Question

The effective atomic number of $Cr$ (At. no. $=24$) in $\left[Cr\right(N{H}_3{)}_6]C{l}_3$ is:

• A. $35$
• B. $27$
• C. $33$
• D. $36$

Answer 1

The correct option is C $33$

Here, atomic number of $Cr$ = $24$.

Coordination number of $Cr$ = $6$

Here, oxidation number of $Cr$ = $+3$

Effective atomic number of $Cr$ = Atomic number - oxidation number + 2(coordination number)

So, effective atomic number of $Cr$ = 24 - 3 + 2(6) = 33

So, correct answer is option C.