- AThe spin magnetic moment of $KO_2$ is $1.73\,BM$
- BThe hybridisation of $B$ in $H_3BO_3$ is $sp^2$
- CIn $H_2S_2O_8\, S-S$ linkage is present
- ✓In $ClF_3$ axial and Equatorial bonds are not identical
The valence shell electron configuration of first excited state of sulphur is $3 \mathrm{s}^{2} 3 \mathrm{p}^{3} 3 \mathrm{d}^{1}$
The valence shell electron configuration of second excited state of sulphur is $3 \mathrm{s}^{1} 3 \mathrm{p}^{2} 3 \mathrm{d}^{2}$.
This is followed by sp ${ }^{3} \mathrm{d}^{2}$ hybridization to form $\mathrm{SF}_{6}$.
Thus $\mathrm{SF}_{6}$ is formed in the second excitation state of sulphur atom.
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$(en = NH_2CH_2CH_2NH_2)$
Reason : Grignard reagents react with hydroxyl group
Method $1\,:\,$ $RBr\xrightarrow[diethyl\,\,ether]{Mg}RMgBr\xrightarrow[2.\,{{H}_{3}}{{O}^{+}}]{1.\,C{{O}_{2}}}RC{{O}_{2}}H$
Method $2\,:\,$ $RBr\xrightarrow{NaCN}RCN\xrightarrow[heat]{{{H}_{2}}O,HCl}RC{{O}_{2}}H$
Which one of the following statements correctly describes this conversion ?