Telangana Board Solutions for Chapter: d- and f- Block Elements and Coordination Compounds, Exercise 13: Intext Questions

Explain on the basis of valence bond theory that $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2–}$ ion with square planar structure is diamagnetic and the ${\left[{\mathrm{NiCl}}_4\right]}^{2–}$ ion with tetrahedral geometry is paramagnetic.

${\left[{\mathrm{NiCl}}_4\right]}^{2–}$ is paramagnetic while $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$ is diamagnetic though both are tetrahedral. Why?

$[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$ is strongly paramagnetic whereas $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{3–}$ is weakly paramagnetic. Explain.

Explain $[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6{]}^{3+}$ is an inner orbital complex whereas $[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+}$ is an outer orbital complex.

Predict the number of unpaired electrons in the square planar $[\mathrm{Pt}{\left(\mathrm{CN}\right)}_4{]}^{2–}$ ion.

The hexaquo manganese $\left(\mathrm{II}\right)$ ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.