In the electronic spectrum of $\left[\mathrm{IrBr}_{6}\right]^{2-}$, the number of charge transfer band(s) and their origin are respectively
(a)Two ligand $\rightarrow$ metal $\left(\sigma \rightarrow \mathrm{t}_{2 \mathrm{g}}\right.$ and $\left.\sigma \rightarrow \mathrm{a}_{1 \mathrm{g}}{ }^{*}\right)$
(b)One ligand → metal $\left(\sigma \rightarrow \mathrm{e}_{\mathrm{g}}\right)$
(c)Two ligand → metal $\left(\sigma \rightarrow \mathrm{t}_{2 \mathrm{g}}\right.$ and $\left.\sigma \rightarrow \mathrm{e}_{\mathrm{g}}\right)$
(d)One ligand → metal $\left(\sigma \rightarrow \mathrm{t}_{2 \mathrm{g}}\right)$
Answer
Answer: C ✓ checked by 4AB · confidence medium
Explanation
[IrBr₆]²⁻ is Ir(IV) t₂g⁵, so it has one hole in t₂g and an empty e_g set. Ligand-to-metal charge transfer can go to either, giving two LMCT bands: σ/π→t₂g and σ→e_g (c).