Answer
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Hint: First, we need to understand the Effective Atomic Number i.e., EAN rule to solve this question. It refers to the total number of electrons that surround the nucleus of an atom in a metal ion.
The formula of Effective Atomic Number = $Z - X + 2 \times Y$
Where,
Z is the atomic number of the metal
X is the oxidation number
And Y is the coordination number.
Complete step by step answer:
We need to understand that the oxidation number is the total number of electrons that are lost in iron formation and coordination number is the total number of electrons that are gained from the donor atoms.
So, now we will calculate to see the exception of the EAN rule.
Case 1:
$[Pt{(N{H_{3{)_6}}}]^{4 + }}$
We can see that the oxidation state of the metal ion = +4
And we know the atomic number = 78 and the coordination number = 6
Therefore, we can substitute the value in EAN rule .
Thus, EAN rule = \[\left( {78 - 4} \right){\text{ }} + {\text{ }}(6 \times 2){\text{ }} = {\text{ }}86\]
We can see that the atomic number of the nearest noble gas Rn (86).
Case 2:
${[Fe{(CN)_6}]^{4 - }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 26 and the coordination number = 6
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule = \[\left( {78 - 4} \right){\text{ }} + {\text{ }}(6 \times 2){\text{ }} = {\text{ }}86\]
We can see that the atomic number of the nearest noble gas Kr (36).
Case 3:
${[Zn{(N{H_3})_4}]^{2 + }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 30 and the coordination number = 4
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule = \[\left( {30 - 2} \right){\text{ }} + {\text{ }}(4 \times 2){\text{ }} = {\text{ }}36\]
We can see that the atomic number of the nearest noble gas Kr (36).
Case 4:
${[Cu{(N{H_3})_4}]^{2 + }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 29 and the coordination number = 4
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule \[ = {\text{ }}\left( {29 - 2} \right){\text{ }} + {\text{ }}(4 \times 2){\text{ }} = {\text{ }}35\]
We can see that it is an exception to the EAN rule.
$\therefore $Option (D) is correct.
Note:
We must know that Sidgwick introduces rules, depending on the effective atomic number of the central metal atom to explain the stability of coordination complexes, known as effective atomic number rule or EAN rule. The effective atomic number is numerically equal to the atomic number of a noble-gas found in the same period of the metal.
The formula of Effective Atomic Number = $Z - X + 2 \times Y$
Where,
Z is the atomic number of the metal
X is the oxidation number
And Y is the coordination number.
Complete step by step answer:
We need to understand that the oxidation number is the total number of electrons that are lost in iron formation and coordination number is the total number of electrons that are gained from the donor atoms.
So, now we will calculate to see the exception of the EAN rule.
Case 1:
$[Pt{(N{H_{3{)_6}}}]^{4 + }}$
We can see that the oxidation state of the metal ion = +4
And we know the atomic number = 78 and the coordination number = 6
Therefore, we can substitute the value in EAN rule .
Thus, EAN rule = \[\left( {78 - 4} \right){\text{ }} + {\text{ }}(6 \times 2){\text{ }} = {\text{ }}86\]
We can see that the atomic number of the nearest noble gas Rn (86).
Case 2:
${[Fe{(CN)_6}]^{4 - }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 26 and the coordination number = 6
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule = \[\left( {78 - 4} \right){\text{ }} + {\text{ }}(6 \times 2){\text{ }} = {\text{ }}86\]
We can see that the atomic number of the nearest noble gas Kr (36).
Case 3:
${[Zn{(N{H_3})_4}]^{2 + }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 30 and the coordination number = 4
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule = \[\left( {30 - 2} \right){\text{ }} + {\text{ }}(4 \times 2){\text{ }} = {\text{ }}36\]
We can see that the atomic number of the nearest noble gas Kr (36).
Case 4:
${[Cu{(N{H_3})_4}]^{2 + }}$
We can see that the oxidation state of the metal ion = +2
And we know the atomic number = 29 and the coordination number = 4
Therefore, we can substitute the value in EAN rule.
Thus, EAN rule \[ = {\text{ }}\left( {29 - 2} \right){\text{ }} + {\text{ }}(4 \times 2){\text{ }} = {\text{ }}35\]
We can see that it is an exception to the EAN rule.
$\therefore $Option (D) is correct.
Note:
We must know that Sidgwick introduces rules, depending on the effective atomic number of the central metal atom to explain the stability of coordination complexes, known as effective atomic number rule or EAN rule. The effective atomic number is numerically equal to the atomic number of a noble-gas found in the same period of the metal.
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