Download App

p -Block elements - ll — Chemistry STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceChemistryp -Block elements - ll1 Mark
MCQ
Assertion : Fluorine exists only in $-\,1$ oxidation state.

Reason : Fluorine has $2s^2\,2p^5$ configuration.

  • A
    If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
  • B
    If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
  • ✓
    If the Assertion is correct but Reason is incorrect.
  • D
    If both the Assertion and Reason are incorrect.
✓

Answer

Correct option: C.
If the Assertion is correct but Reason is incorrect.
c
Highest occupied molecular orbital ($HOMO$) $\Rightarrow \pi^*$ Lowest unoccupied molecular orbital ($LUMO$) $\Rightarrow \sigma^*$ On descending the group gap between $\pi^*$ and $\sigma^*$ decreases.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from p -Block elements - ll→p -Block elements - ll — all question types→Chemistry STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Matrix is defined as:
→
Among the following outermost configurations of transition metals, which shows the highest oxidation state
→
Which of these do not contain $ - COOH$ group
→
${P_4} + 3NaOH + 3{H_2}O \to A + 3Na{H_2}P{O_2}$ here, $'A'$ is
→
When $KBr$ is treated with concentrated ${H_2}S{O_4}$ redish brown gas evolved, gas is
→
$\begin{array}{*{20}{c}}
  {C{H_3} - C - C{H_2} - COOH} \\ 
  {||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$ $\xrightarrow{\Delta }X\xrightarrow[{\mathop O\limits^\Theta  H}]{{HCN}}Y$,

$Y$ is

→
Arrange the following in order of their decreasing acidic strength?
→
The reaction of $N{a_2}{S_2}{O_3}$ with iodine gives
→
Number of cross products in the given reaction :

$CH_3COOC_2H_5 + C_6H_5 -CH_2 -COOC_2H_5$ $\mathop {\xrightarrow{{{C_2}{H_5}ONa}}}\limits_{{C_2}{H_5}OH} $

→
A trinitro compound, $1, 3,5$ tris-($4$-nitrophenyl) benzene, on complete reaction with an excess of $\mathrm{Sn} / \mathrm{HCl}$ gives major product, which on treatment with an excess of $NaNO_2 /  HCl$ at $0^{\circ} C$ provides $P$ as the product. $P$, upon treatment with excess of $H_2 O$ at room temperature, gives the product $Q$. Bromination of $Q$ in aqueous medium furnishes the product $R$. The compound $P$ upon treatment with an excess of phenol under basic conditions gives the product $S$.

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 ($2$)

→