(B) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion should equal the atomic number of the next noble gas.
$EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number})$.
For $A: [Zn(NH_3)_4]^{2+}: EAN = 30 - 2 + 8 = 36$ (Obeys rule).
For $B: [Cu(NH_3)_4]^{2+}: EAN = 29 - 2 + 8 = 35$ (Does $NOT$ obey rule).
For $C: [Pt(NH_3)_6]^{4+}: EAN = 78 - 4 + 12 = 86$ (Obeys rule).
For $D: [Fe(CN)_6]^{4-}: EAN = 26 - 2 + 12 = 36$ (Obeys rule).
Therefore,the complex that does not obey the $EAN$ rule is $[Cu(NH_3)_4]^{2+}$.

(C) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the total number of electrons donated by ligands.
In $[Zn(NH_3)_4]SO_4$,the oxidation state of $Zn$ is $+2$ (since $NH_3$ is neutral and $SO_4$ is $-2$).
$Z = 30$,$X = 2$,and $Y = 4 \times 2 = 8$.
$EAN = 30 - 2 + 8 = 36$.

(D) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the number of electrons donated by ligands.
For $[Co(NH_3)_6]Cl_3$:
$1$. The oxidation state of $Co$ $(X)$ is $+3$.
$2$. The atomic number of $Co$ $(Z)$ is $27$.
$3$. The number of electrons donated by $6$ $NH_3$ ligands $(Y)$ is $6 \times 2 = 12$.
$EAN = 27 - 3 + 12 = 36$.

(C) Key Idea: The total number of electrons which the central metal ion appears to possess in the complex,including those gained by it in bonding,is called the effective atomic number $(EAN)$ of the central metal ion.
In the given complex ion $[Fe(CN)_6]^{3-}$,the $Fe$ is in $+3$ oxidation state.
Atomic number of $Fe = 26$.
The number of electrons in $Fe^{3+} = 26 - 3 = 23$.
Each of the six $CN^-$ ligands donates a pair of electrons (total $6 \times 2 = 12$ electrons).
Therefore,$EAN = 23 + 12 = 35$.

(D) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion in a complex should equal the atomic number of the next noble gas.
For $[Pt(NH_3)_6]^{4+}$: $EAN = 78 - 4 + 6(2) = 86$ (matches Radon).
For $[Fe(CN)_6]^{4-}$: $EAN = 26 - 2 + 6(2) = 36$ (matches Krypton).
For $[Zn(NH_3)_4]^{2+}$: $EAN = 30 - 2 + 4(2) = 36$ (matches Krypton).
For $[Cu(NH_3)_4]^{2+}$: $EAN = 29 - 2 + 4(2) = 35$ (does not match Krypton,$36$).
Thus,$[Cu(NH_3)_4]^{2+}$ is an exception to the $EAN$ rule.

(B) The oxidation state of $Co$ in $Na_{3}[Co(NO_{2})_{4} Cl_{2}]$ is calculated as follows:
$3(+1) + x + 4(-1) + 2(-1) = 0$
$3 + x - 4 - 2 = 0$
$x - 3 = 0$
$x = +3$
The $EAN$ (Effective Atomic Number) is given by the formula:
$EAN = (\text{Atomic number of metal}) - (\text{Oxidation state of metal}) + 2 \times (\text{Coordination number})$
$EAN = (27 - 3) + 2 \times 6$
$EAN = 24 + 12 = 36$.

(B) The complex $Cr(CO)_x$ follows the $18$-electron rule for stability in metal carbonyls.
$Cr$ (atomic number $24$) has $6$ valence electrons.
Each $CO$ ligand acts as a $2$-electron donor.
According to the $18$-electron rule: $6 + 2x = 18$.
$2x = 12$,which gives $x = 6$.
Thus,the compound is $Cr(CO)_6$,which is a stable octahedral complex.
Hence,the correct option is $(B)$.