MCQ
Iso-electronic species is
- ✓${F^ - },\,\,{O^{ - 2}}$
- B${F^ - },\,\,O$
- C${F^ - },\,\,{O^ + }$
- D${F^ - },\,\,{O^{ + 2}}$
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|
List$-I$ (Species) |
List$-II$ (Hybrid Orbitals) |
| $(a)$ $\mathrm{SF}_{4}$ | $(i)$ $\mathrm{sp}^{3} \mathrm{~d}^{2}$ |
| $(b)$ $\mathrm{IF}_{5}$ | $(ii)$ $\mathrm{d}^{2} \mathrm{sp}^{3}$ |
| $(c)$ $\mathrm{NO}_{2}^{+}$ | $(iii)$ $\mathrm{sp}^{3} \mathrm{~d}$ |
| $(d)$ $\mathrm{NH}_{4}^{+}$ | $(iv)$ $\mathrm{sp}^{3}$ |
| $(v)$ $\mathrm{sp}$ |
Choose the correct answer form the options given below :
$3 HC \equiv CH _{( g )} \rightleftharpoons C _{6} H _{6(\ell)}$
[Given: $\Delta_{f} G ^{\circ}( HC \equiv CH )=-2.04 \times 10^{5}\, J mol ^{-1}$
$\Delta_{f} G ^{\circ}\left( C _{6} H _{6}\right)=-1.24 \times 10^{5}\, J mol ^{-1} ; R =8.314\,\left. J K ^{-1} mol ^{-1}\right]$
Assertion $(A)$ : $\mathrm{NH}_3$ and $\mathrm{NF}_3$ molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of $\mathrm{NH}_3$ is greater than that of $\mathrm{NF}_3$.
Reason $(R)$ : In $\mathrm{NH}_3$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $\mathrm{N}-\mathrm{H}$ bonds. $\mathrm{F}$ is the most electronegative element.
In the light of the above statements, choose the correct answer from the options given below: