Question

write electronic configuration of holmium (Z=67) what is it's stable oxidation state​

Answer 1

Explanation:

Holmium is a chemical element with the symbol Ho and atomic number 67. Part of the lanthanide series, holmium is a rare-earth element.Pronunciation/ˈhoʊlmiəm/ (HOHL-mee-əm)Appearancesilvery whiteStandard atomic weight Ar, std(Ho)164.930328(7)[1]

Answer 2

The electronic configuration of Ho is

${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6} {4s}^{2} {3d}^{10} {4p}^{6} {5s}^{2} {4d}^{10} {5p}^{6} {4f}^{11} {6s}^{2}$

• The symbol of Holmium is Ho. It is a rare earth element. It is a transition element.
• It has an atomic number 67 and its electronic configuration is as follows-
• ${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6} {4s}^{2} {3d}^{10} {4p}^{6} {5s}^{2} {4d}^{10} {5p}^{6} {4f}^{11} {6s}^{2}$
• It has two electrons in its 6s orbital. Thus its outermost shell is 6s.
• It can remove its three electrons to give the most stable oxidation state +3.
• Thus most stable oxidation state of Ho is +3.