MCQ
After filling the $4d-$ orbitals, an electron will enter in :
  • A
    $4p$
  • B
    $4s$
  • $5p$
  • D
    $4f$

Answer

Correct option: C.
$5p$
c
After filling the $4 d$-orbital, an electron will enter in $5 p$ orbital.

The $4 d$ sub-energy level is at a lower energy than the $5 p$ sub-energy level

For $4 d , n =4$ and $1=2$. Hence, $( n +l)=4+2=6$

For $5 p , n =5$ and $1=1$. Hence, $( n +l)=5+1=6$

As the value of $(n+l)$ for $4 d$ orbital is same as that of $5 p$ orbital, $4 d$ orbital is filled before $5 p$ orbital as $4 d$ orbital has lower value of $n$ than $5 p$ orbital.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Final product of the reaction

 $C{H_3} - CH = C{H_2}\xrightarrow[{CC{l_4}}]{{B{r_2}}}A\xrightarrow[{(3\,\,moles)}]{{\mathop N\limits^ \oplus  a\mathop N\limits^\Theta  {H_2}}}B\xrightarrow{{C{H_3} - Br}}C\xrightarrow[{{H_2}}]{{Pd + BaS{O_4}}}D$

In weak electrolytic solution, degree of ionization
The solubility product of $Mg{(OH)_2}$ is $1.2 \times {10^{ - 11}}$. The solubility of this compound in gram per $100\,c{m^3}$ of solution is
The number of $N$ atoms is $681 \,g$ of $C _{7} H _{5} N _{3} O _{6}$ is $x \times 10^{21}$. The value of $x$ is $.....$ $\left( N _{ A }=6.02 \times\right.$ $10^{23}\, mol ^{-1}$ ) (Nearest Integer)
$8\,g$ of $NaOH$ is dissolved in $18\,g$ of $H_2O.$ Mole fraction of $NaOH$ in solution and molality (in $mol\,kg^{-1}$ ) of the solution respectively are
The correct statement/s about Hydrogen bonding is/are:

$A$. Hydrogen bonding exists when $\mathrm{H}$ is covalently bonded to the highly electro negative atom.

$B$. Intermolecular $\mathrm{H}$ bonding is present in o-nitro phenol

$C$. Intramolecular $\mathrm{H}$ bonding is present in $\mathrm{HF}$.

$D$. The magnitude of $\mathrm{H}$ bonding depends on the physical state of the compound.

$E$. H-bonding has powerful effect on the structure and properties of compounds.

Choose the correct answer from the options given below :

The compound $\begin{matrix}
   C{{H}_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}  \\
   |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{3}}-C=CH-C{{H}_{3}}  \\
\end{matrix}$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives :-
The major products $\mathrm{A}$ and $\mathrm{B}$ formed in the following reaction sequence are :
In the given reaction $'A'$ is

$\left( A \right)\xrightarrow{{\operatorname{Re}duction}}\left( B \right)\xrightarrow{{HN{O_2}}}{C_2}{H_5}OH$

In nucleuphilic substitution reaction, order of halogens as incoming (attacking) nucleophile is:
$I^-> Br^ - > Cl^-$
The order of halogens as departing nucleophile should be