MCQ
Krypton ${(_{36}}Kr)$ has the electronic configuration ${(_{18}}Ar)$ $4{s^2},3{d^{10}},4{p^6}$. The ${37^{th}}$ electron will go into which one of the following sub-levels
- A$4f$
- B$4d$
- C$3p$
- ✓$5s$
✓
Answer
Correct option: D.
$5s$
d
The $37^{th}$ electron will go into $5 s$ sub-level.
The $37^{th}$ electron will go into $5 s$ sub-level.
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 $5 s$ 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 $5 s$ orbital as $4 p$ orbital has lower value of $n$ than $5 s$ orbital.
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
In the reaction
→$B\xleftarrow{{Pd + CaC{O_3}}}R - C \equiv C - R\xrightarrow{{liq.N{H_3} + Na}}A$
$A$ and $B$ are geometrical Isomers. Then
For the given reaction
The function of fluorspar in the electrolytic reduction of alumina dissolved in fused cryolite $(Na_3AlF_6)$ is :
→Predict the major product
Non-polar solvent is
For the reactions, $C + {O_2} \to C{O_2};\Delta H = - 393\,J$
→$2Zn + {O_2} \to 2ZnO;\Delta H = - 412\,J$
The most polar bond is
For the gaseous reaction, ${N_2}{O_4} \to 2N{O_2}$
→The total energy of the electron of $H-$ atom in the second quantum state is $-E_2$ . The total energy of the $He^+$ atom in the third quantum state is
→The $pH$ of solution having $[O{H^ - }] = {10^{ - 7}}$ is
→