Question

Consider the following metal complexes:
$\left[Co\right(N{H}_3{)}_6{]}^{3+}$
$\left[CoCl\right(N{H}_3{)}_5{]}^{2+}$
$\left[Co\right(CN{)}_6{]}^{3–}$
$\left[Co\right(N{H}_3{)}_5\left(H_{2}O\right){]}^{3+}$
The spin-only magnetic moment value of the
complex that absorbs light with shortest wavelength
is B.M. (Nearest integer)

Answer 1

In all complexes, Co is present in $+3$ oxidation state and all complexes are low spin or inner orbital complex.
Stronger the ligand, higher the crystal field splitting.
So, order of crystal field splitting is
$\left[Co\right(CN{)}_6{]}^{3–}>\left[Co\right(N{H}_3{)}_6{]}^{3+}>\left[Co\right(N{H}_3{)}_5\left(H_{2}O\right){]}^{3+}>\left[CoCl\right(N{H}_3{)}_5{]}^{2+}$
Shortest wavelength is shown by complex having
maximum crystal field splitting.



Spin only magnetic moment$=\sqrt{0(0+2)}=0B.M$