- ✓Low spin complex
- BParamagnetic
- CHigh spin
- D$sp ^{3} d ^{2}$ hybridized
$Co ^{3+}$ shows $d ^{2} sp ^{3}$ hybridization hence all the electrons are paired and diamagnetic in nature. Thus, it is a low spin complex.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(I)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_3} - C{H_2} - O - S - C{F_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}
\end{array}$
$(II)$ $CH_3-CH_2-O-TsCl$
$(III)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}} \\
{|\,\,\,\,} \\
{OH\,}
\end{array}$
$(IV)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - OH} \\
| \\
{{C_6}{H_5}}
\end{array}$
The molar mass difference between compounds $Q$ and $R$ is $474 \mathrm{~mol}^{-1}$ and between compounds $P$ and $S$ is $172.5 \mathrm{~g} \mathrm{~mol}^{-1}$.
($1$)The number of heteroatoms present in one molecule of $R$ is. . . . . .
[Use: Molar mass (in g mol ${ }^{-1}$ ): $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms]
($2$)The total number of carbon atoms and heteroatoms present in one molecule of $S$. . . . . .
[Use: Molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ ]: $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms
Give the answer quetion ($1$) and ($2$)
[Atomic masses (in $u$) $Mn =55 ; Cl =35.5 ; O =16, I =127, Na =23, K =39, S =32]$
$[\Lambda_{\mathrm{H}^{+}}^{\circ}=350 \,\mathrm{~S}\, \mathrm{~cm}^{2}\, \mathrm{~mol}^{-1},\Lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\circ}=50\, \mathrm{~S}\, \mathrm{~cm}^{2}\, \mathrm{~mol}^{-1}]$