Some of the properties of the two species,NO_ | vedclass

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(A) For $NO_{3}^{-}$,the number of hybrid orbitals $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Thus,the central atom $N$ is $sp^{2}$ hybridized and the spe

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Dissimilar in hybridization for the central atom with different structures.

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(A) For $NO_{3}^{-}$,the number of hybrid orbitals $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Thus,the central atom $N$ is $sp^{2}$ hybridized and the species has a trigonal planar geometry.For $H_{3}O^{+}$,the number of hybrid orbitals $H = \frac{1}{2} [6 + 3 - 1 + 0] = 4$. Thus,the central atom $O$ is $sp^{3}$ hybridized and the species has a pyramidal geometry due to the presence of one lone pair.Since they have different hybridization states ($sp^{2}$ vs $sp^{3}$) and different geometries (trigonal planar vs pyramidal),they are dissimilar in both hybridization and structure.

List-$I$ (Molecule Species)	List-$II$ (Property/Shape)
$A$. $SO_2Cl_2$	$I$. Paramagnetic
$B$. $NO$	$II$. Diamagnetic
$C$. $NO_2^-$	$III$. Tetrahedral
$D$. $I_3^-$	$IV$. Linear

Some of the properties of the two species,$NO_{3}^{-}$ and $H_{3}O^{+}$ are described below. Which one of them is correct?

Similar Questions

The option$(s)$ in which at least three molecules follow the Octet Rule is(are):
$(A)$ $CO_2, C_2H_4, NO$ and $HCl$
$(B)$ $NO_2, O_3, HCl$ and $H_2SO_4$
$(C)$ $BCl_3, NO, NO_2$ and $H_2SO_4$
$(D)$ $CO_2, BCl_3, O_3$ and $C_2H_4$

Which of the following statements is correct?

In the reaction $LiH + AlH_3 \to LiAlH_4$,$AlH_3$ and $LiH$ act as:

What is formal charge? Explain by example.

Match List-$I$ with List-$II$ :

List-$I$ (Molecule Species) List-$II$ (Property/Shape)

$A$. $SO_2Cl_2$ $I$. Paramagnetic

$B$. $NO$ $II$. Diamagnetic

$C$. $NO_2^-$ $III$. Tetrahedral

$D$. $I_3^-$ $IV$. Linear

Choose the correct answer from the options given below :

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