MCQ
The first ionisation potentials $(eV)$ of $Be$ and $B$ respectively are :-
- A$8.29\, eV, 9.32\, eV$
- B$9.32\, eV, 9.32\, eV$
- C$8.29\, eV, 8.29\, eV$
- ✓$9.32\, eV, 8.29\, eV$
According to theory as we move from lest to right in a period the ionization potential should increase but in case of $B e$ and $B$ Beryllium is having more ionization potential than Boran due to completely filled s orbital in Beryllium $\left(1 s^{2} 2 s^{2}\right)$.
Hence option D is correct.
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| List$-I ($Compound$)$ | List$-II ($Colour$)$ | ||
| $A.$ | $Fe_4[Fe(CN)_6]_3.xH_2O$ | $I.$ | Violet |
| $B.$ | $[Fe(CN)_5NOS]^{4–}$ | $II.$ | Blood Red |
| $C.$ | $[Fe(SCN)]^{2+}$ | $III.$ | Prussian Blue |
| $D.$ | $(NH_4)_3PO_4.12MoO_3$ | $IV.$ | Yellow |

$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\, M)\,|\,Cu,$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\, M$ and that of $CuSO_4$ changed to $0.01\, M,$ the $emf$ changes to $E_2.$ Fromthe followings, which one is the relationship between $E_1$ and $E_2$ ? (Given, $RT/F = 0.059$)