- A$sp. sp^3$ and $sp^2$
- B$sp^2, sp^3$ and $sp$
- ✓$sp, sp^2$ and $sp^3$
- D$sp^2, sp$ and $sp^3$
$1)$ Count and add all the electrons in valence shell
$2)$ Divide it by $8$
$3)$ On dividing by $8$ you will get a remainder and quotient, add quotient $+$ (remainder $/ 2$)
$4)$ Now get the hybridization corresponding to the number what you got
If $2$ its $sp$, if $3$ its $sp ^2$, if $4$ its $sp ^3$, if $5$ its $sp ^3 d$ and so on.
Ex: $NO ^{2+}$
$\rightarrow$ Total electrons in valence shell : $5+6 \times 2-1=16$.
$\rightarrow 16 / 8=2$. Quotient $=2$ and Remainder $=0$
$\rightarrow 2+0 / 2=2$ i.e $sp$
$N H ^{4+}$
$\rightarrow$ Total electrons in valence shell : $5+7 \times 4-1=32$.
$\rightarrow 32 / 8=4$. Quotient $=4$ and Remainder $=0$
$\rightarrow 4+0 / 2=4$ i.e $sp ^3$
$NO ^{3-}$
$\rightarrow$ Total electrons in valence shell : $5+6 \times 3+1=24$.
$\rightarrow 24 / 8=3$. Quotient $=3$ and Remainder $=0$
$\rightarrow 3+0 / 2=3$ i.e $sp ^2$
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$(CH_3)_3 \, \overline C , \overline C Cl_3 ,(CH_3 )_2 \overline C H ,C_6 H_5 \overline C H_2$ in order of the irdecreasing stability :
$2NOB{r_{(g)\,}}\, \rightleftharpoons \,2N{O_{(g)}}\, + \,B{r_{2(g)}}$
if ${P_{B{r_2}}}$ is $\frac {P}{4}$ At equilibrium and $P$ is total pressure then calculate $\frac {P}{K_P}.$
$STATEMENT$-$2$: In water, orthoboric acid acts as a proton donor.