MCQ
Amongst the following element with their electronic configuration given below, which one is having highest ionisation energy ?
- 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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) \quad \Delta H _1^\theta=- x\,kJ\,mol { }^{-1}$
$(B)$ $C$ (graphite) $+ O _2$ (g) $\rightarrow CO _2$ (g) $\Delta H _2^\theta=- y\,kJ\,mol -1$ The $\Delta H ^\theta$ for the reaction $......$.$C ($ graphite $)+\frac{1}{2} O _2( g ) \rightarrow CO ( g )$ is
(Given : Atomic Number: F :9, $Cl : 17, Na =11$, $Mg =12, Al =13, K =19, Ca =20, Sc =21$ )