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
266.7k+ 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
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Types of Solutions in Chemistry: Explained Simply

Difference Between Alcohol and Phenol: Structure, Tests & Uses

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Sign up for JEE Main 2026 Live Classes - Vedantu

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Class 12 Chemistry Question Paper 2026 PDF Download (All Sets) with Answer Key

NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2025-26

