MCQ
Correct configuration of $F{e^{ + 3}}$ $[26] $ is
- ✓$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}$
- B$1{s^2},2{s^2}s{p^6},3{s^2}3{p^6}3{d^3},4{s^2}$
- C$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}$
- D$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}$
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$(I)$ As wavelength decreases, the lines in the series converge
$(II)$ The integer $n_{1}$ is equal to $2$
$(III)$ The lines of longest wavelength corresponds to $\mathrm{n}_{2}=3$
$(IV)$ The ionization energy of hydrogen can be calculated from wave number of these lines
$\mathrm{A} \stackrel{700 \mathrm{K}}{\rightarrow}$ Product
$\mathrm{A}\xrightarrow[\text { catalyst }]{500 \mathrm{K}} $ Product
it was found that $\mathrm{E}_{\mathrm{a}}$ is decreased by $30 \;\mathrm{kJ} / \mathrm{mol}$ in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same $):$
$(I)$ $\left[ {{{\left( {P{h_3}P} \right)}_2}PdC{l_2}PdC{l_2}} \right]$
$(II)$ $\left[ {NiBrCl\left( {en} \right)} \right]$
$(III)$ $N{a_4}\left[ {Fe{{\left( {CN} \right)}_5}NOS} \right]$
$(IV)$ $Cr{(CO)_3}{\left( {NO} \right)_2}$
$(I)$ - $(II)$ - $(III)$ - $(IV)$