MCQ
Compound $(P)$ is
- A
- B
- C
- ✓
✓
Answer
Correct option: D.
d
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
Which is the weakest base
If $5$ moles of $BaCl _2$ is mixed with $2$ moles of $Na _3 PO _4$, the maximum number of moles of $Ba _3\left( PO _4\right)_2$ formed is $............$(Nearest integer)
→Which of the following molecules is/are chiral ?
For the reaction : ${H_{2(g)}} + C{O_{2(g)}}$ $\rightleftharpoons$ $C{O_{(g)}} + {H_2}{O_{(g)}}$, if the initial concentration of $[{H_2}] = [C{O_2}]$ and $x$ moles/litre of hydrogen is consumed at equilibrium, the correct expression of ${K_p}$ is
→$ ? + \mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{O}$
→The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy ($BDE$) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :
→$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 kcal mol ^{-1} \\ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 kcal mol ^{-1} \\ H - Cl ( g ) \quad \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 kcal mol ^{-1} \\\end{array}$
($1$) Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their BDE in Column $K$ is
|
Column $J$ Molecule |
Column $K$ $\operatorname{BDE}( kcal mol -1)$ |
| $(P)$ $H - C H \left( CH _3\right)_2$ | ${ (i) } 132$ |
| $(Q)$ $H - CH _2 Ph$ | ${ (ii) } 110$ |
| $(R)$ $H - C H = CH _2$ | ${ (iii) } 95$ |
| $(S)$ $H - C \equiv CH$ | ${ (iv) } 88$ |
$(A)$ $P - iii, Q - iv, R - ii, S - i$
$(B)$ $P - i, Q - ii, R - iii, S - iv$
$(C)$ $P - iii, Q - ii, R - i, S - iv$
$(D)$ $P - ii, Q - i, R - iv, S - iii$
($2$) For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 kcal mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 kcal mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 kcal mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 kcal mol ^{-1}$.
Degree of ionisation does not depend on:
Number of unpaired electrons in inert gas is
The number of structural isomers for $\mathrm{C}_6 \mathrm{H}_{14}$ is
→$sp^3d^2$ hybrid orbitals have
→