A) II, III and IV
B) I, II and III
C) I, II and IV
D) II and I
Correct Answer: A
Solution :
[a] No of electrons in \[\overset{\oplus }{\mathop{C}}\,{{H}_{3}}\](I) =6+3-1=8 |
No. of electrons in \[{{\overset{\odot }{\mathop{NH}}\,}_{2}}\](II) =7+2+1=10 |
No. of electrons in \[{{\overset{\oplus }{\mathop{NH}}\,}_{4}}\](III) =7+4+-1=10 |
No. of electrons in \[N{{H}_{3}}\] (IV) =7+3=10 |
II, III, and IV are iso-electronic |
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