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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($1$) The compound $R$ is
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($2$) The compound $S$ is
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Give the answer or quetion ($1$) and ($2$)
$(I)$ gauche conformation of $1, 2$ -dibromoethane
$(II)$ anti conformation of $1, 2$ -dibromoethane
$(III)$ trans- $1, 4$ -dibromocyclohexane
$(IV)$ cis- $1, 4$ -dibromocyclohexane
$(V)$ tetrabromomethane
$(VI)$ $1, 1$ -dibromocyclohexane