MCQ
Krypton $(_{36}Kr)$ has the electronic configuration $[Ar]4s^2 ,3d^{10} ,4p^6 .$ The $37^{th}$ electron will go into which of following subshells
- A$4f$
- B$4d$
- C$3p$
- ✓$5s$
According to aufbau principle, the correct order of energy of $4 p$ and $5 s$-orbitals is $4 p \,<\,5 s$
Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.
The $4 p$ sub-energy level is at a lower energy than the $5$ sub-energy level
For $4 p , n =4$ and $1=1$. Hence, $( n +1)=4+1=5$
For $5 s , n =5$ and $1=0$. Hence, $( n +1)=5+0=5$
As the value of $( n +1)$ for $4 p$ orbital is same as that of $5 s$ orbital, $4 p$ orbital is filled before $5$ orbital as $4 p$ orbital has lower value of $n$ than $5 s$ orbital.
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$2 X + Y \xrightarrow{i} P$
the rate of reaction is $\frac{ d [ P ]}{ dt }=k[ X ]$. Two moles of $X$ are mixed with one mole of $Y$ to make $1.0 L$ of solution. At $50 s , 0.5$ mole of $Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use: $\ln 2=0.693$ )
$(A)$ The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} s ^{-1}$.
$(B)$ Half-life of $X$ is $50 s$.
$(C)$ At $50 s ,-\frac{ d [ X ]}{ dt }=13.86 \times 10^{-3} mol L ^{-1} s ^{-1}$.
$(D)$ At $100 s ,-\frac{ d [ Y ]}{ dt }=3.46 \times 10^{-3} mol L ^{-1} s ^{-1}$.
Given $\Delta H$
$(i)\, Fe_2O_{3(s)}+3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ $492\, kJ/mol$
$(ii)\, FeO_{(s)}+C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ $156\, kJ/mol$
$(iii)\, C_{(graphite)} + O_{2(g)} \to CO_{2(g)}$ $-393 \,kJ/mol$
$(iv)\, CO_{(g)} + \frac{1}{2}\, O_{2(g)} \to CO_{2(g)}$ $-283\, kJ/mol$
$\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
N\equiv C-C{{H}_{2}}-C-C{{H}_{2}}-CH=C{{H}_{2}}\to \\
\end{matrix}$ $\begin{matrix}
OH\,\,\,\,\,\,\,\,\, \\
|\,\,\,\,\,\,\,\,\, \,\,\,\, \\
N\equiv C-C{{H}_{2}}-\underset{H}{\mathop{C}}\,-C{{H}_{2}}-CH=C{{H}_{2}} \\
\end{matrix}$