MCQ
Pyramidal shape would be of
- A$NO_3^ - $
- B${H_2}O$
- ✓${H_3}{O^ + }$
- D$NH_4^ + $
$H _2 O$ has $sp ^3$ hybridisation $H _2 O$ is bent shaped molecule .
$H _3 O ^{+}$ has $sp ^3$ hybridisation and $H _3 O ^{+}$ is pyramidal
$NH _4^{+}$ has $sp ^3$ hybridisation and the shape of $NH _4^{+}$ is tetrahedral.
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| Column $-I$ (various solutions) |
Column $-II$ (Their freezing point ) |
||
| $a$ | $0.1\,M$ $BaCl_2$ solution | $p$ | $271\,K$ |
| $b$ | $0.1\,M$ $NaCl$ solution | $q$ | $270\,K$ |
| $c$ | $0.1\,M\, K_3 [Fe(CN)_6]$ solution | $r$ | $268\,K$ |
| $d$ | $0.1\,M\, Al_2 (SO_4)_3$ solution | $s$ | $269\,K$ |
Given : Freezing point of $0.1\,M$ sucrose solution $= 272\,K$ and $F.pt.$ of water $= 273\,K$
Which of the following option show correct matches ?
$\mathrm{H}_2 \mathrm{O}_2, \mathrm{ClO}_3^{-}, \mathrm{P}_4, \mathrm{Cl}_2, \mathrm{Ag}, \mathrm{Cu}^{+1}, \mathrm{~F}_2, \mathrm{NO}_2, \mathrm{~K}^{+}$