- A$4p$
- B$4s$
- ✓$5p$
- D$4f$
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.
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$C{H_3} - CH = C{H_2}\xrightarrow[{CC{l_4}}]{{B{r_2}}}A\xrightarrow[{(3\,\,moles)}]{{\mathop N\limits^ \oplus a\mathop N\limits^\Theta {H_2}}}B\xrightarrow{{C{H_3} - Br}}C\xrightarrow[{{H_2}}]{{Pd + BaS{O_4}}}D$
$A$. 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 :

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