MCQ
The first ionisation potential is maximum for
- A$B$
- ✓$N$
- C$O$
- D$Be$
It is due to the more stable (half-filled orbitals) configurations of $N$.
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$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron
$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$
$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.
$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$
(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )
The correct statements are
$2H_2(g) + O_2(g) \to 2H_2O(l)$ ; $\Delta _fH^o_{298}(H_2O(l)) = -285.5\, kJ/mol$
What is $\Delta S^o_{298}$ for the given fuel cell reaction ?
Given $: O_2(g) + 4H^+(aq) + 4e^- \to 2H_2O(l)$ $E^o = 1.23\, V$