
In $[Ni{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}$, the E.A.N of Ni is
A. $34$
B. $35$
C. $36$
D. $37$
Answer
161.4k+ views
Hint: E.A.N is the effective atomic number of any coordination compound that can be determined by knowing the atomic number, oxidation number, and coordination number of the central metal atom. Putting these values in the E.A.N formula, one can easily calculate the effective atomic number of the central metal atom.
Formula used: Effective atomic number can be calculated by using a formula which is shown below:
$E.A.N = Z-O.N+2\times (C.N)$
Here, z = nuclear charge number or atomic number of the central metal atom.
O.N = oxidation number of central metal atoms.
C.N = coordination number of central metal atoms.
Complete step by step solution:
The effective atomic number is the total number of electrons surrounding the central nucleus of a metal ion. Therefore, the number of electrons present on the central metal ion and all ligands surrounding the metal ion is required to calculate the E.A.N. Generally, the E.A.N of the metal atom is numerically equal to the nearest noble gas configuration. Kr($Z=36$), Xe(Z=54), and Rn($Z=86$) are three noble gases that fall under the consideration of the effective atomic number rule. Not every coordination compound follows this rule.
We have the complex $[Ni{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}$, the central metal ion $N{{i}^{2+}}$, and the oxidation number,$O.N=2$. Atomic number or central metal ion, $Z=28$.
There are $4$$N{{H}_{3}}$ ligands, So, $C.N=4$and each$N{{H}_{3}}$ ligand donates $2$electrons to the metal ion.
Now,$E.A.N=Z-O.N+2\times (C.N)$
$E.A.N=28-2+2\times 4$
$E.A.N=34$
So, the effective atomic number (E.A.N) of $[Ni{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}$is $34$.
Thus, option (A) is correct.
Note: It is necessary to remember the atomic number of d-block elements at least 1st row from Sc ($Z=21$) to Zn($Z=30$) in the periodic table. As it will be helpful to deal with frequently asked questions about the effective atomic number,$18$ electrons rule, and so on in coordination chemistry
Formula used: Effective atomic number can be calculated by using a formula which is shown below:
$E.A.N = Z-O.N+2\times (C.N)$
Here, z = nuclear charge number or atomic number of the central metal atom.
O.N = oxidation number of central metal atoms.
C.N = coordination number of central metal atoms.
Complete step by step solution:
The effective atomic number is the total number of electrons surrounding the central nucleus of a metal ion. Therefore, the number of electrons present on the central metal ion and all ligands surrounding the metal ion is required to calculate the E.A.N. Generally, the E.A.N of the metal atom is numerically equal to the nearest noble gas configuration. Kr($Z=36$), Xe(Z=54), and Rn($Z=86$) are three noble gases that fall under the consideration of the effective atomic number rule. Not every coordination compound follows this rule.
We have the complex $[Ni{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}$, the central metal ion $N{{i}^{2+}}$, and the oxidation number,$O.N=2$. Atomic number or central metal ion, $Z=28$.
There are $4$$N{{H}_{3}}$ ligands, So, $C.N=4$and each$N{{H}_{3}}$ ligand donates $2$electrons to the metal ion.
Now,$E.A.N=Z-O.N+2\times (C.N)$
$E.A.N=28-2+2\times 4$
$E.A.N=34$
So, the effective atomic number (E.A.N) of $[Ni{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}$is $34$.
Thus, option (A) is correct.
Note: It is necessary to remember the atomic number of d-block elements at least 1st row from Sc ($Z=21$) to Zn($Z=30$) in the periodic table. As it will be helpful to deal with frequently asked questions about the effective atomic number,$18$ electrons rule, and so on in coordination chemistry
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