Question
Correct order of Ist ionisation potential $$(IP)$$ among following elements $$Be, B, C, N, O$$ is
A.
$$B < Be < C < O < N$$
B.
$$B < Be < C < N < O$$
C.
$$Be < B < C < N < O$$
D.
$$Be < B < C < O < N$$
Answer :
$$B < Be < C < O < N$$
Solution :
An atom has electronic configuration
$$1{s^2},2{s^2},2{p^6}\,3{s^2}3{p^6}3{d^3},4{s^2}$$
It is a member of $$d$$- block element because the last electron is filled in $$d$$- subshell as $$3{d^3}$$ and the following electronic configuration is possible for $$d$$-subshell as $$\left( {n - 1} \right){d^{\left( {1\,to\,10} \right)}}$$
\[\underset{\begin{smallmatrix}
\\
\\
n{{s}^{2}}\left( n-1 \right){{s}^{2}}{{p}^{6}}
\end{smallmatrix}}{\mathop{\text{Group}\,\,\text{number}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
3 \\
{{d}^{1}}
\end{smallmatrix}}{\mathop{\text{III}\,B}}\,\,\,\underset{\begin{smallmatrix}
4 \\
{{d}^{2}}
\end{smallmatrix}}{\mathop{\,\,\,\text{IV}B}}\,\] \[\underset{\begin{smallmatrix}
5 \\
{{d}^{3}}
\end{smallmatrix}}{\mathop{\text{V}B}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
6 \\
{{d}^{4}}
\end{smallmatrix}}{\mathop{\text{VI}\,B}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
7 \\
{{d}^{5}}
\end{smallmatrix}}{\mathop{\text{VII}\,B\,\,}}\,\,\,\,\underset{\begin{smallmatrix}
8 \\
{{d}^{6}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\] \[\underset{\begin{smallmatrix}
9 \\
{{d}^{7}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
10 \\
{{d}^{8}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
11 \\
{{d}^{9}}
\end{smallmatrix}}{\mathop{\text{IB}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
12 \\
{{d}^{10}}
\end{smallmatrix}}{\mathop{\text{IIB}}}\,\]
Hence, it is member of third group