MCQ
The atomic number of an element having the valency shell electronic configuration $4{s^2}4{p^6}$ is
- A$35$
- ✓$36$
- C$37$
- D$38$
Hence no. of ${e^ - } = $ no. of protons $ = 36 = Z$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)\, [Co(en)_2(Ox)]^+$
$(ii)\, [Fe(Trien)(Ox)]^+$
$(iii)\, [Ca(EDTA)]^{-2}$
$(iv)\, [Pt(Ox)_2 (H_2O)_2]$