MCQ
The electronic configuration of gadolinium (atomic no. $64$) is
- A$[Xe]4{f^8}5{d^9}6{s^2}$
- ✓$[Xe]4{f^7}5{d^1}6{s^2}$
- C$[Xe]4{f^3}5{d^5}6{s^2}$
- D$[Xe]4{f^6}5{d^2}6{s^2}$
Therefore, further addition of electrons occurs in the lower energy $4 f$ - orbital till it is exactly half filled at $Eu \,(Z=63)$
Thus, the electronic configuration of $Eu$ is $[ Xe ] \,4 f ^7 \,6 s ^2$, thereafter, addition of next electron does not occur in the more stable exactly half filled $4 f^7$ shell but occurs in the little higher energy $5 d$ - orbital.
Thus, the electronic configuration of $Gd (Z=64)$ is $[X e]\, 4 f^7\, 5 d^1\, 6 s^2$
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$2NOCl(g)\, \rightleftharpoons \,2NO(g)\, + \,C{l_2}(g)$
[Given : $K_P = 8 \times10^{12}\,atm$ at $500\,K$ use $R = 0.08\,L\,atm\,mol^{-1}\,K^{-1}$ ]
$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 kcal mol ^{-1} \\ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 kcal mol ^{-1} \\ H - Cl ( g ) \quad \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 kcal mol ^{-1} \\\end{array}$
($1$) 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 $\Delta H ^{\circ}=-25 kcal mol ^{-1}$.