Question

Consider the following complex: $\left[Co\right(C{O}_3\left)\right(N{H}_3{)}_5]Cl{O}_4$
Mark the correct option:
Coordination no., Oxidation no., No. of d-electrons, unpaired d-electrons

• A. $6,3,6,0$
• B. $7,2,7,1$
• C. $7,1,6,4$
• D. $6,2,7,3$

Answer 1

The correct option is A $6,3,6,0$

In $\left[Co\right(N{H}_3{)}_{5}C{O}_3]Cl{O}_4$, five ammonia ligands and one carbonate ligand is attached to the central cobalt atom. Hence, the coordination number is six.
Let $X$ be the oxidation number of cobalt. $X+(-2)+(-1)=0$ or $X=3$. Hence, the oxidation number is 3.
The outer electronic configuration of cobalt (atomic number 27) is $3d^{7}4s^2$. The outer electronic configuration of $C{o}^{+3}$ will be $3d^{6}4s^0$. Thus it contains six d electrons.
$\left[Co\right(N{H}_3{)}_{5}C{O}_3]Cl{O}_4$ is inner orbital or low spin complexes. all the electrons are paired due to $d^{2}s{p}^3$ hybridization of $C{o}^{+3}$ ion. Hence, the number of unpaired d electrons is zero.