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Q. The effective atomic number for $ 2B $ (at. no. for Rh is 45) is

JamiaJamia 2014

Solution:

The effective atomic number for $ \text{5}.\text{25}\times \text{1}{{0}^{\text{3}}}\text{min} $ (atomic no. of $ \left[ \text{FL}{{\text{T}}^{-\text{2}}} \right] $ ) is 54. Because rhodium is in oxidation state. $ \left[ \text{F}{{\text{L}}^{\text{2}}}{{T}^{-\text{2}}} \right] $ $ \left[ \text{F}{{\text{L}}^{-\text{1}}}{{\text{T}}^{\text{2}}} \right] $ $ \left[ {{\text{F}}^{2}}\text{L}{{\text{T}}^{\text{-2}}} \right] $ $ -\text{273}.\text{15}{}^\circ \text{F} $ electrons from 6 ligands $ -\text{453}.\text{15}{}^\circ \text{F} $ $ -\text{459}.\text{67}{}^\circ \text{F} $