MCQ
Which of the order for ionization energy is correct?
- A$Be < B < C < N < O$
- ✓$B < Be < C < O < N$
- C$Be > B > C > N > O$
- D$B < Be < N < C < O$
On moving from left to right in a period, with increase in the atomic number, the ionisation enthalpy increases.
However there are few exceptions
$(1)$ Ionization energy of $B\,<\,$ ionization energy of $B e$. This is because less amount of energy is required to remove a p electron than s electron from the same principal quantum shell.
$(2)$ Ionization energy of $O\, <\,$ ionization energy of $N$.
Nitrogen has half filled $2 p$ subshell which is very stable. More energy is required for removal of electron from half filled, stable subshell.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
A for precipitate formation reaction.
B for precipitate dissolution reaction.
C for precipitate exchange reaction.
D for no reaction.
${P_4} + NaOH \longrightarrow P{H_3} \uparrow + Na{H_2}P{O_2}$
$\mathop {CH_3^ + }\limits_{\rm{I}} $ $\mathop {{H_3}{O^ + }}\limits_{{\rm{II}}} $ $\mathop {N{H_3}}\limits_{{\rm{III}}} $ $\mathop {CH_3^ - }\limits_{{\rm{IV}}} $
$(CH_3)_3 \, \overline C , \overline C Cl_3 ,(CH_3 )_2 \overline C H ,C_6 H_5 \overline C H_2$ in order of the irdecreasing stability :