Question

The effective atomic number of Cr (atomic number = 24) in ${\text{[Cr(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{]C}}{{\text{l}}_{\text{3}}}$ is?
A.35
B.27
C.33
D.36

Answer 1

Hint: The EAN rule is often referred to as the “18-electron rule”. The compounds that obey this rule are said to be stable complexes. It is calculated from the formula EAN= z - c + e where ‘z’ is the atomic number of the metal and ‘c’ is the charge on the central atom, and ‘e’ is the number of electrons acquired by metal from the ligands.

Complete step by step answer:
The English chemist Nevil V. Sidgwick made the observation, since known as the EAN rule, states that in a number of metal complexes the metal atom tends to surround itself with sufficient ligands that the resulting effective atomic number is numerically equal to the atomic number of the noble-gas element found in the same period in which the metal is situated.
This rule seems to hold for most of the metal complexes with carbon monoxide, the metal carbonyls as well as many organometallic compounds. By using this rule, it is possible to predict the number of ligands in these types of compounds and also the products of their reactions.
The effective atomic number, abbreviated as EAN can be calculated from the below formula:
EAN= z - c + e
Where ‘z’ is the atomic number of the metal and ‘c’ is the charge on the central atom, and ‘e’ is the number of electrons acquired by metal from the ligands.
In ${\text{[Cr(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{]C}}{{\text{l}}_{\text{3}}}$ complex, the Chromium is in +3 oxidation state which means it loses 3 electrons to form the complex and the ligand ammonia contributes 2 electrons each.
\[
   \Rightarrow {\text{EAN = 24-3 + }}\left( {{\text{6x2}}} \right){\text{ }} \\
   \Rightarrow {\text{EAN = 24 - 3 + 12 = 33}} \\
\]
Therefore, the effective atomic number of Cr in ${\text{[Cr(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{]C}}{{\text{l}}_{\text{3}}}$ complex is 33.

So, the correct option is C.

Note:
The given complex ${\text{[Cr(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{]C}}{{\text{l}}_{\text{3}}}$ is unstable since it does not have the effective atomic number equal to noble gas atomic number. In other words, the given complex does not obey the 18 electron rule.