MCQ
Chromyl chloride test is for
- A$Cl^-$
- ✓$N{ O_3}^-$
- C$CrO_4^{2 - }$
- D$(A)$ and $(C)$
$NH _3+3 Cl _2 \rightarrow NCl _3+3 HCl$
$X$ is $NH _3$ and $Y$ is $NCl _3$.
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$(I)\, (CH_3)_2P(CF_3)_3$ is non-polar and $(CH_3)_3P(CF_3)_2$ is polar molecule
$(II)\, CH_3 \widehat{P} CH_3$ bond angles are equal in $(CH_3)_3P(CF_3)_2$ molecule
$(III)\,PF_3$ will be more soluble in polar solvent than $SiF_4$