MCQ
Fluorine is a better oxidising agent than $B{r_2}$. It is due to
- ASmall size of fluorine
- BMore electron repulsion in fluorine
- CMore electronegativity of fluorine
- ✓Non-metallic nature of fluorine
✓
Answer
Correct option: D.
Non-metallic nature of fluorine
d
$F$ is more electronegative than $Br$. Therefore it can easily accept $e ^{-}$ and oxidise others.
$F$ is more electronegative than $Br$. Therefore it can easily accept $e ^{-}$ and oxidise others.
Therefore $F _2$ is better oxidising agent than $Br _2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The crystal lattice of ice is mostly formed by:
For $0.1\,M $ solution, the colligative property will follow the order
→Lowest Oxidation number of an atom in a compound $\mathrm{A}_2 \mathrm{~B}$ is -$2$ . The number of an electron in its valence shell is
→The major products obtained from the following sequence of reactions are
→$(CH_3)_2CHCH_2N(CH_2CH_3)_2 \xrightarrow{C{{H}_{3}}I}\xrightarrow[{{H}_{2}}O]{A{{g}_{2}}O}\xrightarrow{heat}$ products
Salts of transition metals are generally coloured, because :
The major organic product formed from the following reaction
The number of unpaired electrons is maximum in (Atomic no. : $ Ti = 22; V= 23; Cr = 24; Fe = 26$)
→Match items of Row $I$ with those of Row $II$.
→Row $I-Image$
Row $II:$
$(i)$ $\alpha- D -(-)$ Fructofuranose.
$(ii)$ $\beta$-D-(-) Fructofuranose
$(iii)$ $\alpha$-D-(-) Glucopyranose.
$(iv)$ $\beta- D -(-)$ Glucopyranose
Correct match is
Which equation is correct for first order reactions
The degree of dissociation of first order reaction is