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Q.
The total number of electrons present in all the s orbitals, all the p orbitals and all the d orbitals of cesium ion are respectively :
EAMCETEAMCET 2003
Solution:
The atomic number of cesium is 55. The electronic configuration of cesium atom is $ {{\,}_{55}}Cs=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}4{{s}^{2}},3{{d}^{10}} $ $ 4{{p}^{6}},5{{s}^{2}}4{{d}^{10}},5{{p}^{6}},6{{s}^{1}} $ $ =1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}, $ $ 4{{s}^{2}}4{{p}^{6}}4{{d}^{10}},5{{s}^{2}}5{{p}^{6}},6{{s}^{1}} $ The electronic configuration of cesium ion will be $ C{{s}^{+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}, $ $ 4{{s}^{2}}4{{p}^{6}}4{{d}^{10}},5{{s}^{2}}5{{p}^{6}},6{{s}^{0}} $ So, the total number of s electrons = 10 the total number of p electrons = 24 Total number of d electrons = 20