- A$[Ne] 3s^23p^1$
- ✓$[Ne] 3s^23p^3$
- C$[Ne] 3s^23p^2$
- D$[Ar] 3d^{10}4s^24p^3$
A) $[N e] 3 s^{2} 3 p^{1}$ - can lose the $p$ electron with relative ease.
B) $[N e] 3 s^{2} 3 p^{3}$ - since it is half filled, it is difficult to remove an electron. Hene ionization energy
will be high in this case.
C) $[N e] 3 s^{2} 3 p^{2}$ - Can lose $p$ electron with relative ease.
D) $[A r] 3 d^{10}, 4 s^{2} 4 p^{3}$ - Half filled. Difficult to remove electron.
Hence, we are left with two options, either $B$ or $D$. But in option $D$ the half filled configuration is in $4 p$ whereas it is in $3 p$ in option $B$. It is relatively easier to remove an electron from $4 p$ than $3 p .$
Hence, option $B$ is the right answer.
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$CO ( g )+ H _2 O ( g ) \rightleftharpoons CO _2( g )+ H _2( g )$ The equilibrium constant $K _{ C } \times 10^2$ for the reaction is $.......$ (Nearest integer)
(Molecular mass of $CHCl_3 = 119.5 \,u$ and molecular mass of $CH_2Cl_2 = 85\, u$)
