MCQ
Which one is the electronic configuration of $F{e^{ + 2}}$
- ✓$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6}$
- B$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^4},4{s^2}$
- C$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}$
- DNone of these
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$A.$ For 1 s orbital, the probability density is maximum at the nucleus.
$B.$ For $2 s$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.
$D.$ $p$ and d-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of p-orbital is zero at the nucleus.
$\begin{array}{|l|l|} \hline \,\,List\,\,\,\,I\,\,(Compound) & \,\,List\,\,II\,\,(Oxidation\,\,state\,\,of\,\,N) \\ \hline (A)\,\,N{{O}_{2}} & (1)\,\,\,\,+\,\,5 \\ \hline (B)\,\,HNO & (2)\,\,-\,\,3 \\ \hline (C)\,\,N{{H}_{3}} & (3)\,\,+\,\,4 \\ \hline (D)\,\,{{N}_{2}}{{O}_{5}} & (4)\,\,+\,\,1 \\ \hline \end{array}$
code : $A \,\, B \,\,C \,\,D$